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(196)
In view of Lemma 8we then have
/bardblI/an}bracketri}ht/an}bracketri}ht=1
κd2/summationdisplay
r=1lr/bardblLr/an}bracketri}ht/an}bracketri}ht (197)
Since
Tr(A) =d2/summationdisplay
r=1arlr=κ/an}bracketle{t/an}bracketle{tI/bardblA/an}bracketri}ht/an}bracketri}ht (198)
we have that A∈sl(d,C) if and only if /an}bracketle{t/an}bracketle{tI/bardblA/an}bracketri}ht/an}bracketri}ht= 0.
Now observe that it follows from Lemma 8and the definition of Mthat
M/bardblI/an}bracketri}ht/an}bracketri}ht=κ/bardblI/an}bracketri}ht/an}bracketri}ht (199)
IfMis a multiple of the identity we have Mrs=κδrsand the lemma is proved.
OtherwiseMhas at least one more eigenvalue, βsay. Let Ebe the corresponding
eigenspace. Since Eis orthogonal to /bardblI/an}bracketri}ht/an}bracketri}htit follows from Eq. ( 198) thatE⊆sl(d,C).
SinceMcommutes with every adjoint representation matrix we have
adAE⊆E (200)
for allA∈sl(d,C). SoEis an ideal of sl( d,C). However sl( d,C) is a simple Lie
algebra, meaning it has no proper ideals [ 52–55]. So we must have E= sl(d,C). It
follows that if we define
˜Lr=Lr−lr
dI (201)
then
M/bardblLr/an}bracketri}ht/an}bracketri}ht=lr
dM/bardblI/an}bracketri}ht/an}bracketri}ht+M/bardbl˜Lr/an}bracketri}ht/an}bracketri}ht (202)
=κlr
d/bardblI/an}bracketri}ht/an}bracketri}ht+β/bardbl˜Lr/an}bracketri}ht/an}bracketri}ht (203)
=d2/summationdisplay
s=1(βδrs+γlrls)/bardblLs/an}bracketri}ht/an}bracketri}ht (204)
whereγ=1
d/parenleftig
1−β
κ/parenrightig
. Eqs. (186), (187) and (188) are now immediate (in view of
Lemma8).27
It remains to establish the bounds on β,γ. LetA=/summationtextd2
r=1arLrbe any non-zero
element of sl( d,C). Then/summationtextd2
r=1arlr= 0, so in view of Eq. ( 186) we have
0<Tr(A2) =βd2/summationdisplay
r=1a2
r (205)
It follows that β >0. Also, using Lemma 8once more, we find
lr=1
κd2/summationdisplay
s=1lsTr(LrLs)
=βlr
κ+γlr
κd2/summationdisplay
s=1l2
s
=lr/parenleftbiggβ
κ+dγ/parenrightbigg
(206)
Since thelrcannot all be zero this implies
β
κ= 1−dγ (207)
Sinceβ
κ>0 we deduce that γ <1
d. /square
Eq. (186) only depends on the Crbeing Hermitian. If we make the assumption
that theCrhave theQ-QTproperty we get a stronger statement:
Corollary 10. LetLr,CrstandCrbe as defined in the statement of Theorem 7.
Suppose that the Crhave the spectral decomposition
Cr=Pr−PT
r (208)
wherePris a rankd−1projector which is orthogonal to its own transpose. Then
(1)For allr
Tr(Lr) =ǫ′
rl (209)
(2)For allr,s
Tr(LrLs) =d
d+1δrs+ǫ′
rǫ′
s
d/parenleftbigg
l2−1
d+1/parenrightbigg
(210)
(3)
d2/summationdisplay
r=1ǫ′
rLr=dlI (211)
for some real constant l>0and signsǫ′
r=±1.
Proof.The proof relies on the fact that the Killing form for gl( d,C) is related to
the Hilbert-Schmidt inner product by [ 55]
Tr(adAadB) = 2dTr(AB)−2Tr(A)Tr(B) (212)
Specializing to the case A=B=Lrand making use of the Q-QTproperty we find
d−1 =dTr(L2
r)−l2
r (213)
Using Lemma 9we deduce
l2
r=dβ−d+1
1−dγ(214)28
It follows that
lr=ǫ′
rl (215)
for some real constant l≥0 and signs ǫ′
r=±1. The fact that the Lrare a basis