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r=ξrΠ′
r+1−ξr
dI (235)
Ifξr= +1 (respectively −1) for allrthen Eq. ( 227) holds with Π r= Π′
randǫ= +1
(respectively −1). Also, if d= 2 thenL′
ris a rank-1 projector irrespective of the
value ofξr, so Eq. ( 227) holds with Π r=L′
randǫ= +1. The problem therefore
reduces to showing that if d >2 it cannot happen that ξr= +1 for some values
ofrand−1 for others. We will do this by assuming the contrary and deducing a
contradiction.
Letmbe the number of values of rfor whichξr= +1. We are assuming that
mis in the range 1 ≤m≤d2−1. We may also assume, without loss of generality,30
that the labelling is such that ξr= +1 for the first mvalues ofr, and−1 for the
rest. So
L′
r=/braceleftigg
Π′
r ifr≤m
2
dI−Π′
rifr>m(236)
Now define
˜Trst= Tr/parenleftbig
L′
rL′
sL′
t/parenrightbig
(237)
Eqs. (230) and (231) mean that the same argument which led to Eq. ( 9) can be
used to deduce
L′
rL′
s=d+1
d
d2/summationdisplay
t=1˜TrstL′
t
−K2
rsI (238)
SinceL′
1is a projector it follows that
L′
1L′
s=/parenleftbig
L′
1/parenrightbig2L′
s=d+1
d
d2/summationdisplay
t=1˜T1stL′
1L′
t
−K2
1sL′
1 (239)
By essentially the same argument which led to Eq. ( 118) we can use this to infer
/parenleftbig˜T′
1/parenrightbig2=d
d+1˜T1+2d2
(d+1)2/bardble1/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{te1/bardbl (240)
where˜T′
1is the matrix with matrix elements ˜T′
1rsand/bardble1/an}bracketri}ht/an}bracketri}htis the vector defined by
Eq.(116). Asbefore /bardble1/an}bracketri}ht/an}bracketri}htisaneigenvectorof ˜T′
1witheigenvalue2d
d+1. Consequently
the matrix
˜Q1=d+1
d˜T′
1−2/bardble1/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{te1/bardbl (241)
is a projector. But that means Tr( ˜Q1) must be an integer. We now use this to
derive a contradiction.
It follows from Eq. ( 236) that
(L′
r)2=/braceleftigg
L′
r r≤m
2(d−2)
d2I−d−4
dL′
rr>m(242)
Consequently
˜T1rr=/braceleftigg
K2
1r r≤m
2(d−2)
d2−d−4
dK2
1rr>m(243)
and so
Tr(˜Q1) =d+1
dd2/summationdisplay
r=1˜T1rr−2
=d+1−4d2+2m(d−2)
d3(244)
So if Tr( ˜Q1) is an integer/parenleftbig
4d2+2n(d−2)/parenrightbig
/d3must also be an integer. But the