16 P. DEIFT, L. C. LI, AND C. TOMEI
Suppose A = ^2j Ajzi £ £* We compute -K+A. For I g j ,
(Tt*+A,X) = (A,Tr+X)
= J tr A(z)(v+X)(z)^
= tt(ir+A)(0)(ir+X)(0) + (ir-A,X)- tr (v.A)(oo)(ir-X)(oo) ,(2.23)
where we have used Cauchy's theorem repeatedly. But for any constant matrix C,
i tr CX{z),_~~^_ = - tr C(n+X)(0) (2.24)
'(z-l)z
and
J tr CX{z)^±^ = tr C(TT_A')(CO) . (2.25)
Applying these formulae to (2.23), we find
A =
n_A.(jI±^B+^-A)^)
z - \ z - \
A0
(2.26)
Now
z - l
A,
suppose A = —^— G £* ,. Then a similar calculation shows that
2 - 1 4V
*+A = r - ^ T + 7(-4p)diag^i , (2.27)
where (Ap)diag is the diagonal part of Ap and ! i is the delta function at z = 1,
/ B(z)6l(z)az = B(l).
J\z\=l
Note that as X(l) is diagonal, only the diagonal part of Ap plays a role at 2 = 1 in (2.27).
We have proved the following result:
L e m m a 2.28. For A = Ap/(z - 1 ) 4 - ^reg £in ,
=
^ - ( ^ r e g ) o
+
_
+
h^
Sl
(2.29)
2: 1 4
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