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)