Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth projective geometrically connected variety over a field $k$.

Let the cyclic group $G=\{e,a\}$ with two elements act on $X \times X$ via $a\cdot (x_1,x_2) = (x_2,x_1)$.

When is the quotient $X\times X/ G$ nonsingular?

I wrote down the case of $X=\mathbf{A}^1$. The answer is that $X\times X/G$ is given by nonsingular scheme $\mathrm{Spec} (k[xy,x+y]) \cong \mathbf{A}^2$, where $\mathbf{A}^1_x\times \mathbf{A}^1_y = X\times X = \mathrm{Spec} (k[x]\otimes k[y])$. (The subscript $x$ and $y$ indicate the coordinate used.)

I'm very interested in the case $\dim X = 1$.

share|cite|improve this question
Notice that your example is not of a projective variety! :-) – Mariano Suárez-Alvarez Jul 19 '12 at 7:08
Ha! You're right. But can't this calculation be used to show that the 2-symmetric product of $\mathbf{P}^1$ is nonsingular? – Harry Jul 19 '12 at 7:19
up vote 4 down vote accepted

Over an algebraically closed field:

The quotient $X^{(2)}=X\times X/C_2$ is called the symmetric square of $X$.

If $X$ is an irreducible (quasi)projective nonsingular curve, then $X^{(2)}$ is always smooth and, in fact, this generalizes to $X^{(n)}=X\times\cdots\times X/S_n$, the obvious quotient by the symmetric group on $n$ letters.

If $X$ is an irreducible (quasi)projective nonsingular surface, then $X^{(n)}$ is singular for all $n\geq2$.

Lothar Göttsche wrote notes for a short course he gave in ICTP a few years ago on the subject of Hilbert schemes of points, with title Hilbert schemes: local properties and Hilbert scheme of points. There you should find proofs of all the above statements. (The Hilbert scheme in all this is a «less singular» replacement for what I wrote $X^{(n)}$ which, in the case of smooth surfaces, turns out to be itself a desingularization of the symmetric power; in general, it is also singular, though)

share|cite|improve this answer
That's great! Could you tell me how one proves this? Is this easy? Or should I look it up in some book somewhere? – Harry Jul 19 '12 at 7:20
Thanks for the edit! I found the answer in Proposition 4.2 (page 22) and Theorem 3.2 (page 20) of Gottsche's notes. – Harry Jul 19 '12 at 7:33
Do add a link to the notes if they are online. – Mariano Suárez-Alvarez Jul 19 '12 at 7:35
I found the notes on his website : – Harry Jul 19 '12 at 7:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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