Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $k$, let $U$ be its enveloping algebra, and let $M$ be a $\mathfrak{g}$-module (not necessarily finite dimensional). Call the invariant dimension of $M$ the largest $i$ such that $\operatorname{Ext}^i_U(k,M)\neq 0$. This is the same as the degree of the largest non-vanishing Lie algebra cohomology group.

Here are two equivalent statements of my question.

  1. If $M$ is a cyclic $\mathfrak{g}$-module (that is, its generated by a single element as a $U$-module), then is the number of relations of $M$ always greater than the invariant dimension minus $1$?
  2. If $I$ is a left ideal in $U$, then is the number of generators of I always greater than the invariant dimension?

The reason why I would suspect such a thing is that it is true in the case of an abelian Lie algebra; that is, when $U$ is a polynomial ring. In this case, the invariant dimension of an ideal $I$ coincides with the height of $I$, and so (2) becomes Krull's height theorem (and (1) follows immediately from (2)).


2 Answers 2


This looks like the kind of thing that one might be able to prove by filtering $U$ by word-length over $g$ and then passing to the associated graded ring. The idea is that this would reduce the problem to the case that $g$ is abelian, since the associated graded ring is a polynomial ring in $\dim g$ variables and the number of generators of $I$ as a left ideal will be at least as big as the number of generators of $\mathrm{gr}(I)$ its associated graded ideal.

I'm not sure whether the invariant dimension will be preserved when you pass to the associated graded though. You might expect it to be as a similar invariant, the grade of the module M, j(M) is preserved when you do this. (Recall that j(M) is the smallest integer j such that $Ext^j_U(M,U)\neq 0$)

Edit: Expansion of strategy:

Proposition 7.1 of Auslander-Gorenstein Rings by Ajitabh, Smith and Zhang tells us that $\mathrm{Ext}^j_U(M,k)$ is isomorphic to the $k$-vectorspace dual of $\mathrm{Ext}^{d-j}_U(k,M)$ where $d=\dim g$. Thus invariant dimension is $d-min(j|\mathrm{Ext}^j_U(M,k)\neq 0)$. If one can relate the non-vanishing of $\mathrm{Ext}^j_U(M,U)$ and $\mathrm{Ext}^j(M,k)$ then it might be possible to reduce the problem to something relating the grade of $M$ to the number of relations of $M$. This can be done by passing to the associated graded ring.

  • $\begingroup$ Hmmmn. My first go at LaTeX on this seems to have failed. $\endgroup$ Nov 19, 2009 at 20:57
  • $\begingroup$ I don't think there is support for the mathfrak font - if you change this it should compile properly. $\endgroup$ Nov 19, 2009 at 21:02

This is FALSE as stated, even for the augmentation ideal. For example, consider the 3-dimensional Heisenberg Lie algebra g and the trivial module k. Then it's easy to see that the third (=top) cohomology is non-zero, but due to the noncommutativity of g, the left ideal I=annihilator of k may be generated by only two elements. The inequality proposed by Simon, in fact, goes the other way: the number of generators for I is at most the number of generators of gr I.

Admittedly, this is a cheap shot, since the right notion of the number of generators for a left ideal I of U(g) is the minimal dimension of an ad(g)-invariant subspace of U(g) generating I - in particular, this number survives the passage to the associated graded. I don't know whether the modified statement is true.


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