3
$\begingroup$

Setting

Let $I\subseteq\mathbb C[x_0,\ldots,x_n]=:S$ be a homogeneous ideal and $X\subseteq\mathbb P^n$ the scheme defined by $I$. Consider the action of the symmetric group $\mathfrak S_{n+1}$ on $S$ by permuting the variables. Assume that $I$ is invariant under under some subgroup $G\subseteq\mathfrak S_{n+1}$. Assume furthermore that $G$ acts transitively and freely on the irreducible components of $X$. In other words, all irreducible components of $X$ are isomorphic and we have one component for each permutation in $G$. You can obtain something like this by picking any (possibly open) point of $\mathbb P^n$ which has trivial stabilizer in $G$ and taking (the closure of) its $G$-orbit.

Question

I am in such a situation and I want to figure out whether each component of $X$ is nonsingular. I thought it might be a good idea to consider the quotient $X/G$, which is defined as $\operatorname{Proj}(S^G/I^G)$. Here, I denote by $S^G := \{ f\in S \mid G.f=\{f\}\}$ the $G$-invariants in $S$. My question is whether the following is true:

The irreducible components of $X$ are nonsingular if and only if $X/G$ is nonsingular.

Thoughts so far

My intuition tells me that $X/G$ should be isomorphic to each component of $X$ (in the general case, also including its embedded points). If $X$ is normal, then this is easily true: Restricting the projection $\pi:X\twoheadrightarrow X/G$ to any component of $X$ yields a surjective morphism between normal varieties whose fibers generically contain one element, so this morphism is bijective. Because it maps between normal varieties, it is an isomorphism. I am not sure how to treat this in the general case, though.

I went through the examples $I=(x,yz)$ and $I=(x^2,xz,yx,yz)$ in $\Bbb C[x,y,z]$ with $G$ generated by the transposition of $y$ and $z$ only. It behaves as I expected, but I gained no insights.

$\endgroup$
3
$\begingroup$

Here is an example where $X/G$ is singular and the components of $X$ are smooth.
In $\mathbb{P}^3$ with coordinates $x,y,z,w$, consider the smooth conics
$$\begin{array}{lll}X_1:& x^2-y^2=yw,&z=w\\ X_2:& z^2-w^2=yw,&x=y.\end{array}$$ They are exchanged by the involution $\sigma:(x:y:z:w)\mapsto(z:w:x:y)$ and meet only at the points $(0:0:1:1)$ and $(1:1:0:0)$ which are themselves exchanged by $\sigma$. In particular, $\sigma$ acts freely on $X:=X_1\cup X_2$, so the projection $X\to X/\langle\sigma\rangle$ is étale. Hence $X/\langle\sigma\rangle$ is singular (because $X$ is; in fact $X/\langle\sigma\rangle$ is a rational curve with one node) while $X_1$ and $X_2$ are smooth.

$\endgroup$
  • $\begingroup$ Thanks. This also made me realise that I made a mistake in my question. I would require $X$ itself to be normal in order to deduce that the components are isomorphic to the quotient, it doesn't suffice that the components are normal individually. This is a great example, exactly what I was looking for. $\endgroup$ – Jesko Hüttenhain Apr 17 '16 at 14:01
2
$\begingroup$
  1. You apparently mean the action on the set of components is free, not just faithful.

  2. For $C$ a component of $X$, the composite map $C \to X \to X//G$ is a finite map, and (by the freeness assumption) birational. From there you need only that $X//G$ is normal to infer that this map is an isomorphism. (You don't need any assumption on $C$ nor do you need smoothness of $X//G$.)

I'm trying to involve the square $$\begin{matrix} C \times G &\to& X \\ \downarrow &&\downarrow \\ C &\to& X//G\end{matrix}$$ where down-arrows divide by the $G$-action on the right (as indicated by the notation $X//G$) and right-arrows by the diagonal interior action $(c,g)\sim (ch, h^{-1}g)$. Your questions should be about comparing the two horizontal (birational) maps.

$\endgroup$
  • $\begingroup$ Thanks and +1, I accepted @Laurent Moret-Bailly's answer below simply by coinflip, I'd really like to accept both answers. $\endgroup$ – Jesko Hüttenhain Apr 17 '16 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.