What does this particular geometric quotient locally look like? - MathOverflow

What does this particular geometric quotient locally look like?

2012-11-07T19:33:32Z

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 $T$).

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 $R$.

(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.)

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.

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?

Answer by Will Sawin for What does this particular geometric quotient locally look like?

2012-11-07T22:03:02Z

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.

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.

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.