What does this particular geometric quotient locally look like? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:01:29Z http://mathoverflow.net/feeds/question/111751 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111751/what-does-this-particular-geometric-quotient-locally-look-like What does this particular geometric quotient locally look like? Florian 2012-11-07T19:33:32Z 2012-11-08T02:53:35Z <p>Let \$k\$ be a field and consider the algebraic group \$GL_n\$ over \$Spec(k)\$. It has as a closed (but not normal) algebraic subgroup the group \$M\$ of monomial matrices, i.e. matrices having exactly one nonzero entry in each row and each column (this is the normalizer \$T\rtimes\Sigma_n\$ of the diagonal matrices \$T\$).</p> <p>The geometric quotient \$GL_n/M\$ of the canonical action of \$M\$ on \$GL_n\$ exists (if I checked everything correctly) and it is the affine scheme associated to the ring of invariants \$R\$.</p> <p>(Intuitively, \$GL_n/M\$ should be some open subset of \$(\mathbb{P}^{n-1})^n\$ of \$n\$ lines spanning the whole space with permutations identified.)</p> <p>This ring of invariants \$R\$ is finitely generated as a \$k\$-algebra: \$M\$ is reductive, by Mumford’s Conjecture geometrically reductive and hence finitely generated by Nagata’s Theorem (it's possibly easier to see this directly in this example). I do not think, that I need an infinite field somewhere.</p> <p>Is there a Zariski open covering of \$GL_n/M\$ by nice affine schemes \$Spec(k[x_1,\ldots,x_m]/I])\$ which I can explicitly write down?</p> http://mathoverflow.net/questions/111751/what-does-this-particular-geometric-quotient-locally-look-like/111759#111759 Answer by Will Sawin for What does this particular geometric quotient locally look like? Will Sawin 2012-11-07T22:03:02Z 2012-11-07T22:03:02Z <p>First we compute \$GL_n\$ mod the group of diagonal matrices. \$GL_n\$ embeds into \$(\mathbb A^n-0)^n\$, so you are correct that the quotient by \$(\mathbb G_m)^n\$ is \$(\mathbb P^{n-1})^n\$. The only points we have to remove are those with determinant \$0\$, a hypersurface.</p> <p>The coordinate ring is thus generated by functions which are a product of one coefficient in each row, divided by the determinant. There are \$n^n\$ of these. They satisfy one relation coming from the fact that the determinant over the determinant is \$1\$, and the rest of the relations are toric: one product of generators is equal to another product of generators because each coefficient shows up the same number of times.</p> <p>Computing the quotient of this ring by the symmetric group action is more subtle. It is easy to find a lot of elements: just take any product of generators and add together all the \$S^n\$ conjugates. I don't think it's too hard to find a basis of these, but I don't know how to find generators and relations.</p>