Let $X$ be a (smooth complex algebraic) surface. Suppose $\theta$ is an automorphism of order $2$ of $X$, such that its fixed locus is a disjoint union of smooth curves. I am trying to prove that the quotient $$Y=X/\langle\theta\rangle$$ is in fact a smooth surface. (First of all: is this true/does it follow from some general result?)

My attempt:

Since the quotient is clearly smooth away from the fixed points, we can localize the question around a fixed point $p=(0,0)\in \Bbb{C}^2$. Up to a chart change we can suppose that the fixed curve is the $z$-axis and $$\theta(z,w)=(z,-w)$$ $\Bbb{C}[z,w]^\theta$ has two generators $Z=z^2$ and $W=w$. Now the smoothness of the quotient amounts to show that in fact there are no relations between $Z$ and $W$ and thus the quotient around $p$ is locally the spec of $\Bbb{C}[z_1,z_2]$. But I have no idea how to do this.

A secondary question is: is it always $\rho(Y)=\rho(X)$ (Picard number) ?

  • 2
    $\begingroup$ $\mathbb{C}[z^2, w] \cong \mathbb{C}[z^2][w]$. It is obvious that $\mathbb{C}[z^2]$ is a polynomial ring, so $\mathbb{C}[z^2, w]$ is a polynomial ring, too. Hence $\textrm{Spec}\,(\mathbb{C}[z^2, w])$ is smooth. $\endgroup$ – Francesco Polizzi Nov 29 '13 at 16:11
  • $\begingroup$ Any nontrivial relation between $X$ and $Y$ would, by substitution, immediately give a nontrivial relation between $z$ and $w$, which does not exist. $\endgroup$ – Will Sawin Nov 29 '13 at 17:30
  • $\begingroup$ Thanks @WillSawin. You also made me notice the terrible mistake I made by mixing notation for $X$ and $Y$ (now $Z$ and $W$). Let me edit and fix that. $\endgroup$ – Heitor Nov 29 '13 at 17:37

Francesco gave the answer to your first question. As for the second one, look at my answer to this post.

  • $\begingroup$ Dear abx, if I understand correctly what you are saying is that in general $S(\theta)=\pi^\ast S_Y$, where $S_Y$ is the Picard group of $Y$ and $S(\theta)$ are the $\theta$-fixed elements of $S_X$. Could you please explain me why is this true? $\endgroup$ – Heitor Nov 29 '13 at 17:32
  • $\begingroup$ This is true is up to multiplication by 2 : there is a homomorphism $Nm: \mathrm{Pic}(X)\rightarrow \mathrm{Pic}(Y)$ such that $\pi ^*Nm(L)=L\otimes \theta ^*L$. Thus if $L$ is $\theta $-invariant, $L^2$ belongs to $\pi ^*S_Y$. $\endgroup$ – abx Nov 29 '13 at 17:48

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.