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rsQsr=QrsQT
sr= 0. Consequently
RrsRsrRrs=QrsQsrQrs+QT
rsQT
srQT
rs
=d
d+1Qrs+d
d+1QT
rs
=1
d+1Rrs (386)
In view of Lemma 15it follows that RrsandRsrare uniformly inclined at angle
cos−1/parenleftbig1√d+1/parenrightbig
.
Further Identities. We conclude this section with another set of identities in-
volving the vectors /bardblfrs/an}bracketri}ht/an}bracketri}ht,/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht,/bardblgrs/an}bracketri}ht/an}bracketri}htand/bardbl¯grs/an}bracketri}ht/an}bracketri}ht.
Define
/bardbl¯er/an}bracketri}ht/an}bracketri}ht=/radicalbigg
2d
d−1/bardbler/an}bracketri}ht/an}bracketri}ht−/radicalbigg
d+1
d−1/bardblv0/an}bracketri}ht/an}bracketri}ht (387)
where/bardblv0/an}bracketri}ht/an}bracketri}htis the vector defined by Eq. ( 286). It is readily verified that
/an}bracketle{t/an}bracketle{t¯er/bardbl¯er/an}bracketri}ht/an}bracketri}ht= 0 and /an}bracketle{t/an}bracketle{t¯er/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (388)
So/bardbl¯er/an}bracketri}ht/an}bracketri}ht,/bardblv0/an}bracketri}ht/an}bracketri}htis an orthonormal basis for the 2-dimensional subspace spanned b y
/bardbler/an}bracketri}ht/an}bracketri}ht,/bardblv0/an}bracketri}ht/an}bracketri}ht. Note that
Qr/bardbl¯er/an}bracketri}ht/an}bracketri}ht=QT
r/bardbl¯er/an}bracketri}ht/an}bracketri}ht=¯Rr/bardbl¯er/an}bracketri}ht/an}bracketri}ht= 0 (389)
We then have
Theorem 20. For allr
1
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl=Qr (390)
1
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗
rs/bardbl=QT
r (391)
2
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl=¯Rr (392)
2
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardbl=¯Rr (393)
and
1
d−1d2/summationdisplay
s=1
(s/negationslash=r)/bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardbl=QT
r+/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+1
d2−1/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(394)47
1
d−1d2/summationdisplay
s=1
(s/negationslash=r)/bardblf∗
sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗
sr/bardbl=Qr+/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+1
d2−1/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(395)
2
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardblgsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgsr/bardbl=¯Rr (396)
2
d−3d2/summationdisplay
s=1
(s/negationslash=r)/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯gsr/bardbl=¯Rr+4(d−1)
d−3/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+4
(d+1)(d−3)/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(397)
Proof.It follows from the definition of /bardblfrs/an}bracketri}ht/an}bracketri}htthat
1
d+1d2/summationdisplay
s=1
(s/negationslash=r)/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl=d2/summationdisplay
s=1
(s/negationslash=r)Qr/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardblQr
=Qr
d2/summationdisplay
s=1/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardbl
Qr
=Qr (398)
where in the second step we used the fact that Qr/bardblr/an}bracketri}ht/an}bracketri}ht= 0 (as can be seen by setting
r=sin Eq. (121)). Eq. (391) is obtained by taking the complex conjugate on both
sides.
We also have
1
d+1d2/summationdisplay
s=1