Why can I divide an affine variety by the action of the general linear group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:26:39Z http://mathoverflow.net/feeds/question/94303 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94303/why-can-i-divide-an-affine-variety-by-the-action-of-the-general-linear-group Why can I divide an affine variety by the action of the general linear group? Jesko Hüttenhain 2012-04-17T16:44:22Z 2012-04-17T19:52:14Z <p>Let $G\subseteq\mathrm{Gl}_n(\mathbb{C})$ be a subgroup of the general linear group and assume that $\rho:G\to\mathrm{Gl}(V)$ is a representation. Understand the complex vector space $V$ as an affine algebraic variety. Then, it appears to be well-known that the quotient $V/G$ has the structure of an algebraic variety such that the quotient map $\pi:V\to V/G$ is a morphism of varieties. However, I cannot find a proof for this statement. There is an abundance of proofs for the case where $G$ is finite, using the Reynolds operator and corollaries of Hilbert's basis theorem, but I would like to see a proof in the general case. Thanks very much in advance!</p> <p><b>Edit:</b> I forgot to mention that I assume $G$ to be reductive and the action on $V$ is regular. You are free to assume even more about $G$ if that allows you to provide a comprehensible reference for a proof.</p> http://mathoverflow.net/questions/94303/why-can-i-divide-an-affine-variety-by-the-action-of-the-general-linear-group/94304#94304 Answer by Angelo for Why can I divide an affine variety by the action of the general linear group? Angelo 2012-04-17T16:50:32Z 2012-04-17T16:50:32Z <p>If by $V/G$ you mean the space of orbits, this is not true. Consider $\mathbb C^*$ acting on the affine space $\mathbb A^1$ by multiplication; the space of orbits has two points, but the only variety with two points is disconnected, while $\mathbb A^1$ is connected.</p> http://mathoverflow.net/questions/94303/why-can-i-divide-an-affine-variety-by-the-action-of-the-general-linear-group/94305#94305 Answer by Ben McKay for Why can I divide an affine variety by the action of the general linear group? Ben McKay 2012-04-17T16:50:53Z 2012-04-17T19:52:14Z <p>If $G$ is reductive, try looking at <a href="http://books.google.ie/books?id=dFlv3zn_2-gC&amp;pg=PR3&amp;lpg=PR3&amp;dq=fogarty+kirwan&amp;source=bl&amp;ots=nR5HzpIUeY&amp;sig=8Uxe2guLjSavpLeBdA1swNQctL8&amp;hl=en&amp;sa=X&amp;ei=LJ-NT6SzL8fW0QX7k4XsDA&amp;ved=0CCAQ6AEwAA#v=onepage&amp;q=fogarty%2520kirwan&amp;f=false" rel="nofollow">Fogarty, Kirwan, Mumford, Geometric Invariant Theory, p. 27</a></p> http://mathoverflow.net/questions/94303/why-can-i-divide-an-affine-variety-by-the-action-of-the-general-linear-group/94311#94311 Answer by Jim Humphreys for Why can I divide an affine variety by the action of the general linear group? Jim Humphreys 2012-04-17T18:01:13Z 2012-04-17T18:01:13Z <p>To amplify what others have pointed out, it doesn't make sense to get an affine variety here as a "quotient" unless all orbits of <em>G</em> are closed (a condition not met even by the natural action of the general linear group) and the variety itself has regular functions given by a finitely generated algebra: the fixed points of <em>G</em> on the algebra of polynomials in <em>n</em> variables if <code>$n=\dim V$</code>. At the least you need <em>G</em> to be a Zariski-closed subgroup of the general linear group, but the natural condition for finite generation of polynomial invariants is typically satisfied in characteristic 0 by focusing just on <em>reductive</em> algebraic groups. </p> <p>All of this goes back to classical invariant theory and Hilbert's 14th problem, but by now it has become part of the broader study of GIT. Thanks to the proof by Haboush of Mumford's Conjecture, reductive groups over an algebraically closed field of any characteristic yield finitely generated algebras here, but more is needed to get a genuine quotient. </p>