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We next observe that if the Crhave theQ-QTproperty they must, in particular,
be Hermitian. It turns out that that is, by itself, already a very str ong constraint.
Before stating the result it may be helpful if we explain the essential idea on
which it depends. Although we have not done so before, and will not d o so again, it
will be convenient to make use of the covariant/contravariantinde x notation which
is often used to describe the structure constants. Define the me tric tensor
Mrs= Tr(LrLs) (179)
and letMrsbe its inverse. So
d2/summationdisplay
t=1MrtMts=Mr
s=/braceleftigg
1r=s
0r/ne}ationslash=s(180)
We can use these tensors to raise and lower indices (we use the Hilber t-Schmidt
inner product for this purpose because the fact that gl( d,C) is not semi-simple
means that its Killing form is degenerate [ 52–55]). In particular, the matrices
Lr=d2/summationdisplay
t=1MrsLs (181)
are the basis dual to the Lr:
Tr(LrLs) =Mr
s (182)
Suppose we now define structure constants ˜Crstby
[Lr,Ls] =d2/summationdisplay
t=1˜CrstLt(183)
(so in terms of the Crstwe have ˜Ct
rs=Crst). It follows from the relation
˜Crst= Tr/parenleftbig
[Lr,Ls]Lt/parenrightbig
= Tr/parenleftbig
Lr[Ls,Lt]/parenrightbig
(184)
that the ˜Crstare completely antisymmetric for any choice of the Lr. If we now
require that the matrices Crbe Hermitian it means that, not only the ˜Crst, but
also theCrstmust be completely antisymmetric. Since the two quantities are
related by
˜Crst=d2/summationdisplay
u=1CrsuMut (185)25
this is a very strong requirement. It means that the Lrmust, in a certain sense,
be close to orthonormal (relative to the Hilbert-Schmidt inner prod uct). More
precisely, it means we have the following lemma:
Lemma 9. LetLr,CrstandCrbe defined as in the statement of Theorem 7, and
letlr= Tr(Lr). Then the Crare Hermitian if and only if
Tr(LrLs) =βδrs+γlrls (186)
whereβ,γare real constants such that β >0andγ <1
d.
If this condition is satisfied we also have
d2/summationdisplay
r=1lrLr=β
1−dγI (187)
d2/summationdisplay
r=1l2
r=dβ
1−dγ(188)
Proof.To prove sufficiency observe that, in view of Eq. ( 185), the condition means
˜Crst=βCrst+γltd2/summationdisplay
u=1Crsulu (189)
In view of Lemma 8, and the fact that β/ne}ationslash= 0, this implies
Crst=1
β˜Crst (190)
Since the ˜Crstare completely antisymmetric we conclude that the Crstmust be
also. It follows that the Crare Hermitian.
To prove necessity let ˜Cr(respectively M) be the matrix whose matrix elements
are˜Crst(respectively Mst). Then Eq. ( 185) can be written
˜Cr=CrM (191)
Taking the transpose (or, equivalently, the Hermitian conjugate) on both sides of
this equation we find
˜Cr=MCr (192)
implying
[M,Cr] = 0 (193)
for allr. Since the Lrare a basis for gl( d,C) we deduce
[M,adA] = 0 (194)
for allA∈gl(d,C). Eq. (186) is a straightforward consequence of this, the fact
that gl(d,C) has the direct sum decomposition CI⊕sl(d,C), the fact that sl( d,C)
is simple, and Schur’s lemma [ 52–55]. However, for the benefit of the reader who is
not so familiar with the theory of Lie algebras we will give the argument in a little
more detail.26
Given arbitrary A=/summationtextd2
r=1arLr, let/bardblA/an}bracketri}ht/an}bracketri}htdenote the column vector
/bardblA/an}bracketri}ht/an}bracketri}ht=
a1
a2
...
ad2
(195)
So
/bardblLr/an}bracketri}ht/an}bracketri}ht=
1
0
...
0
/bardblL2/an}bracketri}ht/an}bracketri}ht=
0
1
...
0
/bardblLd2/an}bracketri}ht/an}bracketri}ht=
0
0
...
1