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2d(d+1)
d2Tsrr+dd2/summationdisplay
v=1Tsrv+dd2/summationdisplay
u=1Tsur+d2/summationdisplay
u,v=1Tsuv
=d
2(d+1)/parenleftbig
3K2
rs+1/parenrightbig
=d(3dδrs+d+4)
2(d+1)2(280)35
and
/an}bracketle{t/an}bracketle{ter/bardbles/an}bracketri}ht/an}bracketri}ht=d+1
2dd2/summationdisplay
u=1K2
ruK2
su
=dδrs+d+2
2(d+1)(281)
Using these results in the expressions
Tr/parenleftbig
QrQs/parenrightbig
= Tr/parenleftigg/parenleftbiggd+1
dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg/parenleftbiggd+1
dTs−2/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardbl/parenrightbigg/parenrightigg
(282)
and
Tr/parenleftbig
QrQT
s/parenrightbig
= Tr/parenleftigg/parenleftbiggd+1
dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg/parenleftbiggd+1
dTT
s−2/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardbl/parenrightbigg/parenrightigg
(283)
the first two statements follow. The remaining statements are imme diate conse-
quences of these and the fact that
Jr=Qr−QT
r (284)
¯Rr=Qr+QT
r (285)
/square
Now define
/bardblv0/an}bracketri}ht/an}bracketri}ht=1
dd2/summationdisplay
r=1/bardblr/an}bracketri}ht/an}bracketri}ht (286)
where/bardblr/an}bracketri}ht/an}bracketri}htis the basis defined in Eq. ( 117). The following result shows (among
other things) that the subspaces onto which the Qr(respectively QT
r,Rr) project
span the orthogonal complement of /bardblv0/an}bracketri}ht/an}bracketri}ht.
Theorem 14. For allr
Qr/bardblv0/an}bracketri}ht/an}bracketri}ht=QT
r/bardblv0/an}bracketri}ht/an}bracketri}ht=Jr/bardblv0/an}bracketri}ht/an}bracketri}ht=Rr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (287)
Moreover
d2/summationdisplay
r=1Qr=d2/summationdisplay
r=1QT
r=d2
d+1/parenleftbig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig
(288)
d2/summationdisplay
r=1Jr= 0 (289)
d2/summationdisplay
r=1¯Rr=2d2
d+1/parenleftbig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig
(290)
Proof.Some of this is a straightforward consequence of the fact that Jris the
adjoint representative of Π r. Since
d2/summationdisplay
s=1Πs=dI (291)36
we must have
d2/summationdisplay
s,t=1JrstΠt=d2/summationdisplay
s=1adΠrΠs= 0 (292)
In view of the antisymmetry of the Jrstit follows that
d2/summationdisplay
r=1Jr= 0 (293)
and
Jr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (294)
Using the relations
Qr=1
2Jr(Jr+I) (295)
QT
r=1
2Jr(Jr−I) (296)
¯Rr=J2
r (297)
we deduce
Qr/bardblv0/an}bracketri}ht/an}bracketri}ht=QT
r/bardblv0/an}bracketri}ht/an}bracketri}ht=¯Rr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (298)
It remains to prove Eqs. ( 288) and (290). It follows from Eq. ( 120) that
d2/summationdisplay
r=1Qrst=d+1
dd2/summationdisplay
r=1Trst−2d2/summationdisplay
r=1/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardblt/an}bracketri}ht/an}bracketri}ht