Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:11:40Z http://mathoverflow.net/feeds/question/4135 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4135/relation-between-lie-algebra-cohomology-and-number-of-relations-of-a-cyclic-modul Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module? Greg Muller 2009-11-04T18:58:15Z 2010-05-01T12:10:28Z <p>Let <img src="http://latex.mathoverflow.net/png?%5Cmathfrak%7Bg%7D" alt="\mathfrak{g}" title="" /> be a finite dimensional Lie algebra over k, let U be its enveloping algebra, and let M be a <img src="http://latex.mathoverflow.net/png?%5Cmathfrak%7Bg%7D" alt="\mathfrak{g}" title="" />-module (not necessarily finite dimensional). Call the <em>invariant dimension</em> of M the largest i such that <img src="http://latex.mathoverflow.net/png?Ext%5Ei%5FU%28k%2CM%29%5Cneq%200" alt="Ext^i_U(k,M)\neq 0" title="" />. This is the same as the degree of the largest non-vanishing Lie algebra cohomology group.</p> <p>Here are two equivalent statements of my question.</p> <ol> <li>If M is a cyclic <img src="http://latex.mathoverflow.net/png?%5Cmathfrak%7Bg%7D" alt="\mathfrak{g}" title="" />-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?</li> <li>If I is a left ideal in U, then is the number of generators of I always greater than the invariant dimension?</li> </ol> <p>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)).</p> http://mathoverflow.net/questions/4135/relation-between-lie-algebra-cohomology-and-number-of-relations-of-a-cyclic-modul/6170#6170 Answer by Simon Wadsley for Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module? Simon Wadsley 2009-11-19T20:55:58Z 2009-11-25T14:11:54Z <p>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.</p> <p>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 $Ext^j_U(M,U)\neq 0$)</p> <p>Edit: Expansion of strategy:</p> <p>Proposition 7.1 of <a href="http://www.informaworld.com/smpp/content~db=all?content=10.1080/00927879808826267" rel="nofollow"> Auslander-Gorenstein Rings by Ajitabh, Smith and Zhang</a> 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.</p> http://mathoverflow.net/questions/4135/relation-between-lie-algebra-cohomology-and-number-of-relations-of-a-cyclic-modul/23173#23173 Answer by Victor Protsak for Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module? Victor Protsak 2010-05-01T12:10:28Z 2010-05-01T12:10:28Z <p>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 <em>at most</em> the number of generators of gr I.</p> <p>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 <em>ad(g)-invariant</em> 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.</p>