Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 $M$ be a smooth (or complex) manifold. Let $M^{(n)}$ its $n$-th Cartesian product. Let $\Sigma_n$ be the symmetric group of dimension $n$, which acts on $M^n$ in the usual way.

Question: is it possible to give $M^{(n)} / \Sigma_n$ a smooth (or complex) structure which is, in some sense, compatible with the original structure on $M$?

share|cite|improve this question
Is this even a manifold? Take the simplest example $\mathbb{R}^2/\Sigma_2$, this is most naturally a manifold with boundary (the part of $\mathbb{R}^2$ above and including the diagonal). For higher $n$, the analogue would be a manifold with corners. I take it you are only worried about the boundary? Since the "interior", where preimage in $M^n$ of your quotient space has $n$ distinct points, inherits the structure directly from the product structure of $M^n$. – Willie Wong Apr 25 '11 at 17:07
@Willie: Yes, I'm "only" worried about the "diagonal" points. Points like $(p_1,\dots,p_n)$, with $p_i \in M$ all distinct have local charts inherited by the product structure. – Raziel Apr 25 '11 at 18:55
I don't know if this may be of any help, but I'm trying to figure out a complex structure on the set of positive divisors of degree $d$ on a Riemann surface $M$. As a set, the latter is clearly equivalent to $M^{(n)} / \Sigma_n$, (that is the set of unordered n-tuples of points of $M$). I need the explicit complex structure on such a space in order to compute the push-forward of the Abel map, and compute its rank. – Raziel Apr 25 '11 at 19:55
As others have explained, the symmetric powers are usually singular objects. You can consider the "Hilbert scheme of $n$ points" on your $M$, which has more chances of being regular. In the complex case, I think there are $S_n$-equivariant desingularizations of the symmetric powers, which give you something smooth to repcace $M^{(n)}$ with. – Mariano Suárez-Alvarez Apr 25 '11 at 21:44
up vote 7 down vote accepted

If $M$ is a Riemann surface / complex surface then $M^{(n)}$ is a smooth manifold. It's a fairly standard argument. But as Willie Wong mentions, generally $M^{(n)}$ isn't a manifold unless you assume more of $M$. Interestingly enough, $(S^1)^{(3)}$ is a manifold and it's a fun exercise to figure out which one it is.

For the Riemann surface case, first consider $\mathbb C^{(n)}$. This is the space of n-tuples of points in $\mathbb C$ but with the ordering forgotten. As a space, it's homeomorphic to the space of monic complex polynomials of degree $n$ -- since monic complex polynomials have $n$ roots up to multiplicity -- the bijection is given in terms of the roots of the polynomials. So you can use a fundamental domain for the Riemann surface (or some other similar argument) to show $M^{(n)}$ is a manifold when $M$ is a Riemann surface.

FYI: I didn't just invent the above argument. It's a standard argument used in setting up Heegaard-Floer theory for 3-manifolds.

share|cite|improve this answer
Thanks! I do not know Heegaard-Floer theory, but you gave me a precious hint on how to build up the analytic structure by employing the correspondence with complex polynomials! – Raziel Apr 26 '11 at 9:34
After some thinking, it seems like I can't figure out the details for the general case of $M^{(n)}$ ($M$ is a Riemann surface). The case $\mathbb{C}^{(n)}$ is ok though (this case is easy, because you have one global chart). Can you please be so kind and provide some reference for the general case? I'm mostly interested in explicit local charts for $M^{(n)}$, then any complex-manifold-like approach is preferred. Thanks in advance! – Raziel Apr 28 '11 at 8:56
MR0151460 (27 #1445) Macdonald, I. G. Symmetric products of an algebraic curve. Topology 1 1962 319–343. 14.20 – Ryan Budney Apr 28 '11 at 17:45

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.