INTRODUCTION

3

0.4. Combining Theorem 0.3 with a non-commutative version of the First Funda-

mental Theorem of Invariant Theory ([Hoi, Theorems 2 and 7]) gives:

COROLLARY . T{Xk) is a simple Noetherian domain, generated by differential op-

erators of degree 2. Moreover, these generators are the "obvious" elements of V{Xk)

- see (IV, 1.9) for the precise statement.

We remark that tp : U(#) — » V{Xk) is still defined when k is not sufficiently small.

However, in this case V(Xk)is a Weyl algebra, and hence by (IV, Remark 1.5) i/ cannot

be surjective. Indeed, in this case V(Xk) will not even be finitely generated as a U(Q)-

module. This provides the following, rather curious dichotomy: The map ip is nice (that

is, surjective) if and only if Xk is bad (that is, singular).

0.5. In the course of the proof of Theorem 0.3 we prove a number of results about

the Lie algebra 0, of which the most significant are the following. For simplicity, we

state these results here under the assumption that k is sufficiently small, although we

do in fact prove analogous results for all values of k.

0.5.1. J(k) = Ann(L(Xk -f /?)), where L(\k -f p) is a simple highest weight module,

whose highest weight Xk can be explicitly described. Under the isomorphism of Theorem

0.3, L(Xk+p) becomes the standard V{Xk) -module 0{Xk). (See (II, 2.7, 3.7 and 4.7)).

0.5.2. Write k = k in Cases A and B, but k = 2k in Case C. Let Ok be the nilpotent

orbit {£ E 0 : £2 = 0 and rank£ = &}, with Zariski closure Ok • Then the associated

variety V( J{k)) of the ideal J(k) is equal to O* (see (II, 6.4)).

0.5.3. Let g = n~ 0 [} © n + be the usual triangular decomposition of 0 . Then Xk

is an irreducible component of Ok f! n+ (see (II, Proposition 6.3)).

0.6. Let G' = 0(k) act on X = M

M

( C ) as in Case B of (0.2). This induces an

action of SO(k) on X and hence on O(X). We will also study the ring of differential

operators V(A), for A = G(X)so^ . The natural way to approach this ring is to note

that Z/2Z = 0{k)/SO(k) acts on both A and V(A) and so one should try and relate

V(A) both to V(A)Z/2Z and to V(AZI2Z) = V(0(X)0^). In this way we obtain:

T H E O R E M . (V, Theorems 2.6 and 3.11). Let A = Q(X)so^ . If k n then

D(A) is a simple Noetherian ring, and is finitely generated as a module over the subring

R = U(sp(2n))/J(k). Moreover V(A)ZI2Z = R.