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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?

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It seems interesting that $GL_n/M$ embeds in the Grassmannian of $n$-planes in ${\mathfrak gl}_n$, as the conjugation-orbit of the point corresponding to the diagonal matrices. –  Allen Knutson Nov 8 '12 at 3:08
I don't understand what that means. What point corresponds to the diagonal matrices? –  Will Sawin Nov 8 '12 at 4:55
The space of diagonal matrices is an $n$-dimensional subspace of ${\mathfrak gl}_n$. Hence it defines a point of the corresponding Grassmannian. $GL_n$ acts on that Grassmannian. –  Allen Knutson Nov 9 '12 at 3:47

1 Answer 1

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.

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