24 P. DEIFT, L. C. LI, AND C. TOMEI
and we can consider each pair (M, A ) and {R,S) separately.
First observe that
flk(M, A ) = fik{K) = tr A* , 42k(M, A ) = tr MKk , k = 1 , . . . , N , (2.64)
are coadjoint invariants. Indeed
tr{gMg~l + [u,gKg-l]){gKg-l)k
= tr MKk + tr(u[gKg-1,gKkg-1})= tr MKk .
Lemm a 2.65. The functions pik,t2k, 1 k N, are independent at (M,K) if K
is invertible with distinct eigenvalues. If det A' = 0, then the functions are dependent at
(M,K).
Proof: Suppose that £ t i U - ^ + £ L i bk^- = 0, £ ?
= 1
ak^-+T,ti h ^ =
0 for suitable constants ak,bk. Then as dj\k/dM = 0, we obtain Ylk=i bk(KT)k = 0, or
]Cfc=i ^)kKk~1 0, as A is invertible. But this is impossible as A* has distinct eigenvalues,
unless bk = 0, k = l , . . . , n . The first equation now implies X^fc=i UkkKk~l = 0, and
again ak = 0, k = 1,.. . ,n. On the other hand, if det K = 0, then PK{K) = 0, where
-PA'(^) = Ej=o ^J^3 1S ^ e characteristic polynomial of A. But CFQ = det A = 0, and so
({cij 0}^_j, {bj = aj}^!) ^ 0 solves the above gradient equations for fik and (f2k- This
completes the proof of the lemma. D
Definition 2.66. We say that a pair (M, A") is generic if K is invertible with distinct
eigenvalues.
Clearly generic pairs (M, K) form a dense, open set in R2N . Given a generic pair
(Mo,Ko), consider
5(M0,A'o) = {(M,K) : 4ik{M,K) = plk(M0,K0) , 1 * 2, 1 k N} . (2.67)
Then S(Mo,i0) is a submanifold of dimension 2Ar2 - 2iV in R2N . Indeed if (M,K)
^(Mo.Ko)' then spec A = spec Ao and so (M, A") is generic and hence dtik are independent
at (M,K). Clearly
0(Afo,/v0) C S(Mo,K0) ' (2.68)
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