diff --git "a/8tE4T4oBgHgl3EQfdQys/content/tmp_files/2301.05090v1.pdf.txt" "b/8tE4T4oBgHgl3EQfdQys/content/tmp_files/2301.05090v1.pdf.txt" new file mode 100644--- /dev/null +++ "b/8tE4T4oBgHgl3EQfdQys/content/tmp_files/2301.05090v1.pdf.txt" @@ -0,0 +1,4510 @@ +arXiv:2301.05090v1 [math.MG] 12 Jan 2023 +Divide and Conquer: A Distributed Approach +to Five Point Energy Minimization +Richard Evan Schwartz +January 13, 2023 +1 +Introduction +The purpose of this work is to rigorously verify the phase-transition for 5 +point energy minimization first observed in [MKS], in 1977, by T. W. Mel- +nyk, O, Knop, and W. R. Smith. +Our results contain, as special cases, +solutions to Thomson’s 5-electron problem and Polya’s 5-point problem. +This work is an updated version of my monograph from 6 years ago. I +simplified the proof significantly and also I wrote this version in an experi- +mental style designed to facilitate the verification process. This work is just +over half as long as the original. +I wrote the proof in a tree-like form. +Thus, the Main Theorem is an +immediate consequence of Lemma A, Lemma B, and Lemma C. These three +Lemmas are independent from each other. Lemma A is an immediate conse- +quence of Lemma A1 and Lemma A2. And so on. All the “ends” of the tree, +such as Lemma B21121, either have short and straightforward proofs or are +computer calculations which I will describe in enough detail that a compe- +tent programmer could reproduce them. At the same time, all my computer +programs are available to download and use. Figures 0 and 01 below map +out the complete logical structure of the proof of the Main Theorem. +The rest of this introduction states the results and explains how to divide +the verification of the proof into small pieces. Following this, §2 contains a +discussion of the history and context of the results, a high-level discussion of +the ideas in the proof, a discussion of the computer experiments I did, and a +guide to the relevant software I wrote. Following this we get to the proof. +1 + +Results: Let S2 be the unit sphere in R3. Given a configuration {pi} ⊂ S2 +of N distinct points and a function F : (0, 2] → R, define the F-potential of +the configuration: +EF(P) = +� +1≤i 0. +(2) +This is also called the s-potential and the power law potential. +The Triangular Bi-Pyramid (TBP) is the 5 point configuration having one +point at the north pole, one point at the south pole, and 3 points arranged +in an equilateral triangle on the equator. A Four Pyramid (FP) is a 5-point +configuration having one point at the north pole and 4 points arranged in a +square equidistant from the north pole. +Define +15+ = 15 + 24 +512, +15++ = 15 + 25 +512. +(3) +Theorem 1.1 (Main) There existש∈(15+,15++) such that: +1. For s ∈ (0,ש) the TPB is the unique minimizer for Rs. +2. For s =ש the TBP and some FP are the two minimizers for Rs. +3. For each s ∈ (ש,15++) some FP is the unique minimizer for Rs. +In Statement 3, the FP presumably depends on s. The numberש is a new +constant of nature. Its decimal expansion startsש=15.0480773927797... +In §3.5 I will also explain the main details the following theorem. +Theorem 1.2 (Auxiliary) The TBP is the unique minimizer for the Fejes- +Toth potential F(r) = −r−s for all s ∈ (−2, 0). +The original monograph bundled these results together. Here I separate +the results and concentrate on the Main Theorem. +2 + +Lemma Trees: Figure 0 shows most of the lemmas proved in this mono- +graph and how they contribute to the proof of the main theorem. The only +thing missing are the trees of implication for Lemmas B and C1, which we +show in Figure 01 below. The color coding indicates different independent +parts of the monograph. There are 7 colors. Each color (so to speak) may +be read independently from the others. I have indicated the nature of each +colored part, as well as the page numbers containing it, with tags. In the +interest of space I have not given the full name of every lemma. Thus, the +rightmost branch of the tree ends at Lemma C313. + + + + +E +5 +1 +3 +2 +1 +error estimate +41-52 +interpolation +53-61 +local analysis +36-40 +outline +16-21 +A135 +3 +main calculation +22-35 +Figure 0: The tree of implications. +3 + +VThe starred lemmas are the divide-and-conquer computer calculations. +The big one, for A13, is done with interval arithmetic. The others are done +with exact integer arithmetic, though I use floating point operations, in a way +that does not interfere with the logic of the proof, to guide the algorithm. The +general logic is that I’ll compute something with floating point calculations +to see if it is a good candidate for formal verification and then, if so, I’ll +formally verify it with exact calculations. The square nodes indicate sizeable +calculations done with exact evaluations of polynomials in Mathematica. +B3 +1 +1 +Figure 01: The trees of implication for Lemma B and Lemma C1. +Figure 01 shows the trees of implication for Lemma B and for Lemma +C1. This part of the monograph is 20 pages long. I have indicated how it +may be divided up into 4 independent parts, each of which could be verified +separately from the others. The various parts all require the material on +pp 65-66, which explains the computational methods. The proofs of Lemma +B and C1 also involve computer-assisted calculations, but these are done +exactly, in Mathematica, by analyzing the properties of the coefficients of in- +teger polynomials. The square nodes of the tree indicate the lemmas which +rely on such calculations. +4 + +DDtria70厂 +53PDVerification: As Figure 0 indicates, a team of 7 readers could check the +mathematical part of the proof, with the team-leader reading the 6-page +outline and the rest of the people reading self-contained sections at most +20 pages long. (Some readers might also want to consult the discussion on +pp 6-15 to get insight into where the ideas come from.) Figure 01 indicates +how the 20-page portion on symmetrization could be further divided into 4 +independent pieces, each no more than 8 pages. +At the same time, I wrote the computer code in such a way that the pro- +grams for each part are independent from the programs for each other part. +It is probably easier in each case to reproduce the code rather than check +that mine is correct. Each reader could team up with a strong computer +programmer who could reproduce the relevant code. This would be a serious +job only for the calculation in Lemma A13. The rest of the code could be +reproduced, in each case, in a day. I’d say that the code for Lemma A13 +could be reproduced in a few days by one person. +Acknowledgements: I thank Henry Cohn, Doug Hardin, John Hughes, +Abhinav Kumar, Curtis McMullen, Stephen D. Miller, Jill Pipher, Ed Saff, +Sergei Tabachnikov, and Alexander Tumanov for discussions related to this +monograph. I thank I.C.E.R.M. for facilitating this project. My interest +in this problem was rekindled during discussions about point configuration +problems around the time of the I.C.E.R.M. Science Advisory Board meeting +in Nov. 2015. After writing the original version of the monograph around +2016, I let the thing languish on my website for some years. After giving the +Lewis Lectures at Rutgers University in Nov, 2022, I got inspired to re-write +my monograph and try again to make it publishable. Finally, I thank the +National Science Foundation for their continued support, currently in the +form of NSF grant DMS-2102802. +5 + +2 +Discussion +This chapter discusses the history and context for the result, and the high +level ideas in the proof. Following this, I explain some of my experimental +methods and discuss the computer parts of the proof. None of this is logically +needed for the proof, but it will shed light on why the proof looks like it does. +2.1 +History and Context +We take up the question discussed in the introduction: Which configurations +of points on the sphere minimize a given potential function F : (0, 2] → R. +The classic choice for this question is F = Rs, the Riesz potential, defined +in Equation 2. Again, Rs(d) = d−s. The Riesz potential is defined when +s > 0. When s < 0 the corresponding function Rs(d) = −d−s is called the +Fejes-Toth potential. The main difference is the minus sign out in front. +The case s = 1 is specially called the Coulomb potential or the electro- +static potential. This case of the energy minimization problem is known as +Thomson’s problem. See [Th]. The case of s = −1, in which one tries to +maximize the sum of the distances, is known as Polya’s problem. +There is a large literature on the energy minimization problem. See [F¨o] +and [C] for some early local results. See [MKS] for a definitive numerical +study on the minimizers of the Riesz potential for n relatively small. The +website [CCD] has a compilation of experimental results which stretches all +the way up to about n = 1000. The paper [SK] gives a nice survey of results, +with an emphasis on the case when n is large. See also [RSZ]. The paper +[BBCGKS] gives a survey of results, both theoretical and experimental, +about highly symmetric configurations in higher dimensions. +When n = 2, 3 the problem is fairly trivial. In [KY] it is shown that when +n = 4, 6, 12, the most symmetric configurations – i.e. vertices of the relevant +Platonic solids – are the unique minimizers for all Rs with s ∈ (−2, ∞)−{0}. +See [A] for just the case n = 12 and see [Y] for an earler, partial result +in the case n = 4, 6. The result in [KY] is contained in the much more +general and powerful result [CK, Theorem 1.2] concerning the so-called sharp +configurations. +The case n = 5 has been notoriously intractable. There is a general feeling +that for a wide range of energy choices, and in particular for the power law +potentials (when s > −2) the global minimizer is either the TBP or an FP. +Here is a run-down on what is known so far: +6 + +• The paper [HS] has a rigorous computer-assisted proof that the TBP is +the unique minimizer for the potential F(r) = −r. (Polya’s problem). +• My paper [S1] has a rigorous computer-assisted proof that the TBP +is the unique minimizer for R1 (Thomson’s problem) and R2. Again +Rs(d) = d−s. +• The paper [DLT] gives a traditional proof that the TBP is the unique +minimizer for the logarithmic potential. +• In [BHS, Theorem 7] it is shown that, as s → ∞, any sequence of +5-point minimizers w.r.t. Rs must converge (up to rotations) to the +FP having one point at the north pole and the other 4 points on the +equator. In particular, the TBP is not a minimizer w.r.t Rs when s is +sufficiently large. +• In 1977, T. W. Melnyk, O. Knop, and W. R. Smith, [MKS] conjectured +the existence of the phase transition constant, around s = 15.04808, at +which point the TBP ceases to be the minimizer w.r.t. Rs. This is the +phase transition which our Main Theorem estabishes. +• Define +Gk(r) = (4 − r2)k, +k = 1, 2, 3, ... +(4) +In [T], A. Tumanov proves that the TBP is the unique minimizer for +G2. The minimizers for G1 are those configurations whose center of +mass is the origin. The TBP is included amongst these. +Tumanov points out that the G2 potential does not have an obvious ge- +ometric interpretation, but it is amenable to a traditional analysis. He also +mentions that his result might be a step towards proving that the TBP min- +imizes a range of power law potentials. Inspired by similar material in [CK], +he observes that if the TBP is the unique minimizer for G2, G3 and G5, then +the TBP is the unique minimizer for Rs provided that s ∈ (0, 2]. +We will establish implications like this during the course of our proof of +the Main Theorem. The family of potentials {Gk} behaves somewhat like the +Riesz potentials. The TBP is the unique minimizer for G3, G4, G5, G6 (as a +consequence of our work here) but not a minimizer for any of G7, G8, G9, G10. +I am sure the pattern continues, but I did not formally check or prove it. +7 + +2.2 +Ideas in the Proof +Here are the three ingredients in the proof of the Main Theorem. +• The divide-and-conquer approach taken in [S1]. +• Elaboration of Tumanov’s observation. +• A symmetrization trick that works on a small domain. +Divide and Conquer: For certain choices of F, we are interested in search- +ing through the moduli space of all 5-point configurations and eliminating +those which have higher F-potential than the TBP. We win if we eliminate +everything but the TBP. For the functions we consider, most of the configu- +rations have much higher energy than the TBP and we can eliminate most +of the configuration space just by crude calculations. What is left is just a +small neighborhood Ω0 of the TBP. The TBP is a critical point for EF, and +(it turns out) that the function EF is convex in Ω0. In this case, we can say +that the TBP must be the unique global minimizer. +To implement this, we normalize so that (0, 0, 1) is a point of the con- +figuration, and then we map the other 4 points into R2 using stereographic +projection: +Σ(x, y, z) = +� +x +1 − z, +y +1 − z +� +. +(5) +We call the 4-point planar configuration the avatar. We use crude a priori +estimates to produce a subset Ω of a 7-dimensional rectangular solid that (up +to symmetry) contains all avatars that could have lower potential than the +TBP for all the relevant functions. Inside Ω the divide-and-conquer algorithm +is easy to manage. Our basic object is a block, a rectangular solid subset of Ω. +The main mathematical feature of the paper is a result which gives a lower +bound on the energy of any configuration in a block based on the potentials +of the configurations corresponding to the vertices, and an error term. +Having an efficient error term makes the difference between a feasible +calculation and one which would outlast the universe. +Our error term is +fairly sharp, and also the error term is a rational function of the vertices of +the block. For the potentials we end up using, we could run all our com- +puter programs using exact integer arithmetic. Such integer calculations are +too slow (in this century). I implemented the big calculations using interval +arithmetic. Since everything in sight is rational, our calculations only involve +the operations plus, minus, times, divide, min, and max. +8 + +Elaborations of Tumanov’s Observation: So far we have discussed one +function at a time, but we are interested in a 1-parameter family of power +laws and we can only run our program finitely many times. Using the divide +and conquer approach we show that the TBP is the unique global minimizer +for Gk when k = 3, 4, 5, 6 and also for the wierd energy hybrids G5 − 25G1 +and G♯♯ +10 = G10+28G5+102G2. Converting these results to statements about +the power law potentials comes down to variants of Tumanov’s observation. +After a lot of experimenting I found variants which cover large ranges of +exponents. The results for the potentials above combine to prove that the +TBP is the unique minimizer for Rs as long as s ∈ (−2, 0) ∪ (0, 13]. +Symmetrization: The methods above cannot be sharp enough to arrive +at the exact statement of our Main Theorem, because of the phase transi- +tion. We get around this problem as follows. First, we use the divide and +conquer approach to identify a small subset Υ of the configuration space such +that every configuration not in Υ, and not the TBP, has higher G♯ +10-energy, +where G♯ +10 = G10 + 13G5 + 68G2. This combines with the previous calcula- +tions to show that every configuration not in Υ, and not the TBP, has higher +s potential than the TBP whenever s ∈ [13, 15++]. The configurations in Υ +very nearly have 4-fold symmetry. +To analyze configurations in Υ we use a symmetrization operation which +maps Υ to the subset Υ4 ⊂ Υ consisting of configurations having 4-fold +symmetry. This retraction turns out to reduce the s-potential for exponent +values s ∈ [13, 15++]. Finally (and slightly simplifying) we produce a re- +traction from Υ4 to a subset Υ8 ⊂ Υ4 consisting entirely of FPs. This new +retraction reduces the s-potential when s ∈ [15, 15++]. Now we are left with +an analysis of the s-potential on a 1-dimensional set. +Extending the Range: My techniques run out just past the valueש. The +obvious conjecture, already observed by Melnyk, Knop, and Smith, is some +FP is the minimizer for any Riesz potential with exponent larger thanש. The +original version of my monograph contained a proof that some FP beat the +TBP for all exponents up to 100. I omitted this material because it didn’t +seem like such a strong result and mostly it was just a tedious calculation. +I would say that the main bottleneck to proving that some FP is the mini- +mizer for Rs for all s >ש is the delicacy of the symmetrization process which +I discuss above and also below. +9 + +2.3 +Experimentation +The proof of the Main Theorem is mostly just a verification of the things I +discovered experimentally using the software I created. I will follow 3 main +lines of experimental investigation in this discussion. +Experiments with Interpolation: This discussion has to do with what +I called “Tumanov’s observation” in the preceding section. These kinds of +methods go under the name of interpolation. +For the purpose of giving results about the Riesz potentials, the functions +Gk lose their usefulness at k = 7 because the TBP is not a minimizer for +G7, G8, ... At the same time, the general method requires Gk for k large in +order to extend all the way to the phase transition, a phenomenon that occurs +atש=15.04... +I built a graphical user interface which allows me to explore combinations +of the form � ckGk and see whether various lists of these energy hybrids +produce the desired results. The computer program takes a quadruple of +hybrids, Γ1, Γ2, Γ3, Γ4, and then solves a linear algebra problem to find a +linear combination +Λs = a0 + +4 +� +i=1 +ai(s)Γi +(6) +which matches the values of Rs at the values +√ +2, +√ +3, +√ +4, the distances in- +volved in the TBP. (I will usually write 2 as +√ +4 because then the distances +involved in the TBP are easier to remember.) +As an aside, let me say how I knew to use quadruples rather than, say, +triples or quintuples. If you match the values at +√ +2, +√ +3, +√ +4 and the deriva- +tives, at +√ +2, +√ +3 you get a 5 variable problem with 5 unknowns. (You don’t +have to worry about matching the derivative at +√ +4 because this is an end- +point to the domain of distances between points on the unit sphere.) +Concerning Equation 6, what we need for the quadruple to “work” on the +interval (s0, s1) is that the functions a1(s), a2(s), a3(s), a4(s) are nonnegative +for s ∈ (s0, s1) and that simultaneously the comparison function +1 − Λs +Rs +is positive on (0, 2) − { +√ +2, +√ +3, +√ +4}. +So, my computer program lets you +manipulate the coefficients defining the energy hybrids and then see plots of +the functions just mentioned. +10 + +At the same time as this, my program computes the energy hybrid eval- +uated on the space of FPs to see how it compares to the value on the TBP. +I call this the TBP/FP competition. On intervals (s0, s1) ⊂ (0,ש) we want +the TBP to win the competition, as judged by the given energy hybrids. +Repeatedly running these competitions and looking at the plots of the co- +efficients and the comparison function, I eventually arrived at the energy +hybrids mentioned in the previous section. +You can use this program too. If you actually get my Java program to +run on your computer, you can get the same intuition I eventually got about +what works and what doesn’t. If you don’t play around with the software, +then choices like +G♭ +5 = G5 − 25G1, +G♯♯ +10 = G10 + 28G5 + 102G2 +will just seem like random lucky guesses. In fact they are practically the +unique (at most 3 term) energy hybrids which do the job! +To extend all the way toש, I had to accept an energy hybrid for which +the TBP would lose the TBP/FP competition. At the same time, the TBP +would still do well in the overall competition, beating most of the other +configurations. Eventually I hit upon the energy hybrid G♯ +10 and the small +neighborhood Υ mentioned above and defined precisely in the next chapter. +The quadruple (G1, G2, G♭ +5, G♯ +10) extends a bit pastש, up to 15++, and G♯ +10 +is a pretty kind judge: With respect to this judge, the TBP wins against all +configurations outside the tiny Υ. +The intuition I came away with is that you need to use some Gk for fairly +large k, to get enough extension, and then you need to tune it by sharp- +ening and flattening. To sharpen means to add in more of the lower Gks. +To flatten means to do the opposite. When you sharpen, you get an energy +hybrid which is a kinder but less extensive judge: It works better but on a +smaller range of exponents. When you flatten, you get a harsher but more +extensive judge. The miraculous quadruple (G1, G2, G♭ +5, G♯ +10) extends to the +neighborhood [13, 15++]. The TBP is the minimizer for the first 3 potentials, +and for G♯ +10 the TBP wins outside of Υ. +Experiments with Symmetrization: Most successful energy minimiza- +tion results are about symmetry. The work culminating in that of Cohn- +Kumar [CK] shows how to exploit the extreme symmetry of some special +configurations, like the Leech cell, to show that they are the energy mini- +mizers with respect to a wide range of potentials. These methods only work +11 + +for very special numbers of points. The number N = 5 is not special in this +way, because there are no Platonic solids with 5 points. +For N = 5 the TBP and the FPs are competitors for the most symmetric +configurations. +They have different symmetries. +However, they do have +one thing in common: 4 fold dihedral symmetry. One dream for proving the +Main Theorem is to use a kind of symmetrization operation which replaces an +arbitrary configuration with one having 4-fold dihedral symmetry and lower +potential energy. This would reduce the overall problem to an exploration of +a 2-dimensional moduli space and would possibly bring the result within the +range of rather ordinary calculus. Symmetrization operations are extremely +prevalent in some areas of mathematics – e.g., in proofs of isoperimetric type +results. +Such a symmetrization operation in general will surely fail due to the vast +range of possible configurations. However, certain operations might work well +in very specific parts of the configuration space and for very specific func- +tions. Fortunately, the divide-and-conquer-plus-interpolation method rules +out everything of interest except the magical domain Υ and the exponent +range [13, 15++]. What I did is test various symmetrizations and various +choices of Υ until I found a pair that worked. +Let me say a word about the relation between symmetrization and Υ. +The smaller the choice of Υ, the more likely symmetrization is to work. On +the other hand, the smaller Υ is, the more computation we have to use to +eliminate all the competitors outside Υ. The domain I finally settled on was +large enough to make this elimination process a feasible computation but +small enough for the symmetrization to work. +Once I found a symmetrization operation which worked, the question +became: How to prove it? Proving that symmetrization lowers the energy +seems to involve studying what happens on the tiny but still 7-dimensional +moduli space Υ. The secret to the proof is that, within Υ, the symmetrization +operation is so good that it reduces the energy in pieces. What I mean is +that the energy of a configuration is a 10 term sum e1 + ... + e10. What I +show is that one can write +e1 + .... + e10 = (e1 + e2) + (e3 + e4) + (e5 + e6 + e7) + (e8 + e9 + e10) +so that the symmetrization operation decreases each bracketed sum sepa- +rately. This replaces one big verification by a bunch of smaller ones, con- +ducted over lower dimensional configuration spaces. +12 + +Not only did I have to experiment to find the symmetrization operation +itself, I had to experiment with how to break up the expression for the energy +to make for a provable result. +As it is, showing that an expression like +(e5 + e6 + e7) decreases amounts to showing that a polynomial in 5-variables, +with thousands of terms, is positive on the unit cube. I have a positivity +certificate I use, which I call positive dominance, which works like magic on +the relevant polynomials. Positive dominance kills these big monsters with +one thrust of the sword for each monster – but only so because I tuned the +definition of Υ so that the verification process produced killable monsters. I +got lucky in that such a useful domain Υ exists at all. +I should also mention that I use a second symmetrization which improves +a configuration with 4-fold symmetry to one with 8-fold symmetry. +This +symmetrization, though rather simple, is extremely delicate. It works on a +tiny domain �Ψ4 ⊂ Υ and only for power laws with potential greater than +about 13.53. At the same time, this symmetrization has what I would call +miraculous algebraic properties when restricted to the tiny domain where I +use it. I found this operation, once again, by experimentation, and then the +algebraic properties took me by surprise. In the first version of the this work, +the 180 page monograph, I hadn’t noticed the good algebra. +Experiments with Local Analysis: Another part of the monograph deals +with configurations that are very near the TBP. Here we are fortunate be- +cause the functions EF are convex near the TBP. Put another way, their +Hessians are positive definite near the TBP. That means that the TBP is +the unique minimizer in a small neighborhood around the TBP. I want to +emphasize that what we need is not just a calculation at the TBP. In order to +use this information effectively in a computational proof, we need an explicit +neighborhood of convexity. Proving this sets up a recursive problem. +Consider the simpler situation where we would like to show that some +function f is positive on some interval I = [0, ǫ]. Let’s say that we have +free access to the values f(0), f ′(0), f ′′(0), ... and we can also look at the +explicit expressions for f and its derivatives. If we had some information +about maxI |f ′| we could combine it with information about f(0) to perhaps +complete the job. But how do we get information about maxI |f ′|? Well, if +we had information about maxI |f ′′| we could combine it with information +about f ′(0) to perhaps complete the job. And so on. +This is the situation we find ourselves in. We can compute all the partial +derivatives of EF at the TBP, though we have a function of 7 variables, and +13 + +so eventually it gets expensive to compute them all. However, no matter +how many derivatives we compute, it seems that we need to compute more +of them to get the bounds we need. +There is something that saves us: The error multiplier in Taylor’s The- +orem with Remainder. This multiplier is essentially ǫN/N!, a number that +becomes tiny as N increases. If we can get any kind of reasonable bounds on +high derivatives of our function, then we get pretty good bounds when we +multiply through by the tiny number. +I eventually found a combinatorial trick for bounding the derivatives of +EF, at least for the relevant choices of F, without having to evaluate them +anywhere. The magic formula is Equation 47. After a lot of experimentation +I found that I could get a reasonably sized neighborhood of convexity for +EF by explicitly evaluating the kth partial derivatives up to k = 6 and then +using Equation 47 for a global bound on the 7th partials. +2.4 +Guide to the Software +The software for my proof can be dowloaded from +http//www.math.brown.edu/ ∼ res/Java/TBP.tar +Once you untar this program, you get a directory with a suite of smaller +programs. There are 5 subdirectories, corresponding to 5 of the 6 parts of +the monograph. The part having to do with the error estimate does not rely +on any computer assists. The reader who is interested in verifying any part +of the monograph need only look at the programs for that part. +Main: This does the interval arithmetic calculation for the main divide- +and-conquer result, Lemma A135. +This program is quite extensive, and +spread out in about 20 Java files, but almost all the length comes from the +visual/experimental part. I show the calculations in action, allow the user +to experiment with fairly arbitrary energy hybrids, and also give detailed +written instructions on the operation of the program. +The reason for the extensive program is debugging. The big infrastructure +is designed to prevent errors in the actual computation. I have also included +a stripped down version that runs without all the bells and whistles. I try to +explain the main computation in §5 in enough detail that a reasonably good +programmer would be able to reproduce it. +14 + +Interpolation: The main program in this section, contained in the direc- +tory JavaMain, does all the experimentation with the energy hybrids dis- +cussed above, and also formally proves that the given energy hybrids extend +to their advertised ranges. However, the code I actually use in this version +of the proof is different. The subdirectory Proof has this shorter method. +Why both? I only discovered the method in Proof recently, and the method +in JavaMain is more robust. It doesn’t require the kind of ad hoc argument +I give in §11.2 and also (a very small part of) it is still needed logically for +the Auxiliary Theorem. I include a PDF file which explains the method in +JavaMain. +Independent from all this, I also include 5 Mathematica files, LemmaA221.m +and LemmaA222.m and LemmaA231.m and Lemma A232.m and LemmaA233.m, +which generate all the plots for the corresponding lemmas. One can compare +the Mathematica and Java plots and see that they are the same. +Local Analysis: This directory has 3 Mathematica files, LemmaL21.m and +LemmaL22.m and Lemma23.m, which perform the straightforward and exact +calculations needed for these lemmas. The calculations involve manipulating +rational polynomials and evaluating them at special points. These files can +easily be reproduced by anyone who knows how to manipulate polynomials in +Mathematica. Of course, one could also reproduce the programs in e.g. Sage. +Symmetrization: This directory has 6 Mathematica files, with names like +Lemma B3522.m. +These do the calculations for the corresponding lemmas +in this part of the monograph. These short files essentially just manipulate +rational polynomials using standard operations in Mathematica. They all +could be reproduced by anyone who knows how to manipulate polynomials +in Mathematica. +Endgame: This directly contains a program which is a baby version of +the main divide-and-conquer program. This program does the calculation +for Lemma C2. It is not a very complicated program, and I explain it in +detail in §18. The calculation is done with exact integer arithmetic over a 3- +dimensional space. It only takes about 2 minutes to run. Without the exact +integer arithmetic – meaning with floating point calculations – the program +is about 1000 times faster and and nearly instantaneous. +15 + +3 +Main Theorem: Proof Outline +3.1 +Preliminaries +Stereographic Projection: Let S2 ⊂ R3 be the unit 2-sphere. +Stere- +ographic projection is the map Σ : S2 → R2 ∪ ∞ given by the following +formula. +Σ(x, y, z) = +� +x +1 − z, +y +1 − z +� +. +(7) +Here is the inverse map: +Σ−1(x, y) = +� +2x +1 + x2 + y2, +2y +1 + x2 + y2, 1 − +2 +1 + x2 + y2 +� +. +(8) +Σ−1 maps circles in R2 to circles in S2 and Σ−1(∞) = (0, 0, 1). +Avatars: Stereographic projection gives us a correspondence between 5- +point configurations on S2 having (0, 0, 1) as the last point and planar con- +figurations: +V0, V1, V2, V3, (0, 0, 1) ∈ S2 +⇐⇒ p0, p1, p2, p3 ∈ R2, +pk = Σ(Vk). +(9) +We call the planar configuration the avatar of the corresponding configura- +tion in S2. By a slight abuse of notation we write EF(p1, p2, p3, p4) when we +mean the F-potential of the corresponding 5-point configuration. +Figure 3.1 shows the two possible avatars (up to rotations) of the trian- +gular bi-pyramid, first separately and then superimposed. We call the one +on the left the even avatar, and the one in the middle the odd avatar. The +points for the even avatar are (±1, 0) and (0, ± +√ +3/3). When we superimpose +the two avatars we see some extra geometric structure that is not relevant +for our proof but worth mentioning. The two circles respectively have radii +1/2 and 1 and the 6 segments shown are tangent to the inner one. +0 +1 +2 +3 +0 +2 +3 +1 +0 +2 +even +odd +both +Figure 3.1: Even and odd avatars of the TBP. +16 + +The Special Domain: We let Υ ⊂ (R2)4 denote those planar configurations +p0, p1, p2, p3 such that +1. ∥p0∥ ≥ ∥pk∥ for k = 1, 2, 3. +2. 512p0 ∈ [433, 498] × [0, 0]. (That is, p0 ∈ [433/512, 498/512] × {0}.) +3. 512p1 ∈ [−16, 16] × [−464, −349]. +4. 512p2 ∈ [−498, −400] × [0, 24]. +5. 512p3 ∈ [−16, 16] × [349, 464]. +We discuss the significance of Υ extensively in §2.3. The set Υ contains the +avatars that compete with the TBP near the exponentש. +p0 +p1 +p2 +p3 +Figure 3.2: The sets defining Υ compared with two TBP avatars. +Symmetrization: Let (p0, p1, p2, p3) be a planar configuration with p0 ̸= p2. +We define +d02 = 2∥p0 − p2∥, +d13 = 2∥π02(p1 − p3)∥. +(10) +Here π02 is the projection onto the subspace perpendicular to the vector +p0 − p2. Finally, we define +p∗ +0 = (d02, 0), +p∗ +1 = (0, −d13), +p∗ +2 = (−d02, 0), +p∗ +3 = (0, d13). +(11) +The avatar p∗ +1, p∗ +2, p∗ +3, p∗ +4 is invariant under reflections in the coordinate axes. +17 + +3.2 +Reduction to Three Lemmas +We now reduce the Main Theorem to Lemmas A, B, C. Let +15+ = 15 + 24 +512, +15++ = 15 + 25 +512 +as in the Main Theorem. Let Υ be as in §3.1. Recall that Rs is the Riesz +potential. +Lemma 3.1 (A) For s ∈ (0, 13] the TBP uniquely minimizes the Rs-potential. +For s ∈ (13, 15++], any Rs-potential minimizer is either the TBP or else iso- +metric to a configuration whose avatar lies in Υ. +Lemma A focuses our attention on the small domain Υ and the parameter +range [13, 15++]. Now we bring in the symmetrization operation from §3.1. +Lemma 3.2 (B) Let s ∈ [12, 15++] and (p0, p1, p2, p3) ∈ Υ. Then +ERs(p∗ +0, p∗ +1, p∗ +2, p∗ +3) ≤ ERs(p0, p1, p2, p3) +with equality if and only if the two configurations are equal. +Let Υ4 denote the subset of Υ consisting of configurations which are +invariant under reflections in the coordinate axes. +Lemma B (which re- +dundantly works for s ∈ [12, 13]) focuses our attention on the same small +parameter range [13, 15++] and on the symmetric configurations living in +Υ4. +Lemma 3.3 (C) Let ξ0 denote a planar avatar of the TPB. There exist +ש∈(15+,15++) such that the following is true. +1. For s ∈ (13,ש) we have Es(ξ0) < Es(ξ) for all ξ ∈ Υ4. +2. For s ∈ (ש,15++) we have Es(ξ0) > Es(ξ) for some ξ ∈ Υ4. +Also, for s ∈ [15+, 15++] the restriction of Es to Υ4 has a unique minimum, +and this minimum represents an FP. +The Main Theorem is an obvious consequence of Lemma A (§3.3), Lemma +B (§12), and Lemma C (§3.4.) As a matter of convention we will point the +reader to where the proof of the given lemma starts. +Thus, the proof of +Lemma A starts in §3.3. +18 + +3.3 +Proof of Lemma A +Define +Gk(r) = (4 − r2)k. +(12) +Also define +G♭ +5 = G5 − 25G1, +G♯♯ +10 = G10 + 28G5 + 102G2, +G♯ +10 = G10 + 13G5 + 68G2 +(13) +Lemma 3.4 (A1) The following is true. +1. The TBP is the unique minimizer for G4, G♭ +5, G6. +2. The TBP is the unique minimizer for G♯ +10 among configurations which +are not isometric to ones which have avatars in Υ. +3. The TBP is the unique minimizer for G♯♯ +10 among configurations which +have avatars in Υ. +We note two implications of Lemma A1: +• Since G5 is a positive combination of G♭ +5 and G1, Lemma A1 immedi- +ately implies that the TBP is the unique minimizer for G5. +• Since G♯♯ +10 is a positive combination of G♯ +10 and G5 and G2, Lemma A1 +immediately implies that the TBP is the unique minimizer for G♯♯ +10. +Forcing: Let T0 be the TBP. We say that a pair (Γ3, Γ4) of functions forces +the interval I if the following is true: If T is another configuration such that +Γk(T0) < Γk(T) for k = 3, 4 then Es(T0) < Es(T) for all s ∈ I. +Lemma 3.5 (A2) The following is true. +1. The pair (G4, G6) forces (0, 6]. +2. The pair (G5, G♯♯ +10) forces [6, 13]. +3. The pair (G♭ +5, G♯ +10) forces [13, 15++]. +Lemma A is an immediate consequence of Lemma A1 (§4) and Lemma +A2 (§10). +19 + +3.4 +Proof of Lemma C +Let Ψ4 denote the set of planar configurations of the form +(x, 0), +(0, −y), +(−x, 0), +(0, y), +64(x, y) ∈ [43, 64]. +(14) +This is a 2-dimensional domain consisting of avatars having 4-fold dihedral +symmetry. We have Υ4 ⊂ Ψ4. We work with Ψ4 because it is more symmetric +than Υ4. We identify Ψ4 with the square [43/64, 1]2 and we think of ERs as a +function on this square. We usually write Es = ERs. Again, the point (a, b) +corresponds to the planar configuration with points −p2 = p0 = (a, 0) and +−p1 = p3 = (0, b). Though the TBP does not lie in Ψ4, it corresponds in the +same way to the point (1, +√ +3/3). +The FP configurations in Ψ4 lie along the main diagonal. We call this +diagonal Ψ8. We define an even smaller square +�Ψ4 = +�55 +56, 55 +56 +�2 +⊂ Ψ4. +(15) +We think of �Ψ4 as the sweet spot, the place where all the action happens. +We now define another symmetrization. +σ(x, y) = (z, z), +z = x + y + (x − y)2 +2 +. +(16) +We have σ : �Ψ4 → Ψ8. Here is the key result. +Lemma 3.6 (C1) If s ∈ [14, 16] and p ∈ �Ψ4 Then Es(σ(p)) ≤ Es(p) with +equality if and only if σ(p) = p. +Remarks: +(1) The operation σ is extremely delicate. If we take the exponent s = 13, +the operation actually seems to increase the energy for all points of �Υ4 − �Υ8. +The magic only kicks in around exponent 13.53. +(2) Lemma C1 has an algebraic proof, using the Positive Dominance cer- +tificate. See §12.2. It turns out that we get miraculously good algebraic +behavior for s ∈ [14, 16] and for configurations in �Ψ4. +(3) Lemma C1 is false if we use Υ4 or Ψ4 in place of �Ψ4. This is why we +restrict our attention to a very small region. The same proof works on the +somewhat larger domain [27/32, 29/32]2 but that is about as far as we can go. +20 + +Our next result eliminates all those configurations and exponents not +in �Ψ4 × [15+, 15++]. This result is an explicit calculation similar in spirit +to Lemma A2 but easier. Here we are just dealing with functions on a 2 +dimensional configuration space. Let �Ψ8 be the main diagonal of �Ψ4. +Lemma 3.7 (C2) Let ξ0 be the point in the plane representing the TBP. +1. If s ∈ [13, 15+] and p ∈ Ψ4 then Es(ξ0) < Es(ξ). +2. If s ∈ [15+, 15++] and p ∈ Ψ4 − �Ψ4 then Es(ξ0) < Es(ξ). +3. If s ∈ [15+, 15++] the restriction of Es to �Ψ8 has a unique minimum. +Statements 1 and 2 of Lemma C2 imply that for any s ∈ [13, 15++], any +minimizer ξ of Es, not equal to the TBP avatar, lies in �Ψ4. Furthermore, +such a ξ can only exist when s ∈ [15+, 15++]. Lemma C1 now says that ξ in +fact lies in �Ψ8. Statement 3 of Lemma C2 adds the information that ξ is the +unique minimizer in �Ψ8. The final lemma finishes the proof. +Lemma 3.8 (C3) There existש∈(15+,15++) such that: +1. For s ∈ (15+,ש) we have Es(ξ0) < Es(ξ) for all ξ ∈ �Ψ8. +2. For s ∈ (ש,15++) we have Es(ξ0) > Es(ξ) for some ξ ∈ �Ψ8. +Lemma C is a consequence of Lemma C1 (§12, §17), Lemma C2 (§18), +and Lemma C3 (§20). +3.5 +The Polya Case +Here I explain the main details of the proof of the Auxiliary Theorem: the +TBP is the unique minimizer for Fs(r) = −r−s for any s ∈ (−2, 0). I call +this the Polya case because the well-known Polya problem concerns the case +s = 1. All we need here is Lemma A for the interval (−2, 0). +Lemma A1 also applies to G3. +The only differences in the proof are +discussed in the remarks at the end of §4.5 and §6.4. Once these details are +in place, our software gives a computational proof of Lemma A1 for G3 in +the same way it does for the other potentials. +A variant of Lemma A2, with the same kind of proof, shows that the +pair (G3, G5) forces (−2, 0) with respect to Fs. The main extra detail needed +here is the matrix of power combos given in §10.5 . The reader can also +see pictures of the Polya case using my software, and my software gives a +rigorous positivity case in the Polya case just as for the other cases. +21 + +4 +Main Theorem: Proof of Lemma A1 +4.1 +Odd and Even Planar Configurations +We call a pair of points �p, �q ∈ S2 far if ∥�p − �q∥ ≥ 4/ +√ +5. Note that (�p, �q) is +a far pair if and only if (�q, �p) is a far pair. Our rather strange definition has +a more natural interpretation in terms of the planar avatars. If we rotate S2 +so that �p = (0, 0, 1) then q = Σ(�q) lies in the disk of radius 1/2 centered at +the origin if and only if (�p, �q) is a far pair. +We say that a point in a 5-point configuration is odd or even according +to the parity of the number of far pairs it makes with the other points in +the configuration. +Correspondingly, define the parity of the avatar to be +the parity of the number of points which are contained in the closed disk of +radius 1/2 about the origin. This definition extends our definition for the +TBP avatars. Here we repeat Figure 3.1 for convenience. +0 +1 +2 +3 +0 +2 +3 +1 +0 +2 +even +odd +both +Figure 3.1: Even and odd avatars of the TBP. +We call 2 planar configurations isomorphic if they are the avatars of +isometric configurations on the sphere. Thus, for instance, the odd and even +avatars of the TBP are isomorphic. Every planar configuration is isomorphic +to an even configuration. To see this, we form a subgraph of the complete +graph by joining two points in a 5-point configuration by an edge if and only +if they make a far pair. As for any graph, the sum of the degrees is even. +Hence there is some vertex having even degree. When we rotate so that this +vertex is (0, 0, 1), the avatar is even. +The definition of even planar configurations (or something similar which +would play the same role) is an important part of our computational proof +of Lemma A1. By focusing on the even planar configurations, and further +using symmetry, we arrive at a configuration space where there is just one +avatar of the TBP. +22 + +4.2 +The Domains +Now we will use the concept of an even planar configuration to define the +main domains we will use in our calculation. +The Relevant Domains: Given a planar configuration ξ = (p0, p1, p2, p3), +we write pk = (pk1, pk2). We define a domain Ω ⊂ R7 to be the set of planar +configurations ξ satisfying the following conditions. +1. ξ is an even configuration. +2. ∥p0∥ ≥ max(∥p1∥, ∥p2∥, ∥p3∥). +3. p12 ≤ p22 ≤ p32 and p22 ≥ 0. +4. p01 ∈ [0, 2] and p01 = 0. +5. pj ∈ [−3/2, 3/2]2 for j = 1, 2, 3. +6. min(p1k, p2k, p3k) ≤ 0 for k = 1, 2. +We define Ω♭ (to be used specially with G♭ +5) by the same conditions except +that we leave off Condition 6. +Closed Versus Open Conditions: If we want to check for inclusion in +the interior of Ω or Ω♭, all the inequalities above must be strict. We find it +useful to work with the interior of Ω and Ω♭ because we won’t need to spe- +cially treat some boundary cases. This will make for a cleaner calculation. +A Tiny Cube: We cannot make a finite calculation if we wish to treat +configurations arbitrarily close to the TBP avatar ξ0 ∈ Ω. We have to make +a cutoff and deal separately with points very near ξ0. When we string out +the points of ξ0, the coordinates we get are +1, 0 +0, − +√ +3/3, +−1, 0, +0, +√ +3/3. +(17) +Here p0 = (1, 0) and p1 = (0, − +√ +3/3), etc. Again, see Figure 3.1. We let Ω0 +denote the cube of side-length 2−17 centered at ξ0. This is our cutoff. +23 + +4.3 +Reduction to Simpler Lemmas +Recall that we mean EF(ξ) to be the F-potential of the 5-point configuration +on the sphere corresponding to a planar avatar ξ. Lemma A1 makes 3 claims: +1. When F is any of G4, G♭ +5, G6, the TBP avatars are the unique minimizer +for EF. +2. When F = G♯ +10, the TBP avatars are the unique minimizers for EF +among configurations which are not isomorphic to ones in Υ. +3. When F = G♯♯ +10, the TBP avatars have smaller EF value than all con- +figurations in Υ. +Lemma 4.1 (A11) Let F be any of G4, G♭ +5, G6, G♯ +10. Then ξ0 is the unique +minimizer for EF inside Ω0. +Lemma 4.2 (A12) The following is true: +1. Let F = G4, G6, G♯ +10. If ξ is not equivalent to any planar configuration +in Ω then then ξ does not minimize EF. +2. Let F = G♭ +5. If ξ is not equivalent to any planar configuration in Ω♭ +then then ξ does not minimize EF. +Let [F] be the EF value of the TBP avatars. +Lemma 4.3 (A13) The following is true. +1. The infimum of EG4 on interior(Ω) − Ω0 is at least [G4] + 2−50. +2. The infimum of EG6 on interior(Ω) − Ω0 at at least [G6] + 2−50. +3. The infimum of EG♭ +5 on interior(Ω) − Ω0 is at least [G♭ +5] + 2−50. +4. The infimum of EG♯ +10 on interior(Ω) − Υ − Ω0 is at least [G♯ +10] + 2−50. +5. The infimum of EG♯♯ +10 on Υ is at least [G♯♯ +10] + 2−50. +Lemma A13 is the main calculation. +It follows from continuity that +Lemma A13 remains true if we replace the interior of Ω by Ω itself. But +then Lemma A1 follows immediately from Lemma A11 (§4.4), Lemma A12 +(§4.5), and Lemma A13 (§5). Our choice of 2−50 is somewhat arbitrary. +24 + +4.4 +Proof of Lemma A11 +Recall that Ω0 is the cube of side length 2−17 centered at ξ0. For all our +choices of F, the function EF is a smooth function on R7. +Lemma 4.4 (A111) The gradient of EF vanishes at ξ0. +Proof: We make a direct calculation in all cases. ♠ +Lemma 4.5 (A112) The Hessian of EF, meaning the matrix of second par- +tial derivatives, is positive definite at every point of Ω0. +Let ξ ∈ Ω0 be other than ξ0. Lemmas A111 and A112 (§6) imply that +the restriction of EF to the line segment γ joining ξ0 to ξ is convex and has +0 derivative at ξ0. Hence EF(ξ) > EF(ξ0). This proves Lemma A11 +4.5 +Proof of Lemma A12 +Recall that ξ0 is the avatar of the TBP. Let [F] = EF(ξ0). Since the TBP +has 6 bonds of length +√ +2, and 3 of length +√ +3, and 1 of length +√ +4, we have +[Gk] = 6 × 2k + 3. Using this result, and Equation 13, we compute +[G4] = 99, +[G6] = 387, +[G♭ +5] = −180, +[G♯ +10] = 10518. +(18) +Let ξ = p0, p1, p2, p3 some other planar configuration. +Lemma 4.6 (A121) Let F = G4, G6, G♯ +10. +If ∥p0∥ > 2 then ξ does not +minimize EF. Also, if ∥p0∥ > 3/2 then ξ does not mininizer EG♭ +5. +Proof: Let rj = ∥Σ−1(pj), (0, 0, 1)∥. If ∥p0∥ > 2 then r0 < d = 2/ +√ +5. We +compute that +G4(d) > 104 > [G4], +G6(d) > 1073 > [G6], +G♯ +10(d) > 117642 > [G♯ +10]. +Since F is monotone decreasing and non-negative, just the one term in EF(ξ) +is larger than [F]. +We have [G♭ +5] = −180. Using 3/2 in place of 2 we get r0 > d′ = 4/ +√ +13 in +place of d. We check that G♭ +5(r) > 93 for r ∈ [0, d′] and G♭ +5 > −30 in [0, 2]. +Hence EG♭ +5(P) > 93 − 270 > [G♭ +5]. ♠ +25 + +Lemma 4.7 (A122) Let F = G4, G♭ +5, G6, G♯ +10. +Suppose ∥p0∥ ≥ 3/2 and +∥pj∥ > 3/2 for some j = 1, 2, 3. Then ξ does not minimize EF. +Proof: +We use the same notation from the previous proof. +We already +proved a stronger result for G♭ +5, so we let F be one of the other functions +listed. If ∥p0∥, ∥pj∥ > 3/2 then we have r0, rj < d′ = 4/ +√ +13. We compute +2G4(d′) > 117 > [G4], +2G6(d′) > 901 > [G6], +2G♯ +10(d′) > 58319 > [G♯ +10]. +Since F is monotone decreasing and non-negative, just these two terms in +EF(ξ) together are larger than [F]. ♠ +Lemma 4.8 (A123) Let F be any strictly monotone decreasing potential. +If min(p1k, p2k, p3k) > 0 for one of k = 1, 2 then ξ does not minimize EF. +Proof: The conditions imply that the 5-point configuration in S2 is con- +tained in a hemisphere H, and at least 3 of the points are in the interior of +H. If we take one of these interior points and reflect it across ∂H then we +increase at least 2 of the distances in the configuration and keep the rest the +same. ♠ +Assume ξ is a minimizer for EF. As we discussed in §4.1, we can normal- +ize so that ξ is an even configuration. Reordering p0, p1, p2, p3 and rotating, +about the origin, we make ∥p0∥ ≥ ∥pi∥ for i = 1, 2, 3 and we move p0 into the +positive x-axis. Reflecting in the x-axis if necessary and reordering the points +p1, p2, p3 if necessary, we arrange that p12 ≤ p22 ≤ p32 and p22 ≥ 0. Lemmas +A121 and A122 tell us that, in all cases, p01 ∈ [0, 2] and pj ∈ [−3/2, 3/2]2 +for j = 1, 2, 3. +We have also arranged that p02 = 0. +For F = G♭ +5 we +have nothing left to check. Otherwise, Lemma A123 shows that ξ satisfies +min(p1k, p2k, p3k) ≤ 0 for k = 1, 2, 3. +Remark: (The Polya Case) For G3, we need to take a number a bit larger +than 2 in Lemma A121. The constant 4 works easily. We specially sepa- +rate out the case of G3 in our code so that we are sure to use the larger +domain. Lemma A122 also works for G3 but we need to work harder: We +have [G5] = 51 but 2G3(d′) ∈ [42, 43]. However, the G3-potential of the 4 +point configuration on S2 not involving (0, 0, 1) is at least that of the regular +tetrahedron, 14+ 2 +9. Since 14+42 > 51 we get the same conclusion as Lemma +A122 for G3. +26 + +5 +Main Calculation: Lemma A13 +5.1 +Blocks +We first list the ingredients in our main calculation and then explain the +calculation itself. +Dyadic Subdivision: The dyadic subdivision of a D-dimensional cube is +the list of 2D cubes obtained by cutting the cube in half in all directions. We +sometimes blur this notation and say that any one of these 2D smaller cubes +is a dyadic subdivision of the big cube. +Blocks: We define a block to be a product of the form +B = Q0 × Q1 × Q2 × Q3 ⊂ □ := [0, 2] × [−2, 2]2 × [−2, 2]2 × [−2, 2]2. (19) +where Q0 is a segment and Q1, Q2, Q3 are squares, each obtained by iterated +dyadic subdivision of [0, 2] or [−2, 2]2. +We call B acceptable if Q0 has length at most 1 and Q1, Q2, Q3 have +sidelength at most 2. If B is not acceptable we let the offending index be +the lowest index where the condition fails. +The kth subdivision of a block amounts to performing dyadic subdivi- +sion to the kth factor and leaving the others alone. We call these operations +S0, S1, S2, S3. Thus S0 cuts B into two pieces and each other Sk cuts B into 4 +pieces. We let Sk(B) denote the list of the blocks obtained by performing Sk +on B. All the blocks our algorithm produces come from iterated subdivision +of □. +Rational Block Calculations: We say that a rational block computation +is a finite calculation, only involving the arithmetic operations and min and +max. The output of a rational block computation will be one of two things: +yes, or an integer. A return of an integer is a statement that the computa- +tion does not definitively answer to the question asked of it. If the integer +is −1 then there is no more information to be learned. If the integer lies +in {0, 1, 2, 3} we use this integer as a guide in our algorithm. For example, +we might ask if the block is acceptable. If not, then we would return the of- +fending index, and our algorithm would subdivide the block along this index. +27 + +5.2 +The Main Calculation +Recall that +ξ0 = (1, 0, − +√ +3/3, −1, 0, 0, +√ +3/3) ∈ Ω +is the avatar of the TBP and Ω0 is the cube of side length 2−17 around ξ0. +Recall also that Υ is the special domain defined in §3.1. +Lemma 5.1 (A131) There exists a rational block computation C1 such that +an output of yes for a block B implies that B ⊂ Ω0. +Lemma 5.2 (A132) There exists a rational block computation C3 such that +an output of yes for an acceptable block B implies that B is disjoint from +the interior of Ω. The same goes for Ω♭. +Lemma 5.3 (A133) There exists a rational block computation C♯ +3 such that +an output of yes for a block B implies that B ⊂ Υ. Likewise, there exists +a rational block computation C♯♯ +3 such that an output of yes for a block B +implies that B is disjoint from Υ. +The proofs of the Lemmas A131, A132, A133, given below, just amount to +checking the conditions in a fairly straightforward way. The final ingredient +is the main ingredient. It is much more involved. All the energy potentials +we consider are what we call energy hybrids. They have the form +F = +m +� +k=1 +ckGk, +Gk(r) = (4 − r2)k, +c1 ∈ Q, +c2, ..., ck ∈ Q+. +(20) +With some modification of Lemma E below we could also handle the case +when some of c2, ..., ck are negative. See Remark (1) after the statement of +Lemma E. +Lemma 5.4 (A134) For any function F given by Equation 20, there exists +a rational block computation C3,F such that an output of yes for an acceptable +block B implies that the minimum of EF on B is at least EF(ξ0) + 2−50. +Otherwise C3,F(B) is an integer in {0, 1, 2, 3}. +Here is the main calculation. +1. We start with the list L = {□}. +28 + +2. If L = ∅ then HALT. Otherwise let B = Q0 × Q1 × Q2 × Q3 be the +last block of L. +3. If B is not acceptable we delete B from L and append to L the subdi- +vision of B along the offending index. We then return to Step 2. Any +blocks considered beyond this step are acceptable. +4. If C1(B) = yes or C2(B) = yes we remove B from L and go to Step +2. Here we are eliminating blocks disjoint from the interior of Ω or else +contained in Ω0. +5. If F = G♯ +10 and C♯ +3(B) = yes we remove B from L and go to Step 2. If +F = G♯♯ +10 and C♯♯ +3 (B) = yes we remove B from L and go to Step 2. +6. If C4,F(B) = yes then we remove B from L and go to Step 2. Here we +have verified that the F-energy of any avatar in B exceeds [F] + 2−50. +7. If C4,F(B) = k ∈ {0, 1, 2, 3} then we delete B from L and append to L +the blocks of the subdivision Sk(B) and return to step 2. +If the algorithm reaches the HALT state for a given choice of F, this +constitutes a proof that the corresponding statement of Lemma A13 is true. +Lemma 5.5 (A135) The Main Computation reaches the HALT state for +each choice of F listed in Lemma A13. +Lemma A13 follows from Lemma A131 (§5.3), Lemma A132 (§5.4), Lemma +A133 (§5.5), Lemma A134 (§5.6) and Lemma A135 (§5.8). +5.3 +Proof of Lemma A131 +Define intervals I0, I1, I√ +3/3 such that +I0 = [−2−17, 2−17], +I1 = [1 − 2−17, 1 + 2−17] +230I√ +3/3 = [619916940, 619933323] (21) +I√ +3/3 is a rational interval that is just barely contained inside the interval +of length 2−17 centered at +√ +3/3. Define +Ω00 = (I1 × {0}) × (I0 × −I√ +3/3) × (−I1 × I0) × (I0 × I√ +3/3). +(22) +We have Ω00 ⊂ Ω0, though just barely. There are 128 vertices of B. We +simply check whether each of these vertices is contained in Ω00. If so then +we return yes. In practice our program scales up all the coordinates by 230 +so that this test just involves integer comparisons. +29 + +5.4 +Proof of Lemma A132 +Let B = Q0×Q1×Q2×Q3 be an acceptable block. These blocks are such that +the squares Q1, Q2, Q3 do not cross the coordinate axes. For such squares, +the minimum and maximum norm of a point in the square is realized at a +vertex. Thus, we check that a square lies inside (respectively outside) a disk +of radius r centered at the origin by checking that the square norms of each +vertex is at most (respectively at least) r2. +We check whether there is an index j ∈ {1, 2, 3} such that all vertices of +Qj have norm at least max Q0. We return yes if this happens, because then +all configurations in the interior of B will have some pj with ∥pj∥ > ∥p0∥. +We check whether there is an index j ∈ {1, 2, 3} such that all vertices +of Qj have norm at least 3/2. If so, we return yes. If this happens then +∥p0∥, ∥pj∥ > 3/2 for all configurations in the interior of B. +We count the number a of indices j such that the vertices of Qj all have +norm at most 1/2. We then count the number b of indices j such that all +vertices of Qj have norm at least 1/2. We return yes if a is odd and a+b = 4. +In this case, every configuration in the interior of B is even. +We write I ≤ J to indicate that all values in an interval I are less or +equal to all values in an interval J. We also allow I and J to be single points +in this notation. For each j = 0, 1, 2, 3 we let Qjk be the projection of Qj +onto the kth factor. Thus Qj1 and Qj2 are both line segments in R. +We return yes for any of the following reasons. +• If Qjk ≤ −3/2 or Qjk ≥ 3/2 for any j = 1, 2, 3 and k = 1, 2. +• Q12 ≥ Q22 or Q12 ≥ Q32 or Q22 ≥ Q32 or Q22 ≤ 0. +• Qj1 ≥ 0 for j = 1, 2, 3 or Qj2 ≥ 0 for j = 1, 2, 3. +If any of these things happens, all configurations in Q violate some condition +for membership in the interior of Ω. We don’t check the last item for Ω♭. ♠ +5.5 +Proof of Lemma A133 +For C♯ +3 we return yes if all the vertices of B lie in Υ. For C♯♯ +3 we return +yes if one of the factors of B is disjoint from the corresponding factor of Υ. +This amounts to checking whether a pair of rational squares in the plane are +disjoint. We do this using the projections defined for Lemma A132. +30 + +5.6 +Proof of Lemma A134 +We say that an acceptable block B = Q0 × Q1 × Q2 × Q3 is good if we +have Qj ∈ [−3/2, 3/2]2 for all j = 1, 2, 3. We first test whether B is a good +block. If not, we return the lowest index i such that Qi has a vertex outside +[−3/2, 3/2]2. Otherwise we proceed with the main calculation. +We let Q denote the set of components of good blocks – either segments +or squares. We also let {∞} be a member of Q. We first define some mea- +surements we take of members in Q. +0. The Flat Approximation: Let Σ−1 be inverse stereographic projection, +as in Equation 8. Given Q ∈ Q we define +Q• = Convex Hull(Σ−1(v(Q)). +(23) +The set Q• is either the point (0, 0, 1), a chord of S2 or else a convex planar +quadrilateral with vertices in S2 that is inscribed in a circle. We let d• be +the diameter of Q•. The quantity d2 +• is a rational function of the vertices of Q. +1. The Hull Approximation Constant: We think of Q• as the linear +approximation to +�Q = Σ−1(Q). +(24) +The constant we define here turns out to measure the distance between �Q +and Q•. When Q = {∞} we define δ(Q) = 0. Otherwise, let +χ(D, d) = d2 +4D + (d2)2 +4D3 . +(25) +This wierd function turns out to be an upper bound to a more geometrically +meaningful non-rational function that computes the distance between an +chord of length d of a circle of radius D and the arc of the circle it subtends. +When Q is a dyadic segment we define +δ(Q) = χ(2, ∥�q1 − �q2∥). +(26) +Here q1, q2 are the endpoints of Q. When Q is a dyadic square we define +δ(Q) = max(s0, s2) + max(s1, s3), +sj = χ(1, ∥qj − qj+1∥). +(27) +Here q1, q2, q3, q4 are the vertices of Q and the indices are taken cyclically. +These are rational computations because χ(2, d) is a polynomial in d2. +31 + +2. +The Dot Product Estimator: By way of motivation, we point out +that if V1, V2 ∈ S2 then Gk(∥V1 − V2∥) = (2 + 2V1 · V2)k. +Now suppose that Q1 and Q2 are two dyadic squares. We set δj = δ(Qj). +Given any p ∈ R2 ∪ ∞ let �p = Σ−1(p). Define +Q1 · Q2 = max +i,j (�q1i · �q2j) + (τ) × (δ1 + δ2 + δ1δ2). +(28) +Here {q1i} and {q2j} respectively are the vertices of Q1 and Q2. The constant +τ is 0 if one of Q1 or Q2 is {∞} and otherwise τ = 1. Finally, we define +T(Q1, Q2) = 2 + 2(Q1 · Q2). +(29) +3. The Local Error Term: For Q1, Q2 ∈ Q and k ≥ 1 we define +ǫk(Q1, Q2) = 1 +2k(k − 1)T k−2d2 +1 + 2kT k−1δ1, +(30) +where +d1 = d•(Q1), +δ1 = δ(Q1), +T = T(Q1, Q2). +One of the terms in the error estimate comes from the analysis of the flat +approximation and the second term comes from the analysis of the difference +between the flat approximation and the actual subset of the sphere. The +quantity is not symmetric in the arguments and ǫk({∞}, Q2) = 0. +4. +The Global Error Estimate: Given a block Q0 × Q1 × Q2 × Q3 +we define +ERRk(B) = +N +� +i=0 +ERRk(B, i), +ERRk(B, i) = +� +j̸=i +ǫ(Qi, Qj). +(31) +More generally, when F = � ckGk is as in Equation 20, we define +ERRF(B) = +N +� +k=0 +ERRF(B, i), +ERRF(B, i) = +� +|ck| ERRk(B, i) +(32) +Now we state the main error estimate, proved in §7. For the most part +we only care about the (+) case of the lemma. We only need the (−) case +when we deal with the potential G5 − 25G1. +32 + +Lemma 5.6 (E) Let B be a good block. Let F = Gk for any k ≥ 1 or +F = −G1. Then +min +p∈B EF(v) ≥ min +p∈v(B) Ek(v) − ERRk(B) +Proof of Lemma A134: Now we can prove lemma A134. Let B be an +acceptable block. Once again, we mention that we immediately return an +integer if our block B is not a good block. So, assume B is a good block. +Let F be an energy hybrid. Let [F] denote the F-potential of the TBP. If +min +p∈v(B) EF(v) − ERRk(B) ≥ [F] + 2−50 +(33) +we return yes. Otherwise we return the index i such that ERRF(B, i) is the +largest. (In the case of a tie, which probably never happens, we would pick +the lowest such index.) ♠ +Remarks: (1) Lemma E is true more generally for F = ±Gk but we do not +need the general result and so we ignore it. Keeping track of the minus sign +all the time greatly increases the chances of making a sign error and we don’t +want to fool around with this. +(2) The integer we return is designed to be a recommendation for the subdivi- +sion that is most likely to speed the computation along. We try to subdivide +in such a way as to decrease the error term as fast as possible. +5.7 +Discussion of the Implementation +Representing Blocks: We represent the coordinates of blocks by longs, +which have 31 digits of accuracy. What we list are 230 times the coordinates. +Our algorithm never does so many subdivisions that it defeats this method +of representation. In all but the main step (Lemma A134) in the algorithm +below we compute with exact integers. When the calculation (such as squar- +ing a long) could cause an overflow error, we first recast the longs as a +BigIntegers in Java and then do the calculations. +Interval Arithmetic: For the main step of the algorithm we use inter- +val arithmetic. We use the same implementation as we did in [S1], where we +explain it in detail. Here is how it works in brief. If we have a calculation +involving numbers r1, ..., rn, and we produce intervals I1, ..., In with dyadic +33 + +rational numbers represented exactly by the computer such that ri ∈ Ii for +i = 1, ..., n. We then perform the usual arithmetic operations on the inter- +vals, rounding outward at each step. The final output of the calculation, in +interval, contains the result of the actual calculation. +In our situation here, the numbers r1, ..., rn are, with one exception, +dyadic rationals. (The exception is that the coordinates of the point rep- +resenting the TBP are quadratic irrationals.) In principle we could do the +entire computation, save for this one small exception, with expicit integer +arithmetic. However, the complexity of the rationals involved, meaning the +sizes of their numerators and denominators, qets quite large this way and the +calculation is too slow. +One way to think about the difference between our explicitly defined ex- +act integer arithmetic and interval arithmetic is that the integer arithmetic +interrupts the calculation at each step and rounds outward so as to keep the +complexity of the rational numbers from growing too large. +Guess and Check: Here is how we speed up the calculation. When we +do Steps 6-7, we first do the calculation C4,F using floating point operations. +If the floating version returns an integer, we use this integer to subdivide the +box and return to step 2. If C4,F says yes then we retest the box using the +interval arithmetic. In this way, we only pass a box for which the interval +version says yes. This way of doing things keeps the calculation rigorous but +speeds it up by using the interval arithmetic as sparingly as possible. +Parallelization: We also make our calculation more flexible using some +parallelization. +We classify each block B = Q0 × Q1 × Q2 × Q3 with a +number in {0, ..., 7} according to the formula +type(B) = σ(c01 − 1) + 2σ(c11) + 4σ(c31) ∈ {0, ..., 7}. +Here cj1 is the first coordinate of the center of Bj and σ(x) is 0 if x < 0 and +1 if x > 0. Step 3 of our algorithm guarantees that σ(·) is always applied to +nonzero numbers. +We wrote our program so that we can select any subset S ⊂ {0, ..., 7} we +like and then (after Step 3) automatically pass any block whose type is not +in S. Running the algorithm in parallel over sets which partition {0, ..., 7} is +logically the same as running the basic algorithm without any parallelization. +To be able to do the big calculations in pieces, we run the program for various +subsets of {0, ..., j}, sometimes in parallel. +34 + +5.8 +Proof of Lemma A135 +Here I give an account of one time I ran the computations to completion +during January 2023. I used a 2017 iMac Pro with a 3.2 GHz Intel Zeon W +processor, running the Mojave operating system. I ran the programs using +Java 8 Update 201. (The Java version I use is not the latest one. The graph- +ical parts of my program use some methods in the Applet class in a very +minor but somehow essential way that I find hard to eliminate.) In listing +the calculations I will give the approximate time and the exact number of +blocks passed. Since we use floating point calculations to guide the algo- +rithm, the sizes of the partitions can vary slightly with each run. +For G4 : 2 hrs 14 min, 10848537 blocks. +For G6: 5 hr 11 min, 25159337 blocks. +For G♭ +5 types 1&2: 2 hr 31 min, 6668864 blocks. +For G♭ +5 types 3&4: 1 hr 55 min, 4787489 blocks. +For G♭ +5 types 5&6: 5 hr 33 min, 14160332 blocks. +For G♭ +5 types 7&8: 3 hr 49 min, 9219550 blocks. +For G♯ +10 type 1: 4 hr 23 min, 6885912 blocks. +For G♯ +10 type 2: 9 hr 47 min, 15982122 blocks. +For G♯ +10 type 3: 3 hr 47 min, 5872029 blocks. +For G♯ +10 type 4: 7 hr 59 min, 13475260 blocks. +For G♯ +10 type 5: 8 hr 30 min, 13313492 blocks. +For G♯ +10 type 6: 15 hr 16 min, 24110457 blocks. +For G♯ +10 type 7: 5 hr 19 min, 7862780 blocks. +For G♯ +10 type 8: 8 hr 33 min, 13478467 blocks. +For G♯♯ +10 (on the domain Υ): 28 minutes, 805242 blocks. +35 + +6 +Local Analysis: Proof of Lemma A112 +6.1 +Reduction to Simpler Statements +We set L=A122, so that we are trying to prove Lemma L. We consider F to +be any of the 4 functions +G4, +G6, +G♭ +5 = G5 − 25G1, +2−5G♯ +10 = 2−5(G10 + 13G5 + 68G2). +Scaling the last function by 2−5 makes our estimates more uniform. +Recall that Ω0 is the cube of side length 2−17 centered at the point +ξ0 = +� +1, 0, −1 +√ +3 +, −1, 0, 0, 1 +√ +3 +� +∈ R7 +(34) +In general, the point (x1, ..., x7) represents the planar avatar +p0 = (x1, 0), p1 = (x2, x3), p2 = (x4, x5), p3 = (x6, x7). +(35) +The quantity EF(x1, ..., x7) is the F-potential of the 5-point configuration +associated to the planar avatar under inverse stereographic projection Σ−1. +EF(x1, ..., x7) = +� +i 0 for all +points ξ ∈ Ω0. +Lemma 6.2 (L2) M3(EF) < 212λ(HEF(ξ0))) in all cases. +36 + +6.2 +Proof of Lemma L1 +Let +H0 = HEF(ξ0), +H = HEF(ξ), +∆ = H − H0. +(38) +For any real symmetric matrix X define the L2 matrix norm: +∥X∥2 = +�� +ij +X2 +ij = sup +∥v∥=1 +∥Xv∥. +(39) +Given a unit vector v ∈ R7 we have H0v · v ≥ λ. Hence +Hv · v = (H0v + ∆v) · v ≥ H0v · v − |∆v · v| ≥ λ − ∥∆v∥ ≥ λ − ∥∆∥2 > 0. +So, to prove Lemma L1 we just need to establish the implication +M3 < 212λ(H0) +=⇒ +∥∆∥2 < λ(H0). +Let t → γ(t) be the unit speed parametrized line segment connecting p0 to +p in Ω0. Note that γ has length L ≤ +√ +7×2−18. We write γ = (γ1, ..., γ7). Let +Ht denote the Hessian of EF evaluated at γ(t). Let Dt denote the directional +derivative along γ. +Now ∥Dt(Ht)∥2 is the speed of the path t → Ht in R49, and ∥∆∥2 is the +Euclidean distance between the endpoints of this path. Therefore +∥∆∥2 ≤ +� L +0 +∥Dt(Ht)∥2 dt. +(40) +Let (Ht)ij denote the ijth entry of Ht. From the definition of directional +derivatives, and from the Cauchy-Schwarz inequality, we have +(DtHt)2 +ij = +� +7 +� +k=1 +dγk +dt +∂Hij +∂k +�2 +≤ 7M2 +3. +∥Dt(Ht)∥2 ≤ 73/2M3. +(41) +The second inequality follows from summing the first one over all 72 pairs +(i, j) and taking the square root. Equation 40 now gives +∥∆∥2 ≤ L × 73/2M3 = 49 × 2−18M3 < 2−12M3 < λ(H0). +(42) +This completes the proof of Lemma L1. +37 + +6.3 +Proof of Lemma L2 +Let F be any of our functions. Let H0 = HEF(ξ0). +Lemma 6.3 (L21) λ(H0) > 39. +Proof: Let χ be the characteristic polynomial of H0. This turns out to be +a rational polynomial. We check in Mathematica that the signs of the coef- +ficients of χ(t + 39) alternate. Hence χ(t + 39) has no negative roots. The +file we use is LemmaL21.m. ♠ +Recalling that ξ0 ∈ R7 is the point representing the TBP, we define +µN(EF) = sup +|I|=N +|∂IEF(ξ0)|. +(43) +Lemma 6.4 (L22) For any of our functions we have the bound +µ3 < 45893, +(7 × 2−18)j +j! +µj+3 < 38, +j = 1, 2, 3. +(44) +Proof: We compute this in Mathematica. The file we use is LemmaL22.m. ♠ +Lemma 6.5 (L23) For any of our functions we have the bound +(7 × 2−18)4 +4! +M7 < 2354. +Lemma 6.6 (L24) We have +M3 ≤ µ3 + +3 +� +j=1 +(7 × 2−18)j +j! +µj+3 + (7 × 2−18)4 +4! +M7 +(45) +Proof: Choose any multi-index J with |J| = 3. Let γ be the line segment +connecting ξ0 to any ξ ∈ Ω. We parametrize γ by unit speed and furthermore +set γ(0) = ξ0. Let +f(t) = ∂JEF ◦ γ(t). +38 + +The bound for |MJ| follows from Taylor’s Theorem with remainder once we +notice that +0 ≤ t ≤ +√ +7 × 2−18, +���∂nf(0) +∂tn +��� ≤ ( +√ +7)nµn +���∂nf +∂tn +��� ≤ ( +√ +7)nMn. +Since this works for all J with |J| = 3 we get the same bound for M3. ♠ +Lemmas L21 - L23 and Equation 44 imply +M3 < 45893 + 3 × 38 + 2354 ≤ 65536 = 216 ≤ 212λ(H0). +This completes the proof of Lemma L2. +6.4 +Proof of Lemma L23 +Now we come to the interesting part of the proof, the one place where we +need to go beyond specific evaluations of our functions. When r, s ≥ 0 and +r + s ≤ 2d we have +sup +(x,y)∈R +2 +xrys +(1 + x2 + y2)d ≤ (1/2)min(r,s). +(46) +One can prove Equation 46 by factoring the expression into pieces with +quadratic denominators. Here is a more general version. Say that a function +φ : R4 → R is nice if it has the form +� +i +Ciaαibβicγidδi +(1 + a2 + b2)ui(1 + c2 + d2)vi , αi, βi, γi, δi ≥ 0, +αi+βi ≤ 2ui, +γi+δi ≤ 2vi. +It follows from Equation 46 that +sup +R +4 |φ| ≤ ⟨φ⟩, +⟨φ⟩ = +� +i +|Ci|(1/2)min(αi,βi)+min(γi,δi). +(47) +Equation 47 is useful to us because it allows us to bound certain kinds of +functions without having to evaluate then anywhere. We also note that if +φ is nice, then so is any iterated partial derivative of φ. Indeed, the nice +functions form a ring that is invariant under partial differentiation. This fact +makes it easy to identify nice functions. +39 + +For any φ : Rn → R we define +M 7(ψ) = sup +|J|=7 +M J(ψ), +M J(ψ) = sup +ξ∈R +n |∂J(φ)|. +(48) +We obviously have +M7(EF) ≤ M 7(EF). +(49) +Recall that �p = Σ−1(p), the inverse stereographic image of p. Define +f(a, b) = 4 − ∥� +(a, b) − (0, 0, 1)∥2 = 4(a2 + b2) +1 + a2 + b2. +(50) +g(a, b, c, d) = 4 − ∥� +(a, b) − � +(c, d)∥2 = 4(1 + 2ac + 2bd + (a2 + b2)(c2 + d2)) +(1 + a2 + b2)(1 + c2 + d2) +. (51) +Notice that g is nice. Hence gk is nice and ∂Igk is nice for any multi-index. +That means we can apply Equation 47 to ∂Igk. +EGk is a 10-term expression involving 4 instances of f k and 6 of gk. How- +ever, each variable appears in at most 4 terms. So, as soon as we take a +partial derivative, at least 6 of the terms vanish. Moreover, ∂If is a limiting +case of ∂Ig for any multi-index I. From these considerations, we see that +M7(EGk) ≤ 4 × M 7(gk). +(52) +The function ∂I(gk) is nice in the sense of Equation 47. Therefore +4 × M 7(gk) ≤ 4 × max +|I|=7 ⟨∂Igk⟩. +(53) +Using this estimate, and the Mathematica file LemmaL23.m, we get +max +k∈{1,2,3,4,5,6} +(7 × 2−18)4 +4! +× 4 × M 7(gk) ≤ +1 +1000. +2−5 × (7 × 2−18)4 +4! +× 4 × M 7(g10) ≤ 2353. +(54) +The bounds in Lemma L23 follow directly from Equation 52 and from the +definitions of our functions. +Remark: The Polya Case. +The analysis above works easily for G3. +In +this case, the minimum eigenvalue satisfies λ0 > 14. +The bounds satisfy +µ3 ≤ 316 and µj < 1 for j = 4, 5, 6. Again, M7 < 1/1000 in this case. +40 + +7 +Error Estimate: Proof of Lemma E +7.1 +Guide to the Proof +Lemma E is stated in §5.6. It is the main error estimate that feeds into +Lemma A134, which in turn feeds into Lemma A13, our main computation. +Our proof of Lemma E splits into two halves, an algebraic part and a +geometric part. +The algebraic part, which we do in this chapter, has no +geometric content. It simply promotes a “local” result to a “global result”. +The geometric part, done in the next chapter, explains the meaning of the +local error term ǫk(Q1, Q2) for Q1, Q2 ∈ Q. Here Q is the space of components +of good blocks, and also the point ∞. +The algebraic part promotes the +information about ǫk(Q1, Q2), which we use as a black box in this chapter, +to the information about the global sum given in Lemma E. +The secret to the algebraic part is what we call an averaging system. For +these purposes, we will treat every member of Q as a quadrilateral by the +trick of repeating vertices. Thus, if we have a dyadic segment with vertices +q1, q2 we will list them as q1, q1, q2, q2. For the point {∞} we will list the +single vertex q1 = ∞ as q1, q1, q1, q1. We do this so as to give a uniform +description without having to stop each time and say what we are doing in +each case. +We say that an averaging system for a member of Q is a collection of +maps λ1, λ2, λ3, λ4 : Q → [0, 1] such that +4 +� +i=1 +λi(z) = 1, +∀ z ∈ Q. +One might also call this a partition of unity. The functions need not vary +continously. In case Q is a segment, we would have λ1 = λ2 and λ3 = λ4. In +case Q = {∞} we would have λj = 1/4 for j = 1, 2, 3, 4. +We say that an averaging system for Q is a choice of averaging system +for each member Q of Q. The averaging systems for different members need +not have anything to do with each other. In this chapter we will posit some +additional properties of an averaging system and then prove Lemma E under +the assumption such such an averaging system exists. In the next chapter +we will use geometric considerations to prove the existence of the desired +averaging system. +41 + +7.2 +Reduction to a Local Result +We fix the function F = Gk for some k ≥ 1 or else F = −G1. Lemma E1 +is true more generally for F = ±Gk but we don’t need the result in this +generality. +We write E = EF. We let ǫ = ǫk, as in Equation 30. +Our algebraic +argument would work for any choice of F, but we need F = Gk to actually +get the averaging system we need. Let q1,1, q1,2, q1,3, q1,4 be the vertices of Q1. +Lemma 7.1 (E1) There exists an averaging system on Q with the following +property: Let Q1, Q2 be distinct members of Q. +Given any z1 ∈ Q1 and +z2 ∈ Q2 we have +4 +� +i=1 +λi(z1)F(∥�q1,i − �z2∥) − F(∥�z1 − �z2∥) ≤ ǫ(Q1, Q2). +(55) +See §8 for the proof. +We call the averaging system in Lemma E1 an +efficient averaging system. We now explore the consequences of having an +efficient averaging system. +We suppose that we have the good dyadic block B = Q0 × ... × QN. The +vertices of B are indexed by a multi-index +I = (i0, ..., in) ∈ {1, 2, 3, 4}N+1. +Given such a multi-index, which amounts to a choice of vertex of in each +component member of the block. We define the energy of the corresponding +vertex configuration: +E(I) = E(q0,i0, ..., qN,iN) +(56) +Here is one more piece of notation. Given z = (z0, ..., zn) ∈ B and a +multi-index I we define +λI(z) = +N +� +i=0 +λij(zj). +(57) +Here λij is defined relative to the averaging system on Qj. +Now we are ready to state our main global result. The global result uses +the existence of an efficient averaging system. That is, it relies on Lemma +E1. +42 + +Lemma 7.2 (E2) Let z = (z0, ..., zN) ∈ B. Then +� +I +λI(z)E(I) − E(z) ≤ +N +� +i=0 +N +� +j=0 +ǫ(Qi, Qj). +(58) +The sum is taken over all multi-indices with the convention that ǫ(Qi, Qi) = 0 +for all i. +Now let us deduce Lemma E from Lemma E2. +Lemma 7.3 (E3) Lemma E1 implies Lemma E. +Proof: Notice that +� +I +λI(z) = +N +� +j=0 +� +4 +� +a=1 +λa(zj) +� += 1. +(59) +The minimum is always less or equal to a convex average. Hence +min +p∈v(B) E(v) ≤ +� +I +λI(z)E(I). +(60) +Choose some (z1, ..., zN) ∈ B which minimizes E. We have +0 ≤ min +p∈v(B) E(v) − min +v∈B E(v) = min +p∈v(B) E(v) − E(z) ≤∗ +� +I +λI(z)E(I) − E(z) ≤ +N +� +i=0 +N +� +j=0 +ǫ(Qi, Qj). +(61) +The starred inequality is Equation 60. The last expression is ERR(B) when +N = 4 and Q4 = ∞. ♠ +43 + +7.3 +From Local to Global +Now we deduce the global Lemma E2 from the local Lemma E1. +Lemma 7.4 (E4) Lemma E2 holds when N = 1. +Proof: In this case, we have a block B = Q0 × Q1. Setting ǫij = ǫ(Qi, Qj), +Lemma E1 gives us +F(∥z0 − z1∥) ≥ +4 +� +α=1 +λα(z0)F(∥q0α − z1∥) − ǫ01. +(62) +Applying Lemma E1 to the pair of points (z1, q0α) ∈ Q1 × Q0 we have +F(∥z1 − q0α∥) ≥ +4 +� +β=1 +λβ(z1)F(∥q1β − q0α∥) − ǫ10. +(63) +Plugging the second equation into the first and using � λα(z0) = 1, we have +F(∥z0 − z1∥) ≥ +� +α,β +λα(z0)[λβ(z1)F(∥q1β − q0α∥) − ǫ10] − ǫ01 = +� +α,β +λα(z0)λβ(z1)F(∥q1β − q0α∥) − (ǫ10 + ǫ01). +(64) +Equation 64 is equivalent to Equation 58 when N = 1. ♠ +Now we do the general case. +Lemma 7.5 (E5) Lemma E2 holds when N ≥ 2. +Proof: We rewrite Equation 64 as follows: +F(∥z0 − z1∥) ≥ +� +A +λA0(z0)λA1(z1) F(∥q0A0 − q1A1∥) − (ǫ01 + ǫ10). +(65) +The sum is taken over multi-indices A of length 2. +We also observe that +� +I′ +λI′(z′) = 1, +z′ = (z2, ..., zN). +(66) +44 + +The sum is taken over all multi-indices I′ = (i2, ..., iN). Therefore, if we hold +A = (A0, A1) fixed, we have +λA0(z0)λA1(z1) = +� +I′′ +λI′′(z). +(67) +The sum is taken over all multi-indices of length N + 1 which have I0 = A0 +and I1 = A1. Combining these equations, we have +F(∥z0 − z1∥) ≥ +� +I +λI(z)F(∥q0I0 − q1I1∥) − (ǫ01 + ǫ10). +(68) +The same argument works for other pairs of indices, giving +F(∥zi − zj∥) ≥ +� +I +λI(z)F(∥qiIi − qjIj∥) − (ǫij + ǫji). +(69) +Let us restate this as +Xij − Yij ≥ Zij, +where +Xij = +� +I +λI(z)F(∥qiIi − qjIj∥), +Yij = F(∥zi − zj∥), +Zij = ǫij + ǫji. +When we sum Yij over all i < j we get the second term in Equation 58. +When we sum Zij over all i < j we get the third term in Equation 58. When +we sum Xij over all i < j we get +� +i A∗ +when d = 1/2. Hence A > A∗ on (0, 1). This implies the inequality. ♠ +51 + +9.3 +Proof of Lemma E112 +Let Q be a dyadic square and let z ∈ Q be a point. Let L be the vertical line +through x and let z01, z23 be the endpoints of the segment L ∩ Q. We label +the vertices of Q (in cyclic order) so that z01 lies on the edge joining q0 to q1 +and z23 lies on the edge joining q2 to q3. +Lemma 9.4 (E1121) If M is any horizontal or vertical line intersecting Q +then the circle Σ−1(M ∪ ∞) has diameter at least 1. +Proof: Since Q is a component of a good block, and M is horizontal or verti- +cal, M contains a point p with ∥p∥ ≤ 3/2. At the same time, M ∪∞ contains +∞. An easy calculation shows that ∥Σ−1(p)−(0, 0, 1)∥ ≥ 4/ +√ +13 > 1. Hence, +our circle contains two points which are greater than 1 unit apart. ♠ +Define +dj = ∥�pj − �pj+1∥. +The length of the segment σ joining the endpoints of Σ−1(L∩Q) varies mono- +tonically with the position of L. Hence, σ has length at most max(d1, d3). +At the same time, Σ−1(L ∩ Q) is contained in a circle of diameter at least 1. +The same argument as in the proof of Lemma E111 now shows that there is +a point z∗ ∈ σ which is within +t13 = χ(1, max(d1, d3)) = max(χ(1, d1), χ(1, d3)) +of �z. +The endpoints of σ respectively are on the spherical arcs obtained by +mapping the top and bottom edge of Q onto S2 via Σ−1. Hence, one endpoint +of σ is within χ(1, d0) of a point on the corresponding edge of ∂Q• and the +other endpoint of σ is within χ(1, d2) of a point on the opposite edge of ∂Q•. +But that means that either endpoint of σ is within +t02 = max(χ(1, d0), χ(1, d2)) +of a point in Q•. But then every point of σ is within t02 of some point of the +line segment joining these two points of Q•. In particular, there is a point +z• ∈ Q• which is within t of z∗. +We have shown that �z is within t13 of z∗ and z∗ is within t02 of z•. Hence +�z is within t02 + t13 = δ(Q) of �z. This completes the proof of Lemma E112. +52 + +10 +Interpolation: Proof of Lemma A2 +10.1 +Statement of the Result +The goal of this part of the monograph is to prove Lemma A2. Here we +repeat all the needed definitions. Rs(r) = r−s is the Riesz potential. The +other potentials we consider are: +Gk(d) = (4 − r2)k. +(90) +Also define +G♭ +5 = G5 − 25G1, +G♯♯ +10 = G10 + 28G5 + 102G2, +G♯ +10 = G10 + 13G5 + 68G2 +(91) +Given any function F, and a configuration p0, p1, p2, p3, p4 of 5 distinct +points on S2, we define +EF(P) = +� +i a0 + +4 +� +j=1 +ajEΓj(T0) = EΛ(T0) = Es(T0). +The central inequality is strict because a1, a2, a3, a4 ≥ 0 and at least one is +nonzero. ♠ +Lemma 10.3 (A22) The pair (G4, G6) specially forces (0, 6]. +Lemma 10.4 (A23) The following is true. +1. The pair (G5, G## +10 ) specially forces [6, 13]. +2. The pair (G♭ +5, G# +5 ) specially forces [13, 15++]. +54 + +10.3 +Proof of Lemma A22 +Before we start the proofs, we direct the reader’s attention to §2.4, which +explains how the reader can see plots of all relevant functions. Using the +software here is like eating a cooked meal rather than just reading a recipe. +Here is how we find the coefficients for the pair (Γ3, Γ4) = (G4, G6). The +same method works for the other cases. Let Γj = Gj for j = 1, 2. We define +Λs = a0(s) + +4 +� +j=1 +aj(s)Γj. +(94) +We then solve the equations +Λs(√m) = Rs(√m), +m = 2, 3, 4, +Λ′ +s(√m) = R′ +s(√m), +m = 2, 3. (95) +Here f ′ denotes the derivative of f, a function defined on (0, 2]. We don’t need +to constrain f ′(2). For each s this gives us a linear system with 5 variables +and 5 equations. In all cases, our solutions have the following structure +(a0, a1, a2, a3, a4) = M(2−s/2, 3−s/2, 4−s/2, s2−s/2, s3−s/2) +(96) +Solving this system of equations, we get +792M = + + +0 +0 +792 +0 +0 +792 +1152 +−1944 +−54 +−288 +−1254 +−96 +1350 +87 +376 +528 +−312 +−216 +−39 +−98 +−66 +48 +18 +6 +10 + + +(97) +Lemma 10.5 (A221) aj(s) > 0 for s ∈ (0, 6] and j = 1, 2, 3, 4. +We define the comparison function +Hs = 1 − Λs +Rs +. +(98) +We want to show that Hs > 0 on (0, 2) − { +√ +2, +√ +3}. We have +H′ +s(r) = rs−1(sΛs(r) + rΛ′ +s(r)). +(99) +55 + +The extrema of Hs occur at the same places as the roots of H′ +s. The roots +of H′ +s coincide with the roots of sΛs(r) + rΛ′ +s(r). Using the general equation +rG′ +k(r) = 2kGk(r) − 8kGk−1(r), +(100) +we see that sΛs(r) + rΛ′ +s(r) is a polynomial in 4 − r2. Hence, the roots of H′ +s +in (0, 2) are in bijection with the roots in (0, 4) of +ψs(t) = (sΛ(r) + rΛ′(r)) +��� +r=√4−t = t6 − +48 +12 + st5 + ... +(101) +Lemma 10.6 (A222) ψs(0) > 0 and ψs(4) > 0 for all s ∈ (0, 6]. +Lemma 10.7 (A223) The only minina for Hs occur at r = +√ +2 and r = +√ +3. +Proof: Since ψs has degree 6 we conclude that ψs has at most N = 6 roots, +counting multiplicity. By construction Hs(√m) = H′ +s(√m) = 0 for m = 2, 3 +and Hs( +√ +4) = 0. This means that Hs has extrema at r2 = +√ +2 and r3 = +√ +3 +and at points r23 ∈ ( +√ +2, +√ +3) and r34 ∈ ( +√ +3, +√ +4). Correspondingly ψs has +roots t1 = 1 and t2 = 2 and t01 ∈ (0, 1) and t12 ∈ (1, 2). The sum of all the +roots of ψs is 48/(12 + s) < 4. Since t1 + t2 + t01 + t12 > 4 we see that not all +roots can be positive. Hence N < 6. Since ψs is positive at t = 0, 4 we see +that N is even. Hence N = 4. This means that the only roots of ψs in (0, 4) +are the ones we already know about. Hence the only extrema of Hs in (0, 2) +are ones we already have mentioned. As s varies the nature of these extrema +cannot change. So, we check at s = 2 that +√ +2 and +√ +3 are minima and then +this fact persists for all a ∈ (0, 6]. ♠ +Lemma A22 is an immediate consequence of Lemmas A221, A222, A223 +10.4 +Proof of Lemma A23 +We find the coefficients for the case (G5, G♯♯ +10) just as we did for the case +(G4, G6). This time the matrix M is given below. To get an integer matrix, +we list 268536M, + + +0 +0 +268536 +0 +0 +88440 +503040 +−591480 +−4254 +−65728 +−77586 +−249648 +327234 +2361 +65896 +41808 +−19440 +−22368 +−2430 +−9076 +−402 +264 +138 +33 +68 + + +(102) +56 + +Lemma 10.8 (A231) With respect to the pair (G5, G♯♯ +10) the coefficients +a1(s), ..., a4(s) are positive for s ∈ [6, 13]. +Similarly, for the pair (G♭ +5, G♯ +10) we get the following matrix M, which we +list as 268536M + + +0 +0 +268536 +0 +0 +0 +982890 +116040 +−1098930 +−52629 +−267128 +0 +−91254 +−240672 +331926 +3483 +68208 +0 +35778 +−15480 +−20298 +−1935 +−8056 +0 +−402 +264 +138 +33 +68 +0 + + +(103) +Lemma 10.9 (A232) With respect to the pair (G♭ +5, G♯ +10) the coefficients +a1(s), ..., a4(s) are positive for s ∈ [13, 15++]. +Now we proceed as in the previous section. It turns out that Λs (and +hence Hs) is the same with respect to either pair (G5, G♯♯ +10) and (G♭ +5, G♯ +10). +This is not an accident. The reason is that for both pairs we are simply using +the four functions G1, G2, G5, G10 to under-approximate the power functions. +We derive the polynomial ψs exactly as we did for the pair (G4, G6). +Lemma 10.10 (A233) The function ψs has at most 4 roots in [0, 4]. +Lemma A233 shows that ψs only has the 4 roots we already know about +– the same ones enumerated for the pair (G4, G5). Lemma A233 then shows +that these are the only roots. As s varies, the nature of these roots cannot +change. When s = 6 we confirm that the roots 1 and 2 are the two roots +corresponding to minima. Hence, this is the case for all s ∈ [6, 16]. Hence, +√ +2 and +√ +3 are the only local minima for Hs for all s ∈ [6, 16]. This proves +Lemma A23. +10.5 +The Polya Case +For the pair (G3, G5), and with respect to the potentials −r−s, the matrix is: +1 +144 + + +0 +0 +−144 +0 +0 +−312 +−96 +408 +24 +80 +684 +−288 +−396 +−54 +−144 +−402 +264 +138 +33 +68 +30 +−24 +−6 +−3 +−4 + + +. +Rows 2,3,4,5 give positive power combos for s ∈ (−2, 0). +57 + +11 +Interpolation: Auxiliary Lemmas +11.1 +Proof of Lemmas A231 and A232 +All the expressions that arise in Lemma A231 and A232 (and Lemmas A221 +and A222) have the following form. +F(s) = +� +cistibs/2 +i +, +(104) +where bi, ci ∈ Q and ti ∈ Z and bi > 0. +For each summand we com- +pute a floating point value, xi. We then consider the floor and ceiling of +232xi and divide by 232. This gives us rational numbers xi0 and xi1 such +that xi0 ≤ xi ≤ xi1. Since we don’t want to trust floating point operations +without proof, we formally check these inequalities with what we call the +expanding out method. +Expanding Out Method: Suppose we want to establish an inequality like +( a +b) +p +q < c +d, where every number involved is a positive integer. This inequality +is true iff bpcq − apdq > 0. We check this using exact integer arithmetic. The +same idea works with (>) in place of (<). +To check the positivity of F on some interval [s0, s1] we produce, for each +term, the 4 rationals xi00, xi10, xi01, xi01. Where xijk is the approximation +computed with respect to sk. We then let yi be the minimum of these ex- +pressions. The sum � yi is a lower bound for Equation 104 for all s ∈ [s0, s1]. +On any interval exponent I where we want to show that Equation 104 is pos- +itive, we pick the smallest dyadic interval [0, 2k] that contains I and then run +the following subdivision algorithm. +1. Start with a list L of intervals. Initially L = {[0, 2k]}. +2. If L is empty, then HALT. Otherwise let Q be the last member of L. +3. If either Q ∩ I = ∅ or the method above shows that Equation 104 is +positive on Q we delete Q from L and go to Step 2. +4. Otherwise we delete Q from L and append to L the 2 intervals obtained +by cutting Q in half. Then we ago to to Step 2. +If this algorithm halts then it constitutes a proof that F(s) > 0 for all s ∈ I. +We check that the algorithm halts for all 8 quantities associated to Lemmas +A231 and A232, respectively for the intervals [6, 13] and [13, 15++]. +58 + +11.2 +Proof of Lemmas A221 and A222 +The 6 quantities associated to Lemma A221 and A222 all have the form given +in Equation 104. Using the same method as in the previous section we show +that all these quantities are positive on [1/4, 6]. To analyze what happens +on the interval (0, 1/4] we take Taylor series expansions. Up to constants +and factors of s the 6 expressions all have the form Y · V (s), where Y is an +integer vector and V (s) = (2−s/2, ..., s3−s/2). For Lemma A221, the various +choices of Y are the rows of the matrix in Equation 97. For Lemma A222: +11ψs(0) = + + +−88 +−128 ++216 ++6 ++32 ++11 + + +· + + +2−s/2 +3−s/2 +4−s/2 +s2−s/2 +s3−s/2 +s4−s/2 + + +, +11 +s ψs(4) = + + +−2112 ++1664 ++459 ++219 +288 +0 + + +· + + +2−s/2 +3−s/2 +4−s/2 +s2−s/2 +s3−s/2 +s4−s/2 + + +We work with the expressions on the right hand sides of these equations, so +that they look more like what we have in Lemma A221. Note that +sup +m=2,3,4 sup +s∈[0,1] +��� ∂6 +∂s6m−s/2��� < 1 +8. +(105) +Moreover the sum of the absolute values of the coefficients in each of the Y +vectors is at most 5000. This means that, when we take the 5th order Taylor +series expansion for Y · V (s), the error term is at most +5000 × 1 +8 × 1 +6! < 1. +We compute each Taylor series, set all non-leading positive terms to 0, and +crudely round down the other terms: +98s − 69s2 + 0s3 − 6s4 + 0s5 − 1s6 +14s − 3s2 − 2s3 + 0s4 − 1s5 − 1s6. +1s + 0s2 − 1s3 + 0s4 + 0s5 − 1s6. +.03s + 0s2 + 0s3 − .01s4 + 0s5 − 1s6. +.08s + 0s2 − .02s3 + 0s4 − .01s5 − 1s6. +11 + 0s + 0s2 − 1s3 − 1s4 + 0s5 − 1s6. +These under-approximations are all positive on (0, 1/4]. For example, if f(s) +is any one of these polynomials them g(u) = f(u/4) is weak positive dominant +in the sense of §12.2. My computer code does these calculations rigorously +with interval arithmetic, but it hardly seems necessary. +59 + +11.3 +Proof of Lemma A233 +I will describe a proof which took me quite a lot of experimentation to find. +We want to show that the degree 10 polynomial ψs(t) has at most 4 roots in +[0, 4] for all s ∈ [6, 16]. The same analysis shows that ψs has roots at 1, 2, +and in (0, 1) and in (1, 2). We just want to see that there are no other roots. +We can factor ψs as (t − 1)(t − 2)βs where βs is a degree 8 polynomial. +Taking derivatives with respect to t, we notice that +Lemma 11.1 (A2331) The following is true: +1. γs = 268536 × 12s/2 × (β′′ +s − β′ +s) is positive for s × t ∈ [6, 16] × [0, 4]. +2. −β′ +s(0) > 0 for all s ∈ [6, 16]. +3. β′ +s(4) > 0 for all s ∈ [6, 16]. +Lemma A2331 Statement 1 shows in particular that β′ +s never has a double +root. This combines with Statements 2 and 3 to show that the number of +roots of β′ +s in [0, 4] is independent of s ∈ [6, 16]. We check explicitly that +β′ +6 has only one root in [0, 4]. Hence β′ +s always has just one root. But this +means that βs has at most 2 roots in [0, 4]. This, in turn, means that ψs has +at most 4 roots in [0, 4]. This completes the proof. +11.4 +Proof of Lemma A2331 +We first give a formula for γs. Define matrices M3, M4, M6 respectively as: + + +−546840 +−1800480 +99720 +−397440 +−234600 +−33120 +173880 +−22080 +18366 +17112 +80766 +24288 +18630 +11592 +4830 +−1104 +0 +0 +0 +0 +0 +0 +0 +0 + + + + +−345600 +−1576320 +−509760 +−760320 +−448800 +−63360 +332640 +−42240 +−199296 +−698784 +75216 +−149376 +−79960 +5856 +94920 +−12992 +7104 +8432 +33960 +11968 +9180 +5712 +2380 +−544 + + + + +892440 +3376800 +410040 +1157760 +683400 +96480 +−506520 +64320 +−73350 +−246888 +−228942 +−165792 +−110370 +−41688 +27510 +−2064 +1473 +4092 +10557 +5808 +4455 +2772 +1155 +−264 + + +60 + +Define 3 polynomials P3, P4, P6 by the formula: +Pk(s, t) = (1, s, s2) · Mk · (1, ..., t7) = +2 +� +i=0 +7 +� +j=0 +(Mk)ijsitj, +k = 3, 4, 6. (106) +We have +γ = P33s/2 + P44s/2 + P66s/2. +(107) +To check the positivity of γs we check that each of the 16 functions +γs(v/4 + 1/4) = av,0 + av,1t + ...av,7t7 +(108) +is positive dominant in the sense of §12.2 for each v = 0, ..., 15. This amounts +to showing that each sum Av,k = av,0 + ... + av,k is positive for all s ∈ [6, 16]. +For each v = 0, ..., 15 and each k = 0, ...., 7 we have a 3 × 3 integer matrix +µv,k such that +Av,k = (1, s, s2) · µv,t · (3s/2, 4s/2, 6s,2). +(109) +This gives 128 matrices to check. We get two more such matrices from the +conditions −β′ +s(0) > 0 and β′ +s(4) > 0. All in all, we have to check that 130 +expressions of the form in Equation 109 are positive for s ∈ [6, 16]. These +expressions are all special cases of Equation 104, and we use the method +discussed above to show positivity in all 130 cases. The program runs in +several hours. +Remark: I found the 130 matrices by manipulating ψs in Mathematica +and then exporting the results to a file which our Java code reads. Our Java +code has an auxiliary debugger which uses the 130 matrices to reconstruct +the values of γs at randomly chosen points. The reader can check the print- +out against the output of our Mathematica file LemmaA233.m, which has γs +defined in a more direct way. In other words, I didn’t make mistakes when +computing these 130 matrices. +61 + +12 +Symmetrization: Preliminaries +In this part of the monograph we prove Lemma B and Lemma C1. In this +preliminary chapter we discuss a few useful lemmas. The reader can read +the proof of Lemma C1 in §17 directly after reading this chapter. +12.1 +Exponential Sums +We begin with two easy and well-known lemmas about exponential sums. +The first is an exercise with Lagrange multipliers. +Lemma 12.1 (Convexity) Suppose that α, β, γ ≥ 0 have the property that +α + β ≥ 2γ. Then αs + βs ≥ 2γs for all s > 1, with equality iff α = β = γ. +Lemma 12.2 (Descartes) Let 0 < r1 ≤ r1... ≤ rn < 1 be a sequence of +positive numbers. Let c1, ..., cn be a sequence of nonzero numbers. Define +E(s) = +n +� +i=1 +ci rs +i . +(110) +Let K denote the number of sign changes in the sequence c1, ..., cn. Then E +changes sign at most K times on R. +Proof: Suppose we have a counterexample. By continuity, perturbation, +and taking mth roots, it suffices to consider a counterexample of the form +� citei where t = rs and r ∈ (0, 1) and e1 > ... > en ∈ N. As s ranges in r, +the variable t ranges in (0, ∞). But P(t) changes sign at most K times on +(0, ∞) by Descartes’ Rule of Signs. This gives us a contradiction. ♠ +12.2 +Positive Dominance +See [S2] and [S3] for more details about the material here. Let G ∈ R[x1, ..., xn] +be a multivariable polynomial: +G = +� +I +cIXI, +XI = +n +� +i=1 +xIi +i . +(111) +62 + +Given two multi-indices I and J, we write I ⪯ J if Ii ≤ Ji for all i. Define +GJ = +� +I⪯J +cI, +G∞ = +� +I +cI. +(112) +We call G weak positive dominant (WPD) if GJ ≥ 0 for all J and G∞ > 0. +We call G positive dominant if GJ > 0 for all J. +Lemma 12.3 (Weak Positive Dominance) If G is weak positive domi- +nant then G > 0 on (0, 1]n. If G is positive dominant then G > 0 on [0, 1]n. +Proof: We prove the first statement. The second one has almost the same +proof. Suppose n = 1. Let P(x) = a0 + a1x + .... Let Ai = a0 + ... + ai. The +proof goes by induction on the degree of P. The case deg(P) = 0 is obvious. +Let x ∈ (0, 1]. We have +P(x) = a0 + a1x + x2x2 + · · · + anxn ≥ +x(A1 + a2x + a3x2 + · · · anxn−1) = xQ(x) > 0 +Here Q(x) is WPD and has degree n − 1. +Now we consider the general case. We write +P = f0 + f1xk + ... + fmxm +k , +fj ∈ R[x1, ..., xn−1]. +(113) +Since P is WBP so are the functions Pj = f0 + ... + fj. By induction on the +number of variables, Pj > 0 on (0, 1]n−1. But then, when we arbitrarily set +the first n − 1 variables to values in (0, 1), the resulting polynomial in xn is +WPD. By the n = 1 case, this polynomial is positive for all xn ∈ (0, 1]. ♠ +Polynomial Subdivision: Let P ∈ R[x1, ..., xn] as above. For any xj and +k ∈ {0, 1} we define +Sxj,k(P)(x1, ..., xn) = P(x1, ..., xj−1, x∗ +j, xj+1, ..., xn), +x∗ +j = k +2 + xj +2 . +(114) +If Sxj,k(P) > 0 on (0, 1]n for k = 0, 1 then we also have P > 0 on (0, 1]n. +Positive Numerator Selection: If f = f1/f2 is a bounded rational func- +tion on [0, 1]n, written in so that f1, f2 have no common factors, we always +choose f2 so that f2(1, ..., 1) > 0. If we then show, one way or another, that +f1 > 0 on (0, 1]n we can conclude that f2 > 0 on (0, 1]n as well. The point +is that f2 cannot change sign because then f blows up. But then we can +conclude that f > 0 on (0, 1]n. We write num+(f) = f1. +63 + +13 +Symmetrization: Proof of Lemma B +13.1 +The Domains +Now we define the domains involved in our proof. Recall that the domain Υ +is defined in §3.1 and shown in Figure 3.1. In this section we describe the +domains that play a role in the proof of Lemma B. The various transforma- +tions we make start in Υ but then move us into slightly different but related +domains. +Let Υ′ denote the domain of planar configurations p′ +0, p′ +1, p′ +2, p′ +3 such that +1. ∥p′ +0∥ ≥ ∥p′ +k∥ for k = 1, 2, 3. +2. 512p′ +0 ∈ [432, 498] × [0, 16]. (Compare [433, 498] × [0, 0].) +3. 512p′ +1 ∈ [−16, 32] × [−465, −348]. (Compare [−16, 16] × [−464, −349].) +4. 512p′ +2 ∈ [−498, −400] × [0, 16]. (Compare [−498, −400] × [0, 24].) +5. 512p′ +3 ∈ [−32, 16] × [348, 465]. (Compare [−16, 16] × [349, 464].) +6. p′ +02 = p′ +22. (Compare p02 = 0.) +The comparison conditions are what we had for Υ. +Up to a very small +perturbation Υ and Υ′ are the same domain. Condition 6 says that p′ +0 and +p′ +2 are on the same horizontal line. +Let Υ′′ denote the domain of configurations p′′ +0, p′′ +1, p′′ +2, p′′ +3 such that +1. 512p′′ +01 ∈ [416, 498] +2. 512p′′ +02 ∈ [0, 16]. +3. 512p′′ +12 ∈ [−465, −348]. +4. 512p′′ +32 ∈ [348, 465]. +5. (p′′ +21, p′′ +22) = (p′′ +01, −p′′ +21). +6. p′′ +11 = p′′ +31 = 0. +Conditions 5,6 say that the configuration is invariant under reflection in the +y-axis. +64 + +13.2 +Reduction to Smaller Steps +Lemma B concerns a map from Υ into the subset K4 of configurations having +4-fold symmetry. Here we describe this map as a 3 step process. We start +with the configuration X having points (p1, p2, p3, p4) ∈ Υ. +Balanced Rotation: We let (p′ +1, p′ +2, p′ +3, p′ +4) be the planar configuration which +is obtained by rotating X about the origin so that p′ +0 and p′ +2 lie on the same +horizontal line, with p′ +0 lying on the right. We call this operation balanced +rotation. Balanced rotation does not change the energy of the configuration. +Horizontal Symmetrization: Given a configuration X′ = (p′ +0, p′ +1, p′ +2, p′ +3), +there is a unique configuration X′′ = (p′′ +0, p′′ +1, p′′ +2, p′′ +3), invariant under under +reflection in the y-axis, such that p′ +j and p′′ +j lie on the same horizontal line for +j = 0, 1, 2, 3 and ∥p′′ +0 − p′′ +2∥ = ∥p′ +0 − p′ +2∥. We call this horizontal symmetriza- +tion. +Vertical Symmetrization: Let Υ′′ denote the union of all configurations +which are obtained from configurations in Υ′ by horizontal symmetrization. +Given a configuration X′′ = (p′′ +0, p′′ +1, p′′ +2, p′′ +3) ∈ Υ′′ there is a unique configura- +tion X′′′ = (p′′′ +0 , p′′′ +1 , p′′′ +2 , p′′′ +3 ) ∈ K4 such that p′′ +j and p′′′ +j lie on the same vertical +line for j = 0, 1, 2, 3. The configuration X′′′ coincides with the configuration +X∗ defined in Lemma B. We call this operation vertical symmetrization. +Lemma 13.1 (B1) Let P ∈ Υ and let P ′ be the balanced rotation of P ′. +Then P ′ ∈ Υ′. +Lemma 13.2 (B2) Let P ′ ∈ Υ′ and let P ′′ be the horizontal symmetrization +of P ′. Then P ′′ ∈ Υ′′ and Es(P ′′) ≤ Es(P ′) for all X′ ∈ Υ′ and all parameters +s ≥ 2. We have equality if and only if P ′ = P ′′. +Lemma 13.3 (B3) Let P ′′ ∈ Υ′ and let P ′′′ be the vertical symmetrization +of P ′′. Then P ′′′ ∈ Ψ4 and Es(P ′′′) ≤ Es(P ′′) for all s ∈ [12, 15++]. We have +equality if and only if P ′′ = P ′′′. +Lemma B follows immediately from Lemma B1 (§13.3), Lemma B2 (§13.4) +and Lemms B3 (§13.5) +Remark: Our proof of Lemma B2 is robust while our proof of Lemma B3 +is delicate. That accounts for the differing range of exponents. +65 + +13.3 +Proof of Lemma B1 +Let P ∈ Υ and let P ′ be the balanced rotation of P. Rotation about the +origin does not change the norms, so P ′ satisfies Condition 1. Moreover, +Condition 6 holds by construction. Now we verify the other properties. Let +ρθ denote the counterclockwise rotation through the angle θ. +Since p0 lies on the x axis and p2 lies on or above it, we have to rotate +by a small amount counterclockwise to get p′ +0 and p′ +2 on the same horizontal +line. That is, the rotation moves the right point up and the left one down. +Hence θ ≥ 0. This angle is maximized when p0 is an endpoint of its segment +of constraint and p2 is one of the two upper vertices of rectangle of constaint. +Not thinking too hard which of the 4 possibilities actually realizes the max, +we check for all 4 pairs (p0, p2) that the second coordinate of ρ1/34(p0) is +larger than the second coordinate of ρ1/34(p0). From this we conclude that +θ < 1/34. This yields +512 cos(θ) ∈ [0, 1], +512 sin(θ) ∈ [0, 16]. +(115) +From Equation 115, the map 512p0 → 512p′ +0 changes the first coordinate +by 512δ01 ∈ [0, 16] and 512δ02 ∈ [−1, 0]. This gives Condition 2 for Υ′. At +the same time, we have p′ +21 = p′ +01 and the change 512p2 → 512p′ +2 changes the +second coordinate by 512δ21 ∈ [0, 1]. This gives Condition 4 for Υ′ once we +observe that |p′ +21| ≤ |p′ +01|. +For Condition 3 we just have to check (using the same notation) that +512δ11 ∈ [0, 16] and 512δ12 ∈ [−1, 1]. The first bound comes from the inequal- +ity 512 sin(θ) < 16. For the second bound we note that the angle that p1 +makes with the y-axis is maximized when p1 is at the corners of its constraints +in Υ. That is, +p1 = +�±16 +512 , 349 +512 +� +. +Since tan(1/21) > 16/349 we conclude that this angle is at most 1/21. Hence +|512δ12| ≤ max +|x|≤1/21 +��� cos +� +x + 1 +34 +� +− cos(x) +��� < 1. +This gives Condition 3. The same argument gives Condition 5. +66 + +13.4 +Proof of Lemma B2 +Each planar configuration corresponds to a 5-point configuration on S2 via +stereographic projection. The energy of the 5 point configuration involves 10 +pairs of points. A typical term is: +Rs(pi, pj) = +1 +∥Σ−1(pi) − Σ−1(pj)∥s. +(116) +Given a list L of pairs of points in the set {0, 1, 2, 3, 4} we define Es(P, L) to +be the sum of the Rs-potentials just over the pairs in L. Thus, for instance +L = {(0, 2), (0, 4), (2, 4)} =⇒ Es(P, L) = Rs(p0, p2)+Rs(p0, p4)+Rs(p2, p4). +We call the subset L good for the parameter s if we have +Es(P ′′, L) ≤ Es(P ′, L). +(117) +Here P ′ ∈ Υ′ and P ′′ is obtained from P ′ by horizontal symmetrization. We +call L great if L is good and if we get equality in Equation 117 if and only +if the points involved on both sides of the equation coincide. For example, +if {(0, 2), (2, 4)} is great it means that we get equality if and only if p′′ +j = p′ +j +for j = 0, 2. +Lemma 13.4 (B21) {(0, 1), (1, 2)} and {(0, 3), (2, 3)} are good for s ≥ 2. +Lemma 13.5 (B22) {(0, 2)} and {(0, 4), (2, 4)} are great for all s ≥ 2. +Lemma 13.6 (B23) {(1, 3), (1, 4), (3, 4)} is great for all s ≥ 2. +Lemma B2 is an immediate consequence of Lemma B21 (§14), Lemma +B22 (§15), and Lemma B23 (§15.2). +Remark: We could probably derive greatness rather than goodness in Lemma +B21, but we don’t need it and we save some trouble by not worrying about +when precisely we get equality. +67 + +13.5 +Proof of Lemma B3 +We define the notation of goodness with respect to vertical symmetrization +just as we did for horizontal symmetrization. This time we are working with +points in the domain Υ′′ described in §13.1. For convenience we set qk = p′′ +k +and q′ +k = p′′′ +k . Let Q be the configuration q0, q1, q2, q3 and let Q′ be the image +under vertical symmetrization. +Lemma 13.7 (B31) The lists {(0, 4)} and {(2, 4)} are great for s ≥ 2. +Proof: Let Σ−1 denote inverse stereographic projection. Let us assume that +q0 ̸= q′ +0. Vertical symmetrization moves the point q0 = (q01, q21) to the point +q′ +0 = (q01, 0) which is closer to the origin. Therefore Σ−1(q′ +0) is farther from +(0, 0, 1) than is Σ−1(q0). This shows the result for {(0, 4)}. The result for +the list {(2, 4)} now follows from symmetry, namely reflection in the y-axis. ♠ +Lemma 13.8 (B32) {(0, 2)} is great for all s ≥ 2. +Proof: Since this inequality only involves 1 term, it suffices to prove the +result for s = 2. Setting x = q01 = −q21 and y = q02 = q22 we +L = {(0, 2)} +−→ +E2(L, Q) = +�1 + x2 + y2 +4x +�2 +(118) +Vertical symmetrization sets y = 0 and keeps x the same. Therefore, we have +that E(L, Q′) ≤ E2(L, Q) with equality if and only if y = 0. ♠ +Lemma 13.9 (B33) {(1, 3)} is great for all s ≥ 2. +Proof: Since our inequality only involves 1 term, it suffices to prove the +result for s = 2. We keep the notation from the previous lemma, except that +now we set y and h so that q1 = (0, −y + h) and q3 = (0, y + h). All we +need in our proof is that y ∈ (0, 1), which we have for configurations in Υ′′. +Vertical symmetrization sets h = 0 and keeps y the same. +We compute that +L = {(1, 3)} +−→ +E2(L, Q) = (1 + (y − h)2)(1 + (y + h)2) +(4y)2 +. +(119) +68 + +Call this function f(h). We suppress the dependence on y in the notation. +Given that our configuration Q lies Υ′′ we have y ∈ (0, 1). +Setting f ′(h) = 0 and solving for h, we find that the local extrema for +this function occur at h = 0 and at h = ± +� +y2 − 1. Since y2 < 1 the latter +roots are not real. Hence f has a unique local extremum, and this occurs at +h = 0. We compute +f ′′(0) = (1 − y2)/4y2 > 0. +We conclude that h = 0 is the unique global minimizer for f. ♠ +Lemma 13.10 (B34) {(1, 4), (3, 4)} is great for all s ≥ 2. +Proof: Consider first the case s = 2. Keeping the notation from the previous +section, we compute that +L = {(1, 4), (3, 4)} +E2(L, Q) = 1 + y2 + h2 +2 +. +(120) +Now we consider the case s > 2. Set +α = ∥Σ−1(q1), (0, 0, 1)∥−2 +β = ∥Σ−1(q3), (0, 0, 1)∥−2 +γ = ∥Σ−1(q′ +1), (0, 0, 1)∥−2 = ∥Σ−1(q′ +3), (0, 0, 1)∥−2. +We have just proved that α + β ≥ 2γ. The Convexity Lemma now says that +αt + βt ≥ 2γt with equality if and only if α = β = γ. ♠ +Lemma 13.11 (B35) {(0, 1), (0, 3)} is good for all s ≥ 12. +It follows from Lemma B35 and symmetry – namely reflection in the y- +axis – that {(2, 1), (2, 3)} is also good for all s ≥ 12. Finally, we observe that +vertical symmetrization maps Υ′′ into Ψ4 because we get a configurations +invariant under reflections in the coordinate axes which satisfy +p01 ∈ +�416 +512, 498 +512 +� +⊂ +�43 +64, 1 +� +, +p32 ∈ +�348 +512, 465 +512 +� +⊂ +�43 +64, 1 +� +. +Thus Lemma B3 follows from Lemmas B31, B32, B33, B34, and B35 (§16). +69 + +14 +Symmetrization: Proof of Lemma B21 +14.1 +Reduction to a Simpler Statement +Let D denote the set of triples of points (q0, q1, q2) ∈ (R2)3 such that +1. |q21| ≤ q01 and q22 = q02. +2. 512q0 ∈ [432, 498] × [−16, 16]. +3. 512q1 ∈ [−32, 32] × [348, 465]. +4. 512q2 ∈ [−498, −400] × [−16, 16]. +The domain D is a symmetrized version of Υ′ with the following properties. +If p0, p1, p2, p3 ∈ Υ′ then +• (q0, q1, q2) = (p0, p3, p2) ∈ D. +• (q0, q1, q2) = (r(p0), r(p1), r(p2)) ∈ D. +Here r is reflection in the x-axis. Thus, rather than consider the two lists +{(0, 1), (1, 2)} and {(0, 3), (3, 2)} separately, we just consider the symmetriza- +tion operation once, as it is applied to configurations in D. +The symmetrization operation is given by (q0, q1, q2) → (q′ +1, q′ +2, q′ +3), where +q′ +0 = +�q01 − q21 +2 +, q02 +� +, +q′ +1 = (0, q21), +q′ +2 = +�q21 − q01 +2 +, q22 +� +, +(121) +Define +λk = ∥Σ−1(qk) − Σ−1(qk+1)∥−2, +(122) +for k = 0, 1 and likewise define λ′ +k with respect to q′ +0, q′ +1, q′ +2. Note that λ′ +0 = λ′ +1 +by symmetry. To prove Lemma B21 it suffices to show +Lemma 14.1 (B11) With respect to all triples in D we have λ0 +λ1 ≥ 2λ′ +0. +The Convexity Lemma then shows that λs +0 + λs +1 ≥ (2λ′ +0)s for all s > 1, with +equality if and only if λ0 = λ1. This is exactly the statement that the lists +{(0, 1), (1, 2)} and {(0, 3), (3, 2)} are great for all s > 2. +70 + +14.2 +Proof of Lemma B211 +We define +[u, v]t = u(1 − t) + vt. +(123) +The map t → [u, v]t maps [0, 1] to [u, v]. +For all 4 choices of signs we define a map φ±,± : [0, 1]5 → (R2)3 by the +following formula +φ±,±(a, b, c, d, e) = q0(a, d, ±b), q1(±e, c), q2(a, d, ±b), +(124) +where +512q0(a, d, ±b) = ([416, 498]a + 49e, ±16b). +512q1(±d, c) = (±32d, [348 + 465]c) +512q2(a, d, ±b) = ([−416, −498]a + 49e, ±16b). +Lemma 14.2 (B2111) We have +D ⊂ φ+,+([0, 1]5) ∪ φ+,−([0, 1]5) ∪ φ−,+([0, 1]5) ∪ φ−.−([0, 1]5). +Proof: Let Dij denote the set of possible coordinates qij that can arise for +points in D. This, for instance D01 = [−16, 16]/512. Let D∗ +ij denote the set of +possible coordinates qij that can arise from the union of our parametrizations. +By construction Di2 ⊂ D∗ +i2 for i = 0, 1, 2 and D11 ⊂ D∗ +11. +Remembering that we have q01 ≥ |q21|, we see that the set of points pairs +(q01, q21) satisfying all the conditions for inclusion in D lies in the triangle X +with vertices +(498, −498), +(498, −400), +(432, −400) +At the same time, the set of pairs (512)(p∗ +01, p∗ +21) that we can reach with our +parametrization is the rectangle X∗ with vertices +(498, −498), +(416, −416), +(498, −498)+(49, 49), +(416, −416)+(49, 49). +We have X ⊂ X∗ because +(432, −400) = (416, −416)+(16, 16), +(498, −400) = (449, −449)+(49, 49). +This completes the proof. ♠ +71 + +In our coordinates, horizontal symmetrization corresponds to the map +(a, b, c, d, e) → (a, b, c, 0, 0). +(125) +We define F±,±(a, b, c, d, e) = λ0 + λ1, where these quantities are defined +relative to the triple φ±,±(a, bb, c, d, e). We define +Φ±,±(a, b, c, d, e) = num+(F±,±(a, b, c, d, e) − F±,±(a, b, c, 0, 0)). +(126) +Lemma 14.3 (B2112) For any sign choice Φ±,± > 0 on (0, 1)5. +As we discussed in §12.2, Lemma B2112 implies that +F±(a, b, c, d, e) ≥ F±(a, b, c, 0, 0) +. Lemma B211 thus follows from Lemma B2111 and Lemma B2112. +14.3 +Proof of Lemma B2112 +Our argument works the same for any of the 4 sign choices, so we set +Φ = Φ±,±, with the understanding that each statement about Φ is really +a statement about each of the 4 cases. Let Φa = ∂Φ/∂a, etc. +Lemma 14.4 (B21121) The following is true: +• F and Φd and Φe are zero when d = e = 0. +• Φd + 2Φe and Φdd and Φee are non-negative on [0, 1]5. +Proof: +The file LemmaB21121.m does these calculations. +For the second +item, we check that the 3 functions are weak positive dominant (§12.2). ♠ +Let Qd ⊂ [0, 1]5 be the sub-cube where d = 0. Let φ(d) be the restriction +of Φ to a line segment which starts at some point (a, b, c, 0, 0) and moves +parallel to (0, 0, 0, 1, 0). By Lemma B21121 we have φ(0) = φ′(0) and also +φ′′(d) ≥ 0. Hence φ(d) ≥ 0 for d ≥ 0. Hence Φ ≥ 0 on Qd. A similar +argument shows that likewise Φ ≥ 0 on Qe. +Any point in (0, 1)5 can be joined to a point in Qd ∪Qe by a line segment +L which is parallel to the vector (0, 0, 0, 1, 2). By Lemma B21121, Φ increases +along such a line segment as we move out of Qd ∪Qe. Hence Φ ≥ 0 on [0, 1]5. +72 + +15 +Symmetrization: Lemma B22 and B23 +15.1 +Proof of Lemma B22 +The relevant points for Lemma B22 are +p′ +0 = (x + d, y), +p′ +2 = (−x + d, y), +p′′ +0 = (x, y), +p′′ +2 = (−x, y), +(127) +All we will use in our proof is that x2+y2 < 1, which we have for points in the +domain Υ′. Let Σ−1 denote inverse stereographic projection. See Equation +8. +Lemma 15.1 (B221) The list L = {(0, 2)} is great for all s ≥ 2. +Proof: Write λ′ = ∥Σ−1(p′ +0) − Σ−1(p′ +2)∥−2 and likewise define λ′′. We have +λ′ − λ′′ = d2 +x2 × (2 + d2 − 2x2 + 2y2). +Since x2+y2 < 1 we have 2−2x2+2y2 > 0. Hence 0 < λ′′ ≤ λ′, with equality +if and only if d = 0. But then for t = s/2 > 1 we also have (λ′′)t < (λ′)t, +with equality iff d = 0. ♠ +Lemma 15.2 (B222) The list L = {(0, 4), (2, 4)} is great for s ≥ 2. +Proof: By the Convexity Lemma, it suffices to prove this for s = 2. We +have +Σ−1(p′ +4) = Σ−1(p′′ +4) = (0, 0, 1). +(128) +This time we have a beautiful formula: +� +∥Σ−1(p′ +0) − (0, 0, 1)∥−2 + ∥Σ−1(p′ +2) − (0, 0, 1)∥−2� +− +� +∥Σ−1(p′′ +0) − (0, 0, 1)∥−2 + ∥Σ−1(p′′ +2) − (0, 0, 1)∥−2� += d2 +2 , +(129) +This holds identically for all choices of x, y, d. ♠ +Lemma B22 follows immediately from Lemma B221, B222, B223. +73 + +15.2 +Proof of Lemma B23 +In this chapter we prove Lemma B23, The relevant points are q1 = (x1, y1) +and q3 = (x3, y3). Let �q = Σ−1(q) be the inverse stereographic image of q. +Horizontal symmetrization sets x1 = x3 = 0. Define +Fs(x1, x3, y1, y3) = As(x1, y1) + As(x3, y3) + Bs(x1, x3, y1, y3), +As(x, y) = ∥ � +(x, y) − (0, 0, 1)∥−s, +Bs(x1, x3, y1, y3) = ∥�q1 − �q3∥−s. +Lemma B23 says that +Fs(x1, x3, y1, y3) ≥ Fs(0, 0, y1, y3) +(130) +for all relevant choices of variables, with equality iff x1 = x3 = 0. All we need +for the proof (and we have it) is that s ≥ 2 and y1 < − +√ +3/3 and y3 > +√ +3/3. +The file LemmaB23.m has our calculations for this lemma. +Lemma 15.3 (B231) If Equation 130 is true when x1x3 ≤ 0 it is also true +when x1x3 > 0. +Proof: When x1, x3 > 0 the points �q1 and �q3 lie in the same hemisphere on +S2, namely the inverse stereographic image of the right halfplane. Replacing +(x1, y1) with (−x1, y1) increases the B-term and fixed the A-terms. Hence +Fs(x1, x3, y1, y3) > Fs(x1, −x3, y1, y3). The same goes when x1, x3 < 0. ♠ +Lemma 15.4 (B232) Equation 130 holds if x1x3 = 0. We have equality if +and only if x1 = x3 = 0. +Proof: +Without loss of generality we assume x1 = 0. +We analyze the +two terms of Fs separately and show that both decrease. +Just as in the +proof of Lemma B221 it suffices to take s = 2 for each of these terms. Set +y1 = − +√ +3/3 − t1 and y3 = +√ +3/3 + t3. We compute +A2(x3, y3) − A2(0, y3) = x2 +3 +4 > 0. +B2(0, x3, y1, y3) − B2(0, 0, y1, y3) = +3x2 +3(4 + 2 +√ +3t1 + 3t2 +1)(4 +√ +3t1 + 3t2 +1 + 2 +√ +3t3 + 6t1t3) +(4(2 +√ +3 + 3t1 + 3t3)2(4 + 3x2 +3 + 4 +√ +3t1 + 3t2 +1 + 4 +√ +3t3 + 6t1t3 + 3t2 +3) > 0. +When x3 ̸= 0 this has all positive terms because t1, t2 > 0. ♠ +74 + +Lemma 15.5 (B233) If Equation 130 has a counterexample with x1x3 < 0 +it also has a counterexample with x1x3 = 0. +Proof: +Without loss of generality we can assume x1 > 0 and x3 < 0. +The point q1 lies in the (+, −) quadrant and the point q3 lies in the (−, +) +quadrant. +Let ρ be the clockwise rotation about the origin, through the +smallest positive angle, such that one of the points q∗ +1 = ρ(q1) or q∗ +3 = ρ(q3) +lies on the y-axis. As our points move, their vertical distance from the origin +increases. Setting q∗ = (x∗, y∗), we have +y∗ +1 < y1 < − +√ +3/3, +y∗ +3 > y3 > +√ +3/3, +x∗ +1x∗ +3 = 0. +(131) +If (x1, x3, y1, y3) gives a counterexample to Equation 130 we have +Fs(0, 0, y1, y3) > Fs(x1, x3, y1, y3) = Fs(x∗ +1, x∗ +3, y∗ +1, y∗ +3) ≥∗ Fs(0, 0, y∗ +1, y∗ +3). +The middle equality is rotational symmetry. The starred inequality is Lemma +B232. We will get a contradiction by showing Fs(0, 0, y∗ +1, y∗ +3) > Fs(0, 0, y1, y3). +It suffices to prove this result when y1 = y∗ +1 because then we can reverse +the roles of the two points and apply the result twice to get the general case. +The key point in the proof (aside from calculation) is that Σ−1(0, y3) is closer +to (0, 0, 1) than it is to Σ−1(0, y1), and Σ−1(0, y∗ +3) is even closer to (0, 0, 1). +We set y1 = − +√ +3/3 − t1 and y3 = +√ +3/3 + t3. Here t1, t3 > 0. We com- +pute that ∂F2(0, 0, y1, y3)/∂t3 is a rational function of t1, t3 with all positive +coefficients and ∂F−2(0, 0, y1, y3)/∂t3 is a rational function of t1, t3 with all +negative coefficients. From this we conclude that the exponential sum +E(s) = Fs(0, 0, y1, y∗ +3) − Fs(0, 0, y1, y3) +satisfies E(2) > 0 and E(−2) < 0. We also have E(0) = 0. +The motion of the point (0, y3) → (0, y∗ +3) increases one of the terms in- +volved in Fs and decreases the other. From this we see that E(s) has at +most 2 sign changes in the sense of §12.1. So, by Descartes’ Lemma, E(s) +changes sign at most twice. +The first and last term in E(s) are positive +because the motion (0, y3) → (0, y∗ +3) decreases the shortest distance involved +and increases the longest. Hence E(s) > 0 when |s| is sufficiently large. +We now see that E(s) changes sign on (−∞, −2). If E′(0) ̸= 0 then E(s) +also changes sign at s = 0. But then E(s) does not change sign on (0, ∞) +and hence is positive there. If E′(0) = 0 and E(s) changes sign in (0, ∞) we +can perturb the points slightly, reduce to the case when E′(0) ̸= 0, and get +a contradictory situation with 3 sign changes. ♠ +75 + +16 +Symmetrization: Proof of Lemma B35 +For ease of notation set qk = p′′ +k. Let D be the set of configurations (q0, q1, q3) +such that +1. 512q01 ∈ [416, 498] +2. 512q02 ∈ [0, 16]. +3. 512q12 ∈ [−465, −348]. +4. 512q32 ∈ [348, 465]. +5. q11 = q31 = 0. +Lemma B35 does not involve the point p2, so we ignore it. +The subset +D ⊂ (R2)3 denotes the set of triples (q0, q1, q3) which satisfy the conditions +for inclusion in Υ′′. This set is not meant to be confused with the set from +the proof of Lemma B21, though it plays the same role in the proof here. We +let D± ⊂ D denote those configurations with +±(q12 + q32) ≥ 0. +(132) +Obviously D = D+ ∪ D−. +We adopt the convention in Equation 123, namely [u, v]t = u(1 − t) + vt. +We define map φ± : [0, 1]4 → (R2)3 as follows: +φ(a, b, c, d) = (q0(b, d), q1(a, c), q3(a, c)), +(133) +512q0(b, d) = ([416, 498]b, 16d). +512q1(a, c) = (0, −[348, 465]a ± 59c). +512q3(a, c) = (0, +[348, 465]a ± 59c). +In these coordinates, the symmetrization operation is (a, b, c, d) → (a, b, 0, 0). +Lemma 16.1 (B351) D± ⊂ φ±([0, 1]4). +Proof: This is just like the proof of Lemma B2111. The only non-obvious +point is why every pair (p12, p32) is reached by the map φ±. The essential +point is that for configurations in D± we have 512|p12 + p32| ≤ 2 × 59. ♠ +76 + +Following the same idea as in the proof of Lemma B21, we define +Fs,± = +� +∥Σ−1(q0)−Σ−1(q1)∥−s+∥Σ−1(q0)−Σ−1(q3)∥−s� +◦φ±(a, b, c, d) (134) +Here Σ−1 is the inverse of stereographic projection. Next, we define +Φs,±(a, b, c, d) = num+(Fs,±(a, b, c, d) − Fs,±(a, b, 0, 0)). +(135) +We can finish the proof by showing that φ2,+ ≥ 0 and φ12,− ≥ 0 on [0, 1]4. +The Convexity Lemma then takes care of all exponents greater than 2 on D+ +and all exponents greater than 12 on D−. +Lemma 16.2 (B352) Φ2,+ ≥ 0 on [0, 1]4. +Proof: Let Φ = Φ2,+. Let Φ|c=0 denote the polynomial we get by setting +c = 0. We define other such symbols similarly. Let Φc = ∂Φ/∂c, etc. The +Mathematica file LemmaB352.m computes that Φ|c=0 and Φ|d=0 and Φc + Φd +are weak positive dominant. Hence Φ ≥ 0 when c = 0 or d = 0 and the +directional derivative of Φ in the direction (0, 0, 1, 1) is non-negative. This +suffices to show that Φ ≥ 0 on [0, 1]4. ♠ +Lemma 16.3 (B353) Φ12,− ≥ 0 on [0, 1]4. +Proof: The file LemmaB353.m has the calculations for our argument. Let +Φ = Φ12,−. This polynomial is a monster. It has 102218 terms. The first +thing we notice is that most of the terms are much smaller than the biggest +terms. So, we first kill off these terms carefully. +Let M denote the maximum coefficient of Φ. We let Φ∗ be the polynomial +we get by taking each coefficient of c of Φ and replacing it with +floor +�1010c +M +� +. +This has the effect of killing off about half the terms of Φ, namely the posi- +tive terms that are less than 10−10M. The “small” negative coefficients are +changed to −1. The polynomial 1010Φ − MΦ∗ has all non-negative coeffi- +cients. Hence, if Φ∗ ≥ 0 on [0, 1]4 so is MΦ ≥ 0 and so is 1010Φ and finally +so is Φ. +77 + +We want to kill more terms. The polynomial Φ∗ has 37760 monomials in +which the coefficient is −1. We check that each such monomial is divisible +by one of c2 or d2 or cd. We therefore define +Ψ = Φ∗∗ − 37760(c2 + d2 + cd), +where Φ∗∗ is obtained from Φ∗ by setting all the (−1) monomials to 0. We +have Ψ ≤ Φ∗ on [0, 1]4. Hence, if Ψ is non-negative on [0, 1]4 then so is Φ. We +have reduced the problem to showing that Ψ ≥ 0 on [0, 1]4. The polynomial +Ψ has 5743 terms, which is more manageable. +Again we let Fa = ∂F/∂a, etc. +We check that Ψaaa is weak positive +dominant and hence non-negative on [0, 1]4. This massive calculation reduces +us to showing that the restrictions Ψ|a=0 and Ψa|a=0 and Ψaa|a=0 are all non- +negative on [0, 1]3. Letting F be any of these 3 functions, we consider +F|c=0, +F|d=0 +4Fc + Fd, +(136) +We show that all three functions are weak positive dominant for Ψa|a=0 and +Ψaa|a=0. +This shows that Ψa|a=0 and Ψaa|a=0 are non-negative on [0, 1]3. +Concerning the choice F = Ψ|a=0, all that remains is showing (in some other +way) that G = 4Fc + Fd ≥ 0. +We check that Gd is weak positive dominant and hence non-negative on +[0, 1]3. This reduces us to showing that H = G|d=0 is non-negative on [0, 1]2. +Here H is a 2-variable polynomial in b, c. We check that the two subdivi- +sions Sb,0(H) and Sb,1(H) are weak positive dominant. This proves that H +is non-negative on [0, 1]2. ♠ +Remark: The proof of Lemma B35 is pretty crazy. Initially we gave an +alternate, which works for s ≥ 13, based on the following fact: Suppose that +x1, y1, x2, y2 are positive numbers with +x2 = y2, +7x1+8y1 ≥ 7x2+8y2, +3x2 +1+3y2 +1−4x1y1 ≥ 3x2 +2+3y2 +2−4x2y2. +Then xt +1 + yt +1 ≥ xt +2 + yt +2 for all t ≥ 13/2. We leave this as an exercise to +the reader. This leaves us to check that vertical symmetrization does not +increase 7E1 +8E3 and 3E2 +1 +3E2 +3 −4E1E3 on D1. The polynomials involved +are much smaller and similar positive dominance methods work for them. +We have preserved the Mathematica files which do the relevant calculations. +78 + +17 +Symmetrization: Proof of Lemma C1 +17.1 +Reduction to Two Halves +Recall that �Ψ4 and Ψ8 respectively are defined by +64�Ψ4 = [55, 56]2, +64Ψ8 = {(t, t)| t ∈ [43, 64]}. +(137) +We have the symmetrization operation σ : �Ψ4 → Ψ8 given by +σ(x, y) = (z, z), +z = x + y + (x − y)2 +2 +. +(138) +Given (x, y) ∈ �Ψ4 the corresponding planar avatar p0, p1, p2, p4 is given +by −p2 = p0 = (x, 0) and −p1 = p2 = (0, y). The quantity Es(x, y) denotes +the s-potential of the 5-point configuration on S2 corresponding to the above +planar configuration under Σ−1, the inverse stereographic projection. Lemma +C1 says that Es ◦ σ ≤ Es on �Ψ4 for all s ∈ [14, 16], with equality iff x = y. +Define �pj = Σ−1(pj). We have (by symmetry): +Es(x, y) = As(x, y) + Bs(x, y), +As(x, y) = ∥�p0, �p2∥−s + ∥�p1, �p3∥−s, +Bs(x, y) = 2∥�p0, (0, 0, 1)∥−s + 2∥�p1, (0, 0, 1)∥−s + 4∥�p0, �p1∥−s. +(139) +Lemma 17.1 (C11) For (x, y) ∈ �Ψ4 and s ≥ 2 we have As(x, y) ≥ As(z, z). +When s > 2 we get equality if and only if x = y. +Proof: Let φ : [0, 1]2 → �Ψ4 be the affine isomorphism whose linear part is a +positive diagonal matrix. Define F2 : [0, 1]2 → R +F2 = num+(A2 ◦ φ − A2 ◦ σ ◦ φ). +(140) +We check that F2(a, b) = (a − b)2H2, where H2 is weak positive dominant. +Hence H2 > 0 on (0, 1)2. Hence A2(x, y) ≥ A2(z, z) on �Ψ4. Now we apply +the Convexity Lemma from §12.1. ♠ +Lemma 17.2 (C12) For (x, y) ∈ �Ψ4 and s ∈ [14, 16] we have the inequality +Bs(x, y) ≥ Bs(z, z). +Lemma C1 follows immediately from Lemmas C11 and C12. +79 + +17.2 +Proof of Lemma C12 +Let ∆ ⊂ (0, 1)2 be the open triangular subset above the main diagonal. +For integers k = 2, 14, 16 define +Gk = num+(Bk ◦ φ − Bk ◦ σ ◦ φ). +(141) +Lemma 17.3 (C121) The following is true: +1. B2(x, y) ≤ B2(z, z) for all (x, y) ∈ �Ψ4. +2. B14(x, y) ≤ B14(z, z) for all (x, y) ∈ �Ψ4. +3. B16(x, y) ≤ B16(z, z) for all (x, y) ∈ �Ψ4. +We get strict inequalities for points in the interior of �Ψ4 − Ψ8. +Proof: An algebraic miracle happens. We compute that: +1. −G2(a, b) = (a − b)2H2(a, b) and H2 is weak positive dominant. +2. G14(a, b) = (a − b)2H14(a, b) and H14 is weak positive dominant. +3. G16(a, b) = (a − b)2H16(a, b) and H16 is weak positive dominant. +This does it. ♠ +Now suppose there is some (x, y) ∈ �Ψ4 − Ψ8 and some s0 ∈ (14, 16) such +that Bs0(x, y) < Bs0(z, z). +Perturbing, we can assume that (x, y) lies in +the interior of �Ψ4 − Ψ8. Let p0, p1, p2, p3 and p′ +0, p′ +1, p′ +2, p′ +3 respectively be the +configurations corresponding to (x, y) and (z, z). Define +1. r01 = ∥Σ−1(p0) − Σ−1(p1)∥−1. +2. r0 = ∥Σ−1(p0) − (0, 0, 1)∥−1 and r1 = ∥Σ−1(p1) − (0, 0, 1)∥−1. +3. r′ +01 = ∥Σ−1(p′ +0) − Σ−1(p′ +1)∥−1. +4. r′ +0 = ∥Σ−1(p′ +0) − (0, 0, 1)∥−1 = ∥Σ−1(p′ +1) − (0, 0, 1)∥−1. +Replacing (x, y) by (y, x) if necessary, we arrange that r0 < r1. +Lemma 17.4 (C122) r0, r1, r′ +0 < 1/ +√ +2 < r01, r′ +01 +80 + +Proof: Let h = 1/2. We have x, y, z ∈ (0, 1). We compute +h − r2 +0 = 1 − x2 +4 +, +h − r2 +1 = 1 − y2 +4 +, +h − (r′ +0)2 = 1 − z2 +4 +, +(r01)2 − h = (1 − x2)(1 − y2) +4(x2 + y2) +> 0, +(r′ +01)2 − h = (1 − z2)2 +8z2 +> 0. +This does it. ♠ +Lemma 17.5 (C123) r01 < r′ +01. +Proof: We define +B∗ +2 = ∥�p0 − �p1∥−2 = r2 +01 +and then define G∗ +2 in terms of B2 just as we defined G2 in terms of B2 in +Equation 141. We check that G∗ +2(a, b) = −(a − b)2H∗(a, b) where H∗ is weak +positive dominant. Hence G∗ +2 > 0 on (0, 1)2. Hence B∗ +2(z, z) > B∗ +2(x, y). But +this implies that r01 < r′ +01. ♠ +We now deduce Lemma C12 from Lemmas C121, C122, C123. We fix +(x, y) and (z, z) = σ(x, y) and define +β(s) := Bs − Bs ◦ σ = +2rs +0 − 4(r′ +0)s + 2rs +1 + 4rs +01 − 4(r′ +01)s +(142) +Now we make the following observations. +• β(2) < 0 and β(14) > 0. Hence β changes sign in (2, 14). +• β(14) > 0 and β(s0) < 0. Hence β changes sign in (14, s0). +• β(s0) < 0 and β(16) > 0. Hence β changes sign in (s0, 16). +• β(s) < 0 for s sufficiently large because the term −4(r′ +01)s eventually +dominates. Hence β changes sign in (16, ∞). +Hence β vanishes at least 4 times. By Descartes’ Lemma, the sign sequence +must change signs at least 4 times. Combining Lemmas C121 and C122 we +see that the sign sequence must be one of +−, +, +, +, −, ++, −, +, +, −, ++, +, −, +, −. +In no case does it change sign at least 4 time. This is a contradiction. (In +fact, Equation 142 has the terms in the correct order.) +81 + +18 +Endgame: Proof of Lemma C2 +18.1 +The Goal +The goal of this chapter is to prove Lemma C2. We will reformulate the +result because we want to set up a rational calculation. Define squares Ψ4 +and �Ψ4. Here are their definitions: +64Ψ4 = [43, 64], +64�Ψ4 = [55, 56]. +(143) +Also, �Ψ8 is the main diagonal of �Ψ4. A point (x, y) in these domains defines +a planar configutation +p0 = (x, 0), +p1 = (0, −y), +p2(−x, 0), +p3(0, y) +(144) +We define Es(x, y) to be the s-potential of the corresponding avatar. +Recall that the point (1, +√ +3/3) represents the TBP avatar. We define +Θ(s, x, y) = Es(x, y) − E(1, +√ +3/3). +(145) +Lemma C2 is equivalent to the following three statements. +1. Θ > 0 on [13, 15+] × Ψ4. +2. Θ > 0 on [15+, 15++ × (Ψ4 − �Ψ4). +3. If s ∈ [15+, 15++] then Θ has a unique minimum in �Ψ8. +Let us deduce Statement 3 from Statements 1 and 2 and an auxiliary +lemma. Let Θx be the partial derivative of Θ with respect to x, etc. +Lemma 18.1 (C21) For all s ∈ [13, 15++] and (x, y) ∈ Ψ4 the quantities +Θxx, Θyy, Θxy are all positive. +Statement 3 is equivalent to the statement that the single variable func- +tion f(x) = Θ(s, t, t) has only one minimum for s ∈ I. Here 64I = [55, 56]. +By the Chain Rule, +ftt = Θxx + Θyy + 2Θxy > 0. +(146) +Hence f is a convex function on I. Hence f has a unique minimum in I. +This proves Statement 3 of Lemma C2. +82 + +18.2 +Integer Calculations +The calculation for Lemma C2 is relatively small. We work over a 3-parameter +space, and every coordinate we consider will be a dyadic rational. +The Expanding Out Method: This is a repeat of the definition in §11.1. +Suppose we want to establish an inequality like ( a +b) +p +q < c +d, where every num- +ber involved is a positive integer. This inequality is true iff bpcq − apdq > 0. +We check this using exact integer arithmetic. The same idea works with (>) +in place of (<). We call this the expanding out method. +More generally, we will want to verify inequalities like +10 +� +i=1 +b−s +i +− +10 +� +i=1 +a−s/2 +i +> C. +(147) +where all ai belong to the set {2, 3, 4}, and bi, c, s are all rational. +more +specifically s ∈ [13, 15++] will be a dyadic rational and c will be positive. +The expression on the left will be Es(p) − Es(p0) for various choices of p, and +the constant C will be a kind of cushion introduced to get around an error +term. +Here is how we handle expressions like this. For each index i ∈ {1, ..., 10} +we produce rational numbers Ai and Bi such that +As/2 +i +> ai +Bs +i < bi. +(148) +We use the expanding out method to check these inequalities. We then check +that +10 +� +i=1 +Bi − +10 +� +i=1 +Ai > C. +(149) +This last calculation is again done with integer arithmetic. Equations 148 +and 149 together imply Equation 147. +Logically speaking, the way that we produce the rational Ai and Bi does +not matter, but let us explain how we find them in practice. +For Ai we +compute 232a−s/2 +i +and round the result up to the nearest integer Ni. We then +set Ai = Ni/232. We produce Bi in a similar way. +When we have verified Equation 147 in this manner we say that we have +used the rational approximation method to verify Equation 147. We will only +need to make verifications like this on the order of 20000 times. +83 + +18.3 +Main Argument for Lemma C2 +We say that a block is a rectangular solid, having the following form: +X = I × Q ⊂ [0, 16] × [0, 1]2, +(150) +where I is a dyadic interval and Q is a dyadic square. In this context we +mean that I is obtained by repeatedly cutting [0, 16] in half and selecting +one of the halves. Similarly, Q is obtained by repeatedly cutting [0, 1]2 in +quarters and selecting one of the pieces. +The Energy Estimate: +We call the dyadic block X relevant if X ⊂ +[13, 16] × Ψ4. +We work with the larger domain just to have nice initial +conditions for our divide-and-conquer algorithm. +We define +|X|1 = |I|, +|X|2 = |Q|. +(151) +Here |I| is the length of I and |Q| is the side length of Q. +Lemma 18.2 (C22) The following is true for any relevant block X and any +good point: +min +X Θ ≥ min +v(X) Θ − (|X|2 +1/512 + |X|2 +2). +Here v(X) denotes the vertex set of X. Thus, to show that Θ|X > 0 we just +need to show that +Θv(X) > |X|2 +1 +512 + |X|2 +2. +Grading a Block: Given a block X = I × Q ⊂ [0, 16] × [0, 1]2 we perform +the following pass/fail evaluation. Let I = [s0, s1]. +1. If I ⊂ [0, 13] we pass X because X is irrelevant. +2. If I ⊂ [15 + 25/512, 16] we pass X because X is irrelevant. +3. If Q is disjoint from Ψ4 we pass X because X is irrelevant. We test +this by checking that either Q10 ≤ 43/64 or Q01 ≤ 43/64. +4. If s0 ≥ 15 + 24/512 and Q ⊂ �Ψ4 we pass X. +5. s0 < 13 and s1 > 13 we fail X because we don’t want to make any +computations which involve exponents less than 13. +84 + +6. If X has not been passed or failed, we try to use the rational approxi- +mation method to verify that Θ(v) > |X|2 +1/512 − |X|2 +2 for each vertex +v of X. If we succeed at this, then we pass X. Otherwise we fail X. +If we pass X it either means that X is irrelevant or that Lemma C2 holds +for all configurations in X. To prove Lemma C2 it suffices to find a partition +of [0, 16] × [0, 1]2 into blocks, all of which pass the grading step. +Subdivision: We will either divide a block X into two pieces or 4, de- +pending on the dimensions. We call X fat if 16|X|2 > |X|1, and otherwise +thin. If X is fat we subdivide X into 4 equal equal pieces by dyadically +subdividing Q. If X is thin we subdivide X into 2 pieces by dyadically sub- +dividing I. We found by trial and error that this scheme takes advantage of +the lopsided form of Lemma C22 and produces a small partition. +The Main Algorithm: We perform the following algorithm. +1. We start with a list L of blocks. Initially L has the single member +{0, 16} × {0, 1}2. +2. We let B be the last block on L. We grade B. If B passes, we delete +B from L. If L = ∅ then HALT. If B fails, we delete B from L and +append to L the subdivision of B. Then we go back to Step 1. +Proposition 18.3 (C23) When we run the algorithm, it halts with success +after 23213 steps and in about 2 minutes. The partition it produces has 15519 +blocks. +This proves Statements 1 and 2 of Lemma C2. We have already seen +that Statements 1 and 2, and Lemma C1, together imply Statement 3. This +completes the proof of Lemma C2. +85 + +19 +Endgame: Lemmas C21 and C22 +19.1 +Proof of Lemma C21 +We will show that Θxx > 0 and Θxy > 0 on Γ := [13, 15++] × Ψ4. The case +of Θyy follows from the case of Θxx and symmetry. +Setting u = s/2 we compute +Es(x, y) = A(s, x) + A(s, y) + 2B(s, x) + 2B(s, y) + 4C(s, x, y), +(152) +A(x) = a(x)u, +B(x) = b(x)u, +C(x) = c(x)u, +a(x) = (1 + x2)2 +16x2 +b(x) = 1 + x2 +4 +c(x, y) = (1 + x2)(1 + y2) +4(x2 + y2) +From this we compute +Θxx = Axx + 2Bxx + 4Cxx, +Θxy = 4Cxy. +(153) +For each choice of F = A, B, C, and correspondingly f = a, b, c, we have +Fxx = u(u − 1)f u−2f 2 +x + uf u−1fxx +(154) +We compute +axx = 3 + x4 +8x4 , +bxx = 1 +2, +cxx = (1 − y4)(3x2 − y2) +2(x2 + y2)3 +, +These are all positive on [43/64, 1). Hence Fxx ≥ 0 and on Γ and we have +strict inequality unless F is the C function. Hence Θxx > 0 in Γ. +We also have +Cxy = u(u − 1)cu−2cxcy + ucu−1cxy. +(155) +We compute +cx = x(y4 − 1) +2(x2 + y2)2 < 0, +cxy = 2xy(1 + x2y2) +(x2 + y2)3 +. +Likewise cy < 0. Equation 155 now shows that Cxy > 0 in Γ Hence Θxy > 0 +in Γ. +86 + +19.2 +Proof of Lemma C22 +Lemma C22 follows immediately from these three results: +Lemma 19.1 (C221) Lemma C22 is true provided that +1. |Θss(s, x, y)| ≤ 1/64 for all (s, x, y) ∈ Γ. +2. |Θxx(s, x, y)| < 4 and |Θyy(s, x, y)| < 4 for all (s, x, y) ∈ Γ. +Here Θss is the second partial derivative of Θ with respect to s, etc. +Lemma 19.2 (C222) |Θss(s, x, y)| ≤ 1/64 for all (s, x, y) ∈ Γ. +Lemma 19.3 (C223) Θxx(s, x, y), Θyy(s, x, y) ∈ (0, 4) for all (s, x, y) ∈ Γ. +19.3 +Proof of Lemma C221 +We begin with a familiar lemma about functions of one variable. +Lemma 19.4 (C2211) Suppose F : [0, 1] → R is a smooth function. Then +min +[0,1] F ≥ min(F(0), F(1)) − 1 +8 max +[0,1] |F ′′(x)|. +Proof: This is an application of Taylor’s Theorem with Remainder. ♠ +Lemma 19.5 (C2212) Let X ⊂ Rn be a rectangular solid with vertex set +v(X). Let d1, ..., dn denote the side lengths of X. Let G : X → R be a +smooth function. Let Gii = ∂2X/∂x2 +i . Then +min +X G ≥ min +v(X) G − 1 +8 +n +� +i=1 +max +X +|Gii|d2 +i . +Proof: The truth of the result does not change if we translate the domain +and/or range, and/or scale the coordinates separately. Thus, it suffices to +prove the result when X = [0, 1]n. In this case, the result follows in a straight- +forward way from Lemma C2211, the triangle inequality, and induction on +the dimension. ♠ +We apply Lemma C2212 using the bounds +max +X +|G11| ≤ 1/64, +max +X +|G22| ≤ 4 +max +X +|G33| ≤ 4. +This gives the desired bound. +87 + +19.4 +Proof of Lemma C222 +Lemma 19.6 (C2221) The 10 distances associated to a 5-point configura- +tion parametrized by a point in Ψ4 exceed 1.3 and at least 6 of them exceed +√ +2, and at least 2 of them exceed +√ +3. +Proof: Let x0 = 43/64. An exercise in calculus shows that A(−1, x), which +represents the distance between 2 of the pairs of points, exceed +√ +3 on [x0, 1]. +Similarly, B(−1, x) exceeds +√ +2 on [x0, 1]. Finally, C[−1, x, y], restricted to +[d, 1]2, takes its min at (x0, x0) and the value there exceeds 1.3. ♠ +The conclusion of this result also obviously holds the TBP. +Lemma 19.7 (C2222) Let ψ(s, b) = b−s. Let s ≥ 13. +• When b ≥ 1.3 we have ψss(s, b) ∈ (0, 1/440). +• When b ≥ +√ +2 we have ψss(s, b) ∈ (0, 1/753). +• When b ≥ +√ +3 we have ψss(s, b) ∈ (0, 1/4184). +Proof: As a function of s, and for b > 1 fixed, the second derivative +ψss(s, b) = b−s log(b)2 +(156) +is decreasing. Given the monotonicity, it suffices to prove our result when +s = 13. Choose b ≥ 1.3. The equation ψssb(13, b) = 0 has its unique solution +in [1, ∞) at the value b = exp(2/13) < 1.3. Moreover, the function ψss(13, b) +tends to 0 as b → ∞. Hence the restriction of the function b → ψss(13, b) +to [b, ∞) takes its max at b. We get the specific bounds by evaluating at +b = 1.3, +√ +2, +√ +3. ♠ +Now, 10 of the 20 terms comprising Θss(s, x, y) are positive and 10 are +negative. Also, for the terms of the same sign, all 10 of them are less than +1/440, and at least 6 of them are less than 1/753, and at least 2 of them are +less than 1/4184. Hence +|Θss| ≤ +4 +440 + +4 +753 + +2 +4184 < 1 +64. +88 + +19.5 +Proof of Lemma C223 +We already know Θxx > 0 on our domain Γ. We will show Θxx < 4 on Γ. The +case of Γyy follows from symmetry. Equations 153 and 154 give us formulas +for Θxx. We just need one more estimate. +Lemma 19.8 (C2231) We have the following bounds for x ∈ Ψ4. +a ∈ [.2, .3] +ax ∈ [−.33, 0] +axx ∈ [.5, 2] +b ∈ [.3, .5] +bx ∈ [+.33, .5] +bxx ∈ [.5, .5] +c ∈ [.5, .6] +cx ∈ [−.33, 0] +cxx = [0, .5] +Proof: Let g be one of the 9 functions listed. An exercise in calculus shows +that the gradient ∇g does not vanish on the interior of Ψ4. The only mildly +tricky case is cxx. In this case, the resultant of (x4 +y4)cxxx and (x4 +y4)cxxy +is +−7962624x11(1 − 2x2 − 3x4)2(−1 − 2x2 + 3x4)2, +and this has no roots in (43/64, 1). Hence, for all cases, the restriction of g +to Ψ4 achieves its extrema at the vertices of Ψ4. Computing at the vertices +and rounding outward, we get the advertised bounds on g. ♠ +From Lemma C2232 and Equation 154 we see that Θxx ≥ 0 on Γ. Com- +bining Lemmas C2231 and C2232 we get the following bounds. +Axx ≤ (u)(u − 1)(.3)u−2(.11) + (u)(.3)u−1(2) < 0.1 +(157) +Bxx ≤ (u)(u − 1)(.5)u−2(.25) + (u)(.5)u−1(.5) < 0.5 +(158) +Cxx ≤ (u)(u − 1)(.6)u−2(.11) + (u)(.6)u−1(.5) < 0.6 +(159) +For the final inequalities we note that the expressions are maximized when +u is as small as possible, namely u = 13/2. These calculations plug into +Equations 153 and 154 to show that Θxx < 4. This completes the proof of +Lemma C223. +89 + +20 +Endgame: Proof of Lemma C3 +20.1 +Reduction to a Simpler Statement +Let I = [55/64, 56/64]. Recall that �Ψ8 is the set of points of the form (x, x) +with x ∈ I, and that 15+ = 15 + +24 +512 and 15++ = 15 + +25 +512. +The point +p0 = (1, +√ +3/3) represents the TBP. As in the proof of Lemma C2, define +Θ(s, x, y) = Es(x, y) − E(1, +√ +3/3). +(160) +We write Θs = ∂Θ/∂s, etc. We prove the following result below. +Lemma 20.1 (C31) Θstt(15, t, t) < 0 on I. +Lemma 20.2 (C32) Θs < 0 in [15+, 15++] × �Ψ8. +Proof: Lemma C212 tells us that |Θss| ≤ 2−6. We also have 15++−15 < 2−4. +Therefore, to prove this result, it suffices to prove that +Θs(15, t, t) < −2−10 +(161) +for t ∈ I. Let t0 = 55/64, the left endpoint of I. We compute Θst(15, t0, t0) < +0. Lemma C31 now implies that Θst(15, t, t) < 0 for all t ∈ I. Next, we com- +pute Θs(15, t0, t0) < −2−7. Since Θst(15, t, t) < 0 we get Equation 161. ♠ +We already know Θ(15+, ∗) > 0 on Ψ4. Hence Θ(15+, ∗) > 0 on �Ψ8. +We compute that Θ(15++, x, x) < 0 at x = 445/512 ∈ I. Combining this +with Lemma C32, we see that there exists a smallest parameterש such that +Θ(ש, p∗) = 0 for some p∗ ∈ �Ψ8. +For s >ש, Lemma C31 now says that +Θ(s, p∗) < 0. These statements immediately imply Lemma C3. +20.2 +Proof of Lemma C31 +The file LemmaC31.m has the calculations for this lemma. We have +Es(t, t) = 2A(s, t) + 4B(s, t) + 4C(s, t), +(162) +where +A(s, t) = +�(1 + t2)2 +16t2 +�s/2 +, +B(s, t) = +�1 + t2 +4 +�s/2 +, +C(s, t) = +�(1 + t2)2 +8t2 +�s/2 +. (163) +90 + +Because the s-energy of the TBP does not depend on the t-variable, we have +Θstt(15, t, t) = 2Astt|s=15 + 4Bstt|s=15 + 4Cstt|s=15. +(164) +Call the three functions on the right α(t), β(t), and γ(t). To finish the proof, +we just need to see that each of these three terms is negative in I. We note +that I has length 2−6. We write f ∼ f ∗ if +f +f ∗ = 2utv(1 + t2)w(2 + t2 + t−2)x +for exponents u, v, w, x ∈ R. In this case, f and f ∗ have the same sign. +Lemma 20.3 (C311) β < 0 on I. +Proof: Taking (u, v, w, x, y) = (−14, 0, 11/2, 0) we have β ∼ −β∗, +β∗(t) = (−2 + 30 log(2)) + t2(−58 + 420 log(2)) − 15(1 + 14t2) log(1 + t2). +Noting that log(2) = 0.69... we eyeball β∗ and see that it is positive for t ∈ I. +The term +420 log(2)t2 dominates. Hence β < 0 on I. ♠ +Lemma 20.4 (C312) γ < 0 on I. +Proof: Taking (u, v, w, x, y) = (−41/2, −16, 12, 1/2) we have γ ∼ −γ∗, +γ∗(t) = (−31 + 360 log(2)) + t2(56 − 585 log(2)) + t4(−29 + 315 log(2))+ +15(−8 + 13t2 − 7t4) log(2 + t2 + t−2). +We have γ∗(55/64) > 24 and we estimate easily that γ∗ +t > −210 on I. Only +the underlined term has negative derivative in I. These results show that +γ∗ > 0 on I. Hence γ < 0 on I. ♠ +Lemma 20.5 (C313) α < 0 on I. +Proof: Taking (u, v, w, x, y) = (−29, −14, 10, 3/2) we have α ∼ −α∗, +α∗(t) = γ∗(t) + δ∗(t), +δ∗(t) = 15 log 2 × (8 − 13t2 + 7t4). +We see easily that δ∗ > 0 on I. So, from Lemma C312, we have α∗ > 0 on +I. 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