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https://mathoverflow.net/questions/16494 | 1 | I read about two different versions of the disjunction elimination rule.
The first version (<http://www.fecundity.com/logic/>) says that:
* if $\Sigma\vdash\phi\_0\lor\phi\_1$ and $\Sigma\vdash\lnot\phi\_0$, then $\Sigma\vdash\phi\_1$
* if $\Sigma\vdash\phi\_0\lor\phi\_1$ and $\Sigma\vdash\lnot\phi\_1$, then $\Sigma\vdash\phi\_0$
The second version (S. Hedman - A First Course in Logic) says that:
* if $\Sigma\models\phi\_0\lor\phi\_1$, $\Sigma\cup\left\{\phi\_0\right\}\vdash\phi\_2$ and $\Sigma\cup\left\{\phi\_1\right\}\vdash\phi\_2$, then $\Sigma\vdash\phi\_2$
Using the first version of the rule, I can't even demonstrate that if $\Sigma\vdash\phi\lor\phi$ then $\Sigma\vdash\phi$. Perhaps the entire system presented by the first source is not complete, in the sense that you can't prove certain true statements. Of course, it may be my fault, instead.
Thanks.
| https://mathoverflow.net/users/3554 | What does the disjunction elimination rule say? | The first rule is not the regular disjunction elimination rule, but is known as [disjunctive syllogism](http://en.wikipedia.org/wiki/Disjunctive_syllogism), and is essentially the modus tollendo ponens rule of term logic. The two rules are mutually admissible in reasonable formulations of classical logic, but the first rule is strictly weaker in intuitionistic logic.
| 10 | https://mathoverflow.net/users/3154 | 16528 | 11,077 |
https://mathoverflow.net/questions/16532 | 19 | This question is about an issue left unresolved by [Chad
Groft's excellent
question](https://mathoverflow.net/questions/15957) and
[John Stillwell's excellent
answer](https://mathoverflow.net/questions/15957#16122) of
it. Since I find the possibility of an affirmative answer
so tantalizing, I would like to pursue it further here.
For background, [Rice's Theorem](http://en.wikipedia.org/wiki/Rice%27s_theorem)
asserts essentially that no nontrivial question about computably enumerable sets is decidable. If We is the set enumerated by program e, then the theorem states:
**Rice's Theorem.** If A is a collection of computably
enumerable sets and { e |
We ∈ A } is decidable, then either A is
empty or A contains all computably enumerable sets.
In short, one can decide
essentially nothing about a program e, if the answer is to
depend only on what the program computes rather than how it
computes it.
The question here is about the extent to which a similar
phenomenon holds for finitely presented groups, using the
analogy between programs and finite group presentations:
* a program e is like a finite group presentation p
* the set We enumerated by e is like the group
⟨p⟩ presented by p.
According to this analogy, the analogue of Rice's theorem
would state that any decidable collection of finitely
presented groups (closed under isomorphism) should be either trivial or everything.
John Stillwell pointed out in answer to Chad Groft's
question that this is not true, because from a presentation
p we can easily find a presentation of the abelianization of
⟨p⟩, by insisting that all generators commute,
and many nontrivial questions are decidable about finitely
presented abelian groups. Indeed, since the theory of
abelian groups is a decidable theory, there will be many
interesting questions about finitely presented abelian
groups that are decidable from their presentations.
My question is whether this is the only obstacle.
**Question.** Does Rice's theorem hold for finitely
presented groups modulo abelianization?
In other words, if
A is a set of finitely presented groups (closed under
isomorphism) and the corresponding set of presentations { p | ⟨p⟩ ∈ A } is
decidable, then does A completely reduce to a question
about the abelianizations of the groups, in the sense that there is a set B of abelian groups such that G ∈ A iff Ab(G) ∈ B?
Of course, in this case B consists exactly of the abelian groups in A. The question is equivalently asking whether A respects the equivalence of groups having isomorphic abelianizations. In other words, must it be that G ∈ A iff Ab(G) ∈ A?
The question is asking whether every decidable set of finitely presented groups amounts actually
to a decidable set of abelian groups, extended to all
finitely presented groups just by saturating with respect to abelianization.
In particular, the set A should contain either none or all perfect groups.
An affirmative answer would seem to provide a thorough
explanation of the pervasive undecidability phenomenon in
group presentations. But perhaps this may simply be too much to hope for...
In any event, I suppose that there is an equivalence relation on finite group presentations, saying that p ≡ q just in case ⟨p⟩ and ⟨q⟩ have the same answer with repsect to any decidable question about finitely presented groups. The question above asks whether this equivalence relation is just Ab(⟨p⟩) = Ab(⟨q⟩). If this turns out not to be true, then what can be said about ≡?
| https://mathoverflow.net/users/1946 | Does every decidable question about finitely presented groups amount to a question about abelian groups? | The question "Is there a nonzero homomorphism from your group to $A\_5$?" is decidable. (Just write down all ways of sending the generators of your group to $A\_5$, and see whether they satisfy the required relations.) The same is true with $A\_5$ replaced by any finite group. I don't see how to reduce this to questions about the abelianization.
| 23 | https://mathoverflow.net/users/297 | 16539 | 11,084 |
https://mathoverflow.net/questions/16491 | 13 | Actually, my question is a bit more specific: Does every complex semisimple Lie group $G$ admit a faithful finite-dimensional *holomorphic* representation? [As remarked by Brian Conrad, this is enough to prove that $G$ is a matrix group (at least when it's connected) because $G$ can be made into an (affine) algebraic group over $\mathbb{C}$ in unique way which is compatible with its complex Lie group structure, and under which every finite-dimensional holomorphic representation is algebraic. Furthermore, one can show that the image of a faithful representation would then be closed.]
Of course the analogous question for real semisimple Lie groups has a negative answer -- "holomorphic" having been replaced by "continuous", "smooth" or "real analytic" -- with the canonical counterexample being a nontrivial cover of $\mathrm{SL}(2,\mathbb{R})$.
For a *connected* complex semisimple Lie group $G$ I believe the answer is "YES." The idea is to piggy back off a 'sufficiently large' representation of a compact real form $G\_\mathbb{R}$; here by "compact real form" I'm referring specifically to the definition which allows us to uniquely extend continuous finite-dimensional representations of $G\_\mathbb{R}$ to holomorphic representations of $G$. I know (e.g. from the proof of Theorem 27.1 in D. Bump's *Lie Groups*) that such a definition is possible if we require $G$ to be connected (and I'd like to know if it's possible in general).
The details of the argument for connected $G$ are as follows. Consider the adjoint representation $\mathrm{Ad} \colon G \to \mathrm{GL}(\mathfrak{g})$. Since $G$ is semisimple, $\mathrm{Ad}$ has discrete kernel $K$. Consider next the restriction of $\mathrm{Ad}$ to $G\_\mathbb{R}$. Observe that the kernel of this map is also $K$, for otherwise its holomorphic extension is different from the adjoint representation of $G$. Thus $K$ is finite, being a discrete, closed subset of a compact space. So by the Peter-Weyl theorem, we can find a representation $\pi\_0$ of $G\_\mathbb{R}$ that is nonzero on $K$. Extend $\pi\_0$ to a holomorphic representation $\pi$ of $G$ and put $\rho = \pi \oplus \mathrm{Ad}$. Notice that $\rho$ is a holomorphic representation of $G$ with kernel $\ker\pi \cap K = 0$, which is what we were after.
What can we say if $G$ is disconnected?
| https://mathoverflow.net/users/430 | Are complex semisimple Lie groups matrix groups? | As requested by Faisal, I am posting as an answer the observation that if $G$ has more components than the size of the complex numbers then G has no faithful finite-dimensional irreducible representation over the complexes, for cardinality reasons.
To be honest I thought Faisal's response to this would be "what happens if $G$ has only countably infinitely many components"? That is then a different question. Does every countable group have a faithful finite-dimensional complex representation? If this is well-known I'd be happy to hear the answer. If it isn't I'd be happy if someone else asked this as a separate question. If it's true that any countable group has a faithful representation then one might ask about complex Lie groups with countably infinitely many components.
| 9 | https://mathoverflow.net/users/1384 | 16544 | 11,088 |
https://mathoverflow.net/questions/16540 | 4 | Suppose that we are given a smooth projective variety $X$ with a *full exceptional collection* of vector bundles $(F\_1, F\_2, \ldots, F\_k)$ in $D^b(X)$ and two vector bundles $E\_1$, $E\_2$ on $X$. Consider the following statement:
$$ \text{If }H^i(X, E\_1\otimes F\_j) = H^i(X, E\_2\otimes F\_j)\text{ for all }i, j\text{, then } E\_1=E\_2. $$
**Question 1.** Is the above true?
If no,
**Question 2.** What if we assume that the collection is strong?
**Question 3.** What if we consider all twists by the ample sheaf $\mathcal{O}(1)$, i. e.
$H^i(X, E\_1\otimes F\_j(t)) = H^i(X, E\_2\otimes F\_j(t))$ for $i, j$ and all $t$?
**Question 4.** Are there any known cohomological criteria for isomorphism?
If yes,
**Question 5.** Does it suffice to assume that $(F\_1, F\_2, \ldots, F\_k)$ generate the derived category, *without* them forming an exceptional collection?
**Remark.** Take $X=\mathbb{P}^n$ and $F\_i = \mathcal{O}(i)$ or $F\_i = \Omega^i(i)$. Then by the Beilinson spectral sequence, the cohomology groups $H^i(X, E\otimes F\_i)$
*would* determine $E$ unambiguously *if* we knew the differentials.
| https://mathoverflow.net/users/3847 | Exceptional collections and cohomological criteria for isomorphism | It is hard to think of a general way to obtain an isomorphism of vector bundles from several isomorphisms of cohomology spaces. In particular, there can be moduli of vector bundles, I presume, even over projective spaces. But an isomorphism of cohomology only means an equality of dimensions, and dimensions are discrete invariants, taking values in the integers.
It would be an entirely different matter if the question were, is a given morphism of vector bundles an isomorphism provided that it induces isomorphisms of certain cohomology spaces. For this version of your question to have a positive answer, it suffices to require that (Fi) generate the derived category (due to Serre duality).
| 3 | https://mathoverflow.net/users/2106 | 16552 | 11,095 |
https://mathoverflow.net/questions/16554 | 16 | Suppose that $C$ is a 2-category, perhaps $C=\rm{Cat}$, the 2-category of small categories, functors, and natural transformations. Let $T$ be an object in $C$.
I form the new 1-category whose objects are morphisms $f\colon A\rightarrow T$ in $C$, and in which a morphism from $f$ to some $f'\colon A'\rightarrow T$ consists of a pair $(\phi,\phi^\sharp)$ where $\phi\colon A\rightarrow A'$ is a 1-morphism in $C$ and $\phi^\sharp\colon\phi\circ f'\rightarrow f$ is a 2-morphism between arrows $A\rightarrow T$. Call this new category the $(C\Uparrow T)$. An obvious variation comes about by reversing the direction of the 2-morphism, i.e. we could take $\phi^\sharp\colon f\rightarrow\phi\circ f'$; perhaps I might call this variation $(C\Downarrow T)$.
What is the high-brow way to refer to these strange slice-categories? How do you locate them within a good understanding of 2-categories? Where are the properties of such things discussed? What is the relation between these strange slices and the usual 2-categorical slices?
Thanks!
| https://mathoverflow.net/users/2811 | The urge to combine 1- and 2-morphisms in slicing a 2-category. | The second definition looks like the 'lax comma category' $C // T$, where a morphism $f \to f'$ is given by a 2-cell $f \to f'\phi$. The defining universal property is the same as for [comma objects](http://ncatlab.org/nlab/show/comma+object), except that the 2-cells in the squares are [lax](http://ncatlab.org/nlab/show/lax+natural+transformation) natural transformations. Your first definition should be the oplax version.
See Kelly, *On clubs and doctrines*, LNM 420, or Gray, *Adjointness For 2-Categories*, LNM 391, who calls these '2-comma categories'.
In more detail, Gray's 2-comma categories come from (strict, I think) 2-functors $A \overset{F}{\rightarrow} K \overset{G}{\leftarrow} B$. An object is a 1-cell $FA \to GB$, a morphism is a square with a 2-cell in, and a 2-cell is given by a pair of 2-cells in $K$ that fit into a commuting cylinder (it's pretty obvious if you draw a picture). In your example, (what I've called) $C // T$ has 2-cells $(\phi,\phi^\sharp) \Rightarrow (\psi,\psi^\sharp)$ given by 2-cells $\alpha \colon \phi \Rightarrow \psi$ such that $\psi^\sharp \circ f'\alpha = \phi^\sharp$. (Again, pictures make it much clearer!) So your slices are actually 2-categories, coming from $C \overset{1}{\rightarrow} C \overset{T}{\leftarrow} \bullet$.
| 14 | https://mathoverflow.net/users/4262 | 16559 | 11,100 |
https://mathoverflow.net/questions/16522 | 11 | I am trying to prove that a certain class of polynomials have symmetric galois group.
Using the Newton polygon, I have shown that the galois groups of these polynomials are transitive on $k$-sets for all $k$ less than the degree of the polynomial. Beaumont and Peterson proved that this condition holds only for symmetric and alternating groups, along with 4 sporadic exceptions. I have discounted the 4 exceptions, so I am left with the task of showing that the alternating group does not occur.
I am looking for a way to do this. I have tried without success to show that there must be a transposition in the group; likewise showing that the determinant is not a square does not seem to be possible.
Does anybody know of any tricks for showing that one of the above sufficient conditions holds? Alternatively, do you know of any other way to prove that a group is the symmetric group, or not the alternating group?
| https://mathoverflow.net/users/4078 | How to show the galois group of a polynomial is not an alternating group? | There *is* a version of the Chebotarev density theorem for finitely generated fields, or more precisely, after spreading out, for an étale Galois cover of schemes of finite type over a ring of $S$-integers. This is a consequence of work surrounding the Weil conjectures. See Lemma 1.2 in Torsten Ekedahl, An effective version of Hilbert's irreducibility theorem, Séminaire de Théorie des Nombres, Paris 1988-89, Birkhäuser 1990.
But what I would recommend trying first is to use the monodromy groups. If you spread out your Galois extension to a finite branched cover of a dense open subscheme $U$ of $\mathbf{P}^n\_{\mathbf{Q}}$, then the monodromy group (inertia group) associated to any irreducible divisor on $U$ is contained in your Galois group. In particular, if there is a divisor that ramifies in the simplest possible way (ramification indices $2,1,1,\ldots,1$), then you know that your group contains a transposition.
The other approach to try, already mentioned by others in the comments, is specialization, which amounts to spreading out as above and then restricting to the cover above an irreducible closed subscheme. More concretely, you could plug in rational numbers for some or all of $y\_1,\ldots,y\_n$ such that you get a separable polynomial (of the same degree as the original polynomial). Or you could spread out to a scheme of finite type over $\mathbf{Z}$ and then restrict to an irreducible closed subscheme; this includes reduction modulo primes. If specialization results in a separable polynomial of the same degree over the new function field, then the Galois group of the specialization is a subgroup of the original Galois group, so it suffices to show that the specialized Galois group is not contained in $A\_n$.
For more techniques for computing Galois groups, I recommend the book *Topics in Galois theory* by Jean-Pierre Serre.
| 12 | https://mathoverflow.net/users/2757 | 16561 | 11,102 |
https://mathoverflow.net/questions/16566 | 4 | Given a curve $C$. Is there any relation between the etale fundamental group $\pi\_1(C)$ and the first etale cohomology of the constant sheaf , say $Z/nZ$, on $C$ ?
For example, if $C$ is a complex curve, then the singular cohomology $H^1(C,Z)$ is the dual of the topological fundamental group divided by the commutators ( which is the same as Hom$(\pi\_1(C),Z) )$.
So it seems that there should be some relation between Hom$(\pi\_1(C),Z/nZ)$ and $H^1(C,Z/nZ)$ in the etale case, but how?
| https://mathoverflow.net/users/4275 | etale fundamental group and etale cohomology of curves | The two groups you want to compare are canonically isomorphic, so long as C is connected. See Example 11.3 of Milne's notes:
<http://www.jmilne.org/math/CourseNotes/lec.html>
| 9 | https://mathoverflow.net/users/271 | 16568 | 11,105 |
https://mathoverflow.net/questions/16363 | 17 | The original [Kuratowski closure-complement problem](http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem) asks:
>
> How many distinct sets can be obtained by repeatedly applying the set operations of closure and complement to a given starting subset of a topological space?
>
>
>
My question is: **what is known about analogous questions in other settings?**
Here's an example of what I'm thinking of, for rings:
>
> How many distinct ideals can be obtained by repeatedly applying the operations of radical and annihilator to a given starting ideal $I$ of a commutative ring $R$?
>
>
>
Note that $r(r(I))=r(I)$ and $I\subseteq Ann(Ann(I))=\{x\in R: x\cdot Ann(I)=(0)\}$, which are the best analogs I could think of to $\overline{\overline{S}}=\overline{S}$ and $(S^C)^C=S$.
Also: **what is the structure necessary to formulate this kind of question called, and where does it occur naturally?**
It seems like we need at least a poset, but with distinguished idempotent and involution operations to generalize the closure and complement, respectively.
| https://mathoverflow.net/users/1916 | Kuratowski closure-complement problem for other mathematical objects? | Here's a paper that might be of interest:
D. Peleg, A generalized closure and complement phenomenon, Discrete Math., v.50 (1984) pp.285-293.
Other than what's found in the above paper I do not know of any general theory or framework specifically aimed at organizing results similar to the Kuratowski closure-complement problem, i.e., those which involve starting with a seed object (or objects) and repeatedly applying operations to generate further objects of the same type in a given space.
Here's a general sub-question I thought of recently, that might be interesting to study:
"What's the minimum possible cardinality of a seed set that generates the maximum number of sets via the given operations?"
A few years ago I proposed a challenging Monthly problem (11059) that essentially asks this question for the operations of closure, complement, and union in a topological space. It does turn out there's a space containing a singleton that generates infinitely many sets under the three operations, but it's a bit tricky to find. I haven't looked into the question yet for other operations. As far as I know it hasn't been discussed yet in the literature (apart from the specific case addressed by my problem proposal).
| 8 | https://mathoverflow.net/users/4276 | 16574 | 11,110 |
https://mathoverflow.net/questions/16583 | 6 | Let $x = \pi/(2k+1)$, for $k>0$.
Prove that
$$
\cos(x)\cos(2x)\cos(3x)\dots\cos(kx) = \frac{1}{2^k}
$$
I've confirmed this numerically for $n$ from $1$ to $30$.
I'm finding it surprisingly difficult using standard trigonometric formula manipulation.
Even for the case $k = 2$, I needed to actually work out $\cos x$ by other methods to get the result.
Please let me know if you have a neat proof.
| https://mathoverflow.net/users/4279 | An identity for the cosine function | Let
$S(x)=\prod\_{j=1}^k \text{sin}(jx)$
and
$C(x)=\prod\_{j=1}^k \text{cos}(jx)$.
Let x = $\frac{\pi}{2k+1}$.
Then $S(2x) = S(x)$ (from $\text{sin}(\pi-x)=\text{sin}(x)$), and $S(2x)=2^kS(x)C(x)$ (from $\text{sin}(2x)=2\text{sin}(x)\text{cos}(x)$), from which the result follows.
Steve
| 21 | https://mathoverflow.net/users/1446 | 16591 | 11,120 |
https://mathoverflow.net/questions/16587 | 8 | **Topic**: this is a mathematics education question (but applies to other sciences too).
**Assumptions**: my first assumption is that most mathematical concepts used in research are not intrinsically more complicated to grasp than high-school and undergraduate maths, the main difference is the amount of prerequisites (and hence time and experience) involved. My second assumption is that some undergraduate topics currently taught compulsarily are a bit of a burden for someone focussed on a particular topic.
Now of course cognitive development is a constraint, but upon reaching the age of high-school, I would think that a fairly large proportion of the scientifically-enclined students could really understand things usually taught much later and indeed become active at research level within a few years, provided some shortcuts are introduced.
**Early specialization**: I'm wondering if a balanced curriculum already exists (or is planned) to provide such early specialization. What I'm looking for is this: a one-week panorama of maths (or physics, or biology) would be organized at the beginning, and then the students would decide which subtopic to study. For example someone interested by group theory (or quantum optics, or genetics) would thus start with basics at the age 15 or 16, and gradually learn more stuff and skills, but for a few years with a strong emphasis on things directly relevant for the chosen subtopic.
So for example the student specializing in group theory would only learn differential calculus and manifolds in passing in the context of Lie groups, and would skip most undergraduate real and functional analysis until it becomes relevant for his/her research topic, if at all. Of course other general courses would still be taught (history, sciences, programming, foreign languages...), but at least 50% of the student's week would be devoted to the research topic, ensuring satisfying progress.
**Question**: do you know of any active or planned educative curriculum (at a high-school or university, or maybe a specific home-schooling program) as outlined above? As an example of successful early specialization see e.g. the winners of the [Siemens Foundation Prizes](http://www.siemens-foundation.org/en/), but I haven't been able to learn much about their specific curriculum if any.
**Note**: Skipping grades in school to enter university earlier is not the point, I'm really interested in a subtopic-oriented curriculum.
| https://mathoverflow.net/users/469 | Specializing early | As far as getting high school students involved in research by learning rapidly a narrow range of mathematics but in some depth, this is actually done in the mathematics section of the [Research Science Institute](http://en.wikipedia.org/wiki/Research_Science_Institute) program at MIT for students who have completed their junior year. Last year there were four projects in representation theory; I recall that one of them did not now linear algebra until some two weeks before the program (but learned quickly and completed a very successful project).
Sadly, I do not know of many other opportunities- RSI is a small program, and only a portion of it is for mathematics. I believe the PROMYS program supervises some research projects, but it is primarily for learning mathematics. Incidentally, many of the winners of competitions such as Siemens begin their projects at RSI.
Also, alumni of the RSI program do not necessarily end up specializing in the same fields that they did their projects (if they do eventually choose to pursue a career in mathematics, which does not always happen). It does give an exposure to a certain field, though.
| 7 | https://mathoverflow.net/users/344 | 16599 | 11,125 |
https://mathoverflow.net/questions/16600 | 14 | Why do elliptic curves have bad reduction at some point if they are defined over Q, but not necessarily over arbitrary number fields?
| https://mathoverflow.net/users/5730 | bad reduction for elliptic curves | Here are two answers:
(a) If you try to write down an elliptic curve $y^2 = x^3 + a x + b$ with everywhere good reduction, you need to choose $a$ and $b$ such that $4a^3 + 27 b^2 = $ a unit. We can certainly solve this equation over some (lots!) of number fields, say if we set the unit equal to $1$ or $-1$, or a unit in some fixed base number field. But we can't solve it in ${\mathbb Q}$.
[Edit: As Bjorn intimates in his comment below, one has to be a little more careful than I am being here to be sure of good reduction mod primes above 2; the details are left to the interested reader (or, I imagine, can be found in Silverman in the section where he discusses the proof that there are no good reduction elliptic curves over $\mathbb Q$).]
(b) There are many non-trivial everywhere unramified extensions of number fields (e.g. $\mathbb Q(\sqrt{-5}, i)$ over $\mathbb Q(\sqrt{-5})$), but there are no everywhere unramified extensions of the particular number field $\mathbb Q$. The situation with elliptic curves is completely analogous.
| 26 | https://mathoverflow.net/users/2874 | 16602 | 11,127 |
https://mathoverflow.net/questions/16584 | 6 | In the definition of vertex algebra, we call the vertex operator state-field correspondence, does that mean that it is an injective map??
Are there some physical interpretations about state-field correspondence ? Or why we need state-field correspondence in physical viewpoint??
Does it have some relations to highest weight representations?
| https://mathoverflow.net/users/4155 | About state-field correspondence | Yes, the state-field map $v\mapsto Y(v,z)$ is an injective map, since by the axioms of VOA, $Y(v,z)1|\_{z=0}=v$.
The state-field correspondence appears in 2-dimensional field theory because such a field theory attaches an amplitude to a "pair of pants" (a 2-sphere with 3 holes). Namely, if you regard two of the holes as "incoming" and one as "outgoing"
(i.e., you regard the surface as a worldsheet of an interaction of two strings in which they unite into one), then to this surface corresponds an operator $A: H\otimes H\to H$, where $H$ is the Hilbert space of the theory (attached to the circle). This can be viewed as a map $A: H\to {\rm End}(H)$, which is the state-field correspondence. In conformal field theory, the field map $Y$ above is obtained in the limit when the holes become very small (so the surface becomes the Riemann sphere without 3 points, say, $0,z,\infty$). For instance, in the minimal models $H=\oplus\_{i=0}^n V\_i\otimes V\_i^\ast$, where $V\_i$ are all the irreducible unitary highest weight representations of the Virasoro algebra of central charge $c=1-6/(m+2)(m+3)$, where $m\ge 1$ is an integer. The restriction of $A$ to the summand $V\_0\otimes V\_0^\ast$ at all three holes turns out to be a tensor product $Y\otimes Y^\vee$, where $Y=Y(z): V\_0\to {\rm End}(V\_0)[[z,z^{-1}]]$, which equips $V\_0$ with a structure of a vertex operator algebra.
| 15 | https://mathoverflow.net/users/3696 | 16606 | 11,130 |
https://mathoverflow.net/questions/16615 | 8 | Given a plane affine curve $\sum\_{i,j}a\_{i,j}X^iY^j = 0$, its genus can be calculated as the number of integral points of the interior of the convex hull of $\{(i,j) \mid a\_{i,j} \neq 0\}$. (claimed here: <http://lamington.wordpress.com/2009/09/23/how-to-see-the-genus/>)
How can this be proved?
| https://mathoverflow.net/users/nan | calculating the genus of a curve using the Newton polygon | Here are the references I know concerning this:
H. F. Baker, Examples of applications of Newton's polygon to the theory of singular points of algebraic functions, *Trans. Cambridge Phil. Soc.* **15** (1893), 403-450.
A. G. Khovanskii, Newton polyhedra and the genus of complete intersections, *Funct. Anal. i ego pril*. English translation: *Functional Anal. Appl.*, **12** (1978), 38-46.
V. I. Danilov and A. G. Khovanskii, Newton polyhedra and an algorithm for
computing Hodge-Deligne numbers, *Izv. Akad. Nauk SSSR Ser. Mat.* **50** (1986), 925-945; English translation: *Math. USSR-Izv.* **29** (1987), 279-298.
P. Beelen and R. Pellikaan, The Newton polygon of plane curves with many
rational points, *Designs, Codes and Cryptography* **21** (2000), 41-67.
(See Theorem 4.2.)
I think the statement should really be, given an irreducible curve in $\mathbf{G}\_m^2$, a formula for the arithmetic genus of its closure in the 2-dimensional projective toric variety corresponding to the polygon. This way one should not need to impose genericity hypotheses or restrictions on the characteristic. The references above don't quite do all of this, however, so there is still room for a better reference or proof, I think.
| 13 | https://mathoverflow.net/users/2757 | 16617 | 11,136 |
https://mathoverflow.net/questions/16578 | 33 | What the title said. In a slightly more leisurely fashion:-
>
> Let $X$ be a compact, connected subset of $\mathbb{R}^2$ with more than one point, and let $x\in X$. Can $X\smallsetminus\{x\}$ be totally disconnected?
>
>
>
Note that the [Knaster-Kuratowski fan](http://en.wikipedia.org/wiki/Knaster%E2%80%93Kuratowski_fan) shows that, in the absence of the compactness hypothesis, the answer can be 'yes'.
To give credit where it's due, this question was inspired by one that I was asked by Barry Simon.
| https://mathoverflow.net/users/1463 | Can a connected planar compactum minus a point be totally disconnected? | Being planar has nothing to do with the problem. Suppose a totally disconnects $X$ and choose $b$ different from $a$. By passing to a sub continuum, assume that no proper sub continuum contains both $a$ and $b$. Take non empty disjoint open sets $U$ and $V$ whose union is $ X\sim a$. WLOG $b$ is in $U$, and observe that $U\cup \{a\}$ is closed and connected.
| 25 | https://mathoverflow.net/users/2554 | 16630 | 11,144 |
https://mathoverflow.net/questions/16611 | 17 | Take a smooth manifold $M^n$ with a smooth foliation $F$. Consider the sheaf $\cal F$ of $C^{\infty}$ functions on $M^n$, locally constant along the foliation $F$. What is known about Chech cohomology of such a sheaf?
I am pretty sure that such a question was studied (and maybe even has a complete answer), but I don't know a reference.
A more specific question is: what happen when $F$ is 1-dimensional, given by integral trajectories of a non-vanishing vector field? Or even more specifically, suppose $H^1(M^n)=0$ and we consider a Killing vector field $v$ on $M^n$ (i.e. $v$ is preserving a metric). Is it true the the sheaf of functions $\cal F$ locally constant along trajectories of $v$ is acyclic? (we need $H^1(M^n)=0$, otherwise $S^1$ will be an obvious counterexample).
**An example of a foliation.** Consider the unit sphere $S^3$ in $\mathbb C^2$ and conisder the action of $\mathbb R$ via diagonal matrixes : $(z,w)\to (e^{ita}z, e^{itb}w)$ with $\frac{a}{b}$ irrational.
| https://mathoverflow.net/users/943 | Cohomology of a sheaf of functions locally constant along a foliation | Nikita Markarian just explained to me (if there is a mistake below, it is mine), that the last and more specific question about acyclicity has 100% negative answer. Namely, we can consider the case $M^3=S^3$ ($H^1(S^3)=0$) and the foliation is given by the fibers of the Hopf fibration $S^3\to S^2$. In this chase the sheaf of functions locally constant on the fibers has a two-term resolution (by soft sheaves). The first term is given by all functions on $S^3$ and the second by $1$-forms on $S^3$, restricted to fibers. The differential is just the differential along the fibers. In this case it is clear, that the first cohomology is huge, it is parameterised by all functions on the base $S^2$.
So this condition $H^1(M^n)=0$ does not help at all.
It is a good exercise to apply the same reasoning to the other foliation on $S^3$, described in the question.
| 4 | https://mathoverflow.net/users/943 | 16631 | 11,145 |
https://mathoverflow.net/questions/16632 | 23 | An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ **connect sum** $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an *oriented* manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes.
I believe one relation is
>
> $p\_i (\mathrm{mod} 2) \equiv w\_{2i}^2$ (pg. 181 Milnor-Stasheff)
>
>
>
So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w\_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! **This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies!**
This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in [Ilya Grigoriev's response to Solbap's question](https://mathoverflow.net/questions/13813/construction-of-the-stiefel-whitney-and-chern-classes) when he says
>
> On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes?
>
>
>
Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension.
Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.
| https://mathoverflow.net/users/1622 | Stiefel-Whitney Classes over Integers? | *I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically.*
If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f\_0$ and $f\_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)
We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out
[in this recent answer](https://mathoverflow.net/questions/56932/what-characteristic-class-information-comes-from-the-2-torsion-of-hbsonz),
all the torsion is actually 2-torsion.
It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W\_{i+1}=\beta\_2(w\_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w\_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W\_3=0$.
[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p\_1$ is not a square.]
| 23 | https://mathoverflow.net/users/2356 | 16639 | 11,152 |
https://mathoverflow.net/questions/16468 | 29 | Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?
Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}\_m$. The affine Grassmannian case also provides positive evidence.
Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. Regarding putting the classical topology of X((t))(ℂ), one should not be scared of the ind-scheminess. ℂ((t)) has a natural structure of a topological ring, and hence we topologise X(ℂ((t))) in the usual manner, taking the subspace topology using a closed embedding into affine n-space for some n.
[paragraph redacted]
| https://mathoverflow.net/users/425 | Topologists loops versus algebraists loops | Here's an example constructed using moonface's idea without leaving the smooth realm: Take an affine curve $X$ whose smooth projective model $\overline{X}$ has genus $g > 0$. Define $S^1\_a = \mathrm{Spec}(\mathbf{C}((t)))$, and $D^2\_a = \mathrm{Spec}(\mathbf{C}[[t]])$.
Claim: The map $X((t))(\mathbf{C}) \to LX$ is not a homotopy equivalence. In fact, it is not even surjective on $\pi\_0$.
Proof: The (split) fibration $\Omega(X) \to LX \to X$ shows that $\pi\_0(LX) = \pi\_0(\Omega(X)) = \pi\_1(X)$. So it suffices to show that the natural map $X(S\_a^1) \to \pi\_1(X)$ is not surjective. As $\pi\_1(X) \twoheadrightarrow \pi\_1(\overline{X})$, it even suffices to show that not every element in $\pi\_1(\overline{X})$ is realised by a map $f:S^1\_a \to X$. Given such an $f$, the composite map $S^1\_a \to X \to \overline{X}$ factors as a map $S^1\_a \to D^2\_a \to \overline{X}$ by the valuative criterion. In particular, the induced map on fundamental groups is trivial as $D^2\_a$ is simply connected. As $g > 0$, we are done.
[ It seems that $\mathrm{Spec}(\mathbf{C}((t)))$ has a Hodge structure of Tate type and, consequently, cannot detect loops except those of weight $0$, i.e., those that come from removing divisors. Does anyone know if Hodge theory makes sense for such big objects? ]
| 11 | https://mathoverflow.net/users/986 | 16650 | 11,160 |
https://mathoverflow.net/questions/16648 | 5 | Suppose $X\subset \mathbb{P}^n$ is a smooth hypersurface defined over $\mathbb{Q}$. For a "generic" prime $p$, what can be said about the set of hyperplanes $H$ in $\mathbb{P}^n(\mathbb{F}\_p)$ for which $H \cap X$ is smooth over $\mathbb{F}\_p$? For $p$ fixed and $X$ varying, by contrast, the situation can be arbitrarily bad: in fact, every hyperplane section of
${\sum\_{i=1}^{n+1}X\_i X\_{i+n+1}^p=0} \subset \mathbb{P}^{2n+1}$ over $\mathbb{F}\_p$ is singular.
| https://mathoverflow.net/users/1464 | Smoothness of hyperplane sections | Spread out $X$ over some $R=\mathbf{Z}[1/n]$ to a hypersurface $\mathcal{X} \subseteq \mathbf{P}^n\_R$ that is smooth and projective over $R$. The standard proof of the Bertini smoothness theorem (as given in Hartshorne, *Algebraic geometry*, for instance) works over $R$: there is a Zariski dense open subscheme $U$ of the dual projective space $\mathbf{P}^n\_R$ such that for $p \nmid n$, the hyperplanes $H$ in $\mathbf{P}^n\_{\mathbf{F}\_p}$ such that $H \cap \mathcal{X}\_p$ is smooth are exactly those corresponding to $\mathbf{F}\_p$-points of $U$. The complement of $U$ has at most $O(p^{n-1})$ points over $\mathbf{F}\_p$ as $p \to \infty$, but $\#\mathbf{P}^n(\mathbf{F}\_p)= p^n+p^{n-1}+\cdots+1$, so when $p$ is large enough, most hyperplanes over $\mathbf{F}\_p$ will intersect the fiber $\mathcal{X}\_p$ in something smooth.
| 9 | https://mathoverflow.net/users/2757 | 16652 | 11,161 |
https://mathoverflow.net/questions/16562 | 10 | Given the group SU(N) of NxN unitary matrices, does there exist a subgroup S
with a manifold dimension larger than the SU(N-1) manifold dimension and
smaller than the SU(N) one? S should not necessarily have SU(N-1) as subgroup.
| https://mathoverflow.net/users/4274 | subgroup of SU(N) with maximal manifold dimension | Your question is equivalent to the question about maximal dimension of a proper Lie subalgebra
of $su(N)$. Clearly such subalgebra is reductive since it has a positive definite invariant form.
Thus you are looking for a reductive Lie algebra of maximal possible dimension strictly smaller
than $N^2-1$ with
faithful representation $V$ of dimension $N$. We claim that this dimension is $(N-1)^2$ for $N>4$.
Here is a sketch of argument:
1) representation $V$ is irreducible: if it splits into direct sum of subrepresentations of dimensions $a$ and $b$, then dimension of our Lie algebra is less than $a^2+b^2\le (N-1)^2$ (one needs to
consider the special case when $a=1$ separately).
2) representation $V$ is not a tensor product of irreducible representations: if it is a tensor
product of representations of dimensions $a$ and $b$, then our Lie algebra has dimension less than $a^2+b^2\le (N/2)^2+(N/2)^2=N^2/2\le (N-1)^2$.
3) it follows from 2) above that our Lie algebra is in fact simple Lie algebra. The smallest possible dimension of an irreducible representation of any simple Lie algebra is well known
(this is always one of the fundamental representations). Now quick search gives you the result
(and the counterexample pointed out by Somnath).
| 4 | https://mathoverflow.net/users/4158 | 16654 | 11,163 |
https://mathoverflow.net/questions/16651 | 20 | I was reading a series of article from the Corvallis volume. There are couple of questions which came to my mind:
1. Why do we need to consider representation of Weil-Deligne group? That is what is an example of irreducible admissible representation of $ Gl(n,F)$ which does not correspond to a representation of $W\_F$ of dimension $n$ ? An example for $ n=2 $ will be of great help.
2. In the setting of global Langlands conjecture, why extension of $W\_F$ by $G\_a$ or products of $W'\_{F\_v}$ does not work?
Thank you.
| https://mathoverflow.net/users/4291 | Weil group, Weil-Deligne group scheme and conjectural Langlands group | Regarding (1), from the point of view of Galois representations, the point is that continuous Weil group representations on a complex vector space, by their nature,
have finite image on inertia.
On the other hand, while a continuous $\ell$-adic Galois representation of $G\_{\mathbb Q\_p}$ (with $\ell \neq p$ of course) must have finite image on wild inertia, it can have infinite image on tame inertia. The formalism
of Weil--Deligne representations extracts out this possibly infinite image, and encodes it as a nilpotent operator (something that is algebraic, and doesn't refer to the $\ell$-adic topology,
and hence has a chance to be independent of $\ell$).
As for (2): Representations of the Weil group are essentially the same thing as representations
of $G\_{\mathbb Q}$ which, when restricted to some open subgroup, become abelian. Thus
(as one example) if $E$ is an elliptic curve over $\mathbb Q$ that is not CM, its $\ell$-adic Tate module cannot be explained by a representation of the Weil group (or any simple modification thereof). Thus neither can the weight 2 modular form to which it corresponds.
In summary: the difference between the global and local situations is that an $\ell$-adic representation of $G\_{\mathbb Q\_p}$ (or of $G\_E$ for any $p$-adic local field) becomes, after
a finite base-change to kill off the action of wild inertia, a tamely ramified representation,
which can then be described by two matrices, the image of a lift of Frobenius and the image of a generator of tame inertia, satisfying a simple commutation relation.
On the other hand, global Galois representations arising from $\ell$-adic cohomology of varieties over number fields are much more profoundly non-abelian.
Added: Let me also address the question about a product of $W\_{F\_v}'$. Again, it is simplest to think in terms of Galois representations (which roughly correspond to motives,
which, one hopes, roughly correspond to automorphic forms).
So one can reinterpret the question as asking: is giving a representation of $G\_F$ (for a number field $F$) the same as giving representations of each $G\_{F\_v}$ (as $v$ ranges over the places of $F$). Certainly, by Cebotarev, the restriction of the global representation
to the local Galois groups will determine it; but it will *overdetermine* it; so giving a collection of local representations, it is unlikely that they will combine into a global one. ($G\_F$ is very far from being the free product of the $G\_{F\_v}$, as Cebotarev shows.)
To say something on the automorphic side, imagine writing down a random degree 2 Euler product. You can match this with a formal $q$-expansion, which will be a Hecke eigenform, by taking Mellin transforms,
and with a representation of $GL\_2(\mathbb A\_F)$, by writing down a corresponding tensor product of unramified representations of the various $G\_{F\_v}$. But what chance is there
that this object is an automorphic representation? What chance is there that your random formal Hecke eigenform is actually a modular form? What chance is there that your random Euler product is actually an automorphic $L$-function? Basically none.
You have left out some vital global glue, the same glue which describes the interrelations of all the $G\_{F\_v}$ inside $G\_F$. Teasing out the nature of this glue is at the heart of proving the conjectured relationship between automorphic forms and motives; its mysterious nature is what makes the theories of automorphic forms, and of Galois representations, so challenging.
| 28 | https://mathoverflow.net/users/2874 | 16659 | 11,166 |
https://mathoverflow.net/questions/16657 | 10 | Suppose M is an arbitrary smooth manifold and D is its bundle of 1-densities.
On the category of finite-dimensional vector bundles over M and linear differential operators between them
there is a contravariant endofunctor that sends a vector bundle E to E\*⊗D
and a differential operator f: E→F to the adjoint differential operator f\*: F\*⊗D→E\*⊗D.
Applying this endofunctor to the standard de Rham (cochain) complex 0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0
with morphisms being de Rham differentials we obtain another (chain) complex 0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0
with morphisms being codifferentials.
Here Λ^k(M) denotes the bundle of k-polyvectors (kth exterior power of the tangent bundle).
What is the exact relationship between the homology of this complex and the usual singular (co)homology of M?
Using Hodge duality we can rewrite this complex as
0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0,
where W is the orientation bundle.
It looks like the answer should be some standard fact from the 1950s,
therefore any references will be appreciated.
| https://mathoverflow.net/users/402 | De Rham homology | When you dualize the bundle of differential forms and multiply it with the line bundle of top forms, you get differential forms again, and not polyvector fields.
| 12 | https://mathoverflow.net/users/2106 | 16663 | 11,169 |
https://mathoverflow.net/questions/16526 | 9 | Problem statement
-----------------
Let $G=(V,E)$ be an undirected graph whose vertices are either black or white. A *local complementation* of $G$ with respect to a black vertex $v$ consists in:
1. complementing the subgraph induced by $v$ and its neighbours,
2. flipping the colour of each neighbour of $v$ (i.e. black vertices become white and conversely), and finally
3. removing $v$ from $V$.
The goal is to delete the whole graph using only local complementations.
Questions
---------
Given an ordering $\mathcal O$ of the vertices of $V$, can we characterise cases in which $\mathcal O$ allows us (or not) to delete $G$?
Comments
--------
A lot of work on local complementations (or "vertex eliminations" in some papers) concerns itself with algorithmic issues, especially with finding orderings that will work. Note that this differs from my question, since here you don't get to choose an ordering.
Of course, verifying whether an ordering works is easy: keep complementing until you're done or stuck. Finding necessary or sufficient nontrivial structural conditions on $G$ or $\mathcal O$ seems harder. Does this problem ring any bell?
Example
-------
Two different orderings for the same graph; the first one does not work:
<http://homepages.ulb.ac.be/~alabarre/local-complementation-1.png>
The second one does:
<http://homepages.ulb.ac.be/~alabarre/local-complementation-2.png>
References
----------
[Hannenhalli and Pevzner](http://dx.doi.org/10.1145/300515.300516), starting from page 14, and [Hartman and Verbin](http://dx.doi.org/10.1007/11880561_23). All other authors (e.g. [Sabidussi](http://dx.doi.org/10.1016/0012-365X(87)90240-8)) consider variants like using directed graphs, or non-coloured vertices, or complementations which do not modify edges adjacent to $v$. Other authors whose papers I'm currently looking into are Donald J. Rose, Robert Endre Tarjan and François Genest.
| https://mathoverflow.net/users/3356 | Local complementation in undirected graphs | **Warning.** I just realized that my reduction is not good as if a node has two outputs, their will be new edges created between them, so we would need a more complicated gadget. I suspect this to be doable, but as meanwhile the question turned out to have a different motivation (see Parity below), I did not give it much thought.
No, it is not possible to characterize them. Of course it is possible to find some conditions for special cases but do not hope to find any simple/local dependent rule as this problem is [P-complete](http://en.wikipedia.org/wiki/P-complete). Below I give a sketch of how to reduce CVP (Circuit Value Problem) to it.
Before I start one simple observation that you did not mention. For every graph with ordered vertices, there is exactly one coloring that deletes the whole graph. This suggests that in the reduction true variables should correspond to certain edges in our graph rather than to colors.
There will be one main gadget that we use, which I tried to depict here with my humbling artistic laziness.
[alt text http://dcg.epfl.ch/webdav/site/dcg/users/184485/public/gadget.JPG](http://dcg.epfl.ch/webdav/site/dcg/users/184485/public/gadget.JPG)
Now I'll try to sketch the reduction. To every node of the circuit we will have a corresponding pair of vertices, $v\_1$ and $v\_2$, which are next to each other in the order. They will be both black and unconnected before we start deleting them if the node is evaluated as TRUE while $v\_1$ will be black, $v\_2$ white and they will be connected before we start deleting them if the node is evaluated as FALSE. Thus the above gadget is a negation gate.
Note that the color of 3 changes iff our first variable is true. Thus basically we can modify a later part of the graph arbitrarily by using several copies of this gadget. If we combine two gadgets, we can also simulate an AND gate. For this it is enough to show how to have an edge iff two variables are true. Let A, B, C have this order, at the beginning A being black, A being the "4" in two gadgets, the role of "3" played by B and C in the two gadgets. Then after deleting both variables and A, the color of B and C will be unchanged and there will be an edge between iff both variables were true.
I know that this is a lengthy answer and uses complexity instead of some nice combinatorial observations, but I think it helps to show why there can be no simple characterization. Let me know if you have any questions/would like a more detailed exposition.
**Parity.** If a graph satisfies that every black vertex has an odd degree and every white vertex has an even degree, then it is not possible to delete it. The proof is by induction on the number of vertices. If there is only one vertex, it must be white, thus we are struck. Otherwise, whenever you delete a black vertex, the parity of the degree of all of its neighbors will change as well as their color, thus we are done by applying the induction hypothesis to this graph.
| 4 | https://mathoverflow.net/users/955 | 16667 | 11,171 |
https://mathoverflow.net/questions/16640 | 6 | Given a finite CW complex X, there is a filtration of the topological K-theory of X given by setting $K\_n(X) = \ker \left(K(X) \to K(X^{(n-1)})\right)$, where $X^{(n-1)}$ is the (n-1)-skeleton of X. (The choice of indexing here is from Atiyah-Hirzebruch.)
My question is:
How does this filtration interact with the external product $K(X)\times K(Y)\to K(X \times Y)$? I believe the answer should be that $K\_n (X) \cdot K\_m (Y) \subset K\_{n+m} (X\times Y)$.
Just to be clear, and to set notation, this external product is the one induced by sending a pair of vector bundles $V\to X$ and $W\to Y$ to the external tensor product, which I'll write $V\widetilde{\otimes} W = \pi\_1^\* V \otimes \pi\_2^\* W \to X\times Y$.
Of course, if $V\in K\_n (X)$ and $W \in K\_m (Y)$, then $V\widetilde{\otimes} W$ restricts to zero in both $K(X^{(n-1)} \times Y)$ and $K(X \times Y^{(m-1)})$, and $(X\times Y)^{(n+m-1)}$ is contained in the union of these two subsets. Is there some way to deduce from this information that the class $V\widetilde{\otimes} W$ is actually trivial in $K((X\times Y)^{(n+m-1)})$?
Here's the reason I'm asking (which is really a second question, I guess).
In Characters and Cohomology Theories, Atiyah states (without comment) that for the internal product $K(X)\times K(X)\to K(X)$, one has $K\_n (X) \cdot K\_m (X) \subset K\_{n+m} (X)$. In Atiyah-Hirzebruch, they state this formula and say that it "admits a straighforward proof."
I thought I remembered that the straighforward proof was the following:
1. Show that the external product satisfies $K\_n (X) \cdot K\_m (Y) \subset K\_{n+m} (X\times Y)$
2. Observe that if $f:X\to X\times X$ is a cellular approximation to the diagonal $X\to X\times X$, then $f(X^{(n+m-1)}) \subset (X\times X)^{(n+m-1)}$. So for any $V, W\in K(X)$, we have $V\otimes W = f^\*(V\widetilde{\otimes} W)$, and if $V\in K\_n (X)$ and $W\in K\_m (X)$, it then follows from 1. that $V\otimes W\in K\_{n+m} (X)$.
Am I barking up the wrong tree here?
Presumably these questions will turn out to have an easy answer, but I've been thinking about them for a while now and haven't gotten any further. Any suggestions or references would be great! I haven't found any sources other than the two mentioned above that talk about the relation between skeleta and products, and neither of these sources mentions case of external products.
| https://mathoverflow.net/users/4042 | Products and the skeletal filtration in K-theory | Hi Dan, welcome to Math Overflow.
The group you denote $K\_m(X)$ is the image of the relative K-group $K(X,X^{(m-1)})$, which for nice spaces (e.g. finite CW-complexes) consists of equivalence classes of formal differences $V - W$ of vector bundles equipped with an isomorphism $V|\_{X^{(m-1)}} \cong W|\_{X^{(m-1)}}$. The product on K-groups lifts to an exterior pairing
$$
K(X,A) \otimes K(Y,B) \to K(X \times Y,A \times Y \cup X \times B).
$$
In particular, if $X$ and $Y$ are CW then using the standard CW structure on the product we have $$(X \times Y)^{(n+m-1)} \subset (X^{(n-1)} \times Y) \cup (X \times Y^{(m-1)}).$$
This gives us an exterior pairing
$$
K(X,X^{(n-1)}) \otimes K(Y,Y^{(m-1)}) \to K(X \times Y,(X \times Y)^{(n+m-1)})
$$
that lifts the ordinary K-theory product, and implies the result you want about the image of the group $K\_n(X) \times K\_m(Y)$. This answers your part (1), and (2) follows just as you said.
| 5 | https://mathoverflow.net/users/360 | 16677 | 11,177 |
https://mathoverflow.net/questions/16666 | 28 | How many field automorphisms does $\mathbf{C}$ have? If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph\_0}}$, I believe. And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$? Specifically, if you assume that $\mathsf{ZF}$ admits a model, does that imply that $\mathsf{ZF}$ admits a model where $\mathbf{C}$ has no wild automorphisms: $\operatorname{Aut}(\mathbf{C})=\mathbf{Z}/2\mathbf{Z}$?
I suppose if that's true, then the next logical question is to construct models of $\mathsf{ZF}$ where $\operatorname{Aut}(\mathbf{C})$ has cardinality strictly between $2$ and $2^{2^{\aleph\_0}}$--pretty disturbing if you ask me. Which finite groups can you hit?
| https://mathoverflow.net/users/271 | Does $\operatorname{Con}\sf(ZF)$ imply $\operatorname{Con}\sf(ZF + \operatorname{Aut}{\bf C = Z/\mathrm 2Z})$? | The use of inaccessible cardinals is not necessary here, the Baire property works just as well as Lebesgue measure. Shelah (*Can you take Solovay's inaccessible away*, Isr. J. Math. 48, 1984, 1-47) shows that $\mathsf{ZF}$ + $\mathsf{DC}$ + "every subset of $\mathbf{R}$ has the [Baire property](http://en.wikipedia.org/wiki/Property_of_Baire)" is relatively consistent with $\mathsf{ZF}$. (This is also the paper where Shelah also shows that the inaccessible cardinal is necessary for Solovay's result.)
The connection is an old theorem of Banach and Pettis which says that any Baire measurable homomorphism between Polish groups is automatically continuous. This result is provable in $\mathsf{ZF}$ + $\mathsf{DC}$. Since $\mathbf{C}$ is a Polish group under addition, it follows that every additive endomorphism of $\mathbf{C}$ is continuous in Shelah's model. Since the continuous additive endomorphisms of $\mathbf{C}$ are precisely the $\mathbf{R}$-vector space endomorphisms, it follows that the only field automorphisms of $\mathbf{C}$ in Shelah's model are the identity and conjugation.
---
As pointed out by Pete Clark in the comments, the Artin-Schreier Theorem goes through using only the Boolean Prime Ideal Theorem ($\mathsf{PIT}$), which is significantly weaker than full $\mathsf{AC}$. This shows that $\mathsf{AC}$ is not completely necessary to show that there is a unique conjugacy class of elements of order $2$ in $\mathrm{Aut}(\mathbf{C})$ and that these correspond precisely to the finite subgroups of $\mathrm{Aut}(\mathbf{C})$.
Looking at Pete Clark's [Field Theory Notes](http://alpha.math.uga.edu/%7Epete/FieldTheory.pdf), specifically at Steps 4 and 5 of his proof of the Grand Artin-Schreier Theorem on pages 62-63, I think that it is a theorem of $\mathsf{ZF}$ that the only possible order for a nontrivial finite subgroup of $\mathrm{Aut}(\mathbf{C})$ is $2$.
| 34 | https://mathoverflow.net/users/2000 | 16683 | 11,181 |
https://mathoverflow.net/questions/16684 | 14 | This could well be too general a question, but I'd be interested in solutions to special cases too. Say you have some finite set of positive real numbers $x\_i$, when is it the case that $\sum\_i x\_i > \prod\_i x\_i$? And when are they equal?
The special case that prompted this was an argument about whether any number is equal to the sum of its prime factors.
Any references or quick proofs welcome.
| https://mathoverflow.net/users/4076 | When is the product of a set of numbers greater than the sum of them? | If you have a *set* of positive integers (that is, no duplicates are allowed) then the sum is greater than the product if and only if the set is of the form {1,x}. The sum is equal to the product only for singleton sets {x} and the set {1,2,3}.
For, examining the remaining cases:
* If the set is empty the sum is 0 and the product is 1, so sum < product
* If the set has two elements {x,y}, neither of which is 1, then $xy\ge 2\max(x,y)>x+y$.
* If the set has three elements {1,2,x}, with $x>3$, the sum is $x+3$ and the product is the larger number $2x$.
* If the set has any other three elements then its product is at least three times its max and its sum is less than that.
* If the set has {1,2,3,x} then the product is 6x and the sum is x+6, smaller for all $x\ge 4$.
* If the set has any other form with $k>3$ elements then by induction the sum of the smallest $k-1$ items is less than their product. Multiplying or adding the largest item doesn't change the inequality.
| 16 | https://mathoverflow.net/users/440 | 16687 | 11,182 |
https://mathoverflow.net/questions/16672 | 10 | Hello everyone.
My question is concerned with the following statement.
*"Having a grothendieck topology on a category C is equivalent to having a full reflective subcategory Sh(C) in the category PSh(C) of presheaves, whose reflection is left exact."*
What i need is a reference for this containing a proof. I tried google but could not find anything besides citations of this result.
| https://mathoverflow.net/users/1261 | Sheaves as full reflective subcategories | I have seen a reference for this fact, and I *think* it was in Artin's book on Grothendieck Topologies. I have no copy available to check this right now.
Before I found that reference, I wrote up a little treatment for my own benefit; I took the "full reflective subcategory" idea as the *definition* of a Grothendieck topos, then proved that all such come from Grothendieck topologies. It's in section 3.7 of <http://www.math.uiuc.edu/~rezk/homotopy-topos-sketch.pdf>
The proof goes like this. That a Grothendieck topology gives rise to a full reflective subcategory with left-exact reflection is standard. If you're given such a reflective subcategory $D\subseteq Psh(C)$, consider all the *sieves*, i.e., monomorphisms $f:S\to h\_X$ where $h\_X$ is the representable functor determined by $X\in C$. Call $f$ a *covering sieve* if $Lf$ is an isomorphism, where $L: Psh(C)\to D$ is the left adjoint. You then show (i) the collection of covering sieves is a Grothendieck topology $\tau$, and (ii) sheaves for $\tau$ are exactly those presheaves isomorphic to objects of $D$. Both (i) and (ii) require using the fact that $L$ is left exact. (ii) is equivalent to the statement: (ii') for all $f:X\to Y$ in $Psh(C)$, $Lf$ is iso if and only if $L\_\tau f$ is iso (where "$L\_\tau$" is sheafification with respect to $\tau$.) It's convenient to prove (ii') first for monomorphisms $f$, and then for epimorphisms $f$.
| 12 | https://mathoverflow.net/users/437 | 16690 | 11,184 |
https://mathoverflow.net/questions/16673 | 17 | Let $x\_1 < x\_2 < \ldots < x\_n$ and $y\_1 < y\_2 < \ldots < y\_n$ be two sequences
of $n$ real numbers. It is well known that there are polynomials that "interpolate"
in that $f(x\_i)=y\_i$ for all $i$, and the Lagrange interpolating polynomial
even warrants a solution of degree $ < n$. Now, what happens if we want the
polynomial $f$ to be nondecreasing on the interval $[x\_0,x\_n]$ ? Is there always
a solution, and is there a bound on the degree also ?
| https://mathoverflow.net/users/2389 | Finite interpolation by a nondecreasing polynomial | This problem has appeared before in literature and is now well understood, I guess. The general version is when you have no restriction on the $y\_i$'s and you ask for an interpolating polynomial that is monotone on each sub-interval $[x\_ix\_{i+1}]$. The first paper proving the existence of such a polynomial is:
W.Wolibner, "Sur un polynom d'interpolation", Colloq. Math (2) 1951, 136-137
but it is a non-constructive proof, as it uses the Weierstrass approximation theorem much like the answer given by Harald Hanche-Olsen above. Another proof for the case $0=y\_0\le \cdots \le y\_n=1$ is given in "Polynomial Approximations to Finitely Oscillating Functions" by W.J. Kammerer (Theorem 4.1) and the non-constructive aspect of his proof is the use of uniform convergence of appropriate Bernstein polynomials. In "Piecewise monotone polynomial interpolation", S.W. Young proves the same theorem and makes the final remark that the existence of such monotone interpolating polynomial is in fact *equivalent* to the Weierstrass theorem. On the other hand Rubinstein has some papers devoted to proving the existence of interpolating polynomials which are increasing in all of $\mathbb R$.
The first paper which gives bounds on the degrees is, I think,
E. Passow, L. Raymon, "The degree of piecewise monotone interpolation", which is [here](http://www.jstor.org/stable/2040274)
and an improvement is made in "Exact estimates for monotone interpolation" by G.L. Iliev.
Note that the bounds are in terms of
$$A=\max \Delta y\_i=\max (y\_{i}-y\_{i-1}) \qquad B=\min \Delta y\_i \qquad C=\min \Delta x \_i$$
And no uniform bound exists.
| 18 | https://mathoverflow.net/users/2384 | 16697 | 11,186 |
https://mathoverflow.net/questions/16686 | 8 | Did Conway pay the wager for either of the proofs to the [The Angel Problem](http://en.wikipedia.org/wiki/Angel_problem)?
I'd check in on this every now and again when it was an unsolved problem and would like to know how the story ends. Anyone know more details?
| https://mathoverflow.net/users/3623 | The Angel Problem - was the bet paid? | Actually, until very recently, Conway didn't even believe his problem had been solved. (This despite the fact that multiple solutions have been published, some years ago by now, and the solutions had even been exposited at seminars at Princeton.)
Only a few months ago did a few graduate students at Princeton convince him that the problem was solved. He was particularly excited when he heard about the "nice devil" (who never kills a square that could have been visited before).
I have checked with Conway: the bet has not yet been paid. However, it will be soon. I will update this answer if and when it has been paid.
| 15 | https://mathoverflow.net/users/1079 | 16704 | 11,193 |
https://mathoverflow.net/questions/16664 | 10 | In standard calculus it is a well known fact that left-point and mid-point [Riemann sums](http://en.wikipedia.org/wiki/Riemann_sum) do become equal in the limit. When it comes to stochastic integration this is no longer the case.
Taking the left-hand sums renders the [Ito integral](http://en.wikipedia.org/wiki/Ito_calculus) with an extra term, taking the midpoints renders the [Stratonovich integral](http://en.wikipedia.org/wiki/Stratonovich_integral) (see for example: [Higham](http://www.caam.rice.edu/~cox/stoch/dhigham.pdf), p 531).
While the Ito integral is the usual choice in applied math, the Stratonovich integral is frequently used in physics. Unlike the Itō calculus, Stratonovich integrals are defined such that e.g. the chain rule of ordinary calculus holds.
**My question**
1.) What is/are the deeper reason(s) that we have a convergence in ordinary calculus but have a non-convergence here?
2.) Are there examples in non-stochastic calculus where we also have a non-convergence of these limiting cases? If yes, how do they look like (and why)? Could you give a toy example of such a function?
| https://mathoverflow.net/users/1047 | Convergence and non-convergence of left-point and mid-point Riemann sums | The reason that in stochastic calculus the left-hand and right-hand sums give different integrals really all boils down to quadratic variations. Processes such as Brownian motion have non-zero quadratic variation.
Suppose that you are integrating a process X with respect to some other process Y, then choosing a partition 0=t0≤...≤tn=t the approximations using left and right hand sums respectively are,
$$
\int\_0^t X\ dY\approx \sum\_{k=1}^nX\_{t\_{k-1}}(Y\_{t\_k}-Y\_{t\_{k-1}})
$$
$$
\int\_0^t X\ \overleftarrow{d}Y\approx \sum\_{k=1}^nX\_{t\_{k}}(Y\_{t\_k}-Y\_{t\_{k-1}})
$$
The difference between these can be bounded as follows
$$
\sum\_{k=1}^n(X\_{t\_k}-X\_{t\_{k-1}})(Y\_{t\_k}-Y\_{t\_{k-1}}) \le\max\_k\vert X\_{t\_k}-X\_{t\_{k-1}}\vert\sum\_k\vert Y\_{t\_k}-Y\_{t\_{k-1}}\vert
$$
The final term on the right hand side converges to the variation of Y as the mesh of the partition goes to zero and, if X is continuous, the first term goes to zero. So, in standard calculus where integration is always with respect to finite variation functions, it makes no difference whether the left hand or right hand sums are used.
Alternatively, the Cauchy-Schwarz inequality can be applied to get the following bound.
$$
\sum\_{k=1}^n(X\_{t\_k}-X\_{t\_{k-1}})(Y\_{t\_k}-Y\_{t\_{k-1}}) \le\sqrt{\sum\_{k=1}^n(X\_{t\_k}-X\_{t\_{k-1}})^2\sum\_{k=1}^n(Y\_{t\_k}-Y\_{t\_{k-1}})^2}.
$$
As the mesh of the partition goes to zero, the terms inside the square root converge to the quadratic variations of X and Y respectively, denoted by [X] and [Y]. Again, in standard calculus, we use (continuous) finite variation functions, which have zero quadratic variation. However, in stochastic calculus, processes such as Brownian motion have non-zero quadratic variation. Convergence to the quadratic variation along partitions occurs in the sense of convergence in probability - it does not have to converge in the usual sense with any positive probability. A Brownian motion B has [B]t=t, so the left and right hand sums can converge to different numbers.
With the Stratonovich integration the correct thing to use is the average of the left and right hand sums, *not* the mid-point. For Ito processes, which are integrals with respect to Brownian motion and time, it makes no difference. This is because their quadratic variations are absolutely continuous. However, for general continuous semimartingales, the mid-point sums don't actually have to converge to anything. (See [Sur quelques approximations d'intégrales stochastiques](http://www-mathdoc.ujf-grenoble.fr/cgi-bin/spitem?id=1136) by Marc Yor).
So, Stratonovich integration uses the average of the left and right hand sums, and the difference between this and the Ito integral is precisely half of what you get using the right-hand sums.
As to the question of whether this difference shows up in standard (non-stochastic) calculus, the answer is, as far as I know, hardly ever. In fact, given any continuous functions then you can choose a sequence of partitions along which the quadratic variation vanishes as the mesh goes to zero (I'll leave this as an exercise!). So, given any continuous function with non-zero quadratic variation with respect to some sequence of partitions, so that the left and right hand sums converge to different numbers then, there will be other partitions along which the quadratic variation vanishes. So the integral isn't really defined at all in this case.
However, there is one case I have seen where left and right hand sums for deterministic functions converge to different numbers. For this to happen, you have to fix some sequence of partitions with mesh going to zero, and stick to using these to define the integral. Different partitions could lead to different results. Hans Follmer published a paper using this idea in 1981 ([Calcul d'Ito Sans Probabilities](http://www-mathdoc.ujf-grenoble.fr/cgi-bin/spitem?id=1509)). There aren't any natural (and useful) cases that I know of where this occurs, but you can construct some examples.
Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alternatively, you could construct a continuous function along the lines of the [Weierstrass function](http://en.wikipedia.org/wiki/Weierstrass_function) while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer.
For example, let s(t) be the 'sawtooth' function s(t) = 1-|1-2{t/2}| ({t}=fractional part of t). Then, define the following function on the unit interval,
$$
f(t) = s(t)+\sum\_{n=1}^\infty 2^{-(n+1)/2}s(2^nt).
$$
This has quadratic variation 1, calculated along [dyadic](http://en.wikipedia.org/wiki/Dyadic_rational) partitions. The following left-hand, right-hand and `Stratonovich' integrals are easily verified,
$$
\int\_0^1 f df = (f(1)^2-f(0)^2-[f]\_1)/2.
$$
$$
\int\_0^1 f \overleftarrow{d}f = (f(1)^2-f(0)^2+[f]\_1)/2.
$$
$$
\int\_0^1 f \partial f = (f(1)^2-f(0)^2)/2.
$$
| 12 | https://mathoverflow.net/users/1004 | 16706 | 11,195 |
https://mathoverflow.net/questions/16668 | 9 | I wonder how strong the power of Tannaka philosophy is, and if we accept that a tensor category is a generalized bialgebra, what difficulties we will come up against ?
Edit: Whether most tensor categories are representable, or whether for every "good enough" tensor category there exist a bialgebra with its module category isomorphic to the given category?
| https://mathoverflow.net/users/4155 | Does any tensor category correspond to a bialgebra? | I'd like to explain Bruce's answer a bit more. The fusion categories Bruce mentioned have non-integer Frobenius-Perron dimensions, so it is very easy to see that they are not categories of finite dimensional modules over a bialgebra. E.g. one of the simplest of them, the so called Yang-Lie category, has simple objects $1,X$ with $X^2=X+1$. So if $X$ were a finite dimensional representation of a bialgebra, then the dimension of $X$ would be the golden ratio, which is absurd.
This, however, can be fixed if we allow weak bialgebras and weak Hopf algebras. In fact, any fusion category is the category of modules over a finite dimensional weak Hopf algebra, see arXiv.math/0203060.
As to Akhil's example (Deligne's categories), it is also true that they cannot be realized as categories of finite dimensional representations of a bialgebra (or even a weak bialgebra), but for a different reason. Namely, if X is a finite dimensional representation of a bialgebra, then the length of the object $X^{\otimes n}$ is at most ${\rm dim}(X)^n$, where ${\rm dim}$ means the vector space dimension. But in Deligne's categories, the length of $X^{\otimes n}$ grows faster as $n\to \infty$. Actually, in another paper, Deligne shows that if in a symmetric rigid tensor category over an algebraically closed field of characteristic zero, the length of $X^{\otimes n}$ grows at most exponentially, then this is the category of representations of a proalgebraic supergroup, where some fixed central order 2 element acts by parity
(so essentially this IS the category of (co)modules over a bialgebra). This is, however, violently false in characteristic $p$, since if the root of unity $q$ is of order $p$, where $p$ is a prime, the the fusion categories for $U\_q({\mathfrak g})$ mentioned by Bruce admit good reduction to characteristic $p$, which are semisimple symmetric rigid tensor categories with finitely many simple objects and non-integer Frobenius-Perron dimensions.
A third very simple example of a tensor category not coming from a bialgebra is the category of vector spaces graded by a finite group $G$ with associator defined by a nontrivial $3$-cocycle. This category, however, is the category of representatins of a quasibialgebra (and also of a weak bialgebra, as mentioned above).
So the conclusion is as in the previous two answers: tensor categories are more general than bialgebras. More precisely, the existence of a bialgebra for a tensor category is equivalent to the existence of a fiber functor to vector spaces, which is an additional structure that does not always exist. And if it exists, it is often not unique, so you may have many different bialgebras giving rise to the same tensor category.
| 23 | https://mathoverflow.net/users/3696 | 16709 | 11,197 |
https://mathoverflow.net/questions/16681 | 2 | Is 'small enough' ellipse projected on a surface of a sphere convex? By ellipse I mean a set of points 'C' with a constant sum |AC| + |BC|, A and B are the centers. By 'small enough' I mean that the radii fits into 90 degrees (I think it is not convex once you make it large enough, though the limit is probably more like 180 degrees).
It seems to me that it is indeed convex, but is there some simple proof? The mathematics I tried to do usually ends up as f(x)=arccos(a(x)) + arccos(b(x)) and it isn't quite easy to prove that that the function has a reasonable shape when a(x) is decreasing and b(x) is increasing. Is there some easy proof I have overlooked?
By convex I mean that any shortest line connecting the points on the ellipse is 'inside' the ellipse (i.e. the distance |AX| + |XB| is smaller or equal then the distance defining the ellipse for any point X of the line).
Update: I think I eventually found a solution; triangle inequality works for these 'small enough' triangles. Geometrically the problem can be somewhat shuffled, in the end I have to prove that a triangle that is 'inside' another triangle is indeed smaller; triangle inequality combined with the way of computing a distance on a sphere will do the trick.
| https://mathoverflow.net/users/4302 | Is ellipse on a sphere convex? (proof) | Yes it is. After central projection on the plane (Klein model for sphere) you obtain usual ellipse.
Also you can show it using triangle inequality. All proofs from euclidean plane works. For example this one:
Suppose $F\_1$ and $F\_2$ foci of the ellipse. Take any two points $A$ and $B$ inside and reflect $F\_2$ with respect to the line $AB$. New point denote by $F\_2'$.
Take any point $X$ on the segment $AB$. Suppose ray $F\_1X$ intersect the segment $F\_2'A$ (the case $F\_2'B$ is the same) in the point $Y$. We have,
$$F\_1X+F\_2X=F\_1X+F\_2'X< F\_1X+XY+YF\_2'=F\_1Y+YF\_2'< F\_1A+AY+YF\_2'=F\_1A+AF\_2'$$
| 3 | https://mathoverflow.net/users/2158 | 16710 | 11,198 |
https://mathoverflow.net/questions/16711 | 8 | Let $[X, Y]\_0$ denote base point preserving homotopy classes of maps $X\rightarrow Y$. A multiplication on a pointed space $Y$ is a map $\phi: Y\times Y\rightarrow Y.$ From this map, we can define a continuous map for each pointed space $X$, $\phi\_X: [X, Y]\_0\times [X, Y]\_0\rightarrow [X, Y]\_0,$
by the composition $$\phi\_X (\alpha, \beta)(x)=\phi(\alpha(x), \beta(x)).$$
If $([X, Y]\_0, \phi\_X)$ is a group for each $X$, then $(Y, \phi)$ is called a homotopy associative $H$-space.
A $coH$-space is defined from a comultiplication, namely, a map $\psi: X\rightarrow X\vee X.$ Then, for each pointed space $Y$, we can define a function $\psi^Y: [X, Y]\_0\times [X, Y]\_0\rightarrow [X, Y]\_0$ in this way:
$$\psi^Y(\alpha, \beta)=(\alpha\vee\beta)\circ\psi.$$ If $([X, Y]\_0, \psi^Y)$ is a group for each $Y$, then $(X, \psi)$ is called a homotopy associative $coH$-space.
So, as we can see, if we have a homotopy associative $coH$-space $(X, \psi)$ and a homotopy associative $H$-space $(Y, \phi)$, then we can define two group structures on the space $[X, Y]\_0$. My question is: are they "equivalent" in some sense? Obviously, whatever $\phi$ or $\psi$ is, the zero element of the group is the constant map in $[X, Y]\_0.$ However, the two group structures do depend on the choice of $\phi$ and $\psi$, which seems have little relationship with each other.
| https://mathoverflow.net/users/1537 | homotopy associative $H$-space and $coH$-space | I looked at my homotopy theory lecture notes and we had the following similar result:
$X$ H-CoGroup, $Y$ $H$-Group, then both group structures defined on [X,Y] agree.
The proof goes roughly as follows:
Call the upper products $\cdot$, resp. $\*$. Inserting the definitions of those products, one can show the following "distributivity":
$(a\cdot b)\*(c\cdot d)=(a \* c)\cdot(b \* d)$
Then one shows that both products have the same neutral element and finally
$f\*g=(f\cdot 1) \* (1\cdot g)=(f \* 1)\cdot(1 \* g)=f\cdot g$,
gives the result. That's the strategy of the proof in the case of $H$-(co-)groups.
| 8 | https://mathoverflow.net/users/3969 | 16713 | 11,199 |
https://mathoverflow.net/questions/16691 | 12 | Let E be an [ellipse](http://en.wikipedia.org/wiki/Ellipse) centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question:
>
> Can you give a geometric reason for why the length is a+b?
>
>
>
This was originally asked to me by Frank Jones a few years ago.
| https://mathoverflow.net/users/3970 | Geometrically interpreting the answer to a vector calculus question involving tangent line segments to ellipses. | There is a geometric way to show that $n$-gon circumscribed around an ellipse has minimal perimeter if it is inscribed in a confocal ellipse. From Poncelet porism (and generalization of optical property) it follows that we have continuous family of "minimal" polygons.
If we know it, then it is easy to understand that the circumscribed rhomb (from your question) and the circumscribed rectangular (with perimeter $4(a+b)$) are minimal polygons. So, side of the rhomb equals $a+b$.
| 13 | https://mathoverflow.net/users/2158 | 16718 | 11,203 |
https://mathoverflow.net/questions/16721 | 23 | Erdős, Ginzburg and Ziv prove the following:
Let $n \geq 1$ and $a\_1,\ldots, a\_{2n-1}\in \mathbb{Z}$. There exist distinct $i\_1,\ldots , i\_n$
such that
$$
a\_{i\_1} + \cdots + a\_{i\_n} \equiv 0 \pmod{n}.
$$
Is there a proof that doesn't use the Chevalley–Warning theorem (or a variant of its proof)?
| https://mathoverflow.net/users/3958 | EGZ theorem (Erdős-Ginzburg-Ziv) | The original proof used Cauchy-Davenport lemma. Several proofs are given in [this](http://www.tau.ac.il/~nogaa/PDFS/egz1.pdf) article of Alon-Dubiner (The proofs deal only with the case when $n$ is prime, but deducing the general case is straightforward from there). Note that the ideas behind most of these proofs could be interpreted as special cases of the more powerful theorem that is commonly known as "Combinatorial Nullstellensatz" (proven by N. Alon, see [here](http://www.cs.tau.ac.il/~nogaa/PDFS/null2.pdf)). The keyword for results like these is "Zero-sum Ramsey theory".
ETA: You might also find the paper by Olson, "A combinatorial problem in finite abelian groups", Journal of Number Theory (1969) Vol.1 very interesting. It proves a generalization of EGZ theorem for finite abelian p-groups (I think this was one of the first among many other generalizations).
| 13 | https://mathoverflow.net/users/2384 | 16724 | 11,207 |
https://mathoverflow.net/questions/16742 | 2 | An infinite sequence is normal if all strings of equal length occur with equal asymptotic frequency.
Formally, let $\Sigma$ be a finite alphabet of $b$ digits. Let $S$ be an infinite sequence and $\omega$ a finite sequence, both over $\Sigma$, i.e., $S\in\Sigma^\infty$ and $\omega\in\Sigma^\*$. Define $N\_S(\omega, n)$ to be the number of times the string $\omega$ appears as a substring in the first $n$ digits of the sequence $S$. The sequence $S$ is normal if for all finite strings $\omega\in\Sigma^\*$,
$$
\lim\_{n\rightarrow\infty}{N\_S(\omega,n)\over n}={1 \over b^{\left|\omega\right|}}
$$
Or you can find the definition here: <http://en.wikipedia.org/wiki/Normal_number>
The webpage in wiki above also states a property of normal sequences: "A sequence is normal if and only if every block of equal length appears with equal frequency. (A block of length $k$ is a substring of length $k$ appearing at a position in the sequence that is a multiple of k: e.g. the first length-$k$ block in $S$ is $S[1..k]$, the second length-$k$ block is $S[k+1..2k]$, etc.)"
Wiki gives me the reference. More specifically, it says this result was made explicit in the work of Bourke, Hitchcock, and Vinodchandran (2005). But I cannot figure out which theorems in that paper imply this property and how they do.
This "block characterization of normality" seems a natural property and should be easy to prove, but so far I still have difficulties in proving the "only if" direction. That is, how can I show the every block of equal length appears with equal frequency when the sequence is known to be normal? On the other hand this result seems important as it implies several other properties (see wiki). One paper I read about the connections between normal sequences and finite automata also relies on this result.
So I will be grateful if someone can help.
| https://mathoverflow.net/users/4162 | How to show that an infinite sequence is normal if and only if every block of equal length appears with equal frequency? | Earlier, I gave the following sketch:
**Evenly distributed blocks implies evenly distributed sequences:**
Estimate the upper and lower densities of a sequence of length $k$ from the frequencies of blocks of length $L$ much greater than $k$. The difference between the upper and lower estimates is from the sequences which cross the ends of the blocks. As $L$ increases, the difference drops to $0$.
However, you wanted the other direction.
**Evenly distributed sequences implies evenly distributed blocks:**
For sufficiently large $L$ much greater than $k$, most sequences of length $L$ contain about the right number of copies of each subsequence of length $k$ offset by each class mod $k$, so that any translate of $S$ contains about the right number of blocks of each type. That is, we can choose $L\_\epsilon$ so that a random sequence of length $L\_\epsilon$ has probability over $1-\epsilon$ of being $\epsilon$-good, having at least $(1-\epsilon) L\_\epsilon/(k b^k)$ and at most $(1+\epsilon)L\_\epsilon/(k b^k)$ copies of each block with each offset mod $k$. (A weakly dependent law of large numbers suffices.)
That the sequences of length $L\_\epsilon$ are evenly distributed means that for each sequence $B$ of length $k$, the lower density is at least $(1-\epsilon)^2\times$average of pairs of an $\epsilon$-good sequence of length $L\_\epsilon$ and a block it contains with pattern $B$. Thus, the lower density of blocks of pattern $B$ is at least $(1-\epsilon)^2\times$average. Since this is true for all $\epsilon\gt 0$, the density is $1/b^k$.
| 1 | https://mathoverflow.net/users/2954 | 16744 | 11,221 |
https://mathoverflow.net/questions/16745 | 9 | In Milne, Étale cohomology, it is proved that $\mathrm{Br}(X) = H^2(X,\mathbf{G}\_m)$ for $X$ regular of dimension $\leq 2$. Are there in the meantime further results for $X$ regular?
| https://mathoverflow.net/users/nan | When is Br(X) = H^2(X,G_m)? | When $X$ is quasi-projective over an affine scheme (or more generally if $X$ has an ample [**EDIT:** invertible] sheaf), then its Brauer group is isomorphic to the **torsion part** of $H^2(X, {\mathbb G}\_m)$. This is an unpublished result of Gabber, and J. de Jong [wrote down a different proof](http://www.math.columbia.edu/~dejong/papers/2-gabber.pdf).
| 14 | https://mathoverflow.net/users/3485 | 16749 | 11,224 |
https://mathoverflow.net/questions/16752 | 9 | I have twice heard it attributed to Dana Scott that he said something to the effect that the consistency of the lambda-calculus was an accident.
Does anyone have a reasonable-sounding source for this? I find it hard to believe that Scott would talk about the Church-Rosser theorem in this way; I guess that this a mangling of something else he said, or some context is hidden.
| https://mathoverflow.net/users/3154 | Scott on the consistency of the lambda calculus | You could ask [him](http://www.cs.cmu.edu/~scott/) directly, but the story he told me was that he was working on domain theory because he wanted to give a denotational semantics of typed lambda calculus, or more generally typed programming languages. (He had been telling people they should design typed languages, rather than untyped ones, and so he wanted to show how a mathematical theory of typed programming languages would work.) But it turned out that his theory of domains also provides a model of the untyped lambda calculus. In this sense it was an accident.
I also asked him once why he thought it was important to give a model of the untyped lambda calculus when it had been known by the Church-Rosser theorem the calculus was consistent. I cannot reporoduce the exact answer, but in effect he said that it was important to understand what models of a theory looked like, not just that it was consistent. I think this reveals a certain "semantic" view of mathematics.
| 17 | https://mathoverflow.net/users/1176 | 16754 | 11,226 |
https://mathoverflow.net/questions/16760 | 2 | Hello,
I am looking for a reference (if it exists) that makes the link between cohomology of sheaves for sites and Galois cohomology :
quickly said, I would like to see Galois cohomology (at least in the commutative case) as the cohomology of a sheaf over the étale site of extensions of k.
By the way, what is a reference for cohomology of sites ?
Thanks
| https://mathoverflow.net/users/2330 | Cohomology of sheaves for sites and Galois cohomology | The two references from my comment above, now with links!
* Milne, James S. Étale cohomology. Princeton Mathematical Series, 33. Princeton University Press, Princeton, N.J., 1980. xiii+323 pp. [MR0559531](http://www.ams.org/mathscinet-getitem?mr=MR0559531) You can get another set of notes on étale cohomology from his [web page](http://www.jmilne.org/math/CourseNotes/lec.html): «in comparison with my book, the emphasis is on heuristic arguments rather than formal proofs and on varieties rather than schemes».
* Tamme, Günter. Introduction to étale cohomology. Translated from the German by Manfred Kolster. Universitext. Springer-Verlag, Berlin, 1994. x+186 pp. [MR1317816](http://www.ams.org/mathscinet-getitem?mr=MR1317816)
| 6 | https://mathoverflow.net/users/1409 | 16761 | 11,232 |
https://mathoverflow.net/questions/16743 | 12 | **Notation.** Let $p$ be a prime number, $K$ a finite extension of
$\mathbb{Q}\_p$ and $E|K$ an elliptic curve which has *good reduction.*
The discriminant $d\_{E|K}$ of $E|K$ is an element of the
multiplicative group $\mathfrak{o}^\times/\mathfrak{o}^{\times12}$,
where $\mathfrak{o}$ is the ring of integers of $K$.
**Question.** Does the order of $d\_{E|K}$ as an element of
$\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ show up somewhere ? Is it related to some other invariant of $E|K$ ?
**Background.** $E$ can be defined over $K$ by a minimal cubic
$f=y^2+a\_1xy+a\_3y-x^3-a\_2x^2-a\_4x-a\_6=0,\ \ (a\_i\in\mathfrak{o})$;
its discriminant $d\_f$ is in $\mathfrak{o}^\times$ (because $E$ has
good reduction). If we replace $f$ by another minimal cubic $g$
defining $E$, then $d\_f$ gets replaced by $d\_g=u^{12}d\_f$ for some
$u\in\mathfrak{o}^\times$. So the class $d\_{E|K}$ of $d\_f$ in
$\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ depends only on $E|K$,
not on the choice of a minimal cubic defining $E$. It can be shown
that every class in the finite group
$\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ is the discriminant of
some good-reduction elliptic curve.
**Addendum.** As Qing Liu remarks, one may ask, given an elliptic curve $E$ over a finite extension $k|\mathbb{F}\_p$, whether the order of the discriminant $d\_{E|k}\in k^\times/k^{\times12}$ shows up somewhere. When $p\neq2,3$, the two questions are equivalent.
| https://mathoverflow.net/users/2821 | The order of the discriminant of a good-reduction elliptic curve | I will give an intrinsic characterization below for what this unit class modulo 12th powers means, which may be viewed as an answer of sorts: it expresses the obstruction to extracting the 12th root of a certain canonical isomorphism between 12th powers of line bundles (and so one could shift the answer to: where does the need to extract such a 12th root come up?)
For any ring $R$, the group $R^{\times}/(R^{\times})^{12}$ naturally maps into the degree-1 fppf cohomology of $\mu\_{12}$ over ${\rm{Spec}}(R)$, so it classifies isomorphism classes of certain $\mu\_{12}$-torsors for the fppf topology over this base. (Namely, those $\mu\_{12}$-torsors whose pushout to a $\mathbf{G} \_m$-torsor is trivial.)
It is the same to use the etale topology when $12$ is a unit in $R$ (as then $\mu\_{12}$ is etale over $R$). So the issue is to associate to any elliptic curve $f:E \rightarrow {\rm{Spec}}(R)$ over a ring a canonical $\mu\_{12}$-torsor (with the extra property that its pushout to a $\mathbf{G} \_m$-torsor is trivial).
In the theory of Weierstrass planar models for elliptic curves $E$ over a base scheme $S$ (this includes the condition "good reduction") there is an obstruction to the existence of such a model, namely whether or not the line bundle $\omega\_{E/S} = f\_{\ast}(\Omega^1\_{E/S})$ on $S$ admits a global trivialization. The necessity of such triviality is due to the fact that a Weierstrass model produces a trivialization (the ${\rm{dx}}/(2y+\dots)$ thing), and the sufficiency is explained in Chapter 2 of Katz-Mazur (where they use a choice of trivializing section to distinguish some formal parameters along the origin and pass from this to a Weierstrass model via the relationship between global 1-forms, the relative cotangent space ${\rm{Cot}}\_e(E)$ along the identity section $e$, and $\mathcal{O}(ne)/\mathcal{O}((n-1)e) \simeq {\rm{Cot}}\_e(E)^{ \otimes -n}$ for $n = 2, 3$).
That being said, regardless of whether or not the line bundle $\omega\_{E/S}$ is trivial (though it always is when $S$ is local), the line bundle $\omega\_{E/S}^{\otimes 12}$ is canonically trivial (in a manner that is compatible with base change and functorial in isomorphisms of elliptic curves): that is the meaning of the classical fact that the product of $\Delta$ with the 12th power of the section ${\rm{d}}x/(2y+\dots)$ is invariant under choice of Weierstrass model. This also underlies Mumford's calculation (recently revisited by Fulton-Olsson) of the Picard group of the moduli stack of elliptic curves as $\mathbf{Z}/12\mathbf{Z}$, which one could regard as providing a distinguished role to that trivialization. Working with the compactified moduli stack over $\mathbf{Z}$ (so allowing generalized elliptic curves with geometrically irreducible but possibly non-smooth fibers, and hence working with relative dualizing sheaf to generalize $\omega\_{E/S}$ when allowing non-smooth fibers), the trivialization (which we could generously attribute to Ramanujan) is unique up to a sign, which in turn is nailed down by the Tate curve over $\mathbf{Z}[[q]]$ and the isomorphism of its formal group with $\widehat{\mathbf{G}}\_m$. So this trivialization is really a canonical thing, independent of any theory of Weierstrass models.
Letting $\theta\_{E/S}$ denote this intrinsic trivializing section of $\omega \_{E/S}^{\otimes 12}$ as just defined, it is natural to ask if $\theta \_{E/S}$ is the 12th power of a trivializing section of $\omega \_{E/S}$. Note that this is a nontrivial condition even when $\omega \_{E/S}$ is trivial (such as when $S$ is local). Anyway, the functor of such 12th roots is a $\mu \_{12}$-torsor over $S$ for the fppf topology (and etale if 12 is a unit on the base), and as such it corresponds to the inverse of the class of $\Delta$ in the question (for which the base was local). So that is an answer of sorts: it describes the obstruction to extracting a 12th root of the canonical trivialization of $\omega^{\otimes 12}$ obtained by pullback from the trivialization over the moduli space of elliptic curves (up to an issue of signs in the exponent). Now does one ever care to extract such a 12th root? That's another matter...
| 16 | https://mathoverflow.net/users/3927 | 16763 | 11,233 |
https://mathoverflow.net/questions/16764 | 9 | Hi everyone, I got a problem when proving lemmas for some combinatorial problems,
and it is a question about integers.
Let
$\sum\_{k=1}^m a\_k^t = \sum\_{k=1}^n b\_k^t$
be an equation,
where $m, n, t, a\_i, b\_i$ are positive integers, and
$a\_i \neq a\_j$ for all $i, j$,
$b\_i \neq b\_j$ for all $i, j$,
$a\_i \neq b\_j$ for all $i, j$.
>
> Does the equality have no solutions?
>
>
>
For $n \neq m$, it is easy to find solutions for $t=2$ by Pythagorean theorem,
and even for $n = m$, we have solutions like
$1^2 + 4^2 + 6^2 + 7^2 = 2^2 + 3^2 + 5^2 + 8^2$.
For $t > 2$, similar equalities hold:
$1^2 + 4^2 + 6^2 + 7^2 + 10^2 + 11^2 + 13^2 + 16^2 = 2^2 + 3^2 + 5^2 + 8^2 + 9^2 + 12^2 + 14^2 + 15^2$
and
$1^3 + 4^3 + 6^3 + 7^3 + 10^3 + 11^3 + 13^3 + 16^3 = 2^3 + 3^3 + 5^3 + 8^3 + 9^3 + 12^3 + 14^3 + 15^3$,
and we can extend this trick to all $t > 2$.
>
> The question is, if we introduce one more restriction, that is,
> $|a\_i - a\_j| \geq 2$ and $|b\_i - b\_j| \geq 2$ for all $i, j$,
> is it still possible to find solutions for the equation?
>
>
>
For $t = 2$ we can combine two Pythagorean triples, say,
$5^2 + 12^2 + 25^2 = 7^2 + 13^2 + 24^2$,
but how about the cases for $t > 2$ and $n = m$?
| https://mathoverflow.net/users/4248 | Equality of the sum of powers | An even harder problem than $t>2$ and $n=m$ is the Prouhet–Tarry–Escott problem. Now I leave it to you and google to find lots of examples ;-)
<http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_problem>
| 9 | https://mathoverflow.net/users/1384 | 16765 | 11,234 |
https://mathoverflow.net/questions/16751 | 20 | In Mac Lane, there is a definition of an arrow between adjunctions
called a map of adjunctions. In detail, if a functor $F:X\to A$ is left
adjoint to $G:A\to X$ and similarly $F':X'\to A'$ is left adjoint to
$G':A'\to X'$, then a map from the first adjunction to the second is a
pair of functors $K:A\to A'$ and $L:X\to X'$ such that $KF=F'L$,
$LG=G'K$, and $L\eta=\eta'L$, where $\eta$
and $\eta'$ are the units of the first and second adjunction. (The
last condition makes sense because of the first two conditions; also,
there are equivalent conditions in terms of the co-units, or in terms
of the natural bijections of hom-sets).
As far as I can see, after the definition, maps of adjunctions do not
appear anywhere in Mac Lane. Googling, I found this definition also
in the [unapologetic mathematician](http://unapologetic.wordpress.com/2007/07/30/transformations-of-adjoints/),
again with the motivation of being an arrow between adjunctions.
But what is the motivation for defining arrows between adjunctions
in the first place? I find it hard to believe that the only
motivation to define such arrows is, well, to define such arrows...
So my question is: What is the motivation for defining a map of
adjunctions? Where are such maps used?
Besides the unapologetic mathematician, the only places on the web
where I found the term ''map of adjunctions'' were sporadic papers,
from which I was not able to get an answer to my question (perhaps
''map of adjunctions'' is non-standard terminology and I should have
searched with a different name?).
I came to think about this when reading [Emerton's first answer
to a question about completions of metric spaces](https://mathoverflow.net/questions/11622/what-is-the-right-universal-property-of-the-completion-of-a-metric-space).
In that question, $X$ is metric spaces with isometric embeddings, $A$
is complete metric spaces with isometric embeddings, $X'$ is metric
spaces with uniformly continuous maps, $A'$ is complete metric
spaces with uniformly continuous maps, and $G$ and $G'$ are the
inclusions. Now, if I understand the implications of Emerton's answer
correctly, then it
is possible to choose left adjoints $F$ and $F'$ to $G$ and $G'$ such
that the (non-full) inclusions $A\to A'$ and $X\to X'$ form a map of
adjunctions. This made me think whether the fact that we have a map
of adjunctions has any added value. Then I realized that I do not
even know what was the motivation for those maps in the first place.
[EDIT: Corrected a typo pointed out by Theo Johnson-Freyd (thanks!)]
| https://mathoverflow.net/users/2734 | What is the motivation for maps of adjunctions? | One of the applications of adjoint functors is to compose them to get a monad (or comonad, depending on the order in which you compose them). A map of adjoint functors gives rise to a map of monads. So one might ask: what are maps of monads good for? Many algebraic categories (such as abelian groups, rings, modules) can be described as categories of algebras over a monad, others (for example in Arakelov geometry) are most easily described in such a way. A map of monads then gives functors between the categories of algebras over these objects.
Here is a concrete example from topology: Let $E$ be a connective generalized multiplicative homology theory, and let $H = H(-;\pi\_0E)$ be ordinary homology with coefficients in $\pi\_0E$. There exists a map $E \to H$ inducing an isomorphism on $\pi\_0$. For a spectrum $X$, the functor $\underline{E}\colon X \mapsto E \wedge X$ gives rise to a monad, and similarly for $H$, thus we get a morphism of monads $\underline{E} \to \underline{H}$. The completion $X\hat{{}\_E}$ of a spectrum $X$ at $E$ is defined to be the totalization of the cosimplicial spaces obtained by iteratively applying $\underline{E}$ to $X$. The monad map gives a natural map $X\hat{{}\_E} \to X\hat{{}\_H}$ which turns out to be an equivalence for connective $X$.
| 13 | https://mathoverflow.net/users/4183 | 16772 | 11,239 |
https://mathoverflow.net/questions/16773 | 0 | In a computer graphing library, a rectangular region of the Cartesian plane may be defined by {x, y, w, h} (where w,h are width and height).
Intersection (lets say '^') is defined as the overlapping region of two rectangles (and also is a rectangle).
Union of r1 and r2 could be defined as the smallest (ie smallest w, smallest h) rectangle 'u' such that
u ^ r1 = r1 and
u ^ r2 = r2
etc.
My question(s):
* does negative w, h make 'sense',
* if so , what sense does it make?
ie, and in particular, should the result of intersection of non-overlapping rectangles be a rectangle with negative width and/or height?
* can this be used to build a Group over the set of rectangles with the operation of union and/or intersection?
* what is the 'area' of a rectangle with width < 0 and height < 0. I would like it to be negative, but typically it is width x height, thus positive.
Sorry if this sounds more computerish than mathish, but I was once a math major, and non of the other computerish people I might ask know anything about group theory...
More thoughts:
I want to decide if rectangles with negative dimensions makes much sense. I could leave this up to the users of the rectangles, but it is up to me to define what an operation like intersection does.
'neg' rectangles could just mean that the x,y is at the opposite corner than expected. ie they could be 'facing' the wrong way, and by moving x,y to the opposite corner, w,h can be made positive, and the rectangles are 'normalized'.
But when I consider intersection, imagine 2 rectangles moving along the plane, at first intersecting, but moving such that they overlap less and less - they eventually get to the point where they only touch at the corners - and the intersection is then a rectangle with w = h = 0.
Now, If we continue to move the rectangles further apart, you could either say that the result of intersection is still an empty rectangle, or I think it might make more sense to say the result is a 'negative' rectangle (and w,h depicts how far away they are from overlapping).
Given this concept, can it be somehow continued/followed/expanded in a logical/consistent way, and if so, what do we end up with?
Mathematically it would nice if it was a group, and if making it a group required a few other common operations (like union maybe) to be defined in a certain consistent way, I would like to explore that.
I just don't have any colleges nearby that think mathematically...
| https://mathoverflow.net/users/4322 | Intersection/Union of Rectangles as a Group (or Monoid or...?) | I imagine the only useful way of interpreting a rectangle with negative w and/or h should just be as a rectangle with positive width and height, only starting at a different point (it's lower left vertex, it seems?).
With either union or intersection as the operation, you're going to run into a couple of problems if you want the structure to be a group. The only possible choice for the identity under intersection is the entire plane (which, if you are parametrizing the rectangles in the plane using (x,y,w,h), would require you to allow x=y= $-\infty$, w,h=$\infty$), and the only possible choice for the identity under union is an "empty" rectangle (i.e., w=h=0, though you'll also have the problem that there is one of these at each x and y). Of course, given any rectangle which is not the entire plane, you can't take an intersection of it with something and get the entire plane, and similarly, given any rectangle which is non-empty, you can't take the union of it with something and get an empty rectangle - that is, there won't be inverses. However, as you guessed in the title, you still do get a [commutative monoid](http://en.wikipedia.org/wiki/Monoid).
You can make a group (and even ring) structure, similar to what you are describing, out of the collection of all subsets of the plane (or any set), as described [here](http://en.wikipedia.org/wiki/Power_set#Properties), but you'd have to give up your restriction to only rectangles if you want to use this one, because the operation used is the [symmetric difference](http://en.wikipedia.org/wiki/Symmetric_difference).
EDIT: This is to summarize the things you seem to be looking for:
* The set of rectangles in the plane, under the operation of intersection, makes a commutative monoid if you allow for the "infinite rectangle" consisting of the entire plane.
* Unfortunately, under union (as you defined it for rectangles - least rectangle containing both), there is in fact no identity, because given any two distinct "empty" rectangles, let's say the ones at (a,b) and (c,d), their "union" would be the rectangle with opposite corners at (a,b) and (c,d) and so is non-empty, so neither of them can be an identity. Thus, the set of rectangles in the plane, under the operation of "union", is only a commutative semigroup.
* The set of rectangles is not closed under taking symmetric difference or the usual union.
That describes the situation with the main set-theoretic operations, though you may be able to construct another operation on rectangles using some combination of addition and multiplication on the parameters (x,y,w,h) which makes a stronger structure.
| 1 | https://mathoverflow.net/users/1916 | 16776 | 11,240 |
https://mathoverflow.net/questions/16780 | 4 | Given two relatively prime integers a and b, is there an easy characterization for when a^2+b^2 is square free?
Edit: The above question proved to be too general. The problem I had in my mind is as follows: given the two sequences $a\_n$ and $b\_n$ defined by $a\_0=b\_0=1$, $a\_{n+1}=a\_nb\_n,\ b\_{n+1}=a\_n^2+b\_n^2$, are the $b\_n$'s square free and coprime?
| https://mathoverflow.net/users/nan | Square free sum of two squares | No. To the best of my knowledge, it is not even known whether Fermat numbers are squarefree.
| 6 | https://mathoverflow.net/users/3503 | 16781 | 11,241 |
https://mathoverflow.net/questions/16783 | 5 | Suppose I have a compact 3-dimensional submanifold N of S X (0,1) which has one boundary component, where S is a closed surface. Must N be a handlebody?
| https://mathoverflow.net/users/4325 | Are compact submanifolds of "S X (0,1)" with one boundary component handlebodies, where S is a closed surface? | No. N could be homeomorphic to the exterior of any knot in the 3-sphere (i.e. a cube with a knotted hole).
| 10 | https://mathoverflow.net/users/1335 | 16784 | 11,243 |
https://mathoverflow.net/questions/16762 | 3 | Given a permutation $f: \{0,1\}^n \rightarrow \{0,1\}^n$ as $n$ polynomials over $GF(2)$ how to get formulas for the inverse permutation $f^{-1}$?
I am interested in the answer to the previous question, although I would really like to know an answer to a more specific question. Let's consider a restricted permutation $g: \{0,1\}^{n-1} \rightarrow \{0,1\}^{n-1}$ that is obtained from $f(x\_1, \ldots, x\_n)$ if we fix any of its arguments to some constant (for example, $x\_1 = 0$). How does $deg(g^{-1})$ depend on $deg(f^{-1})$ (here $deg(f)$ is a maximum over degrees of polynomials, corresponding to $f$)? My hypothesis is that $deg(g^{-1}) \ge deg(f^{-1}) - 1$ for at least one of the two values we can assign to $x\_1$.
| https://mathoverflow.net/users/2641 | Inverse for a permutation over GF(2) | I very much doubt there is a formula for the inverse of a permutation. As far as degrees are concerned, I expect most permutations and their inverses to have degree $n$. A caveat here: the degree is not well-defined for functions in $GF(2)^n$ as e.g. $x=x^2$, but it is if you require all monomials to be squarefree. Then all functions can be represented by polynomials of degree at most $n$ and most of them have degree exactly $n$.
Your second question does not make sense. If $g(x\_1,\ldots,x\_{n-1})=f(x\_1,\ldots,x\_{n-1},0)$ there is no guarantee that $g(GF(2)^{n-1})$ is contained in the subspace $x\_n=0$ of $GF(2)^n$ (which is where $f$ takes its values).
Edit: I just noticed that, for $n > 1$ a permutation has degree at most $n-1$, so I amend my guess to say that most permutations and their inverses have degree $n-1$.
The degree bound follows from the fact that a coordinate function of a permutation takes both values $0,1$ the same number of times and from the fact that the coefficient of $x\_1 \cdots x\_n$ of such a coordinate function is the sum of its values at all points of the domain.
Edit 2: (in reply to the comment below)
To compute the inverse, interpolation might be reasonable. Another way is to use the above observation on the coefficient of $x\_1 \cdots x\_n$. So, for instance, the coefficient of $x\_2 \cdots x\_n$ in a polynomial $p$ is the sum of the values of $x\_1p$ which is also the sum of the values of $p$ on the points with $x\_1=1$ and so on. I don't see a way to use the fact that we're dealing with permutations to improve on these methods.
As for your second question, if you assume $f$ preserves both $x\_n=0$ and $x\_n=1$ then I think the last coordinate of $f$ is $x\_n$ and the same is true for the inverse. Then the highest degree occurs elsewhere and the inequality you want on degrees is obvious.
| 5 | https://mathoverflow.net/users/2290 | 16785 | 11,244 |
https://mathoverflow.net/questions/16792 | 15 | While playing around with [this question](https://mathoverflow.net/questions/16780/square-free-sum-of-two-squares) (when is the sum of two squares squarefree?), from some experimental computations (and bolstered by the fact that the density of squarefree positive integers is known to exist), I came up with the following conjecture: the asymptotic density of squarefree numbers in the sequence $(k+1^2, k+2^2, k+3^2, \ldots)$, for fixed k, exists and depends on k.
To give an example of what I mean, consider numbers of the form $1 + n^2$. 895 of the numbers $1+1^2, 1+2^2, \ldots, 1+1000^2$ are squarefree; 897 of the next thousand are; 895 of the third thousand; 896 of the fourth thousand. 891 of the numbers $1+1000001^2, \ldots, 1+1001000^2$ are squarefree, as are 895 of the numbers $1+2000001^2, \ldots, 1+2001000^2$. So there seems to be some constant $C\_1$, probably between 0.89 and 0.90, such that $1+k^2$ has ``probability'' $C\_1$ of being squarefree. The same thing seems to happen if we replace 1 with other integers, although with other constants.
Are such constants known to exist? If they are, how can they be computed?
| https://mathoverflow.net/users/143 | What's the probability that k + n^2 is squarefree, for fixed k? | More generally, suppose that $f(x) \in \mathbf{Z}[x]$ has no repeated factors. For each prime $p$, let $c\_p$ be the number of integers $x \in \{0,1,\ldots,p^2-1\}$ satisfying $f(x) \equiv 0 \pmod{p^2}$. Heuristically, the probability that a random integer $x$ is such that $f(x)$ is not divisible by $p^2$ equals $1-c\_p p^{-2}$, and these conditions should be independent by the Chinese remainder theorem, so one would conjecture that the fraction of integers $x$ in $\{1,2,\ldots,N\}$ such that $f(x)$ is squarefree should tend to $\prod\_p (1-c\_p p^{-2})$, where the product is taken over all primes $p$. For large $p$, we have $c\_p \le \deg f$, so this product converges.
This guess has been proved for $\deg f \le 3$ (the case $\deg f=3$ is a nontrivial result of C. Hooley). In particular, this answers your question for $f(x)=x^2+k$. There is no $f$ of degree $4$ or greater for which the density is known to exist (except in cases when the density is $0$ because some $c\_p$ equals $p^2$). On the other hand, A. Granville proved that the $abc$ conjecture implies that the density exists and equals the predicted value for $f$ of any degree. For further references and a generalization to multivariable polynomials, see the papers cited in the references below.
C. Hooley, On the power free values of polynomials, *Mathematika* **14** (1967), 21-26.
A. Granville, $ABC$ allows us to count squarefrees, *Internat. Math. Res. Notices* **1998**, no. 19, 991-1009.
B. Poonen, [Squarefree values of multivariable polynomials](http://www-math.mit.edu/~poonen/papers/sqfree.pdf), *Duke Math. J.* **118** (2003), no. 2, 353-373.
| 25 | https://mathoverflow.net/users/2757 | 16798 | 11,250 |
https://mathoverflow.net/questions/16786 | 8 | Let $X$ be a projective variety. Assume there is an algebraic map $f: X \rightarrow X$ that is a bijection. I am thinking of $X$ as a variety, not a scheme, so by a bijection I mean a bijection on closed points. Most likely I am working over the complex numbers, so if you like I mean a bijection on complex points. Can you conclude that $f$ has an algebraic inverse?
I think this is not immediately obvious, since it is not true that any algebraic bijection between two projective varieties is an isomorphism. For instance, there is an algebraic bijection from ${\Bbb P}^1$ to a cuspidal cubic in ${\Bbb P}^2$ given by $[x,y] \rightarrow [x^3, x^2y, y^3]$. So if this is true one must use the fact that the map is from $X$ to itself.
I am interested in cases where $X$ is both singular and reducible (although is of pure dimension, if that helps), so a complete answer would cover any such case. Alternatively, if it is not true that such a map has an algebraic inverse, I would like an explicit counter example.
| https://mathoverflow.net/users/1799 | Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? | A reference for a proof that a bijective endomorphism of an algebraic variety over a field of characteristic zero is an automorphism (which is a corrected version of Mariano's statement in the comment): see e.g. S. Kaliman, Proc. Amer. Math. Soc. 133 (2005), 975-977, Lemma 1.
| 7 | https://mathoverflow.net/users/3696 | 16803 | 11,254 |
https://mathoverflow.net/questions/16207 | 6 | I've been looking at an example in the non-commutative geometry literature and I'm having trouble figuring out what the classical motivation is. I'll just describe the classical case here: Recall that $\mathbb{CP}^2 = SU(\mathbb{C},3)/U(\mathbb{C},2)$. Recall also that since $T\_\mathbb{C}^\*(\mathbb{CP}^2)$ can be viewed as a vector bundle associated to the $SU(2)$-bundle $SU(\mathbb{C},3) \to \mathbb{CP}^2$, we can view $\Omega\_\mathbb{C}^1(\mathbb{CP}^2)$ as a subset of $\mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2)$. Now in the example, the action of each of the Dolbeault operators $\partial,\overline{\partial}$ is given in terms of two mappings, $\partial\_1,\partial\_2$ and $\overline{\partial\_1},\overline{\partial\_2}$, such that
$$
\partial(f) = (\partial\_1(f),\partial\_2(f)) \in \mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2),
$$
and similarly for $\overline{\partial}$. The mappings ${\partial}\_i, \overline{\partial}\_i$ are constructed using an action of the Lie algebra $\mathfrak{su}(3)$ on $\mathcal{O}(SU(3))$ constructed using the canonical pairing between Lie algebras and coordinate algebras.
I have a feeling this is a complicated incarnation of a simple classical object. Does any of this ring any bells with anyone?
Please ask if you would like more details.
Edit: This question has been superseded by [this question](https://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions) and I am voting to close. I would ask others to do likewise.
| https://mathoverflow.net/users/1648 | Why can the Dolbeault Operators be Realised as Lie Algebra Actions | The Dolbeault operators are usually defined in terms of the de Rham operator and the complex structure (see e.g. Wells' book or Griffith and Harris). The example you outline generalizes to the situation $G\_{\mathbb{C}} / P = G / G\_0$, where $G$ is compact, $G\_{\mathbb{C}}$ is the complexification, $P$ is a parabolic subgroup, and $G\_0 = G \cap P$. The holomorphic tangent bundle is the homogeneous vector bundle $G\_{\mathbb{C}} \times\_P \mathfrak{g}\\_{\mathbb{C}} / \mathfrak{p}$, and the cotangent bundle is a homogeneous vector bundle in similar fashion.
In this case, the Dolbeault complex with coefficients in a homogeneous vector bundle $G\_{\mathbb{C}} \times\_P V$ translates to the Koszul complex for the relative
Lie algebra cohomology $H^\*(\mathfrak{g},\mathfrak{g}\\_0,V \otimes C^{\infty}(G))$. The Dolbeault operator $\overline{\partial}$ translates to the boundary operator for Lie algebra cohomology, which of course involves the action of $\mathfrak{g}$. I haven't worked out what
happens to $\partial$ in this situation, but most likely a similar expression in terms of the Lie algebra can be derived. The translation works by thinking about the smooth sections as elements of $C^{\infty}(G) \otimes V)^{\mathfrak{g}\\_0}$ (for holomorphic sections use $C^{\infty}(G) \otimes V)^{\mathfrak{p}}$).
| 3 | https://mathoverflow.net/users/371 | 16806 | 11,257 |
https://mathoverflow.net/questions/16790 | 3 | Let $X^n$ be an Alexandrov space, and $f: X^n\to \mathbb R^k$ a regular map, does the level set necessary be an Alexandrov space?
In my mind, the intrinsic metric on the level set is 'comparable' to the ambient metric, but is it necessary an Alexandrov space?
| https://mathoverflow.net/users/3922 | Is Level set of Regular functions in Alexandrov spaces again an Alex. space? | The answer is "no" even for regular semiconcave function $f:X\to\mathbb R$
If $f:X\to \mathbb R$ is convex then it is a long-standing open problem.
| 3 | https://mathoverflow.net/users/1441 | 16810 | 11,261 |
https://mathoverflow.net/questions/16795 | -10 | Consider a finite simple graph $G$ with $n$ vertices, presented in two different but equivalent ways:
1. as a logical formula $\Phi= \bigwedge\_{i,j\in[n]} \neg\_{ij}\ Rx\_ix\_j$ with $\neg\_{ij} = \neg$ or $ \neg\neg$
2. as an (unordered) set $\Gamma = \lbrace [n],R \subseteq [n]^2\rbrace$
In each case the complement $G'$ of $G$ is easily presented and is of course *not* isomorphic to $G$ (in the usual sense) generally:
1. $ \Phi' = \bigwedge\_ {i,j} \neg \neg\_{ij}\ R x\_i x\_j $
2. $\Gamma' = \lbrace [n],[n]^2 \setminus R\rbrace$
Let's state for the moment that the presentation as a logical formula is the more "flexible" one: we can easily omit single literals, leaving it open whether $Rx\_ix\_j$ or not. But this can be mimicked for set presentation by making it from a pair to a triple $\lbrace[n],R,\neg R \subseteq [n]^2 \setminus R\rbrace$.
Let's call a presentation *complete*, if it leaves nothing open, i.e. no omitted literal and $\neg R = [n]^2 \setminus R$, resp.
Now, let a graph be given in complete set presentation $\lbrace[n],R,\neg R = [n]^2 \setminus R\rbrace$. Since order in this set should not matter, any sensible definition of "graph isomorphism" should make any graph isomorphic to its complement.
>
> Where and how do I run into trouble when I
> assume - following this line of
> reasoning, contrary to the usual line of thinking - that every (finite) graph is
> isomorphic to its complement?
>
>
>
| https://mathoverflow.net/users/2672 | Isn't a graph to be considered isomorphic to its complement, actually? | It is a strange question, but maybe a useful answer can make it a bit better.
Certainly for many purposes a graph will look totally different from its complement. For instance, a graph and its complement have completely different spectra, diameter, perfect matchings, etc. So that side of the question is kind-of lame, I agree.
On the other hand, for some purposes a graph is much the same as its complement. One obvious case is when you are interested in the automorphism group of a graph, or in the computational problem of graph isomorphism. Then you might as well think of a graph as a bicoloring of the edges of a complete graph. It then has a natural extra automorphism given by switching the two colors. More generally a colored graph with $n$ colors is equivalent, for graph isomorphism and graph automorphism questions, to a complete graph with $n+1$ colors. This viewpoint is more useful than you might first think, because a natural partial algorithm and preparatory step in the graph isomorphism and automorphism problems is to recolor every vertex by its valence, then color every edge by the colors of its vertices, etc., until the recoloring process stabilizes. Many graphs can be completely identified this way in practice. Recognizing the equivalence between a graph and its complement makes it easier to understand what these algorithms are really doing.
Even some specific graphs, such as the Higman-Sims graph, are mainly used for their automorphisms and similar purposes, and it might be better to think of them as colorings of a complete graph.
A much deeper example is the [perfect graph theorem](http://en.wikipedia.org/wiki/Perfect_graph) of Lovasz. The theorem is that a graph is perfect if only if its complement is perfect. For perfect graphs, taking the graph complement is closely related to the dual or polar polytope of a convex polytope.
| 14 | https://mathoverflow.net/users/1450 | 16816 | 11,263 |
https://mathoverflow.net/questions/16799 | 3 | I know that the Heisenberg group { x,y | [x,[x,y]]=[y,[x,y]]=1 } is free nilpotent; what
about the higher dimensional ones? Do the higer dimensional Heisenberg groups have nice presentations? By higher dimensional Heisenberg groups I mean nxn upper triangular matrices with integral entries. Thanks.
| https://mathoverflow.net/users/3804 | Are higher dimensional Heisenberg groups free nilpotent? | Let $U\_n$ be the group of upper triangular integer matrices of size $n$ by $n$ with ones on the diagonal. Then $U\_n$ is generated by the elements $x\_i$, $i=1,...,n-1$, with Serre relations
$$
[x\_i,[x\_i,x\_{i+1}]]=[x\_{i+1},[x\_i,x\_{i+1}]]=1,
$$
and $[x\_i,x\_j]=1$ if $|i-j|\ge 2$ (EDIT: one needs additional relations, see below). Indeed, the group $G\_n$ generated in this way maps surjectively to $U\_n$ by $x\_i\mapsto 1+E\_{i,i+1}$, and one can check that this map is injective (EDIT: with additional relations) by writing every element of $G\_n$ as an ordered product of powers of $x\_{ij}:=[x\_i[x\_{i+1}...x\_j]]$, $i\le j$. The corresponding groups over $\Bbb Z/m\Bbb Z$ are then obtained by adding the relations $x\_{ij}^m=1$.
EDIT: As Jack kindly pointed out, I erroneously ignored the treacherous 2-torsion.
According to Theorem 1 of the paper
D. Biss, S. Dasgupta, "A presentation for the unipotent groups over rings with $1$",
J. of Algebra, v. 237, pp.691-707, 2001,
the above statements are correct over $\Bbb Z/m\Bbb Z$ when $m$ is an odd integer.
In general (i.e. for any integer $m$, including $m=2$ and $m=0$),
it suffices to add the relation
$$
[[x\_i,x\_{i+1}],[x\_{i+1},x\_{i+2}]]=1
$$
for $i=1,...,n-3$.
Moreover, over $\Bbb Z/m\Bbb Z$ with $m\ne 0$ it is enough to impose the relations $x\_i^m=1$ (the relations $x\_{ij}^m=1$ will then follow).
| 3 | https://mathoverflow.net/users/3696 | 16820 | 11,265 |
https://mathoverflow.net/questions/16846 | 6 | Sorry if the question is too vague or if the examples I look for are too boringly well-known: my knowledge of analytic number theory is rather primitive......
So, here it goes: suppose that you want to prove that the set $\Sigma$ of primes satisfying a certain condition $C$ is infinite. Then you may attempt to compute the density
$$
\delta(C)=\lim\_{x\to\infty}\frac{|\text{$p\leq x$ such that $C(p)$ holds}|}{|p\leq x|}.
$$
If $\delta$ turns out to be positive, you're done. But it could as well be that $\delta=0$ and yet $\Sigma$ be infinite.
My questions are: (1) what are the main known examples of this occurrence? (2) in these examples, if any, the proofs of the infiniteness of $\Sigma$ did use ad hoc case-by-case "tricks" or there are somewhat standard techniques than can be employed with the situation? (3) is there a standard reference?
| https://mathoverflow.net/users/3602 | Infinite sets of primes of density 0 | If you can prove any reasonable lower bound for the set of primes which are at most $x$ then it's trivial to find infinite sets of primes with density 0. For example using completely elementary methods one can check that there's always a prime between $n$ and $2n$ (Bertrand's postulate), and hence the number of primes between 1 and $x$ is at least $log\_2(x)$ for $x\geq2$ an integer. So now just build an infinite set $C$ of primes by letting the $n$th element be the smallest prime greater than $2^{2^n}$ (or any function that grows much faster than $2^n$). Does this answer your question or did I misunderstand it?
---
Edit: it appears that the questioner doesn't want arbitrary constructions, but explicit examples of infinite sets of density 0. In this case I would say that there is no "standard technique" (known to me, at least), other than the obvious one of "take a finite set of primes with the property, and construct another one". This is, for example, the technique used by Elkies to prove that there are infinitely many supersingular primes for an elliptic curve over the rationals.
I think that in general if you want to prove that a set of primes is infinite, often you try and compute the rate of growth, or come up with heuristics estimating what its growth should be under suitable "independence hypotheses". I guess that would be another technique. Having positive density is a super-strong condition on a set of primes. Heuristic arguments based on Sato-Tate, for example, tell you that the set of supersingular primes for a non-CM elliptic curve over Q is probably growing something like $O(x^{1/2}/log(x))$. The truth of that statement establishes both that the set is infinite and has density zero all in one stroke. Elkies didn't prove this though, he just took the more naive approach above.
| 5 | https://mathoverflow.net/users/1384 | 16851 | 11,282 |
https://mathoverflow.net/questions/16848 | 23 | Each orientable 3-manifold can be obtained by doing surgery along a framed link in the 3-sphere. Kirby's theorem says that two framed links give homeomorphic manifolds if and only if they are obtained from one another by a sequence of isotopies and Kirby moves.
The original proof by R. Kirby (Inv Math 45, 35-56) uses Morse theory on 5-manifolds and is quite involved. There are two simpler proofs that use Wajnryb's presentations of mapping class groups. One is due to N. Lu (Transactions AMS 331, 143-156) and the other to S. Matveev and M. Polyak (Comm Math Phys 160, 537-556). I would like to ask what other proofs of Kirby's theorem are known. In particular, is there a proof that uses only Morse theory/handle decompositions of 3-manifolds?
| https://mathoverflow.net/users/2349 | Proofs of Kirby's theorem | There is Bob Craggs' 1974 proof, which was never published. It relies on Wall's result, that any two 2-handle cobordisms between S^3 and itself are stably homeomorphic if the associated bilinear forms have the same signature and type. It's very much what you are after. I have a hard-copy in my office. I do not fully follow all aspects of the argument and therefore cannot yet vouch for it. I uploaded a scan of his preprint [HERE](http://www.sumamathematica.com/SR_2nd.pdf) (thanks Ryan for help combining PDF files!).
I'm not positive which proof (Cerf theory or MCG) is really "simpler", because both proofs rely on a lot of background "black boxes". For Kirby's proof, later simplified by Fenn and Rourke, and then later by Justin Roberts in *[by Kirby calculus in manifolds with boundary](http://arxiv.org/abs/math/9812086)*, the only black box is Cerf's theorem. Surely the proof of Cerf's theorem could be tremendously simplified, and after this is eventually done by somebody, my money would be on the Cerf theory proof to be slicker. The MCG proof uses presentations of the mapping class group, which use simple connectedness of the Hatcher-Thurston complex (itself not so easy to prove), and results on buildings (a result of Brown) in order to construct presentations of the group from its action on a simply connected simplicial complex. This is actually a lot of machinery, if you think about it. Again, you can simplify the proof by using Gervais's presentation from *[A finite presentation of the mapping class group of an oriented surface](https://arxiv.org/abs/math/9811162)*, proven directly by Silvia Benvenuti (*[Finite presentations for the mapping class group via the ordered complex of curves](https://eudml.org/doc/121753)*) using an ordered complex of curves, or by Susumu Hirose *[Action of the mapping class group on a complex of curves and a presentation for the mapping class group of a surface](https://arxiv.org/abs/math/0008185)* using a complex of non-separating curves.
When all is said and done, I am personally not satisfied with any of the proofs out there. The Cerf theory proof, while being conceptual, takes you into deep and hard analytic terrain- while there is nothing at all conceptual about the MCG proof, despite it being "easy" (it's definitely much easier than Cerf theory, at least for me). The heart of the proof is to check that it happens to be possible to realize "relations" in your favourite finite presentation of the MCG by Kirby moves on links which generate the Dehn twists. The presentation itself is almost incidental, and the relations don't represent Kirby moves in any obvious way (the proof only goes one way- you could not prove a finite presentation of the MCG from Kirby's theorem).
| 23 | https://mathoverflow.net/users/2051 | 16861 | 11,289 |
https://mathoverflow.net/questions/16132 | 8 | All rings are assumed to be commutative and unital, with all homomorphisms unital as well.
On last week's homework, there was a mistake in one of the questions:
>
>
> >
> > **(2.5)** Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps $R\to S\_{\mathfrak{p}}$ are formally étale for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally étale.
> >
> >
> >
>
>
>
According to our professor, the exercise should have stated additionally that $S$ was finitely presented over $R$ (which allows us to prove that $S$ is in fact étale over $R$ rather than just formally étale). This is not too hard to do and is left as an exercise. (You can also find it in EGA).
However, I'm interested in seeing either a counterexample or a proof for the stronger claim in the grey box.
| https://mathoverflow.net/users/1353 | Formally étale at all primes does not imply formally étale? | **EDIT:** Don't bother reading my partial solution. Brian Conrad pointed out that an easier way to do what I did is to use the equivalent definition of formally unramified in terms of Kähler differentials. And later on, fpqc posted below a *complete* solution passed on by Mel Hochster, who got it from Luc Illusie, who got it from ???.
**OLD ANSWER:**
Here is a half-answer. I'll prove half the conclusion, but on the plus side I'll use only half the hypothesis! Namely, I will prove that if $R \to S\_{\mathfrak{p}}$ is formally unramified for all primes $\mathfrak{p} \subset S$, then $R \to S$ is formally unramified.
Let $A$ be an $R$-algebra, and let $I \subseteq A$ be a nilpotent ideal. Given $R$-algebra homomorphisms $f,g \colon S \to A$ that become equal when composed with $A \to A/I$, we must prove that $f=g$. Fix $\mathfrak{p}\subset S$. Then the localizations $A\_{\mathfrak{p}} := S\_{\mathfrak{p}} \otimes\_{S,f} A$ and $S\_{\mathfrak{p}} \otimes\_{S,g} A$ of $A$ (defined viewing $A$ as an $S$-algebra in the two different ways) are naturally isomorphic, since adjoining the inverse of an $a \in A$ to $A$ automatically makes $a+\epsilon$ invertible for any nilpotent $\epsilon$ (use the geometric series). Now $f$ and $g$ induce $R$-algebra homomorphisms $f\_{\mathfrak{p}},g\_{\mathfrak{p}} \colon S\_{\mathfrak{p}} \to A\_{\mathfrak{p}}$ that become equal when we compose with $A\_{\mathfrak{p}} \to A\_{\mathfrak{p}}/I A\_{\mathfrak{p}}$. Since $R \to S\_{\mathfrak{p}}$ is formally unramified, this means that $f\_{\mathfrak{p}} = g\_{\mathfrak{p}}$. In other words, for every $s \in S$, the difference $f(s)-g(s)$ maps to zero in $A\_{\mathfrak{p}}$ for every $\mathfrak{p}$. An element in an $S$-module that becomes $0$ after localizing at each prime ideal of $S$ is $0$, so $f(s)=g(s)$ for all $s$. So $f=g$.
| 5 | https://mathoverflow.net/users/2757 | 16862 | 11,290 |
https://mathoverflow.net/questions/16858 | 34 | Given a finite group G, I'm interested to know the smallest size of a set X such that G acts faithfully on X. It's easy for abelian groups - decompose into cyclic groups of prime power order and add their sizes. And the non-abelian group of order pq (p, q primes, q = 1 mod p) embeds in the symmetric group of degree q as shown here: www.jstor.org/stable/2306479.
How much is known about this problem in general?
| https://mathoverflow.net/users/4336 | Smallest permutation representation of a finite group? | It is difficult to find this number for arbitrary finite groups, but many families have been solved. A somewhat early paper that has motivated a lot of work in this area is:
Johnson, D. L. "Minimal permutation representations of finite groups."
Amer. J. Math. 93 (1971), 857-866. [MR 316540](http://www.ams.org/mathscinet-getitem?mr=316540) [DOI: 10.2307/2373739](http://dx.doi.org/10.2307/2373739).
This paper classifies those groups for which the regular permutation representation is the minimal faithful permutation representation (cyclic of prime power order, K4, or generalized quaternion) and some results on nilpotent groups (improved in later papers).
The minimal permutation degrees of the finite simple groups are known from:
Cooperstein, Bruce N. "Minimal degree for a permutation representation of a classical group."
Israel J. Math. 30 (1978), no. 3, 213-235. [MR 506701](http://www.ams.org/mathscinet-getitem?mr=506701) [DOI: 10.1007/BF02761072](http://dx.doi.org/10.1007/BF02761072).
This paper only finds the degrees. A fuller description of the permutation representations are given in:
Grechkoseeva, M. A. "On minimal permutation representations of classical simple groups." Siberian Math. J. 44 (2003), no. 3, 443-462 [MR 1984704](http://www.ams.org/mathscinet-getitem?mr=1984704) [DOI: 10.1023/A:1023860730624](http://dx.doi.org/10.1023/A:1023860730624).
There are a great deal of topics associated with minimal permutation degrees. I'll just briefly sketch them below, let me know if any interest you and I can give citations or longer descriptions:
The minimal degree of a subgroup is never larger than the minimal degree of the parent group, but the minimal degree of a quotient group may be much, much larger than that of the original. This poses problems in computational group theory, since quotient groups may be difficult to represent. Some quotients are easy to represent, and this has had a significant impact on CGT in the last 10 years or so.
Finding minimal permutation representations of covering groups can be difficult, and here I think the results are much less complete. Basically what you want are large subgroups not containing normal subgroups. In a covering group Z(G) is contained in really quite a few of the "good choices". This is because it is contained in Φ(G), the Frattini subgroup, the intersection of the maximal subgroups. One has to give up using maximal subgroups (at least if Z(G) is cyclic of prime power order), and so the minimal degree can increase dramatically.
Minimal degrees of primitive permutation groups is a topic with a different flavor (rather than specific families of groups, it is more of the interactions between group properties), but a great deal is known. Similar techniques are used to describe asymptotic behavior of minimal degrees of arbitrary families of finite groups, and quite powerful results are available there.
| 35 | https://mathoverflow.net/users/3710 | 16863 | 11,291 |
https://mathoverflow.net/questions/16818 | 25 | This may well be an open problem, I'm not sure.
In Berger's classification (refined by Simons, Alekseevsky, Bryant,...) of the holonomy representations of irreducible non-symmetric complete simply-connected riemannian manifolds, there are some cases which imply Ricci-flatness: namely, $\mathrm{SU}(n)$ (Calabi-Yau) in dimension $2n$, $\mathrm{Sp}(n)$ (hyperkähler) in dimension $4n$, $G\_2$ in dimension $7$ and $\mathrm{Spin}(7)$ in dimension $8$.
A natural question is the converse: whether Ricci-flatness implies a reduction of the holonomy. The other holonomy representations are known not to be Ricci-flat: $\mathrm{Sp}(n)\cdot \mathrm{Sp}(1)$ (quaternionic kähler) is known to be Einstein with nonzero scalar curvature, and in the case of $\mathrm{U}(n)$ in dimension $2n$ (Kähler) it is known that if a Kähler manifold is Ricci-flat then the holonomy is contained in $\mathrm{SU}(n)$, so that it is Calabi-Yau. So the remaining question is whether there exists any Ricci-flat riemannian manifolds with generic holonomy $\mathrm{SO}(n)$ in dimension $n$.
I would like to know the present status of this question and if it's still open what the experts think: do people expect examples of Ricci-flat riemannian manifolds with generic holonomy?
**Bonus question**: How about if the manifold is pseudoriemannian?
---
**Added**
Thanks to Igor's answer below, here are some further remarks.
The question needs to be refined. The riemannian analogue of the Schwarzschild metric
on $\mathbb{R}^2 \times S^2$ is an example of a complete, simply-connected noncompact Ricci-flat metric with generic holonomy. So the question is about *compact* examples.
In fact, in Berger's 2003 book *A panoramic view of Riemannian geometry* (page 645) one reads at the bottom of the page:
>
> It remains a great mystery that no Ricci flat compact manifolds are known which do not have one of these special holonomy groups.
>
>
>
| https://mathoverflow.net/users/394 | Are there Ricci-flat riemannian manifolds with generic holonomy? | I am not an expert but the question: "Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?"
is a well-known open problem. Note that Schwarzschild metric is a complete Ricci flat metric on $S^2\times\mathbb R^2$ with holonomy $SO(4)$, so the issue is to produce compact examples; I personally think there should be many. The difficulty is that it is hard to solve Einstein equation on compact manifolds. If memory serves me, Berger's book "Panorama of Riemannian geometry" discusses this matter extensively.
| 13 | https://mathoverflow.net/users/1573 | 16864 | 11,292 |
https://mathoverflow.net/questions/16845 | 7 | Currently I'm studying the article *[Moduli of Enriques surfaces and Grothendieck-Riemann-Roch](http://arxiv.org/abs/math/0701546)* by Pappas.
I am particularly interested in how he applies the GRR.
**Q1.** What is meant by a "family of Enriques surfaces"? I was guessing a flat morphism $f:Y\longrightarrow T$ of smooth projective varieties such that each fibre is an Enriques surface, but maybe this is too general?
**Q2.** In the article it says that the higher direct images $R^i f\_\ast O\_Y$ are zero ($i>0$). Is this an application of Grauert's theorem (III.12, Cor. 12.9, Hartshorne)?
**Q3.** How to see easily that $R^0f\_\ast O\_Y = O\_T$ ? I am guessing Grauert's theorem again.
| https://mathoverflow.net/users/4333 | Family of Enriques surfaces and Grothendieck-Riemann-Roch | This is a somewhat technical remark, related to Andrea's answer, which is a bit too big to fit into the comment box.
If $f: Y \rightarrow T$ has connected fibres, to conclude that
$R^0f\_\*\mathcal O\_Y = \mathcal O\_T$, one needs some assumptions beyond just that $f$ is a projective morphism of Noetherian schemes. (Consider these examples: a closed embedding will have connected fibres. To give such an example in which all fibres non-empty, consider a non-reduced $T$, and let $Y$ be the underlying reduced subscheme. Or one could take $T$ to be a cuspidal cubic curve and $Y$ to be its normalization.)
What the theorem on formal functions shows (assuming that $f$ is projective, and that $Y$ and $T$ are Noetherian, so we can apply the result as it is proved in Hartshorne) is that
for any point $P$ in $T$, the $\mathfrak m\_P$-adic completion $(R^0f\_\*\mathcal O\_Y\hat{)}\_P$
is equal to $H^0(\hat{Y}\_P,\mathcal O)$, the global sections of the structure sheaf on the
formal fibre $\hat{Y}\_P$ over $P$.
So if $f$ has connected fibres, and hence connected formal fibres, so that
$H^0(\hat{Y}\\_P,\mathcal O)$ is a local ring, we see that $(R^0f\_\*\mathcal O\_Y\hat{)}\_P$ is a finite local $\hat{\mathcal O}\\_{T,P}$-algebra. In general, one can't do better than this.
But, if $f$ is flat with geometrially connected and reduced fibres (e.g $f$ is smooth with geometrically connected fibres), then base-change for flat maps (Hartshorne III.9.3) shows
that the fibre mod $\mathfrak m\_P$ of $R^0f\_\*\mathcal O\_Y$ is equal to
$H^0(Y\_P,\mathcal O\_P)$ (the actual fibre over $P$, now, not the formal fibre),
which equals $k(P)$ (the residue field at $P$), since $Y\_P$ is projective, geometrically reduced, and geometrically connected over $k(P)$.
So, maintaining these assumptions on $f$, we see that for each point $P$ of $T$,
the stalk $(R^0f\_\*\mathcal O\_Y)\_P$
is a finite $\mathcal O\\_{T,P}$-algebra with the property that its reduction modulo
$\mathfrak m\_P$ is isomorphic to
the residue field $k(P)$ of ${\mathcal O}\\_{T,P}$. This implies (by Nakayama)
that the natural map ${\mathcal O}\\_P \rightarrow (R^0f\_\*\mathcal O\_Y)\\_P$
is surjective. This is true at every $P$, and so we see that
$\mathcal O\_T \rightarrow R^0f\_\*\mathcal O\_Y$ is surjective.
Now one can combine this with the Grauert result to conclude (since a surjection
of invertible sheaves is necessarily an isomorphism) that the natural map
$\mathcal O\_T \rightarrow R^0f\_\*\mathcal O\_Y$ is an isomorphism. (We probably don't
need to use the full force of Grauert here; for example, suppose that $T$ is connected;
a flat map is open, and a projective map is closed, so $f$ is surjective, hence faithfully
flat, and this implies that the map $\mathcal O\_T \rightarrow R^0f\_\*\mathcal O\_Y$ is
injective, I think.)
Added: See Keerthi Madapusi's answer below for a correction to the above discussion
of flat base-change.
| 8 | https://mathoverflow.net/users/2874 | 16867 | 11,295 |
https://mathoverflow.net/questions/16779 | 8 | I recently read [the original paper by Chas-Sullivan on string topology](http://arxiv.org/abs/math/9911159), in which they introduce some operations on homology of free loopspace LM, where M is a compact oriented manifold, giving it the structure of a (Gerstenhaber-)Batalin-Vilkovisky algebra. However, the arguments in this paper rely on some transversality assumptions, and I'm not sure whether these assumptions are justified. I know that the Chas-Sullivan operations have been constructed via homotopy theoretic methods by Cohen, Jones, Voronov (hopefully I'm not missing any names here), but I am wondering whether anybody has managed to construct the Chas-Sullivan operations in a way that more or less follows the original ideas (e.g. without using any homotopy theory).
| https://mathoverflow.net/users/83 | Chas-Sullivan string topology | I would like to point at the Diploma thesis of my student Lennart Meier, who has given various elementary descriptions of the Chaas Sullivan product (for example using my description of singular homology in terms of bordism groups of stratifolds, see: <http://www.hausdorff-research-institute.uni-bonn.de/files/kreck-DA.pdf>). I'm sure he will send you an electronic version of his thesis: lennart@meier-bielefeld.de.
Matthias Kreck
| 11 | https://mathoverflow.net/users/27783 | 16895 | 11,309 |
https://mathoverflow.net/questions/16899 | 4 | First some notation. Given a domain $R$ and $x,a,b \in R$, I write $x=gcd(a,b)\_R$ to mean that $x$ is *one* gcd of $a$ and $b$ in $R$.
I want to find an example of an GCD-domain $R$, a subdomain $S \subseteq R$, and two elements $a, b \in S$ such that there isn't any $x \in S$ such that $x=gcd(a,b)\_R$ and $x=gcd(a,b)\_S$. Notice that it is not enough to find one element $x \in S$ such that $x=gcd(a,b)\_R$ but $x \neq gcd(a,b)\_S$.
I can prove that this is impossible in as little as a Bezout domain, but I cannot prove that this is impossible in a mere GCD-domain. I do not know that many examples of GCD-domains which are not Bezout domains in the first place.
ETA: As suggested below, I also wanted $S$ to be a GCD-domain.
| https://mathoverflow.net/users/3065 | An example where GCD depends on the domain | (Edit: first version was about lcm rather than gcd).
Take $R=k[u,v,w]$, $a=uv$, $b=vw$. Then $gcd\_R(a,b)=v$ (times constant). Now let $S=k[a,b]$. Since $a$ and $b$ are independent, $gcd\_S(a,b)=1$ (times constant). Right?
Edit: here's an even simpler example: $R=k[u,v]$, $a=u$, $b=uv$, $S=k[a,b]$. Then $a|b$ in $R$, but $a$ and $b$ are both irreducible in $S$.
| 14 | https://mathoverflow.net/users/2653 | 16900 | 11,313 |
https://mathoverflow.net/questions/16869 | 9 | Most books I have treat primary decomposition only in the Noetherian case. Atyiah-MacDonald goes a step further and prover the uniqueness theorems of primary decomposition without the Noetherian hypothesis. But it seems to me they get a slight different result from the usual one.
**Definitions**
Recall that a prime $P$ is said to be associated to the $A$-module $M$ if there exists $m \in M$ such that $P = Ann(m)$; equivalently $A/P$ injects into $M$. I denote by $Ass(M)$ the set of associated primes. If $A$ and $M$ are Noetherian, this is always not empty.
For an ideal $I$ we let $Ass(I) = Ass(A/I)$. So a prime $P$ is associated to $I$ if and only if $P$ is of the form $(I : x)$ for some $x \in A$.
For the purposes of this question let me say that $P$ belongs to $I$ if and only if $P$ is of the form $\sqrt{(I : x)}$ for some $x \in A$. We call $Bel(I)$ the set of primes belonging to $I$.
Then the result of Atyiah-MacDonald shows that if $I$ has a (minimal) primary decomposition $I = \bigcap Q\_i$, and if we let $P\_i = \sqrt{Q\_i}$, the set of $P\_i$ which appear is exactly $Bel(I)$. The usual formulation gives instead that for $A$ Noetherian this set is $Ass(I)$.
**The problem**
I want to understand the relationship between $Ass(I)$ and $Bel(I)$. Clearly, since prime ideals are radical, $Ass(I) \subset Bel(I)$. In general I see no reason why the opposite inclusion should be true.
Let us see how to go proving the opposite inclusion in a special case. Assume $I$ is decomposable. Then by the result of Atyiah-MacDonald it is enough to show that if we have a minimal primary decomposition $I = \bigcap Q\_i$, and if we let $P\_i = \sqrt{Q\_i}$, then $P\_i \in Ass(I)$.
Let us do this for $P\_1$ and call $R = Q\_2 \cap \cdots \cap Q\_n$. I also call $P = P\_1$, $Q = Q\_1$, so $I = Q \cap R$.
Then observe that $R/I = R/(R \cap Q) \cong (R + Q) /Q \subset A/Q$. Since $Q$ is $P$-primary, $Ass(A/Q) = P$. So $Ass(R/I) \subset \{ P \}$.
If moreover $A$ is Noetherian this set has to be non empty, so $Ass(R/I) = \{ P \}$ and a fortiori it follows that $P \in Ass(A/I)$.
I don't see how to do this without the Noetherian hypothesis, though.
**Questions**
>
> Is $Ass(I) = Bel(I)$ always, even if $A$ is not Noetherian?
>
>
> Is $Ass(I) = Bel(I)$ if we assume that $A$ is not Noetherian, but at least $I$ is decomposable?
>
>
>
| https://mathoverflow.net/users/828 | Different notions of associated prime (in the non Noetherian case) | Dear Andrea, let $A=K[X\_1,X\_2,\ldots ,X\_n,\ldots]$, the polynomial ring in countably many variables and $I$ be the ideal $I=(X\_1^2, X\_2^2,\ldots )\subset A $. Then
$$\mathcal M=(X\_1,X\_2,\ldots)\in Bel(I) \setminus Ass(I)$$
Indeed, $(I:1)=I$ has as radical $\sqrt I=\mathcal M$, hence $\mathcal M \in Bel(I)$.
But there is no polynomial $x=P(X\_1,X\_2,\ldots ,X\_N)\notin I$ such that $(I:x)=\mathcal M$ because $X\_M$ will not satisfy $X\_M.x\in I$ for $M>N$ [Of course if $x\in I$, we have $(I:x)=A\neq\mathcal M$]
| 4 | https://mathoverflow.net/users/450 | 16903 | 11,314 |
https://mathoverflow.net/questions/16917 | 10 | The question is in the title. The motivation behind the question is as follows: there are plenty of references about profinite groups and profinite completions of groups. It seems that their not exactly a wealth of references about profinite sets and profinite completions of sets.
| https://mathoverflow.net/users/4043 | What is a reference for profinite sets? | Profinite sets are just another name for compact totally disconnected topological spaces. I think this is (essentially) explained somewhere in Bourbaki's books on general topology.
| 10 | https://mathoverflow.net/users/2106 | 16924 | 11,329 |
https://mathoverflow.net/questions/16911 | 2 | Motivation:
Let $G$ be an $\ell$-group (locally profinite group). A map $G\to \mathbb{C}$ is called smooth provided that it is continuous as a map $$G\to \mathbb{C}\_{discrete}.$$This gives us the correct notion of smoothness for $\ell$-groups.
Question: Can we characterize smoothness topologically in other interesting cases, or
is this just a coincidence?
| https://mathoverflow.net/users/1353 | Smoothness as a topological property | The question is a bit awkward, as Pete suggests. First, no need to take an $\ell$-group; an $\ell$-space is the right way to start. Second, you've stated the definition of the word "smooth" in this context. Definitions can't be "correct" -- but the word "smooth" is a good choice in this context, because of some parallels between harmonic analysis on p-adic groups and harmonic analysis on real Lie groups.
I think you should put this question aside for a bit.. look at Ralf Meyer's "Smooth Group Representations on Bornological Vector Spaces" to see an answer to something like your question.
| 6 | https://mathoverflow.net/users/3545 | 16925 | 11,330 |
https://mathoverflow.net/questions/16913 | 6 | Hey,
I'm taking a course in computability theory, but I'm struggling with primitive recursion. More specifically we are often asked to prove that some arbitrary function is primitive recursive, but I really don't know how to go about doing so.
For example:
Let $\pi (x)$ be the number of primes that are $\le x$. Show that $\pi (x)$ is primitive recursive.
How do you go about proving this formally? And, in general, what are the necessary conditions to prove if something is or is not primitive recursive.
Note: this isn't homework, I've pulled the question out from a list of examples we have been given :)
| https://mathoverflow.net/users/4348 | Prove a function is primitive recursive | This is not an exact answer, but it helps to quickly determine in many cases that a given function is primitive recursive. The idea is to use a reasonable programming language in which your function can be expressed more easily than with "raw" arithmetic and primitive recursion. Of course, the programming language must be limited enough to prevent unbounded searches and other constructs that lead to general recursion. I mention here just one easy possibility.
Suppose a function $f : \mathbb{N} \to \mathbb{N}$ is computed by a simple imperative program in polynomial running time (actually, as Grigory points out in his answer, any primitive recursive bound will do). More precisely, the program is allowed to use basic arithmetic operations, boolean values, variables, arrays, loops, and conditional statements. Then the function $f$ is primitive recursive.
The reason that such a program can only define a primitive recursive function is that the entire execution of the program may be encoded by a number $N$ whose bit size is bounded by a polynomial $p(n)$, where $n$ is the input. Because verifying that a number encodes the execution of a simple imperative program is primitive recursive, we can perform a bounded search up to $2^{p(n)}$ to find the number $N$ (such a bounded search is primitive recursive) and extract the answer from it.
Let us apply this to your question. The following Python program computes the function $\pi(n)$ and uses just a couple of loops and basic arithmetic (we could replace the remainder function % with another loop). Its running time is quadratic in $n$, assuming all basic operations are constant time:
```
def pi(n):
'''Computes the number of primes which are less than or equal n.'''
p = 0 # the number of primes found
k = 2 # for each k we test whether it is prime
while k <= n:
j = 1 # possible divisors of k
d = 0 # number of divisors of k found
while j <= n:
if k % j == 0: d = d + 1
j = j + 1
if d == 2: p = p + 1
k = k + 1
return p
```
Therefore your function is primitive recursive.
| 10 | https://mathoverflow.net/users/1176 | 16927 | 11,331 |
https://mathoverflow.net/questions/16901 | 4 | I know that the number 216 + 1 is commonly used for RSA, since 0b 1 0000 0000 0000 0001 only contains two 1 bits. Many sites explain that this makes modular exponentiation faster, but I haven't come across an explanation of why it is faster.
Why is it more efficient to use a number with a lot of zeros for modular exponentiation?
| https://mathoverflow.net/users/4346 | modular exponentation for RSA, why is 2^16 + 1 commonly chosen? | There are a two minor advantages to choosing the exponent 216+1.
The first advantage, as Johannes observed, is that for fixed size exponent, exponentiation to power e using the [basic repeated squaring method](http://en.wikipedia.org/wiki/Exponentiation_by_squaring) is moderately faster when e has lots of zero bits. It is not true that exponents with more one bits are necessarily slower since there are plenty of such numbers with very short [addition chains](http://en.wikipedia.org/wiki/Addition-chain_exponentiation) (though finding such short addition chains is an NP complete problem in general). In any case, e = 3 would be a much better choice than e = 216+1 for the sole purpose of exponentiation.
The second advantage is that 216+1 is a prime number and it is not too small. A requirement of the RSA algorithm is that the exponent e must be relatively prime with φ(pq) = (p-1)(q-1). Since the large primes p and q are chosen randomly, there is always a chance that (p-1)(q-1) is not relatively prime with the (previously chosen) exponent e and the primes p,q must therefore be discarded. So small exponents e are poor choices since about every (e-1)th choice of p and q is a bad one, thus shrinking the overall key space. Choosing e to be a large prime would be best, but too large an e would make exponentiation slow. In the end, e = 216+1 is a nice compromise value.
| 10 | https://mathoverflow.net/users/2000 | 16935 | 11,336 |
https://mathoverflow.net/questions/7732 | 30 | Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (\*)$$
**Comments:**
* This is a modification of a problem of A. Nabutovsky. [Here](https://mathoverflow.net/questions/8534/diameter-of-universal-cover) is yet related question about universal covers.
* You can reformulate it for compact length metric space --- no difference.
* The answer is YES if the cover is [**regular**](http://en.wikipedia.org/wiki/Covering_space#Deck_transformation_group.2C_regular_covers) (but that is not as easy as one might think).
* The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.
* We have equality in $(\*)$ for covers of $S^1$ and for some covers of figure-eight.
| https://mathoverflow.net/users/1441 | Diameter of m-fold cover | I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.
Split $\gamma$ into $m$ arcs $a\_1,\dots,a\_n$ of equal length: $\gamma=a\_1a\_2\dots a\_m$, $L(a\_i)=L(\gamma)/m>diam(M)$. Let $b\_i$ be a shortest path in $M$ connecting the endpoints of $a\_i$. Note that $L(b\_i)\le diam(M)< L(a\_i)$. I want to replace some of the components $a\_i$ of the path $\gamma$ by their "shortcuts" $b\_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.
To switch from $a\_i$ to $b\_i$, you left-multiply $\gamma$ by a loop $l\_i:=a\_1a\_2\dots a\_{i-1}b\_i(a\_1a\_2\dots a\_i)^{-1}$. More precisely, if you replace the arcs $a\_{i\_1},a\_{i\_2},\dots,a\_{i\_k}$, where $i\_1< i\_2<\dots< i\_k$, by their shortcuts, the resulting path is homotopic to the product $l\_{i\_1}l\_{i\_2}\dots l\_{i\_k}\gamma$.
So it suffices to find a product $l\_{i\_1}l\_{i\_2}\dots l\_{i\_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi\_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.
Now consider the following $m+1$ elements of $\pi\_1(M,p)$: $s\_0=e$, $s\_1=l\_1$, $s\_2=l\_1l\_2$, $s\_3=l\_1l\_2l\_3$, ..., $s\_m=l\_1l\_2\dots l\_m$. Two of them, say $s\_i$ and $s\_j$ where $i< j$, are in the same left coset. Then $s\_i^{-1}s\_j=l\_{i+1}l\_{i+2}\dots l\_j\in H$ and we are done.
| 26 | https://mathoverflow.net/users/4354 | 16939 | 11,340 |
https://mathoverflow.net/questions/16892 | 20 | The question
============
Let $a\_1,a\_2,\dots,a\_n$ be a sequence whose entries are +1 or -1. Let t be a parameter. My question is to give an estimate for the number of such sequences so that
>
> $|a\_1+a\_2+\dots a\_k| \le t$, for every $k$, $1 \le k\le n$.
>
>
>
(In other words, the probability that a random sequence will satisfy the above relation.)
I am especially interested in this probability when t is small. Either a constant, or slowly growing (say, it behaves like (log n)^s for some real number s, or slower).
variations
----------
1) I would also like to know what is the situation if you demand that the average value of |a\_1+a\_2+\dots a\_k| is smaller than t, rather than the maximum value.
2) If there are more delicate estimates for the case that t itself is a function of k e.g. t itself grows as (log n)^s I would be very interested as well.
Motivation
==========
This question is relevant to the recent collective effort ([polymath5](http://gowers.wordpress.com/2010/03/02/edp10-a-new-and-very-promising-approach/)) regarding the Erdos Discrepancy Problem (EDP). It particular it is relevant to a [probabilistic heuristic](http://gowers.wordpress.com/2010/02/19/edp8-what-next/#comment-6278) regarding what the answer to EDP, and to several related questions, should be.
It is also relevant to certain probabilistic approaches towards construction of sequences with low discrepancy.
Expectation
===========
I would expect that the answers to the questions above are known. But they are not known to me. It is easy to be convinced, for example, that when t is bounded the number of such sequences is $c\_t^{-n}$, for $c\_t<2$ but I would like to know the dependence of c\_t on t.
| https://mathoverflow.net/users/1532 | The probability for a sequence to have small partial sums | For $t$ fixed, the count is proportional to $\lambda^n$, where $\lambda = 2 \cos \frac\pi{2t+2}$ is the principal eigenvalue of the adjacency matrix of the path with $2t+1$ vertices. The all-positive (Perron-Frobenius) eigenvector corresponding to $\lambda$ is
$$\bigg(\sin \frac{\pi}{2t+2}, \sin \frac{2\pi}{2t+2},\sin \frac{2\pi}{2t+2},\dots,sin \frac{(2t+1)\pi}{2t+2}\bigg).$$
Since $-\lambda$ is also an eigenvalue, the stable behavior of the distribution of endpoints of paths which stay in $[-t,t]$ is an oscillation between the odd entries
$$\bigg(\sin \frac{\pi}{2t+2}, 0,\sin \frac{3\pi}{2t+2},0,\dots,\sin \frac{(2t-1)\pi}{2t+2},0,\sin \frac{(2t+1)\pi}{2t+2}\bigg).$$
and even entries
$$\bigg(0,\sin \frac{2\pi}{2t+2}, 0,\sin \frac{4\pi}{2t+2},0,\cdots ,0,\sin \frac{2t\pi}{2t+2},0\bigg).$$
The exact count of paths staying in $[-t,t]$ is a sum of signed binomial coefficients.
The number of paths from $0$ to $i$ is 0 if $n \not \equiv i ~\mod 2$, and $n \choose (n\pm i)/2$ when $n \equiv i ~\mod 2$.
The number of paths which never leave $[-t,t]$ from $0$ to $i \in [-t,t]$ with $n \equiv i ~\mod 2$ is
$$ \sum\_{j\in \mathbb Z} (-1)^j {n\choose (n +i)/2 + j(t+1)}$$
by the reflection principle applied to the group of isometries of $\mathbb R$ generated by reflecting about $t+1$ and $-t-1$.
If you sum over all $i \in [-t,t]$, then when $n$ is even, you get a signed sum of binomial coefficients with $t+1$ positive signs in a row alternating with $t+1$ negative signs in a row. If $n$ is odd, then you get $t$ positive signs in a row, skip a term (give it a coefficient of $0$ instead of $\pm 1$), then $t$ negative signs in a row, skip a term, etc.
For example, for $n=100, t=2,$ the number of paths is
$$ ... +{100\choose 43} + {100\choose 44} + {100 \choose 45} - {100 \choose 46} - {100 \choose 47} - {100\choose 48} + {100\choose 49} + {100 \choose 50} + {100\choose 51} - ...$$
For $n=101, t=2,$ the number of paths is
$$ ... +{101\choose 44} + {101\choose 45} - {101\choose 47} - {101 \choose 48} + {101\choose 50} + {101\choose 51} - {101\choose 53} - {101\choose 54} + ...$$
These can be summed using the techniques in the answers to the [Binomial distribution parity question](https://mathoverflow.net/questions/16187/binomial-distribution-parity/).
A lot more can be said when $t$ varies, but the answers are more complicated. For $t$ slowly increasing, as $c\sqrt[3]n$, there is enough time for the distribution to stabilize (for each parity) at a given value of $t$, since the ratio between the magnitudes of the largest two eigenvalues and the magnitudes of the next two is about $1+c/t^2$, and the principal eigenvectors have a small $L^1$ distance for adjacent values of $t$. You should pick up a constant factor for each transition. In other words, the number of paths when you spend at least $n\_t \gt c t^2$ steps at a given $t$ should be
$$C \prod\_{t \le t\_{max}} (2 \cos \frac{\pi}{2t+2})^{n\_t}$$
where $C$ is between some functions $f\_{lower}(t\_{max}) \lt C \lt f\_{upper}(t\_{max})$ which does not depend on the values of $n\_t$. I don't think the $n\_t \gt c t^2$ condition is sharp for this behavior. Something like $n\_t \gt c t^2/\log t$ should work, too. The geometry of the eigenvectors for adjacent values of $t$ lets you estimate $f\_{lower}$ and $f\_{upper}$.
For $t$ more rapidly increasing, different behaviors occur. By the law of the iterated logarithm, if $t$ increases as $t(n) = \sqrt {(2-\epsilon) n \log\log n},$ random paths will almost surely violate the constraint. I think there are precise versions of the law of the iterated logarithm which may tell you when a positive proportion of random paths do not violate the constraint. I would guess that if $t(n) = \sqrt{(2+\epsilon) n \log\log n}$ then a positive percentage of random paths won't violate the constraint.
| 12 | https://mathoverflow.net/users/2954 | 16940 | 11,341 |
https://mathoverflow.net/questions/16943 | 5 | The result stated in the title is thoroughly standard - or that's the impression I got.
I seem to remember seeing it stated somewhere in a book I was reading in the library, and then reverse-engineering a proof from the hints given.
For a preprint I'm working on, it would be preferable to give a precise citation from a "standard text", rather than spend time giving the proof "for the reader's convenience". Any suggestions?
---
If anyone's interested, an outline of a proof is as follows: consider an idempotent P in B(H), with H a Hilbert space, and note that we can always decompose H as an orthogonal sum with respect to which P has the block matrix form
$$ P= \left(\begin{matrix} I & R \\\\ 0 & 0 \end{matrix}\right) $$
Then it's not hard to see that conjugating $P$ by the invertible operator
$$ S= \left( \begin{matrix} I & R \\\\ 0 & I \end{matrix} \right) $$
will give
$$ E = \left(\begin{matrix} I & 0 \\\\\ 0 & 0 \end{matrix} \right) $$
Since $S= I+P-E$, it suffices to show that $E$ is in the C\*-algebra generated by I and P (for then S will also lie in that algebra, and then we're done). This follows by messing around with various combinations of P, its adjoint, and their products.
| https://mathoverflow.net/users/763 | Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one | Blackadar's *[K-Theory for operator algebras](http://books.google.com/books?id=214a1Wri63QC&dq=idempotent+projection+similar&client=firefox-a&source=gbs_navlinks_s)* has it, although the way it is done there is perhaps overkill if this is all you need. The result is generalized to local $C^\*$-algebras, and he shows similarity by showing the stronger property of homotopy equivalence. It is [Proposition 4.6.2](http://books.google.com/books?id=214a1Wri63QC&lpg=PA23&dq=idempotent%2520projection%2520similar&client=firefox-a&pg=PA23#v=onepage&q=&f=false) on page 23 of the 2nd edition (1998). ([Proposition 4.3.3](http://books.google.com/books?id=214a1Wri63QC&lpg=PA23&dq=idempotent%2520projection%2520similar&client=firefox-a&pg=PA21#v=onepage&q=idempotent%2520projection%2520similar&f=false) shows that homotopy equivalence is stronger.)
The stronger equivalence (but just for $C^\*$-algebras) is also shown in the K-theory book by Rørdam et al., [Lemma 11.2.7](http://books.google.com/books?id=SMiB8VIB5UIC&lpg=PA192&dq=idempotent%2520projection%2520k-theory&client=firefox-a&pg=PA192#v=onepage&q=&f=false), with a very similar proof.
---
Added after the first two comments:
Kaplansky's *[Rings of operators](http://books.google.com/books?id=hRaoAAAAIAAJ&client=firefox-a&source=gbs_navlinks_s)* has another approach. Since there is no Google preview, I'll outline what is done. Theorem 26 shows that if $A$ is a unital ring with involution $\*$ such that $1+x^\*x$ is invertible for all $x\in A$, then for each idempotent $f\in A$ there is a projection $e\in A$ such that $fA=eA$. (The projection is obtained just as in the previous two sources, $e=ff^\*(1+(f-f^\*)(f^\*-f))^{-1}$.) A previous exercise (4 on page 24) shows that if $f$ and $e$ are idempotents in a unital ring $A$ and $fA=eA$, then $f$ and $e$ are similar.
Here is yet another reference, older and with a different approach, but this time not from a standard text. It is Lemma 16 on page 856 of Kaplansky's "[Modules over operator algebras](http://www.jstor.org/stable/2372552)", American Journal of Mathematics, Vol. 75, No. 4 (Oct., 1953), pp. 839-858. Of the references I have provided, this is the only one that sets out to prove precisely the result in question as opposed to something stronger or more general. The approach is to invoke the theory of polynomial identities to reduce to the case of 2-by-2 scalar matrices. This paper is famous among operator algebraists because of its introduction of what are now called Hilbert $C^\*$-modules (but only over commutative $C^\*$-algebras).
| 6 | https://mathoverflow.net/users/1119 | 16944 | 11,343 |
https://mathoverflow.net/questions/16880 | 4 | This question is related to a previous [question](https://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions) of mine. More specifically, it results from my attempts to understand the simplest incarnation of a phenomenon mentioned therein.
Put a grading on the elements of the coordinate algebra of $SU(2)$ by setting deg$(\hat{a})$=deg$(\hat{c})=1$, and deg$(\hat{b})$=deg$(\hat{d})=-1$; where
$$
\hat{a}\left(\array{a & b \\\ c & d} \right) = a
$$
and so on. We can identify the coordinate algebra of $\mathbb{CP}^1 = SU(2,\mathbb{C})/U(1)$, with the elements of degree $0$, $\Omega^{(0,1)}(\mathbb{CP}^1)$ with the elements of degree $2$. Does there exist an element $X$ of $\mathfrak {su}(2)$ (or its enveloping algebra) such that
$$
X(f) = \overline{\partial}(f) \in \Omega^{(0,1)}(\mathbb{CP}^1),
$$
for all coordinate functions $f$ on $\mathbb{CP}^1$, where $X$ is the canonical action of $\mathfrak {su}(2)$ on the coordinate algebra of $SU(2)$, that is,
$$
X(f)(g) = \frac{d}{dt}\_{t=0}(f(\exp(-tX)g)).
$$
If so, can someone explain in a more general context why so? (Perhaps in the context of spaces of the form $G/T$, where $G$ is a compact Lie group, and $T$ is a maximal torus.)
EDIT: I've been doing some more reading, and it seems that the required element of $\mathfrak{su} (2)$ does exist. Moreover, there exists a formula for extending this to the general $\mathbb{CP}^n$ case, see page 32 of this [paper](http://arxiv.org/abs/0901.4735) by D'Andrea and Dabrowski. What I still don't see though is why.
| https://mathoverflow.net/users/1648 | Dolbeault Operators for $CP^1$ as $\mathfrak{su}(2)$ Actions. | Let $K$ be a compact connected Lie group, and $L$ a Levi subgroup of $K$ (the centralizer of an element of the Lie algebra). Then $X=K/L$ is a complex manifold (a coadjoint orbit).
So one can ask about the representation theoretic interpretation of the Dolbeault operator
on $X$.
We have a direct sum decomposition $Lie(K)\_{\Bbb C}=Lie(L)\_{\Bbb C}\oplus {\mathfrak u}\_+\oplus {\mathfrak u}\_-$,
where ${\mathfrak u}\_\pm$
are the nilpotent subalgebras corresponding to $L$.
The antiholomorphic tangent bundle $T^{0,1}$ is the bundle associated to the representation ${\mathfrak u}\_+$ of $L$ and the principal $L$-bundle $K\to X$.
Now, by the Peter-Weyl theorem, the space of functions (of finite type) on $K$ is
${\rm Fun}(K)=\oplus V\otimes V^\ast$, where $V$ runs over irreducible representations of $K$.
Thus, ${\rm Fun}(X)=\oplus V^L\otimes V^\ast$, and
$\Omega^{0,1}(X)=\oplus ({\mathfrak n}\_+^\ast\otimes V)^L\otimes V^\ast$.
The Dolbeault operator $\bar{\partial}: {\rm Fun(X)}\to \Omega^{0,1}(X)$ is thus an
operator
$\oplus V^L\otimes V^\ast\to \oplus ({\mathfrak n}\_+^\*\otimes V)^L\otimes V^\ast$.
It commutes with the action of $K$, so comes from some operator
$\overline{\partial}\_V: V^L\to ({\mathfrak n}\_+^\*\otimes V)^L$ for each $V$.
I claim that this operator is just the dual to the action map ${\mathfrak n}\_+\otimes V\to V$ (restricted to $L$-invariants).
Indeed, it is well known that the exterior differential of functions on $K$ is given by the formula $df=\sum L\_{a\_i}(f) \omega\_{a\_i^\ast}$, where $a\_i$ is a basis of $Lie(K)\_{\Bbb C}$,
$L\_a$ is the left invariant vector field corresponding to $a$, and
$\omega\_u$ is the left invariant differential 1-form corresponding to $u$.
Choosing a basis corresponding to the decomposition $Lie(K)\_{\Bbb C}=Lie(L)\_{\Bbb C}\oplus {\mathfrak u}\_+\oplus {\mathfrak u}\_-$, and restricting to function on $X=K/L$, we see that the terms corresponding to $Lie(L)\_{\Bbb C}$ are zero, and the terms for ${\mathfrak n}\_+$ and ${\mathfrak n}\_-$ give $\bar\partial$ and $\partial$ respectively.
In your basic example $K=SU(2)$, $L=U(1)$, and the Dolbeault operator is defined by the action of the generator $E$ of $sl(2)$.
| 4 | https://mathoverflow.net/users/3696 | 16952 | 11,350 |
https://mathoverflow.net/questions/16359 | 6 | Select $K$ random binary vectors $Y\_i$ of length $m$ uniformly at random.
Let the following collection of random variables be defined: $X\_{i,j}=w(Y\_i \oplus Y\_j)$ where $w(\cdot)$ denotes the Hamming weight of a binary vector, i.e., the number of the nonzero coordinates in its argument. Define $D\_{min}(Y\_1,\ldots,Y\_K)$ as the smallest of the $X\_{i,j}$ for $i \neq j.$
Thus we have $n=C(K,2)=K(K-1)/2$ non-independent random variables $X\_{i,j}$ with support {$0,1,\ldots,m$} and individual distribution $Bin(m,1/2)$. It seems to me that the random variables $X\_{i,j}$ will be $s$-wise negatively correlated (for $s$ large enough) if distances between pairs chosen from a subcollection of $Y\_{i\_1},Y\_{i\_2},\ldots,Y\_{i\_v}$ where ($v < K$) tincreases then the distances between $Y\_{i\_j}$ and the remaining vectors will tend to decrease. Take $s=v+1.$
It is possible to get a bound on the following quantity. Fix $w$ an integer less than $m/2.$ The Hamming sphere of radius w has "volume", i.e., contains $V\_w(m)=\sum\_{s=0}^w C(m,s)$ vectors and we approximately have to first order in the exponent
$$
V\_w(m) =2^{m H((w+1)/2)}
$$
where $H(\cdot)$ is the binary entropy function. Then, for a random uniform choice of the $Y\_i$ for $i=1,2,\ldots,K$ it is clear that if the Hamming spheres centred at these vectors are disjoint then the minimum distance is at least $2w+1$ thus
$Pr[D\_{min} \geq 2 w+1] \leq \frac{(2^m-V)}{2^m}\frac{ (2^m-2 V) }{2^m} \cdots\frac{ (2^m - (K-1)V)}{ 2^{m}}$
where $V=V\_w(m).$ This means that, by replacing each fraction of the form $(1-x)$ by $exp(-x)$ where $x >0$ but small, we get the approximate upper bound
$Pr[ D\_{min} \geq 2w+1] \leq exp\left[-K(K-1)V^2/(2^{m+1} \right]$ which then expresses this upper bound in terms of the entropy function, which is nice. Unfortunately this upper bound is quite loose.
I will be happy with any pointers to literature or any other suggestions.
| https://mathoverflow.net/users/17773 | Minimum Hamming Distance Distribution in a Random Subset of Binary Vectors+ | Here is a direct application of Theorem 21 from Gabor Lugosi's [concentration of measure notes](http://www.econ.upf.edu/~lugosi/anu.pdf). Your $Y\_i$ corresponds to his $X\_{i,1}^m$ and your $X\_{i,j}$ to his $d(X\_{i,1}^m, X\_{j,1}^m)$. Take his $A$ to be your $\{X\_{i,j}\}\_{i \neq j}$. The birthday problem gives the probability that any two of the $Y\_i$ are exactly the same. That is:
$$\mathbb{P}(0^m \in A) = \mathbb{P}\left(\left\{X\_{i,j} = 0^m : i \neq j\right\}\right) = \mathrm{(omitted\ for\ simplicity)} $$
Now your $D\_{min}$ corresponds to his $d(0^m,A)$. By the Theorem, for any $t > 0$,
$$\mathbb{P}\left(D\_{min} \geq t + \sqrt{\frac{m}{2} \mathrm{log}\frac{1}{\mathbb{P}(0^m \in A)}}\right) \leq e^{-2t^2/m}.$$
This bound may be OK for your needs. If it isn't, see Lugosi's discussion of Talagrand's convex distance inequality, which is a big improvement.
| 5 | https://mathoverflow.net/users/1656 | 16955 | 11,352 |
https://mathoverflow.net/questions/16930 | 8 | I would like to know the standard usage of "lax colimit" and "oplax colimit" in the 2-categorical literature. The nLab does not give an explicit definition of "lax colimit", as far as I can see, and I don't know what the most reliable source is. I think I have seen at least one paper using each convention, but I have not encountered the notion often enough to have a good sense of whether this one of those places where the terminology is not really standardized, or there is general agreement with a few exceptions.
Concretely, given a diagram X : I → C in a 2-category C (for my purposes indexed by a 1-category I), suppose I have a cone (Y, {gi}i∈Ob I, {αf}f∈Mor I), with gi : Xi → Y, and for f : i → j in I, αf : gjf → gi (such that various diagrams commute). Is this a lax colimit cone or a oplax colimit cone?
| https://mathoverflow.net/users/126667 | Terminology: lax vs. oplax colimits | I agree with Finn that the way to derive the correct choice of lax vs oplax is to connect it back to natural transformations. Of course, as Finn pointed out, there is controversy over the choice for natural transformations, but my views on that are clear at the [nlab page](http://ncatlab.org/nlab/show/lax+natural+transformation) so I'll just write using that terminology.
However, in contrast to Finn, I think what you've got there is actually a *lax* cocone, since it is given by a lax natural transformation $\* \to C(X-,Y)$, where $\*:I^{op}\to Cat$ is the functor constant at the point. It's true that, as Finn says, it is also an *oplax* natural transformation from $X$ to $\Delta\_Y$, where $\Delta\_Y$ is the functor $I\to C$ constant at $Y$. But I think it's better to think of a cone as a transformation $\* \to C(X-,Y)$, since this is the version that generalizes to *weighted* limits: for any weight $J:I^{op}\to Cat$, a $J$-weighted cocone is a transformation of the appropriate sort $J\to C(X-,Y)$.
The weighted-limit perspective on lax (co)limits is especially valuable because of the existence of lax morphism classifiers. Namely, for any weight $J$ there is another weight $J^\dagger$ such that lax transformations out of $J$ are the same as strict (or pseudo) transformations out of $J^\dagger$. Thus, lax $J$-weighted limits are the same as ordinary $J^\dagger$-weighted limits, so that lax $J$-weighted cones and limits are the same as ordinary $J^\dagger$-weighted cones and limits. Thus a "lax limit" is really just a particular type of *weighted* limit, whose weight happens to be of the form $J^\dagger$. Similarly, there is an oplax morphism classifier $J^\diamond$.
I think the choice I'm proposing is fairly widespread in Australia. For instance, it's the one used [here](http://arxiv.org/abs/math/0702535) and [here](http://www.maths.usyd.edu.au/u/stevel/papers/talgl.pdf) and [here](http://www.tac.mta.ca/tac/reprints/articles/4/tr4.pdf). Actually, I'm not sure offhand whether I've even ever seen the other choice in print.
| 8 | https://mathoverflow.net/users/49 | 16956 | 11,353 |
https://mathoverflow.net/questions/16953 | 11 | Are all submodules of free modules free? I would like a reference to a proof or counterexample please.
| https://mathoverflow.net/users/3787 | Are submodules of free modules free? | Вот общий пример: неглавной идеал в кольце $A$. Кольцо $A$ -- свободный $A$-модуль. Идеал в кольце -- подмодуль, а он тоже свободный $A$-модуль только в случае, что он главной идеал: ненулевые элементы $a$ и $b$ в кольце удовлетворяют нетривиальному $A$-линейному соотношению $c\_1a + c\_2b = 0$, где $c\_1 = b$ и $c\_2 = -a$, поэтому если существует базис, то мощность является одним.
| 46 | https://mathoverflow.net/users/3272 | 16957 | 11,354 |
https://mathoverflow.net/questions/16946 | 4 | Polya Enumeration Formula gives us 6 equivalence classes of 2-colorings of a square. But, in the Polya coloring, the following 2 colorings belong to 2 different equivalence classes:
```
00
01
```
and
```
11
10
```
(0 and 1 are the 2 colors.)
What is the theory that groups colorings like the above 2 into the **same** equivalence class? The reason to put the above 2 into the same class would be that we can obtain one from the other by a mapping between the colors (0->1 and 1->0). Is there a formula to obtain the number of equivalence classes of colorings with this constraint?
| https://mathoverflow.net/users/4355 | Polya Enumeration Formula with color indifference | There is a paper [A survey of generalizations of Pólya's enumeration theorem](http://alexandria.tue.nl/repository/freearticles/597547.pdf) which discusses a generalization of Polya's theorem involving colors. It gave [Enumerative Combinatorial Problems Concerning Structures](http://alexandria.tue.nl/repository/freearticles/597529.pdf) as a reference for this generalization. I think this is the theorem that you are looking for and and the variant referred to by Yuan in his answer to your question.
| 7 | https://mathoverflow.net/users/1098 | 16958 | 11,355 |
https://mathoverflow.net/questions/16959 | 13 | Suppose we have a curve γ : [0,1] -> ℝn. It is well known that if this curve is Hölder continuous for some exponent α then the Hausdorff dimension of γ[0,1] is bounded above by 1/α.
My question is: Is there a partial converse of the following form. Suppose that the curve γ is such that γ[0,1] has Hausdorff dimension d, then is it true that for any α < 1/d we have that there exists some *reparameterization* of γ so that γ is Hölder continuous with exponent α?
I would like it to be true, although I doubt it is, however I have not been able to find a counterexample. Proofs, references, or counterexamples would be appreciated.
| https://mathoverflow.net/users/4358 | Hausdorff Dimension and Hölder Continuity | Edit:
The short answer is that there are planar curves that cannot be parametrized in a Holder continuous manner. Thus any such curve provides a counterexample for some $d \le 2$.
My original answer, showing that the curve can be chosen even of Hausdorff dimension $d=1$, follows.
I believe the answer to your question is negative. Indeed, I would think you can construct a curve in $\mathbb{R}^2$ that has Hausdorff dimension 1 but cannot be parametrized by a Holder-continuous curve.
The basic idea is that at smaller and smaller scales, the curve oscillates rather wildly in the horizontal direction, which can be done to ensure that the curve cannot be parametrized in a Holder-continuous way. However, if we make sure that the vertical extent of these oscillations is much, much smaller at each step, then the Hausdorff dimension should still be one. (Of course the linear measure of such a curve will be infinite.)
Let me try to be a bit more precise. I hope the idea is suitably clear.
Let us suppose that we construct the curve $\gamma\_k$ at successive scales $\delta\_k = 2^{-k}$. The curve $\gamma\_k$ will consist of a number $M\_k$ of horizontal segments, each of length $\delta\_k$, together with some vertical segments. In the construction, we may ensure that the vertical segments have bounded total length, so we can just ignore them.
To obtain $\gamma\_{k+1}$, we split each horizontal segment in two segments of length $\delta\_{k+1}$, and then replace each of these segments by a curve that oscillates, consisting of some number $m\_{k+1}$ of horizontal segments of length $\delta\_{k+1}$. However, the vertical extent of this oscillation should be some very small number $\varepsilon\_{k+1}$. (Which in particular we may make so small that the total length of vertical segments added in this step is smaller than, say $2^{-k}$, so the total length of vertical segments stays bounded at all times.)
Now the total number of segments at stage $k+1$ is $M\_{k+1} = 2m\_{k+1} M\_k$. Any parametrization of this curve by the unit interval must contain two points that are at distance at most $1/M\_{k+1}$ but have points whose images are at distance at least $\delta\_{k+1}$. So if we choose $m\_{k+1}$ sufficiently large, we can ensure that the limit curve cannot be Holder-parametrized.
On the other hand, the curve can be covered by about
$M\_{k+1}\cdot\frac{\delta\_{k+1}}{\varepsilon\_{k+1}}$
sets of diameter $\varepsilon\_{k+1}$. If we make $\varepsilon\_{k+1}$ small enough, this number will be smaller than, say $\varepsilon\_{k+1}^{-(1+1/k)}$, so the Hausdorff dimension of the limiting curve will be equal to one.
| 5 | https://mathoverflow.net/users/3651 | 16962 | 11,357 |
https://mathoverflow.net/questions/16963 | 0 | Given a $O\_X$ module $\cal F$ whose support is a closed subscheme $Z \subset X$. Under what conditions can we say that $ \cal F$ is an $O\_S$ module ( how far off is $\cal F$ an $O\_S$ module ? )
| https://mathoverflow.net/users/4275 | O_X module with support Z \subset X vs O_S module? | It has to be annihilated by the sheaf of ideals of $Z$. If you are working with a noetherian scheme and a coherent sheaf at least, we can at least filter $\mathcal{F}$ by subsheaves $\mathcal{I}^i \mathcal{F}$ (where $\mathcal{I}$ is the sheaf of ideals of $Z$) whose successive quotients are $O\_S$-modules. Here $\mathcal{I}^i \mathcal{F} = 0$ for $i$ large by the assumption on support in $Y$. Indeed, $Y$ can be defined as $V(\mathcal{a})$ for some ideal $\mathcal{a}$. Then $M\_{\mathfrak{p}} = 0$ if $\mathfrak{p} \not \supset \mathcal{a}$. Therefore, every associated prime must contain $\mathcal{a}$, so there is a filtration $0=M\_0 \subset M\_1 \subset \dots \subset M\_k = M$ whose quotients are isomorphic to the form $A/\mathfrak{p}$ where $\mathfrak{p} \supset \mathcal{a}$ is a prime, which means $M$ is annihilated by $\mathcal{a}^k$.
| 1 | https://mathoverflow.net/users/344 | 16964 | 11,358 |
https://mathoverflow.net/questions/16971 | 13 | Why do algebraic geometers still use the term "quasi-compact" when they almost never deal with Hausdorff spaces? They certainly use "local" rather than "quasi-local" (local = quasi-local + noetherian), so is there any reason other than historical contingency?
Do algebraic geometers who do work in other fields still follow this convention when they write other papers? If they do, do they write at the beginning of the paper something along the lines of "by compact, we mean quasi-compact and Hausdorff"?
| https://mathoverflow.net/users/1353 | Compact and quasi-compact | The condition of quasi-compactness in the Zariski topology bears little resemblance to the condition of compactness in the classical analytic topology: e.g. any variety over a field is quasi-compact in the Zariski topology, but a complex variety is compact in the analytic topology iff it is complete, or better, proper over $\operatorname{Spec} \mathbb{C}$.
I think many algebraic geometers think to themselves that a variety is "compact" if it is proper over the spectrum of a field. I have heard this terminology used and occasionally it shows up in (somewhat informal) writing.
So a perhaps more accurate brief answer is that in algebraic geometry the distinction between quasi-compact and quasi-compact Hausdorff is very important, whereas in other branches of geometry non-Hausdorff spaces turn up more rarely.
Anyway, many mathematicians have been happy with the quasi-compact / compact distinction for about 50 years, so I don't think this usage is going away anytime soon.
To address the last question: when writing for a general mathematical audience, it is a good idea to give an unobtrusive heads up as to your stance on the quasi-compact / compact convention. (The same probably goes for other non-universal conventions in mathematics.) If I were speaking about profinite groups, I would say something like:
"A profinite group is a topological group which can be expressed as an inverse limit of finite discrete groups. Equivalently, a topological group is profinite if it is compact (Hausdorff!) and totally disconnected."
This should let people know what side I'm on, and thus be able to understand me. When writing for students, I might take pains to be more explicit, using a "By compact I mean..." construction as you have indicated above.
| 18 | https://mathoverflow.net/users/1149 | 16976 | 11,365 |
https://mathoverflow.net/questions/454 | 16 | (1) What are some good references for homotopy colimits?
(2) Where can I find a reference for the following concrete construction of a homotopy colimit? Start with a partial ordering, which I will think of as a category and also as a directed graph (objects = vertices, morphisms = edges). Assume we have a functor $F$ from the graph into (say) chain complexes. We will construct a big chain complex (the homotopy colimit) in stages.
Stage 0: direct sum over all vertices $v$ of $F(v)$
Stage 1: direct sum over all edges $e$ of the mapping cylinder of $F(e)$, with the ends of the mapping cylinder identified with the appropriate parts of stage 0.
Stage 2: direct sum over all pairs of composable edges $(e\_1, e\_2)$ of a higher order mapping cylinder, with appropriate identifications to parts of stage 1. This implements a relation between the three stage 1 mapping cylinders corresponding to $e\_1, e\_2$ and $e\_1\*e\_2$.
Stage 3: direct sum over all triples of composable edges $(e\_1, e\_2, e\_3) \dots$
| https://mathoverflow.net/users/284 | References for homotopy colimit | I'm way late on this one, but for the record I'll point out that a nice answer to question 2 can be found in Hatcher's Algebraic Topology book, Section 4.G.
| 5 | https://mathoverflow.net/users/4042 | 16990 | 11,371 |
https://mathoverflow.net/questions/16983 | 6 | Some basic observations lead me to ask the following quesiton
Let $A\_1, \cdots, A\_m$ be $n\times n$ complex matrices. For positive integer $k\ge 1$, show
$$\left(\begin{array}{cccc}Tr\{(A\_1^\*A\_1)^k\}&Tr\{(A\_1^\*A\_2)^k\}&\cdots &Tr\{(A\_1^\*A\_m)^k\}\\Tr\{(A\_2^\*A\_1)^k\}&Tr\{(A\_2^\*A\_2)^k\}&\cdots &Tr\{(A\_2^\*A\_m)^k\}\\\cdots&\cdots&\cdots&\cdots\\Tr\{(A\_m^\*A\_1)^k\}&Tr\{(A\_m^\*A\_2)^k\}&\cdots &Tr\{(A\_m^\*A\_m)^k\}
\end{array}\right)$$
is positive semidefinite.
**Remark**
1). When $m=2$, it suffices to show $|Tr\{(A\_1^\*A\_2)^k\}|^2\le Tr\{(A\_1^\*A\_1)^k\}\cdot Tr\{(A\_2^\*A\_2)^k\}$, which is a consequence of a unitarily invariant norm inequality appeared in p.81 of X.Zhan, Matrix inequalities, Springer, 2002.
2). It is easy to show $$\left(\begin{array}{cccc}(Tr\{A\_1^\*A\_1\})^k&(Tr\{A\_1^\*A\_2\})^k&\cdots &(Tr\{A\_1^\*A\_m\})^k\\(Tr\{A\_2^\*A\_1\})^k&(Tr\{A\_2^\*A\_2\})^k&\cdots &(Tr\{A\_2^\*A\_m\})^k\\\cdots&\cdots&\cdots&\cdots\\(Tr\{A\_m^\*A\_1\})^k&(Tr\{A\_m^\*A\_2\})^k&\cdots &(Tr\{A\_m^\*A\_m\})^k
\end{array}\right)$$
is positive semidefinite, since it is $k$ Hadamard product of a Gram matrix.
| https://mathoverflow.net/users/3818 | On a positivity of a matrix with trace entries. | It seems that this is not true. Here is a counterexample. Consider a regular $d$-gon, where $d\ge 3$ is an odd number. Let $S,T$ be the permutation matrices on the vertices of this $d$-gon, induced by reflections in two adjacent symmetry axes. Let $A\_1=1, A\_2=S, A\_3=T$, which are $d$ by $d$ matrices. We have $S^2=T^2=1$, and $ST$ is a rotation of order $d$
(so $(ST)^2$ has no fixed points).
Let $k=2$. Then the matrix in the question seems to be the following:
$$
\left(\begin{matrix} d & d & d\\ d & d & 0\\ d & 0 & d\\ \end{matrix}\right)
$$
The determinant of this matrix is $-d^3$.
| 9 | https://mathoverflow.net/users/3696 | 16995 | 11,373 |
https://mathoverflow.net/questions/16977 | 16 | I am sitting on my carpet surrounded by books about quantum groups, and the only categorical concept they discuss are the representation categories of quantum groups.
Many notes closer to "Kontsevich stuff" discuss matters in a far more categorical/homotopical way, but they seem not to wish to touch the topic of quantum groups very much....
I know next to nothing about this matter, but it seems tempting to believe one could maybe also phrase the first chapters of the quantum group books in a language, where for example quasitriangulated qHopf algebras would just be particular cases of a general "coweak ${Hopf\_\infty}$-algebra" (does such thing exist?). Probably then there should be some (${\infty}$,1) category around the corner and maybe some other person's inofficial online notes trying to rework such a picture in a dg Hopf algebra model category picture.
My quantum group books don't mention anything in such a direction (in fact ahead of a chapter on braided tensor categories the word 'category' does kind of not really appear at all). So either
1. [ ] I am missing a key point and just proved my stupidity to the public
2. [ ] There is a quantum group book that I have missed, namely ......
3. [ ] Such a picture is boring and/or wrong for the following reason .....
Which is the appropriate box to tick?
| https://mathoverflow.net/users/3888 | Why do my quantum group books avoid homotopical language? | The question is rather rambling and it is more about not so well-defined appetites (do you have a more conrete motivation?).
There is one thing which however makes full sense and deserves the consideration. Namely it has been asked what about higher categorical analogues of (noncommutative noncocommutative) Hopf algebras. This is not a trivial subject, because it is easier to do resolutions of operads than more general properads. Anyway the infinity-bialgebras are much easier than the Hopf counterpart. There is important work of Umble and Saneblidze in this direction (cf. [arxiv/0709.3436](http://arxiv.org/abs/0709.3436)). The motivating examples are however rather different than quantum groups, coming from rational homotopy theory, I think.
Similarly, there is no free Hopf algebra in an obvious sense what makes difficult to naturally interpret deformation complexes for Hopf algebras (there is a notion called [free Hopf algebra](http://ncatlab.org/nlab/show/free+Hopf+algebra), concerning something else). Boris Shoikhet, aided with some help from Kontsevich, as well as Martin Markl have looked into this.
Another relevant issue is to include various higher function algebras on higher categorical groups, enveloping algebras of higher Lie algebras (cf. [baranovsky (pdf)](http://www.mrlonline.org/mrl/2008-015-006/2008-015-006-001.pdf) or [arxiv 0706.1396](http://arxiv.org/abs/0706.1396) version), usual quantum groups, examples of secondary Steenrod algebra of Bauese etc. into a unique natural higher Hopf setting. I have not seen that.
The author of the question might also be interested in a monoidal bicategorical approach to general [Hopf algebroid](http://ncatlab.org/nlab/show/Hopf+algebroid)s by Street and Day.
| 5 | https://mathoverflow.net/users/35833 | 16999 | 11,376 |
https://mathoverflow.net/questions/16991 | 35 | I watched a video that said the probability for Gaussian integers to be relatively prime is an expression in $\pi$, and I also know about $\zeta(2) = \pi^2/6$ but I am wondering what are more connections between $\pi$ and prime numbers?
| https://mathoverflow.net/users/4361 | What are the connections between pi and prime numbers? | Well, first of all, $\pi$ is not just a random real number. Almost every real number is transcendental so how can we make the notion "$\pi$ is special" (in a number-theoretical sense) more precise?
Start by noticing that $$\pi=\int\_{-\infty}^{\infty}\frac{dx}{1+x^2}$$
This already tells us that $\pi$ has something to do with rational numbers. It can be expressed as "a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ given by polynomial inequalities with rational coefficients." Such numbers are called [periods](https://en.wikipedia.org/wiki/Period_(number)).
Coming back to the identity
$$\zeta(2)=\frac{\pi^2}{6}$$
There is a very nice proof of this (that at first seems very unnatural) due to Calabi. It shows that
$$\frac{3\zeta(2)}{4}=\int\_0^1\int\_0^1\frac{dx\,dy}{1-x^2y^2}$$
by expanding the corresponding geometric series, and then evaluates the integral to $\pi^2/8$. (So yes, $\pi^2$ and all other powers of $\pi$ are periods.) But the story doesn't end here as it is believed that there are truly deep connections between values of zeta functions (or [L-functions](https://en.wikipedia.org/wiki/L-function)) and certain evaluations involving periods, such as $\pi$. Another famous problem about primes is Sylvester's problem of which primes can be written as a sum of two rational cubes. So one studies the [elliptic curve](https://en.wikipedia.org/wiki/Elliptic_curve)
$$E\_p: p=x^3+y^3$$ and one wants to know if there is one rational solution, the central value of the corresponding L-function will again involve $\pi$ up to some integer factor and some Gamma factor. Next, periods are also values of multiple zeta functions:
$$\zeta(s\_1,s\_2,\dots,s\_k)=\sum\_{n\_1>n\_2>\cdots>n\_k\geq 1}\frac{1}{n\_1^{s\_1}\cdots n\_k^{s\_k}}$$
And they also appear in other very important conjectures such as the [Birch and Swinnerton-Dyer conjecture](https://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture). But of course all of this is really hard to explain without using appropriate terminology, the language of [motives](https://en.wikipedia.org/wiki/Motive_(algebraic_geometry)) etc. So, though, this answer doesn't mean much, it's trying to show that there is an answer to your question out there, and if you study a lot of modern number theory, it might just be satisfactory :-).
| 61 | https://mathoverflow.net/users/2384 | 17008 | 11,381 |
https://mathoverflow.net/questions/17006 | 52 | Simple linear algebra methods are a surprisingly powerful tool to prove combinatorial results. Some examples of combinatorial theorems with linear algebra proofs are the (weak) [perfect graph theorem](http://mathworld.wolfram.com/PerfectGraphTheorem.html), the [Frankl-Wilson theorem](http://gilkalai.wordpress.com/2009/05/21/extremal-combinatorics-vi-the-frankl-wilson-theorem/), and [Fisher's inequality](http://en.wikipedia.org/wiki/Fisher%27s_inequality).
Are there other good examples?
| https://mathoverflow.net/users/2233 | Linear algebra proofs in combinatorics? | Some other examples are the Erdos-Moser conjecture (see R. Proctor, Solution of two difficult problems with linear algebra, *Amer. Math. Monthly* **89** (1992), 721-734), a few results at
[http://math.mit.edu/~rstan/312/linalg.pdf](http://math.mit.edu/%7Erstan/312/linalg.pdf), and Lovasz's famous result on the Shannon capacity of a 5-cycle and other graphs (*IEEE Trans. Inform. Theory* **25** (1979), 1-7). For a preliminary manuscript of Babai and Frankl on this subject (Linear Algebra Methods in Combinatorics), see [http://people.cs.uchicago.edu/~laci/CLASS/HANDOUTS-COMB/BaFrNew.pdf](http://people.cs.uchicago.edu/%7Elaci/CLASS/HANDOUTS-COMB/BaFrNew.pdf) .
2022 version of Babai & Frankl text: [https://people.cs.uchicago.edu/~laci/babai-frankl-book2022.pdf](https://people.cs.uchicago.edu/%7Elaci/babai-frankl-book2022.pdf)
| 31 | https://mathoverflow.net/users/2807 | 17012 | 11,384 |
https://mathoverflow.net/questions/16992 | 20 | I've read in (abstracts of) papers that there are abelian varieties over fields of positive characteristic that admit no prinicipal polarization. Apparently its not the easiest thing to find an example of, but I was thinking it should be much easier over the complex numbers.
All abelian varieties of dimension 1 are elliptic curves which always have a principal polarization. So any example would have to be at least two dimensional. So my question is given an abelian variety with a polarization, is there a good way of telling if there is or isn't a principal polarization?
Or if that's in general a difficult question, are there some relatively simple examples where you can really see that there are or aren't any principal polarizations?
| https://mathoverflow.net/users/7 | non principally polarized complex abelian varieties | Here is another construction, followed by some comments on how to solve the existence problem in general.
>
> If $A$ is a $g$-dimensional principally polarized abelian variety over $\mathbf{C}$ with $\operatorname{End} A = \mathbf{Z}$, and $G$ is a finite subgroup whose order $n$ is not a $g$-th power, then $B:=A/G$ is an abelian variety that admits no principal polarization.
>
>
>
**Proof:** If $B$ had a principal polarization, its pullback to $A$, given by the composition $A \to B \to B' \to A' \simeq A$ (where $A'$ is the dual of $A$ and so on) would be an endomorphism of degree $n^2$. But this endomorphism is multiplication-by-$m$ for some integer $m$, which has degree $m^{2g}$. So $n$ would have to be a $g$-th power. $\square$
To complete this answer, observe that most abelian varieties $A$ over $\mathbf{C}$ satisfy $\operatorname{End} A=\mathbf{Z}$. An explicit example is the Jacobian of the hyperelliptic curve that is the smooth projective model of the affine curve
$$y^2= a\_{2g+1} x^{2g+1} + \cdots + a\_1 x + a\_0$$
where $a\_{2g+1},\ldots,a\_0 \in \mathbf{C}$ are algebraically independent over $\mathbf{Q}$.
**Remarks:**
1) Of course there is no reason to restrict to $\mathbf{C}$. For instance, one can find examples over $\mathbf{Q}$ by using the fact that the endomorphism ring injects into the endomorphism ring of the reduction modulo any prime of good reduction, and combining this information for several primes.
2) For an arbitrary abelian variety $A$, if you are given a polarization $A \to A'$, then if there is a principal polarization, following the first by the inverse of the second would give you an endomorphism of $A$. So one way of answering the existence question is to determine the endomorphism ring of $A$ and to study those endomorphisms that factor through your given polarization. (That's not quite sufficient, but it gives you an idea of the complexity of the problem since determining the endomorphism ring can be rather difficult.)
| 32 | https://mathoverflow.net/users/2757 | 17014 | 11,386 |
https://mathoverflow.net/questions/16984 | 3 | As I mentioned in my previous post, I am studying the article *[Moduli of Enriques surfaces and Grothendieck-Riemann-Roch](http://arxiv.org/abs/math/0701546)*.
The Grothendieck-Riemann-Roch theorem is applied there to show that, for any family of Enriques surfaces $f:Y\longrightarrow T$, the line bundle $$\mathcal{L} := R^0 f\_\ast \Big ((\Omega^2\_{Y/T})^{\otimes 2}\Big)$$ is a torsion line bundle, i.e., some tensor power $\mathcal{L}^{\otimes n}$ is isomorphic to the structure sheaf on $T$.
To this extent, one applies GRR to the morphism $f$ and the structure sheaf $\mathcal{O}\_Y$. The problem I have now is with the "relative tangent sheaf". I am guessing this is the the quotient sheaf $$f^\ast \mathcal{T}\_T/\mathcal{T}\_Y.$$
**Q1**. Why is this well-defined? That is, why do we have an injection $ \mathcal{T}\_Y \longrightarrow f^\ast \mathcal{T}\_T$?
**Q2**. How can one determine the Chern classes of $\mathcal{T}\_f$ by means of the fibres? That is, can one use the structure on the fibres (Enriques surfaces) to determine $c\_i(\mathcal{T}\_f)$?
**[New questions]**
**Q3**. Let $E$ be a fibre of $f:Y\longrightarrow T$ with injection $i:E\longrightarrow Y$. Is the ringmorphism $i^\ast:A^\cdot Y \longrightarrow A^\cdot E$ injective? If not, is it injective after tensoring with $\mathbf{Q}$?
Let $c\_i=c\_i(T\_f)$.
**Q4**. We have two formulas from the GRR. The first is $1 = \frac{1}{12} f\_\ast(c\_1^2+c\_2).$ This is the degree 0 part. The second comes from the degree 1 part and reads $0 = \frac{1}{24}f\_\ast(c\_1\cdot c\_2).$ Now, why is $f\_\ast(c\_1^2) = 0$ as is suggested by the article?
| https://mathoverflow.net/users/4333 | Family of Enriques surfaces and GRR, Part 2 | **Q1**: It is the other way round. For a smooth family the differential $T\_Y \to f^\ast T\_T$ is surjective and the relative tangent is the kernel, so you have an exact sequence
$0 \to T\_f \to T\_Y \to f^\ast T\_T \to 0$.
In this way the tangent to $f$ actually restricts to the tangent of the fibers.
**Q2**: I don't think that the classes $c\_i(T\_f)$ are determined by the fibers alone; they depend on the family. It does not even make sense to say that $c\_i(T\_f)$ are determined by the fibers since these classes live in $H^{2i}(Y)$ anyway, so you have to know at least the total space.
But since $T\_f$ restricts to the tangent of the fibers, you know, by naturality of the Chern classes, that if $i \colon E \to Y$ is the inclusion of a fiber $i^\ast c\_i(T\_f) = c\_i(T\_E)$.
And these you can compute using the fact that $E$ is Enriques. Namely $2 c\_1(T\_E) = 0$ since twice the canonical is trivial and $c\_2(T\_E) = \chi(E) = 12$.
**Q3**: Surely it is not injective in the top degree, for trivial dimensional reasons. I do not see any reason why it should be in other degrees.
**Q4**: As is written in the article, this follows from $f\_\ast c\_2 = 12$. This is more or less clear in cohomology. In this case $f\_\*$ is the integration along fibers, and since $c\_2(T\_E)$ is $12$ times the fundamental class of $E$ for all fibers $E$ (see **Q2**), that integral is $12$.
To translate this in the Chow language, I think the folllowing will do. Let $D$ be a cycle representing $c\_2(T\_f)$. Since $Y$ is smooth, we can compute the intersection number $D \cdot E = c\_2(T\_f) \cap E = c\_2(T\_E) \cap E = 12$. So $D$ intersectts the generic fiber in $12$ points, and the morphism $D \to T$ has degree $12$. In follows that $f\_\ast D = 12 [T]$, which is what you want.
| 3 | https://mathoverflow.net/users/828 | 17018 | 11,389 |
https://mathoverflow.net/questions/17005 | 6 | Does anyone know of a good method for finding a lower bound of the Hausdorff dimension of a set $G$?
The only method I could find is to find an $\alpha$-Hölder function $f \colon G \to H$ then $\dim\_H(G) \geq \alpha \dim\_H(\operatorname{im}(f))$. Choosing $f$ cleverly will mean that $\operatorname{im}(f)$ will be a set whose Hausdorff dimension is already known (or at least a lower bound for it is known).
| https://mathoverflow.net/users/3121 | Determining a lower bound on the Hausdorff dimension of a set | Upper bound for the Hausdorff dimension is often easy, from the definition.
Lower bound can be harder. One method can be used if you have a measure on your set. Even better, a measure that naturally fits with the structure of the set. Then lower bounds for the Hausdorff dimension come from density computations for that measure.
(Would citing my own book here be considered crass?)
Packing dimension may be opposite. The lower bound is easy from the definition, but the upper bound harder. Again a density with respeact to a measure can help with this upper bound.
**added March 4**
This density theorem is found in: G. Edgar, *Integral, Probability, and Fractal Measures* (Springer 1998) Theorem 1.5.14, p. 52.
*Definitions* ... Let $E \subseteq \mathbb{R}^n$ be a Borel set, let $\mu$ be a nonzero measure
on $E$, let $s>0$ be a real number. Write
$$
B\_r(x) = \{y \in E \colon |y-x|\le r\}
$$
for a closed ball. The **upper** $s$-**density** of $\mu$ at a point
$x \in \mathbb{R}^n$ is
$$
\overline{D}^s\_\mu(x) = \limsup\_{r \to 0} \frac{\mu(B\_r(x))}{(2r)^s} .
$$
A consequence of *Theorem* 1.5.14 is then: If
$\sup\_{x \in E} \overline{D}^s\_\mu(x) < \infty$, then the Hausdorff dimension
of $E$ is at least $s$.
| 3 | https://mathoverflow.net/users/454 | 17021 | 11,391 |
https://mathoverflow.net/questions/17020 | 26 | Today, I heard that people think that if you can prove the Hodge conjecture for abelian varieties, then it should be true in general. Apparently this case is important enough (and hard enough) that Weil wrote up some families of abelian 4-folds that were potential counterexamples to the Hodge conjecture, but I've never heard of another potential counterexample.
Anyway, in short:
1) Does the Hodge Conjecture for abelian varieties imply the full Hodge conjecture?
2) If not, is there an intuitive reason why abelian varieties should be the hardest case?
| https://mathoverflow.net/users/622 | Why do people think that abelian varieties are the hardest case for the Hodge conjecture? | I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost.
Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be *extremely* interesting).
| 46 | https://mathoverflow.net/users/930 | 17025 | 11,393 |
https://mathoverflow.net/questions/17032 | 28 | What is $\mathbb{Q}\_p \cap \overline{\mathbb{Q}}$ ?
For instance, we know that $\mathbb{Q}\_p$ contains the $p-1$st roots of unity, so we might say that $\mathbb{Q}(\zeta) \subset \mathbb{Q}\_p \cap \overline{\mathbb{Q}}$, where $\zeta$ is a primitive $p-1$st root.
As a more specific example, $x^2 - 6$ has 2 solutions in $\mathbb{Q}\_5$, so we could also say that $\mathbb{Q}(\sqrt{6},\sqrt{-1})\subset \mathbb{Q}\_p \cap \overline{\mathbb{Q}}$.
**Edit:** I removed the motivation for this question (which I think stands by itself), as it will be better as a separate question once I think it through a bit better.
| https://mathoverflow.net/users/434 | Which p-adic numbers are also algebraic? | The field $K\_p = \mathbb{Q}\_p \cap \overline{\mathbb{Q}}$ is a very natural and well-studied one. I can throw some terminology at you, but I'm not sure exactly what you want to know about it.
1) It is often called the field of "$p$-adic algebraic numbers". This comes up in model theory: it is a $p$-adically closed field, which is the $p$-adic analogue of a real-closed field. In particular, it is elementarily equivalent to $\mathbb{Q}\_p$.
2) It is the Henselization of $\mathbb{Q}$ with respect to the $p$-adic valuation, or the
fraction field of the Henselization of the ring $\mathbb{Z}\_{(p)}$ -- i.e., $\mathbb{Z}$ localized at the prime ideal $p$.
The idea is that this field is not complete but is Henselian -- it satisfies the conclusion of Hensel's Lemma. Alternately and somewhat more gracefully, Henselian valued fields are characterized by the fact that the valuation extends uniquely to any algebraic field extension.
Roughly speaking, Henselian fields are as good as complete fields for algebraic constructions but are not "big enough" to have the same topological properties. For instance, note that $K\_p$ cannot possibly be complete with respect to the $p$-adic valuation, because it is countably infinite and without isolated points: apply the Baire Category Theorem.
| 28 | https://mathoverflow.net/users/1149 | 17033 | 11,398 |
https://mathoverflow.net/questions/16967 | 20 | $\DeclareMathOperator\Ho{Ho}$Is there a model category $C$ on an additive category such that its homotopy category $\Ho(C)$ is the stable homotopy category of spectra and the additive structure on $\Ho(C)$ is induced from that on $C$.
Basically I want to add and subtract maps in $C$ without going to its homotopy category.
I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure.
As John Palmieri pointed out I should really say what structure I want the equivalence (between $\Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible.
| https://mathoverflow.net/users/3557 | Is there an additive model of the stable homotopy category? | The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care).
Using the main result of
S. Schwede, [The stable homotopy category is rigid](http://www.math.uni-bonn.de/%7Eschwede/rigid.pdf), Annals of Mathematics 166 (2007), 837-863
your question is equivalent to the following: does there exist a model category $C$, which is additive, and such that $C$ is *Quillen equivalent* to the usual model category of spectra?
In particular, we might ask: does there exist an additive category $C$, endowed with a Quillen stable model category structure, such that the corresponding stable $(\infty,1)$-category is equivalent to the stable $(\infty,1)$-category of spectra?
Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of
[Ken Brown](http://pi.math.cornell.edu/%7Ekbrown/scan/1973.0186.pdf)), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(\infty,1)$-category (obtained by inverting weak equivalence of $C$ in the sense of $(\infty,1)$-categories) is equivalent to the stable $(\infty,1)$-category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category).
So, assume there is an additive category of cofibrant objects $C$, with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C\_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$. Then $Ho(C\_f)\simeq S\_f$, where, by abuse of notations, $Ho(C\_f)$ is the $(\infty,1)$-category obtained from $C\_f$ by inverting weak equivalences, while $S\_f$ stands for the stable $(\infty,1)$-category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(\infty,1)$-category of $A$" (that is the $(\infty,1)$-category obtained from the category of bounded complexes of $A$, by inverting the chain homotopy equivalences). Then, the canonical functor $A\to K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$), has the following universal property: given a stable $(\infty,1)$-category $T$, any functor $A\to T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)\to T$. In particular, the functor $C\_f\to Ho(C\_f)\simeq S\_f$ extends uniquely to a finite colimit preserving functor $F:K(C\_f)\to S\_f$. Let $Ker(F)$ be the full $(\infty,1)$-subcategory of $K(C\_f)$ spanned by objects which are sent to zero in $S\_f$. Then the induced functor
$$K(C\_f)/Ker(F)\to S\_f$$
is an equivalence of (stable) $(\infty,1)$-categories (to see this, you may use the universal property of $S\_f$: given a stable $(\infty,1)$-category $T$, a finite colimit preserving functor $S\_f\to T$ is the same as an object of $T$; see Corollary 10.16 in [DAG I](https://people.math.harvard.edu/%7Elurie/papers/DAG-I.pdf)). This implies that, for any object $X$ of $S\_f$, if $X/n$ denotes the cone of the map $n:X\to X$ (multiplication by an integer $n$), then $n.X/n\simeq 0$ (see Proposition 1 in Schwede's paper [Algebraic versus topological triangulated categories](https://www.math.uni-bonn.de/people/schwede/algebraic_topological.pdf)). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in *loc. cit.*). Hence there isn't such a $C$...
| 32 | https://mathoverflow.net/users/1017 | 17034 | 11,399 |
https://mathoverflow.net/questions/16988 | 2 | Given any affine algebraic group $G$ over an algebraically closed field $\mathbb{F}$ of characteristic $0$ with a faithfull representation in $GL\_n(\mathbb{F})$ . If one knows the generators of the corresponding ideal, what can be said about the generators of $G^u$. Here $G^u$ shall denote the group generated by all unipotent elements of $G$. (Unlike the case where $G$ is irreducible and solvable, this group is not necessarily unipotent).
I am particular interested in bounds on the degrees of the generators; also any reference, which deals with unipotent generated groups is welcome.
| https://mathoverflow.net/users/4363 | generators of the ideal of an unipotent-generated algebraic group | Suppose we are over an algebraically closed field, and $G$ is connected. Then, we have an exact sequence
$$
1\to U\to G\to G\_r\to 1,
$$
where $U$ is the unipotent radical of $G$, and $G\_r$ is a reductive group.
Since a semisimple or unipotent group is generated by unipotent elements, this implies that $G^u$ is the intersection of the kernels
of all the characters of $G$. Characters of $G$ are grouplike elements of the Hopf algebra
${\mathcal O}(G)$. So the additional relations are that some grouplike elements $g\_j\in {\mathcal O}(G)$ generating the group of characters of $G$ are equal to $1$.
| 2 | https://mathoverflow.net/users/3696 | 17049 | 11,410 |
https://mathoverflow.net/questions/17023 | 6 | Background
----------
Let $\mathfrak{A}$, $\mathfrak{B}$, and $\mathfrak{C}$ be subcategories of the category of Banach spaces (over $\mathbb{R}$). Suppose we have a functor $\lambda:\mathfrak{A}^{op}\times\mathfrak{B}\to \mathfrak{C}$.
Let $f:E'\to E$ be a morphism belonging to $\mathfrak{A}$, and let $g:F\to F'$ be a morphism belonging to $\mathfrak{B}$. (Note: These are morphisms of topological vector spaces).
Then we have a map $$\matrix{Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))\\
(f,g)\mapsto\lambda(f,g)}$$
We say $\lambda$ is of class $C^p$ if for all manifolds $U$, and any two $C^p$ morphisms $U\to Hom(E',E)$ and $U\to Hom(F,F')$, the composition $$U\to Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))$$
is also of class $C^p$. (Note: We can replace $\mathfrak{A}$ and $\mathfrak{B}$ with categories of tuples to generalize this to several variables. In fact, this is what we do below.)
It is not hard to show that this induces a unique functor $$\lambda\_X:VB(X, \mathfrak{A})^{op}\times VB(X,\mathfrak{B})\to VB(X,\mathfrak{C}).$$ on vector bundles taking values in the appropriate vector bundle categories over $X$.
We define a **tensor bundle of type $\mathbf{\lambda}$** on $X$ to be $\lambda\_X(TX)=\lambda\_X((TX,\dots,TX),(TX,\dots,TX))$, where $TX$ is the tangent bundle.
However, this doesn't agree with the definition given on Wikipedia or anywhere else I've looked.
Questions
---------
* Is this terminology nonstandard?
* Is the notion itself nonstandard?
* If the terminology is nonstandard, but the notion is standard, does it have a different name?
* Is this definition useful?
* Does this include more vector bundles as tensor bundles than the standard definition?
| https://mathoverflow.net/users/1353 | Is Lang's definition of a tensor bundle nonstandard? | Is the notion non-standard? As Emerton says, the answer is no, except perhaps in minor details.
Is the terminology non-standard (and more permissive than the usual notion of tensor)? Yes, I'd say it is.
Because Lang allows arbitrary categories of Banach spaces, his notion is very general; by taking the categories small, one needn't have much functoriality at all. For instance, in the case where $\lambda$ has just one, covariant input, we can take $\mathfrak{B}$ to be the category with one object, $\mathbb{R}^n$, and morphisms $GL(n)$. Then $\lambda(\mathbb{R}^n)$ is a representation $V$ of $GL(n)$. On an $n$-manifold $X$, the resulting bundle $\lambda\_X(TX)$ is usually known as an associated bundle. Form the principal $GL(n)$-bundle $Fr\_X$ of frames (consisting of pairs of $x\in X$ with an isomorphism $T\_x X \to \mathbb{R}^n$). Then $\lambda\_X(TX)= Fr\_X \times\_{GL(n)}V$.
The notion of associated bundle is one that differential geometers often find more convenient than Lang's, because it allows one to say exactly which structure groups of interest (here, $GL(n,\mathbb{R})$) and exactly which representations.
A restrictive definition of tensor bundles allows those associated bundles where $V$ is a tensor product of copies of $\mathbb{R}^n$ and its dual. A broader (and I suspect fairly standard) definition allows subquotient representations of such - in particular, symmetric tensors and differential forms. But Lang apparently allows other things, e.g. $r$-densities (take the representation of $GL(n)\to \mathbb{R}^\times$ given by $g\mapsto |\det g|^r$).
Is the greater generality (compared to the standard notion of tensor) useful? In examples (e.g. densities) yes; in the abstract, not really - the interesting geometry is attached to specific classes of structure groups and their representations.
| 5 | https://mathoverflow.net/users/2356 | 17052 | 11,413 |
https://mathoverflow.net/questions/17072 | 42 | This question is inspired by the recent question ["The finite subgroups of SL(2,C)"](https://mathoverflow.net/questions/16026/the-finite-subgroups-of-sl2-c). While reading the answers there I remembered reading once that identifying the finite subgroups of SU(3) is still an open problem. I have tried to check this and it seems it was at least still open in the Eighties.
Can anyone confirm or deny that the finite subgroups of SU(3) are not all known? And if this is true, then what is the source of the difficulty?
Secondly, what is known of the finite subgroups of SU(n) for n > 3?
**UPDATE:** Thanks to those below who have corrected my ignorance! It seems that I may have been tricked by some particularly sensationalised abstracts (or perhaps just misunderstood them.)
| https://mathoverflow.net/users/3623 | The finite subgroups of SU(n) | There is an algorithm due to Zassenhaus which, in principle, lists all conjugacy classes of finite subgroups of compact Lie groups. I believe that the algorithm was used for $\mathrm{SO}(n)$ for at least $n=6$ if not higher. I believe it is expensive to run, which means that in practice it is only useful for low dimension.
---
**Added**
Now that I'm in my office I have my orbifold folder with me and I can list some relevant links:
1. Zassenhaus's original paper (in German) [Über einen Algorithmus zur Bestimmung der Raumgruppen](https://doi.org/10.1007/BF02568029)
2. There is a book by RLE Schwarzenberger *N-dimensional crystallography* with lots of references
3. There are a couple of papers in *Acta Cryst.* by Neubüser, Wondratschek and Bülow titled *On crystallography in higher dimensions*
4. There is a sequence of papers in *Math. Comp.* by Plesken and Pohst titled *On maximal finite irreducible subgroups of GL(n,Z)* which I remember were relevant.
---
Independent of this algorithm, there is some work on $\mathrm{SU}(n)$ from the physics community motivated by elementary particle physics and more modern considerations of the use of orbifolds in the gauge/gravity correspondence.
The case of $\mathrm{SU}(3)$ was done in the mid 1960s and is contained in the paper [Finite and Disconnected Subgroups of SU(3) and their Application to the Elementary-Particle Spectrum](http://link.aip.org/link/?JMAPAQ/5/1038/1) by Fairbairn, Fulton and Klink. For the case of $\mathrm{SU}(4)$ there is a more recent paper [A Monograph on the Classification of the Discrete Subgroups of SU(4)](https://arxiv.org/abs/hep-th/9905212) by Hanany and He, and references therein.
---
**Further edit**
The paper [Non-abelian finite gauge theories](https://arxiv.org/abs/hep-th/9811183) by Hanany and He have the correct list of finite subgroups of SU(3), based on Yau and Yu's paper [Gorenstein quotient singularities in dimension three](https://mathscinet.ams.org/mathscinet-getitem?mr=1169227).
| 43 | https://mathoverflow.net/users/394 | 17074 | 11,425 |
https://mathoverflow.net/questions/17077 | 2 | Let's start with some family of algebraic structures of the same type indexed by the natural numbers, say the symmetric group $S\_n$. Suppose that the axioms of this algebraic structure (in this case, groups) can be stated within the framework of first-order logic. In this way, we can consider a structure $M$ defined as follows: it contains $\mathbb{N}$, as well as all the $S\_n$. Consider the complete theory $Th(M)$ in first-order logic (i.e. containing all the symbols of $M$ and a symbol for each relation on $M$ of any arity).
In this theory, there are relations $m, e, i$ that correspond to the relations of multiplication, identity, and inverse, and a relation $P$ such that $P(a,b)$ iff $a \in S\_n$ and $b = n$. There are a bunch of axioms that are satisfied, e.g. if $P(a,b)$ then $P( i(a), b)$ and $m( a, i(a)) = e(b)$, etc., that correspond to the group laws. There is also a relation $isParameter$ that tells you whether the object in question is a parameter (i.e. a natural number).
So, in this way (am I misunderstanding this?) one can use an ultraproduct construction (or the compactness theorem applied to $Th(M)$ together with the collection of sentences $isParameter(c) \wedge c> 1 + \dots + 1$ where $c$ is some new constant symbol) to embed $M$ in a bigger structure $M'$ where there are parameters greater than all the standard elements of $\mathbb{N}$, i.e. where the parameters are (some version of) the hypernatural numbers. Since the group $S\_n$ is nonempty for each standard $n$, it should follow by transfer that $S\_n$ is also defined for infinite $n$. It must be a group by transfer.
>
> **Question:** Is $S\_{n}$ for infinite $n$ in any way related to the set of
> permutations of the interval from $1$
> to $n$? If not, what can we say about it?
>
>
>
My hunch is that this probably isn't the case, because "every permutation is contained in $S\_n$" sounds like a second-order statement and this is first-order logic we're dealing with. Nevertheless, I'm curious about what we can say about $S\_{n}$ for infinite $n$, and whether we can deduce additional properties about $S\_n$ for infinite $n$ from the known theory for finite $n$.
| https://mathoverflow.net/users/344 | Algebraic structures at hypernatural parameters | You can build your group directly as an ultraproduct without fussing about the particular language. Namely, let U be any nonprincipal ultrafilter on ω, and let S be the ultraproduct Πn Sn/U. That is, one defines f ≡ g in the product Π Sn if and only if { n | f(n) = g(n) } ∈ U. This is an equivalence relation, and the ultraproduct is the collection of equivalence classes [f], where the algebraic structure is well-defined coordinate-wise. Los's theorem then states that S will satisfy a statement about [f] in the language of groups if and only if the set of n for which f(n) has the property in Sn has the property is in U. For example, S will have elements of infinite order, since we can let f(n) select an element of Sn of order n, and then observe that for any fixed k, the set of n for which f(n) has order at least k is co-finite and hence in U. As Gerald observed, S is not itself a full symmetric group, since one will never be able to define the standard/nonstandard cut.
But let me describe a more general way to accomplish what you want, for all kinds of structures simultaneously. There is little reason when taking ultraproducts to restrict to any particular structure. Rather, one should simply consider the ultrapower of the entire set-theoretic universe V. This results in a new set-theoretic universe W = Vω/U that satisfies all the truths about any [f] that are true of f(n) in V on a set in U. In particular, S is the just [⟨ Sn | n ∈ ω ⟩] in the universe W. (What this shows is that ultraproducts of any set structures are elements of the ultrapower of the universe, and in this sense, ultraproducts are a special case of ultrapowers, although one usually hears the converse.) With this construction, we are not limited to the language of group theory when discussing properties of your group S, and we can freely refer to properties involving subgroups, group extensions and whatever else is expressible in set theory. If almost every Sn (that is, on a set in U) has some property expressible in set theory, then S will have this property in W. Thus, the group S is a "finite" permutation group inside W on the nonstandard natural number N represented by the identity function id(n) = n. That is, W thinks that S is just SN, where N = [id]. So any property that you can prove about all Sn, will be true of S in W. Some but not all of these properties are absolute between W and V, and it is often interesting to compare the differences.
What this shows is that there IS a sense in which your group S is a full group of permutations on a set, because it is the group of all permutations on N inside W. That is, it is a full symmetric group inside the alternative set-theoretic universe W, rather than in the original universe V. In particular, Gerald's counterexample permutation does not exist in W, since W also cannot define the standard/nonstandard cut.
| 4 | https://mathoverflow.net/users/1946 | 17082 | 11,431 |
https://mathoverflow.net/questions/17007 | 9 | I'm looking for a (comprehensible) reference for the Frolicher-Nijenhuis bracket, hopefully more down to earth than Michor's books and different from Saunders's book on Jets.
I'm interested in it as this bracket seems the appropriate means to define the curvature of a general Ehresmann connection on a bundle.
Alternatively it would be really great to have a reference (again different from Michor) that defines connections, curvature and holonomy on general bundles, later specializing to principal bundles, vector bundles and maybe even to the riemannian setting.
| https://mathoverflow.net/users/3701 | Reference for the Frolicher-Nijenhuis Bracket | I also only learned about Frolicher-Nijenhuis brackets from Saunders' book on jets but I doubt that there is any other authorative reference besides Michor's book and the original papers. I don't know if this is what the original poster intended, but here's how I understand the link between FN and curvature. Generally, I tend to stay away from the full generality of the FN bracket, and only use those (axiomatic) properties that I need...
Let $\pi: E \rightarrow M$ be a fiber bundle. An Ehresmann connection is a subbundle $H$ of $TE$ such that $H \oplus VE = TE$, where $VE$ is the vertical bundle. Denote the projector from $TE$ onto $H$ by $h$. Saunders (and many other authors) define Ehresmann connections directly in terms of bundle maps $h$, since they are easier to work with, but this doesn't matter.
Now, to introduce curvature, we would like to formalize the idea that curvature is the failure of "parallel transport" to commute, where of course we haven't define parallel transport properly. However, it is only a small step of the imagination to guess that this must be related to the integrability of $H$, i.e.whether $[h(X), h(Y)] \in H$ for arbitrary vector fields $X, Y$. The failure of two horizontal vector fields to be horizontal again is measured by the expression
$$
R(X,Y) = [h(X), h(Y)] - h([h(X), h(Y)]),
$$
and in fact this is nothing but Saunders' definition 3.5.13 of curvature. Note that $R(X, Y) = 0$ if either $X$ or $Y$ is in $VE$.
The Frohlicher-Nijenhuis bracket of two linear bundle maps is in general quite complicated, but if you look at proposition 3.4.15 in Saunders, you get that for a linear bundle map $h: TE \rightarrow TE$,
$$
[h, h] (X,Y) = 2(h([X,Y]) + [h(X), h(Y)] - h([h(X),Y]) - h([X,h(Y)])).
$$
Now note that the right-hand side is zero as soon as either $X$ or $Y$ is in $VE$, just as for $R$. If both $X$ and $Y$ are horizontal, we can apply the above formula to $X = h(X)$ and $Y = h(Y)$, and conclude that $[h, h] (X, Y) = 2( [h(X), h(Y)] - h([h(X), h(Y)]))$ (where we use the identity $h \circ h = h$ liberally), and so
$$
R = \frac{1}{2} [h,h].
$$
If you have a principal fiber bundle, $h$ is related to the connection one-form $\mathcal{A}$, and the above formula gives the curvature of $\mathcal{A}$ in terms of the covariant differential of $\mathcal{A}$. For a symplectic connection, something similar happens.
**Edit:** here's how I think it works for principal fiber bundles. Take a connection one-form $A: TE \to \mathfrak{g}$, and let $\sigma: \mathfrak{g} \rightarrow VE$ be the infinitesimal generator of the $G$-action. The composition $v := \sigma \circ \mathcal{A}$ is then the vertical projector of the connection and $h := 1 - v$ is the horizontal one. Now plug this expression for $h$ into the formula for the curvature:
$$
[h, h] = [1, 1] - [\sigma \circ \mathcal{A}, 1] - [1, \sigma \circ \mathcal{A}]
+ [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}].
$$
The term $[1,1]$ is zero, and you can use Saunders' proposition 3.4.15 to show that term 2 and 3 vanish as. The last term can be written as (again using S3.4.15) as
$$
[\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) =
2( (\sigma \circ \mathcal{A})^2([X, Y]) + [ (\sigma \circ \mathcal{A})(X), (\sigma \circ \mathcal{A})(Y)] - (\sigma \circ \mathcal{A})([(\sigma \circ \mathcal{A})(X), Y)]) - (\sigma \circ \mathcal{A})([X, (\sigma \circ \mathcal{A})(Y)])).
$$
Now, show that this vanishes whenever $X$ or $Y$ is vertical, so that we can take $X$ and $Y$ to be horizontal. In that case, the above simplifies to
$$
[\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 2(\sigma \circ \mathcal{A})([X, Y])
$$
or
$$
R(X, Y) = (\sigma \circ \mathcal{A}) ([X^h, X^h])
$$
where $X^h$ represents the horizontal part of $X$. But $\mathcal{A} ([X^h, X^h])$ is just the negative of the curvature (as a two-form with values in $\mathfrak{g}$) of $\mathcal{A}$, so that
$$
R(X, Y) = -\sigma ( \mathcal{B}(X, Y) ).
$$
| 6 | https://mathoverflow.net/users/3909 | 17086 | 11,434 |
https://mathoverflow.net/questions/17091 | 4 | Edit (first version was incorrectly stated. Thank you Douglas and others for your corrections) Let $B\_n$ be the $n$th Bell number (the number of partitions of a set with $n$ members). For each $n > 3$, I have a set $A\_n$ of size $|A\_n|=B\_n$. I then have a subset $A'\_n \subset A\_n$ where $|A'\_n|=B\_n-B\_{n-1}$. I would like to say something about the size of $A'\_n$ relative to the size of $A\_n$. For instance, it seems that $lim\_{n \to \infty} \frac{B\_{n-1}}{B\_n}=0$. Can I make a stronger statement about the ratio of successive Bell numbers? How can I formalize the statement "for sufficiently large $n$, most of $A\_n$ is in $A'\_n$."
| https://mathoverflow.net/users/4250 | On the Bell Numbers | It's easy to see that $B\_n \ge 2 B\_{n-1}$ since we always have a choice of whether to add $n$ to the same part as $n-1$ or not. Since the number of parts in a typical set partition of size $n-1$ grows, the choices for adding $n$ to a new or existing part grow, so
$$\lim\_{n\to\infty} B\_{n-1}/B\_n = 0.$$
There are asymptotics in the [Wikipedia article on the Bell numbers](http://en.wikipedia.org/wiki/Bell_number), but it may not be obvious how to work with the Lambert $W$-function in that expression, or how to bound $B\_{n-1}/B\_n$. A faster proof that the limit is $0$ can be obtained from Dobiński's formula, that $B\_n$ is the $n$th moment of a Poisson distribution with mean $1$:
For any $c \in \mathbb R$, the Poisson distribution has positive probability of being greater than $c$. So, for large enough $n$, the $n$th moment $B\_n$ is at least $c^n$. By Jensen's inequality, the moments satisfy
$$B\_n^{\frac{n+1}{n}} \le B\_{n+1}$$
$$c \le \sqrt[n]{B\_n} \le \frac {B\_{n+1}}{B\_n}$$
| 8 | https://mathoverflow.net/users/2954 | 17097 | 11,442 |
https://mathoverflow.net/questions/17062 | 28 | A student asked me why $\mathcal{O}\_K$ is the notation used for the ring of integers in a number field $K$ and why $h$ is the notation for class numbers. I was able to tell him the origin of $\mathcal{O}$ (from Dedekind's use of Ordnung, the German word for order, which was taken from taxonomy in the same way the words class and genus had been stolen for math usage before him), but I was stumped by $h$. Does anyone out there know how $h$ got adopted?
I have a copy of Dirichlet's lecture notes on number theory (the ones Dedekind edited with his famous supplements laying out the theory of ideals), and in there he is using $h$. So this convention goes back at least to Dirichlet -- or maybe Dedekind? -- but is that where the notation starts? And even if so, why the letter $h$?
I had jokingly suggested to the student that $h$ was for Hilbert, but I then told him right away it made no historical sense (Hilbert being too late chronologically).
| https://mathoverflow.net/users/3272 | Why is "h" the notation for class numbers? | Gauss, in his Disquisitiones, used ad hoc notation for the class number when he needed it. He did not use h. Dirichlet used h for the class number in 1838 when he proved the class number formula for binary quadratic forms. I somewhat doubt that he was thinking of "Hauptform" in this connection - back then, the group structure was not as omnipresent as it is today, and the result that $Q^h$ is the principal form was known (and written additively), but did not play any role. Kummer, 10 years later, used H for the class number of the field of p-th roots of unity, and h for the class number of a subfield generated by Gaussiam periods (and "proved" that $h \mid H$); in the introduction he quotes Dirichlet's work on forms at length.
| 22 | https://mathoverflow.net/users/3503 | 17098 | 11,443 |
https://mathoverflow.net/questions/16850 | 24 | Here is an updated formulation of the question, which is more precise and I think completely correct:
Suppose $M$ is a Riemannian manifold. Pick a point $p$ in $M$ and let $U$ be a neighborhood of the origin in $T\_p M$ on which $exp\_p$ restricts to a diffeomorphism. Let $X$ and $Y$ be tangent vectors in $T\_p M$, and let $V$ be the intersection of $U$ and the plane spanned by $X$ and $Y$. Let $c(t)$ be a piecewise smooth simple closed curve in $V$. I claim that for any vector $Z$ in $T\_p M$,
$R(X,Y)Z=(P\_c(Z)−Z)Area(c)+o(Area(c))$
where $R$ is the Riemannian curvature tensor, $P\_c$ is parallel transport around the image of $c$ under $exp\_p$, and $Area(c)$ is the area enclosed by the image of $c$ under $exp\_p$.
Can anyone refer me to a proof of this statement or something similar? I am fairly sure the argument has something to do with integrating the curvature 2-form over the embedded surface obtained by restricting $exp\_p$ to the region enclosed by $c$, but I am having trouble with the estimates. Unfortunately I can't find anything in Kobayashi and Nimazu.
Thanks in advance!
Paul
| https://mathoverflow.net/users/4362 | Curvature and Parallel Transport | It appears to me that one reason why nobody has proved the formula yet is that the formula is still wrong. First, the formula has to depend on $X$ and $Y$. If you rescale $X$ and $Y$, the left side of the formula scales but the right side stays constant. That can't be. Second, the two sides of the equation do not scale the same under a constant scaling of the metric.
Warning: I wrote this up very quickly and did not check for typos and errors. It's possible that my final formula is still not right, but I am confident that my argument can be used to obtain a correct formula. I also did not provide every last detail, so, if you're unfamiliar with an argument like this, you need to do a lot of work making sure that everything really works. The key trick is pulling everything back to the unit square, where elementary calculus can be used. I'm sure this trick can be replaced by Stokes' theorem on the manifold itself, but that's too sophisticated for my taste.
[Holonomy calculation](http://www.deaneyang.com/papers/holonomy.pdf)
ADDED:
The correct formula, if you assume $|X\wedge Y| = 1$, is
$P\_\gamma Z - Z = Area(c) R(X,Y)Z$
This scales properly when you rescale the metric by a constant factor. Notice that the left side is invariant under rescaling of the metric.
I recommend looking at papers written by Hermann Karcher, especially the one with Jost on almost linear functions, the one with Heintze on a generalized comparison theorem, and the one on the Riemannian center of mass. I haven't looked at this or anything else in a long time, but I have the impression that I learned a lot about how to work with Jacobi fields and Riemann curvature from these papers.
Finally, don't worry about citing anything I've said or wrote. Just write up your own proof of whatever you need. If it happens to look very similar to what I wrote, that's OK. I consider all of this "standard stuff" that any good Riemannian geometer knows, even if they would say it differently from me.
EVEN MORE: There are similar calculations in my paper with Penny Smith: P. D. Smith and Deane Yang
*Removing Point Singularities of Riemannian Manifolds*, TAMS (333) 203-219, especially in section 7 titled "Radially parallel vector fields". In section 5, we attribute our approach to H. Karcher and cite specific references.
| 22 | https://mathoverflow.net/users/613 | 17099 | 11,444 |
https://mathoverflow.net/questions/17100 | 13 | Let $G$ be a Lie group and let $H$ be a closed subgroup of $G$. Then $G/H$ may not be a group, but it will be a homogeneous space for $G$ with stabilizers conjugate to $H$. Sometimes, this is a variety, for instance, when $G$ is a complex reductive group and $H$ is a complex subgroup, and it will even be projective when $H$ is parabolic (by definition).
However, when we take real Lie groups, the situation seems more subtle. For instance, if $G=Sp(g,\mathbb{R})$ and $H=U(g)$, then $G/H$ is the Siegel upper half space, which is an (analytic) open set in a variety, namely, the variety is the space of symmetric $g\times g$ matrices and the open set is given by the ones with positive imaginary part. Similarly, many constructions in Hodge theory, particularly that of period domains, end up coming from real Lie groups, and so may be varieties, open subsets of varieties, or not varieties at all *a priori*. Clearly, the quotient being even dimensional is a necessary condition, but I'd be surprised if it were sufficient.
So the first part of the question is
>
> When is a homogeneous space (an open subset of) a variety?
>
>
>
Now, additionally, these period domains often can be quotiented by a discrete (I believe Griffiths says arithmetic) subgroup of the original group to actually get a variety. For instance (if I'm understanding right) if we take $\mathfrak{h}\_g=Sp(g)/U(g)$ above, we can quotient further by $Sp(g,\mathbb{Z})$ and this gives us $\mathcal{A}\_g$, the moduli space of abelian varieties.
>
> When is there a discrete subgroup that we can take a further quotient by to get a variety?
>
>
>
| https://mathoverflow.net/users/622 | When is a homogeneous space a variety? | I'll try to answer both questions, though I will change the first question somewhat. Let's work in the setting of a real reductive algebraic group $G$ and a closed subgroup $H \subset G$.
Your first question asks when $G/H$ is an open subset of some (presumably complex) variety. I think that this question should be modified in a few ways.
You can't really say that $G/H$ "is a subset" of a variety, since $G/H$ is not a priori endowed with a complex structure. So you need a bit more data to go with the question -- a complex structure on the homogeneous space $G/H$. Such a complex structure can be given by an embedding of the circle group $U(1)$ as a subgroup of the center of $H$. Let $\phi: U(1) \rightarrow G$ be such an embedding, and let $\iota = \phi(i)$ be the image of $e^{pi i} \in U(1)$ under this map. Such an embedding yields an integrable complex structure on the real manifold $G/H$, I believe (though I haven't seen this stated in this degree of generality).
So now one can ask if $G/H$, endowed with such a complex structure, is an open subset of a complex algebraic variety. But again, I have some objection to this question -- it's not really the right one to ask. Indeed, it's very interesting when one finds that some quotients $\Gamma \backslash G /H$ are (quasiprojective) varieties -- but such quotients are not obtained as quotients in a category of varieties, from $G/H$ to $\Gamma \backslash G / H$. They are complex analytic quotients, but not quotient varieties in any sense that I know.
So what's the point of knowing whether $G/H$ is an open subset of a variety? Really, one needs to know properties of $G/H$ as a Riemannian manifold and complex analytic space (e.g. curvature, whether it's a Stein space). That's the most important thing!
As Kevin Buzzard and his commentators note, under the assumption that $G$ comes from a reductive group over $Q$, *and* under the assumption that $H$ is a maximal compact subgroup of $G$, *and* under the assumption that there is a "Shimura datum" giving the quotient $G/H$ a complex structure, the quotient $G/H$ is a period domain for Hodge structures, and the quotients $\Gamma \backslash G / H$ are quasiprojective varieties when $\Gamma$ is an arithmetic subgroup of $G$.
But these are quite strong conditions, on $G$ and on $H$! I have also wondered about other situations when $X = \Gamma \backslash G / H$ might have a natural structure of a quasiprojective variety. A general technique to prove such a thing is to use a differential-geometric argument. A great theorem along this line is due to Mok-Zhong (Compactifying complete Kähler-Einstein manifolds of finite topological type and bounded curvature, Ann. of Math 1989). The theorem, as quoted from MathSciNet, reads:
"Let $X$ be a complex manifold of finite topological type. Let $g$ be a complete Kähler metric on $X$ of finite volume and negative Ricci curvature. Suppose furthermore that the sectional curvatures are bounded. Then $X$ is biholomorphic to a Zariski-open subset $X'$ of a projective algebraic variety $M$."
Such results can be applied to prove quasiprojectivity of Shimura varieties of Hodge type. I believe I first learned this by reading J. Milne's notes on Shimura varieties.
I tried once to apply this to an arithmetic quotient of $G/H$, where $H$ was a bit smaller than a maximal compact (when $G/H$ was the twistor covering of a quaternionic symmetric space) -- I couldn't prove Mok-Zhong's conditions for quasiprojectivity, and I still don't know whether such quotients are quasiprojective.
| 7 | https://mathoverflow.net/users/3545 | 17113 | 11,452 |
https://mathoverflow.net/questions/17103 | 11 | I'm reading some very old papers (by Birch et al) on quadratic forms and I don't get the following point:
>
> If $f$ is a quadratic form in $X\_1,X\_2,\cdots,X\_n$ over a
> finite field, then one can change variables such that $f$ can be written as $\sum\_{i = 1}^s Y\_{2i - 1}Y\_{2i} + g$, where $g$ is a quadratic form which involves
> variables other than
> $Y\_1,Y\_2,\cdots,Y\_{2s}$ and has order at
> most 2 (i.e. can be written using at
> most two linear forms).
>
>
>
So either this is a well-known result - but I don't find a reference - or either this is easy to see, but in that case I'm just missing the point. By the way, is this really true in characteristic 2?
(And in fact I'm not sure what role is played by the fact that the field is finite...)
| https://mathoverflow.net/users/1107 | Quadratic forms over finite fields | I will sketch below a standard argument to show what you need, because I find it very neat!
Let $V$ be a finite dimensional vector space over a field $k$ and let $q \colon V \to k$ be a quadratic form on $V$. Denote by $b$ the symmetric bilinear form associated to $q$: thus for vectors $v,w \in V$ define $b(v,w) := q(v+w)-q(v)-q(w)$. Suppose that $v$ is a non-singular zero of $q$. Since $v$ is non-singular, it follows that there is a $w' \in W$ such that $\alpha := b (v , w') \neq 0$. Let $w := \frac{1}{\alpha^2} (\alpha w' - q(w') v)$; it is immediate to check that $q(w)=0$ and $b (v , w) = 1$. Observe that the ``orthogonal complement'' of $v,w$ with respect to the form $q$ has codimension two and does not contain the span of $v,w$. Thus, we conclude that we can find a basis of $V$ such that $q(x\_1,\ldots,x\_n) = x\_1 x\_2 + q'$, where $q'$ is a quadratic form over a space of dimension two less than the dimension of $V$.
The statement about finite fields follows at once, since over a finite field, any quadratic form in three or more variables admits a non-trivial zero. This is a consequence of the Chevalley-Warning Theorem. More generally, any field such that quadratic forms in three variables always admit a zero has the property you need, e.g. any $C\_1$-field would work.
| 10 | https://mathoverflow.net/users/4344 | 17119 | 11,455 |
https://mathoverflow.net/questions/17117 | 12 |
>
> Theorem. Fix $\epsilon > 0$; for sufficiently large n, any graph with n vertices and $\epsilon \binom{n}{2}$ edges contains many (nondegenerate) cycles of length 4.
>
>
>
The proof is simple; put an indicator variable $\delta\_{x, y}$ for each pair of vertices corresponding to whether or not there is an edge there; then start with
$n^8 \epsilon^4 = (\sum \delta\_{x, y})^4$
and apply Cauchy-Schwarz twice; finally, note that there are $O(n^3)$ "degenerate 4-cycles".
A basic corollary of this is the following fact:
>
> Corollary. Any graph with girth at least 5 and n vertices has $o(n^2)$ edges.
>
>
>
This seems like it should be possible to prove without resorting to "analytic" machinery like Cauchy-Schwarz; indeed, it seems like it should be weak enough to prove almost by arguing "locally." But none of the obvious lines of reasoning seem to provide a proof.
Is it possible to get a good bound on the density of large-girth graphs without using Cauchy-Schwarz or equivalent?
| https://mathoverflow.net/users/382 | Combinatorial proof that large-girth graphs are sparse? | It's known more specifically that any graph with girth ≥ 5 has $O(n^{3/2})$ edges — see e.g. [Wikipedia on the Zarankiewicz problem](http://en.wikipedia.org/wiki/Zarankiewicz_problem).
Here's a combinatorial proof. Suppose that graph $G$ has $\ge kn^{3/2}$ edges for a sufficiently large constant $k$. As long as there are vertices with degree smaller than some appropriate constant times $\sqrt n$ one can remove them and get a smaller graph with the same property of having at least $kn^{3/2}$ edges, so eventually one can reach a state where every vertex has degree at least $\Omega(\sqrt n)$. Once this happens, there are $O(n^2)$ possible pairs of neighbors that a vertex might have, and each vertex has $\Omega(n)$ pairs of neighbors, so some pair of neighbors appears twice causing the graph to have a 4-cycle.
| 20 | https://mathoverflow.net/users/440 | 17121 | 11,457 |
https://mathoverflow.net/questions/17110 | 10 | If I have a flat family $f \colon X \to T$ such that some fiber is (locally) a complete intersection, does that imply that there is an open set $U$ in $T$ such that the fibers above $U$ are (locally) complete intersections?
In general, what kind of intuition should one have about which properties are "open" with respect to flat families?
| https://mathoverflow.net/users/321 | Complete intersections and flat families | EGA IV$\_4$, 19.3.8 (and 19.3.6); this addresses openness upstairs without properness, and (as an immediate consequence) the openness downstairs if $f$ is proper (which I assume you meant to require).
The general intuition is that openness holds upstairs for many properties, and so then holds downstairs when map is proper. As for proving openness upstairs, the rough idea is to first prove constructibility results, and then refine to openness by using behavior under generization. But it's a long story, since there are many kinds of properties one can imagine wanting to deal with. These sorts of things are developed in an extraordinarily systematic and comprehensive manner in EGA IV$\_3$, sections 9, 11, 12 (especially section 12 for the niftiest stuff).
| 17 | https://mathoverflow.net/users/3927 | 17123 | 11,458 |
https://mathoverflow.net/questions/17128 | 21 | Let $M$ be a topological monoid. How does the homology-formulation of the group completion theorem, namely (see McDuff, Segal: *Homology Fibrations and the "Group-Completion" Theorem*)
>
> If $\pi\_0$ is in the centre of $H\_\*(M)$ then $H\_\*(M)[\pi\_0^{-1}]\cong H\_\*(\Omega BM)$
>
>
>
imply that $M\to \Omega BM$ is a weak homotopy equivalence if $\pi\_0(M)$ is already a group? I don't see the connection to homology. Can one prove the latter (perhaps weaker) statement more easily than the whole group completion theorem?
A topological group completion $G(M)$ of $M$ should transform the monoid $\pi\_0(M)$ into its (standard algebraic) group completion. But a space with this property is not unique. Why is $\Omega BM$ the "right" choice? Perhaps this is clear when I see the connection to the homology-formulation above.
| https://mathoverflow.net/users/4011 | Group completion theorem | The statement that $M \to \Omega BM$ is a weak equivalence when $M$ is a group-like topological monoid is indeed easier: the map $EM = B(M \wr M) \to BM$ is then a quasi-fibration, has geometric fibre $M$ over the basepoint and homotopy fibre $\Omega BM$.
However the homological group-completion theorem also implies this: if $M$ is group-like then $\pi\_0(M)$ already consists of units in $H\_\*(M)$, so it just says that $M \to \Omega BM$ is a homology equivalence. Each of these spaces has homotopy equivalent path components, so it is then enough to observe that the map of 0 components is a homology equivalence between simple spaces, so a weak homotopy equivalence.
However it is perverse to prove the "$M \simeq \Omega BM$" result this way.
| 22 | https://mathoverflow.net/users/318 | 17134 | 11,464 |
https://mathoverflow.net/questions/17105 | 7 | Let $0 < \alpha < 1$ be a constant. The expected number of prime factors of a "random" integer near $n$ which are greater than $n^\alpha$ is $-\log \alpha$.
It's my understanding that (properly formulated) this is a well-known fact in analytic number theory but I cannot find a reference for it. Can anybody provide a reference?
**Edited to add (March 28):**: The asymptotic density of positive integers $n$ with $k$th largest factor smaller than $n^{1/\alpha}$ is $\rho\_k(\alpha)$, where we have $L\_0(\alpha) = [\alpha > 0]$ and
$$ L\_k(\alpha) = [\alpha \ge k] \int\_k^\alpha L\_{k-1}(t-1) \: {dt \over t}, $$
and $1-\rho\_k(\alpha) = \sum\_{n=0}^\infty {-k \choose n} L\_{n+k}(\alpha)$. (See Riesel, p. 162.) The density of positive integers with $k$th largest factor *larger* than $n^{1/\alpha}$ is therefore $1-\rho\_k(\alpha)$, and so the expected number of factors larger than $n^{1/\alpha}$ is $\sum\_{k \ge 1} (1-\rho\_k(\alpha))$. Therefore the expected number of such factors is
$$ \sum\_{k \ge 1} \sum\_{n \ge 0} {-k \choose n} L\_{n+k}(\alpha). $$
Letting $n+k = j$ we can rewrite this sum as
$$ \sum\_{j \ge 1} \sum\_{n=0}^{j-1} {n-j \choose n} L\_j = \sum\_{j \ge 1} L\_j \left( \sum\_{n=-0}^{j-1} (-1)^n {j-1 \choose n} \right) $$
and the inner sum is $0$ except when $j=1$, when it is $1$. So the expected number of factors larger than $n^{1/\alpha}$ is $L\_1(\alpha)$; this is $\log \alpha$.
| https://mathoverflow.net/users/143 | Reference for the expected number of prime factors of n larger than n^alpha is -log alpha | Theorem 5.4 of Riesel, Prime Numbers and Computer Methods for Factorization, says "the number of prime factors $p$ of integers in the interval $[N-x,N+x]$ such that $a<\log\log p< b$ is proportional to
$b-a$ if $b-a$ as well as $x$ are sufficiently large as $N\to\infty$."
| 3 | https://mathoverflow.net/users/3684 | 17143 | 11,469 |
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