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If $X$ is a quasi-projective variety, then for each integer $n\geq 0$ one can define the symmetric product to be the scheme quotient $$\textrm{Sym}^n(X)=X^n/S_n,$$ where $S_n$ is the symmetric group and $X^n$ is $n$-fold power of copies of $X$ over the base field. I am not sure, but I think one can also characterize $\textrm{Sym}^n(X)$ as the scheme representing the functor of "unordered $n$-tuples of points" in $X$, which I should probably refer to as effective zero cycles of degree $n$.

I heard that the construction of $\textrm{Sym}^n(X)$ does not work if $X$ is not necessarily quasi-projective. I suppose this means (both) that there is no scheme quotient $X^n/S_n$ as above, and that the above functor is not representable.

Question. Is there a geometric object (maybe not a scheme anymore) that replaces $\textrm{Sym}^n(X)$ when $X$ is not quasi-projective?

Thank you in advance.

Brenin
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1 Answers1

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The answer is yes! Basically by definition, the quotient $X^n / S_n$ will be an étale algebraic space for an arbitrary variety $X$. Algebraic spaces are not defined by topological space + structure sheaf like schemes, but they are defined by their functor of points: an algebraic space over $k$ is an étale sheaf of sets $E$ on the site of schemes over $k$ that

  1. has a representable diagonal, which means that for any two schemes $S$ and $T$, the fiber product of functors $h_S \times_E h_T$ is representable;

  2. étale locally looks like a scheme (there is an étale surjection from a scheme to $E$, which makes sense once you realise that condition 1 allows you to define étale and surjective for these maps).

Clearly, the functor of points of any scheme is an algebraic space, so this is a generalisation of the notion of a scheme (and once you go down this rabbit hole, you quickly arrive at stacks as a further generalisation of algebraic spaces where a sheaf of sets is replaced by a "sheaf of groupoids").

Taking quotients by actions of finite groups (modulo being careful in characteristic $p$) in the category of algebraic spaces is not a problem (this is the Keel-Mori theorem, though you could prove this by hand for the symmetric product).

Jakob Oesinghaus
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