## Smoothness of Symmetric Powers

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Here's something that's been bothering me, and that's come up again for me recently while reading some stuff about Hilbert schemes of points (Nakajima's lectures, specifically):

Let $C$ be an algebraic curve. Define $S^nC$ to be $C\times\ldots\times C/S_n$, the symmetric power.

Now, over $\mathbb{C}$, I can show that $C$ a complex manifold implies that $S^nC$ is, and that if $X$ is a variety with $S^nX$ smooth, then $X$ is one dimensional, but the argument I have involves looking in analytic open sets and reducing to the case of $\mathbb{C}^n$, and additionally is fairly unhelpful for identifying the total space (ie, that $S^n\mathbb{P}^1\cong\mathbb{P}^n$)

So here's my question: how can we, in a fairly quick and natural way, show that

1. If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth.
2. If $X$ is a smooth variety over an algebraically closed field $k$, and $S^nX$ is smooth, then $X$ is one dimensional.

Now, I don't want any projectivity hypotheses here, and I'm curious, with the more arithmetically inclined, is this still true over an arbitrary field?

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## 4 Answers

Just convince yourself that if $(C,P_1,\dots,P_n)$ and $(D,Q_1,\dots,D_n)$ are analytically isomorphic -- $C$ at the point $P_i$, $D$ at the point $Q_i$ -- then $Sym^n C$ at the point $(P_1,\dots,P_n)$ is analytically isomorphic to $Sym^n D$ at the point $(Q_1,\dots,Q_n)$. For this, you need to see that completion commutes with taking $S_n$-invariants. Let me hedge and say that this is clear if $P_2,\dots,P_n$ are smooth points.

(When I say "analytically isomorphic", I don't mean working over $\mathbb C$. Instead, as customary, I mean that the completion of the corresponding rings are isomorphic, for example $\widehat{\mathcal O}_{C,P_i}\simeq \widehat{\mathcal O}_{D,Q_i}$.)

Thus for 1 you can reduce to $D=\mathbb A^1$, and you are done.

1 is true over any field, without restrictions on characteristic. Indeed, the ring corresponding to $(\mathbb A^1)^n/S_n$ is the ring of invariants $k[x_1,\dots, x_n]^{S_n}$, and it is a polynomial ring on $n$ variables, the elementary symmetric polynomials in $x_i$. That is true over any ring $k$ (commutative, with identity).

Part 2 is reduced to the case of $\mathbb A^2$ by the same trick, and then it is an explicit computation. I will illustrate the $(\mathbb A^2)^n/S_n$ case. We need to find the ring of invariants $k[x_1,y_1,x_2,y_2,\dots,x_n,y_n]^{S_n}$. Now there are $2n$ elementary polynomials in $x_i$, resp. in $y_i$, of course. But there is more. For example $x_1y_1 + \dots + x_ny_n$ is an invariant.

To prove that $X=(\mathbb A^2)^n/S_n$ is singular at the point $P=(0,\dots,0)$ is equivalent to showing that $\dim T_{P,X} = \dim m/m^2 > 2n = \dim X$, where $m$ is the maximal ideal in $\mathcal O_{P,X}$. It is an exercise that the above $2n+1$ polynomials are linearly equivalent in $m/m^2$, which would complete the proof.

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So then after using completions (always forget about those...need to work on that) it reduces to affine space, and then symmetric polynomials prove the first claim for the affine line over any field.  Now, for the second, you're taking the points to all be distinct, but what about on the diagonals? – Charles Siegel Jan 1 2010 at 17:08
I misunderstood question 2. I will fix the solution. The previous version proved that $X$ singular implies $Sym^n X$ singular (which is trivial of course). – VA Jan 1 2010 at 19:56

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Smoothness is an etale local property, and you can check that symmetric power sends etale maps to etale maps. So checking on $\mathbb{A}^n$ is fine, since all smooth varieties over an algebraically closed field is etale locally isomorphic to it.

Of course, this is secretly the same as VA's answer, since stuff which makes sense etale locally is exactly the stuff remembered by the completion of the local ring, but I like the sound of it better.

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I also wanted to mention a `high-technology' answer to (1). Namely, if $C$ is a smooth algebraic curve, its $n$-th symmetric power coincides with the variety of all degree $n$ effective divisors on $C$ (that is, we consider $n$-element subsets of $C$, and then allow points to merge). In a fancier language, we are looking at the Hilbert scheme of length $n$ subschemes of $C$.

Note that this agrees with the explicit choice of coordinates if $C={\mathbb A}^1$ (and also $C={\mathbb P}^1$): namely, if you want to parametrize $n$-element subsets of ${\mathbb A}^1$ with multiplicities, you do so (essentially by unique factorization) by saying that such a subset is the zero locus of a degree $n$ monic polynomial $p$, so it is uniquely determined by coefficients of $p$ (which of course are the elementary symmetric polynomials in terms of original $n$ points).

OK, so why do finite subschemes of $C$ form a smooth variety? Because deformation theory says that at a particular finite subscheme $D\subset C$, the obstructions to smoothness belong to $Ext^2_C(O_D,O_D)$. This vanishes because $C$ is a smooth curve, and so there are no $Ext$'s beyond first (OK, no local $Ext$'s, but $D$ is finite, so it's the same thing).

I think this is a useful point of view. For instance, if $S$ is a smooth surface, $Sym^n S$ is singular, but one can check that the Hilbert scheme $Hilb_n S$ of degree $n$ finite subschemes is still smooth; it provides a resolution of singularities of $Sym^n S$. One can thus argue that the Hilbert scheme is a 'better behaved' object compared to the symmetric power...

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Edit: It seems appropriate to recall here Chevalley-Shephard-Todd's Theorem. It says the quotient of $\mathbb A^n_k$ by a finite linear group $G$ with order prime to the characteristic of $k$ is smooth (i.e. the algebra of invariants is polynomial) if and only if $G$ is generated by pseudo-reflections (codimension one fixed point set).

Once one localizes the problem at a point, as VA and Ben Webster did, this settles both 1 and 2 over arbitrary fields of characteristic zero. Of course VA argument is preferable as it is more direct/elementary.

Original answer. Below is my original answer commented by David Speyer below:

Over the category of smooth real manifolds the symmetric power of smooth curves is not smooth in general. The second symmetric power is a smooth surface with boundary and starting from the third symmetric power what we get are varieties with corners at the boundary.

Symmetric powers of smooth surfaces are still smooth as they are locally diffeomorphic to complex curves.

If nothing else these examples show that smoothness might mean different things for algebraic geometers and differential geometers in algebraic geometry over $\mathbb R$ and in differential geometry.

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1
I disagree. These examples show that "curve" and "surface" mean something different in algebraic and differential geometry, because algebraic geometers usually describe things by their complex dimension. What you are calling surfaces are being called curves in the other answers. But we agree about which objects are smooth. – David Speyer Jan 1 2010 at 14:59
I was thinking on curves over $\mathbb R$, or more precisely on its real points. According to VA's answer the symmetric powers of $\mathbb A^1$ are smooth while the symmetric powers of $\mathbb A^1(\mathbb R)=\mathbb R$, as real manifolds, are not. For $\mathbb A^2$ and $\mathbb A^2(\mathbb R) = \mathbb R^2$ we have the reverse phenomenon. – jvp Jan 2 2010 at 1:03
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