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Let $X$ be a projective variety. Symmetric product of $X$ is the quotient of the product $X^n$ by the action of the symmetric group $\Sigma_n$ permuting the factors.

When does it exist (as an algebraic variety)?

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Always. This is in many elementary algebraic geometry books, including Harris' book. –  J.C. Ottem Oct 20 '10 at 22:42
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Shouldn't it always exist, and always be given by taking $k[X]$ the homogeneous coordinate ring for some embedding, taking $R=k[X]^{\otimes n}$ and then $R^{S_n}$ with the $S_n$ acting by permuting sets of variables? Finite quotients of varieties exist and are varieties, this is (I believe) Chapter 0 (?) of GIT. –  Charles Siegel Oct 20 '10 at 22:42
    
Ah, of course, taking $Proj(R^{S_n})$, I meant. –  Charles Siegel Oct 20 '10 at 22:43
    
Thanks, Charles. I have already realised that I had asked a silly question. I wanted to ask it for a general scheme, but at the last moment flipped out and resorted to a projective one :) –  Dima Sustretov Oct 21 '10 at 17:33
    
@Charles Siegel: "Finite quotients of varieties exist and are varieties". This might depend on your definition of variety. If a variety is just something like an integral scheme of finite type over a field then I don't think this is true: there are proper non-projective varieties with finite groups acting on them such that the quotient only exists as an algebraic space---indeed one of the motivations of enlarging the category of schemes to things like alg spaces and stacks is precisely to make finite quotients always exist. On the other hand if a variety for you is quasi-proj then you're OK. –  Kevin Buzzard Oct 21 '10 at 19:38
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1 Answer

up vote 3 down vote accepted

To fix ideas, let $K$ be a field and $X/K$ be a seperated $K$-scheme of finite type. Let $G$ be a finite group operating on $X$ via $K$-morphisms. The operation is said to be admissible provided every orbit of $G$ is contained in an open affine subset of $X$. If the operation is admissible, then there is a pair $(Y, p)$ consisting of a seperated $K$-scheme $Y$ of finite type and a finite, surjective morphism $p\in Hom(X, Y)^G$, such that the map $$Hom(Y, Z)\to Hom(X, Z)^G,\ f\mapsto f\circ p$$ is bijective for all schemes $Z/K$. Then $(Y, p)$ is said to be the quotient of $X$ mod $G$. (Cf. SGA I, V.1 for such constructions.)

From now on assume that $X/K$ is projective. Then $X$ is a closed subscheme of $P_n$ and every finite subset of $X$ is contained in an open affine subset of $X$. We see: If a finite group $G$ acts on $X/K$ by $K$-morphisms, then the operation is automatically admissible.

Now take a projective $K$-scheme $V/K$. Then $V^n=V\times\cdots\times V$ is projective over $K$ (Segre-embedding) and hence the natural operation of the symmetric group $S_n$ on $V^n$ will be admissible. Consequently the quotient exists in that case.

Aside: It can happen that $V^n/S_n$ is non-smooth, even if $V^n$ is smooth over $K$.

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