Is symmetric power of a manifold a manifold?

Question

A Hausdorff, second-countable space $M$ is called a topological manifold if $M$ is locally Euclidean. Let $SP^n(M): = \left(M \times M \times \cdots \times M \right)/ \Sigma_m$, where product is done $n$-times and $\Sigma_n$ is the symmetric group. I want to prove that $SP^n(M)$ is a manifold (possibly with boundary). I know that $SP^n(M)$ is Hausdorff and second-countable. Is it locally Euclidean also for any $n$? Note that $M$ has no further structures like smoothness or complex structure (I know IG Macdonald's result when $M$ is a complex algebraic curve, but here I'm interested in general topological manifolds).

In a comment to the following question: https://math.stackexchange.com/questions/35045/symmetric-power-of-a-manifold, it has been doubted (or claimed) that $SP^n(M)$ is not locally Euclidean, though no proof/counterexamples are given. In another question: https://math.stackexchange.com/questions/3271422/symmetric-product-of-a-manifold-is-a-manifold, it is suspected in comments that $SP^n(M)$ is a manifold with boundary, though there is no justification. There are other questions and answers as well but no definitive proof if $SP^n(M)$ is locally Euclidean, or a counterexample if $SP^n(M)$ is not locally Euclidean, is given.

Any help will be greatly appreciated, even if it is for $n=2$ case.

Answer 1

$\newcommand{\Cone}{\operatorname{Cone}}$Let $d$ be the dimension of the manifold $M$. For $n \geq 2$, I will prove that the symmetric power $SP^n(M)$ is a manifold with boundary for $d=1$, a manifold for $d=2$, and a non-manifold for $d \geq 3$.

For disjoint open balls $U_1,\ldots,U_k$ in $M$ and positive integers $n_1,\ldots,n_k$ with $n_1+\cdots+n_k=n$, we have an open subset $$\begin{align} V & = SP^{n_1}(U_1) \times \cdots \times SP^{n_k}(U_k) \\ & \cong SP^{n_1}(\mathbb{R}^d) \times \cdots \times SP^{n_k}(\mathbb{R}^d) \end{align}$$ of $SP^n(M)$. These cover $SP^n(M)$. It is thus enough to prove that $V$ is a manifold with boundary for $d=1$ and a manifold for $d=2$, and for some choice of the $n_i$ a non-manifold for $d \geq 3$. In the arguments below, I will identify $SP^m(X)$ with the set of unordered sequences $\{p_1,\ldots,p_m\}$ of not necessarily distinct points in $X$.

We start with $d=1$. In this case, it is enough to prove that $SP^m(\mathbb{R})$ is a manifold with boundary for all $m \geq 1$. In fact, generalizing Ben McKay's answer we have a homeomorphism $$\mathbb{R} \times [0,\infty)^{m-1} \cong SP^m(\mathbb{R})$$ taking $(x,c_1,\ldots,c_{m-1}) \in \mathbb{R} \times [0,\infty)^{m-1}$ to the collection of points $$\{x,x+c_1,x+c_1+c_2,\ldots,x+c_1+c_2+\cdots+c_{m-1}\}$$ in $\mathbb{R}$. The point here is that unordered collections of points in $\mathbb{R}$ actually have a canonical ordering from smallest to largest.

We now handle $d=2$. In this case, it is enough to prove that $SP^m(\mathbb{R}^2)$ is a manifold for all $m \geq 1$. In fact, we have a homeomorphism $$SP^m(\mathbb{R}^2) \cong \mathbb{R}^{2m}.$$ To see this, identify $\mathbb{R}^2$ with $\mathbb{C}$. For $$\underline{a}=(a_1,\ldots,a_m) \in \mathbb{C}^m,$$ let $$f_{\underline{a}}(z) = z^m + a_1 z^{m-1} + \cdots + a_{m-1} z^1 + a_m \in \mathbb{C}[z].$$ The desired homeomorphism $$\mathbb{C}^m \cong SP^m(\mathbb{C}^m)$$ then takes $\underline{a} \in \mathbb{C}^m$ to the unordered set of roots of the monic degree $m$ polynomial $f_{\underline{a}}$.

The remaining case is $d \geq 3$. Here it is enough to find some $V$ as above that is not a manifold. Below I will prove that $$SP^2(\mathbb{R}^d) \cong \mathbb{R}^d \times \Cone(\mathbb{RP}^{d-1}).$$
This will imply that $$\begin{align} & SP^2(U_1) \times SP^1(U_2) \times \cdots \times SP^1(U_k) \\ \cong {} & \mathbb{R}^d \times \Cone(\mathbb{RP}^{d-1}) \times \mathbb{R}^d \times \cdots \times \mathbb{R}^d \end{align} $$ This is not a manifold (for instance, think about local homology). Here we are using the fact that $d \geq 3$ since $$\Cone(\mathbb{RP}^1) \cong \Cone(S^1) \cong \mathbb{R}^2.$$

It remains to prove that $$SP^2(\mathbb{R}^d) \cong \mathbb{R}^d \times \Cone(\mathbb{RP}^{d-1}).$$
This homeomorphism takes an unordered collection $\{p,q\}$ of points in $\mathbb{R}^d$ to the pair $$\left(\frac{p+q}{2},z\right) \in \mathbb{R}^d \times \Cone(\mathbb{RP}^{d-1}),$$ where $z$ is as follows:

• The cone point if $p=q$.
• Using homogeneous coordinates for projective space, the point in the cone that is $\operatorname{dist}(p,q)$ from the cone point in the direction $[p-q] \in \mathbb{RP}^{d-1}$.

Answer 2

Let $M:=\mathbb{R}$. Consider the map $\varphi\colon M\times M\to\mathbb{R}\times[0,\infty)$ given by $(x,y)\mapsto(x+y,(x-y)^2)$. Two points map to the same point just when they have same sum and product, since $(x-y)^2=(x+y)^2-4xy$, so the polynomial $p(t)=(t-x)(t-y)$ has the same roots, so the points are the same up to possible permutation. Hence the map descends to an injective map $SP^2M\to\mathbb{R}\times[0,\infty)$. By the same argument, if we take any numbers $b,c$ with $c\ge 0$, we can construct a polynomial $p(t)=t^2-bt+(b^2-c)/4$, and we find that it has two real roots $x,y$, or a double root $x,y=x$, and therefore is $\varphi(x,y)=(b,c)$. If we bound $b,c$, then $x^2+y^2=c+b^2$ is bounded, so we bound $x,y$. So $\varphi$ is a proper map. So $\varphi$ is onto and proper, so its quotient on $SP^2M$ is one to one, onto and proper. A proper injection to a locally compact Hausdorff space is a homeomorphism to its image (standard topology fact). So $SP^2M\cong \mathbb{R}\times[0,\infty)$. This obviously generalizes using the Viete formulas, so all of the symmetric products of the real number line are given by some inequalities on elementary symmetric polynomials, but I haven't worked them out.

Answer 3

The beginning of this paper describes the orbit type stratification of the the space $M^n/\Sigma_n$, also for manifolds $M$.