# Quotient space of $\Bbb C^n$ obtained by action of $S_n$

Consider the action of $S_n$ on $\mathbb{C^n}$ given by: $$\sigma(x_1, x_2, \cdots,x_n) = (x_{\sigma(1)}, x_{\sigma(2)}, \cdots,x_{\sigma(n)}).$$

What is the quotient space of $\mathbb{C^n}$ obtained by the action of $S_n$?

I considered the case $n =2$ and tried thinking of the problem geometrically, but I didn't get far.

In this case equivalence classes look like $\{(x_1,x_2),(x_2,x_1)\}$ because action is by $S_2$

• This is not an action. You need $\sigma^{-1}$'s on the RHS. – darij grinberg May 5 '15 at 23:24
• @darijgrinberg Although I prefer left actions, I wouldn't go so far as to call a right action "not an action". – Andreas Blass May 5 '15 at 23:41
• @AndreasBlass: but would you go so far as writing a right action on the left? – darij grinberg May 5 '15 at 23:49
• @darijgrinberg Not while I'm awake. – Andreas Blass May 5 '15 at 23:50

The quotient is actually $\mathbb{C}^n$ indeed. This is really an algebraic corollary and the geometric intuition here wouldn't be that helpful, mainly because we are talking about the homeomorphism type of the quotient, so we have quite a lot of flexibility.

For that matter, we are in complex space, where the hyperplane of a reflection has topological codimension 2, so taking the quotient doesn't create boundaries... (the action of $S_2$ on $\mathbb{C}^2$ is equivalent to the action $x\rightarrow -x$ on $\mathbb{C}$, the quotient of which is homeomorphic to $\mathbb{C}$ in the same way $\mathbb{R}P^1$ is homeomorphic to $S^1$ (the circle).

Let me give a description of the quotient and try and persuade you that it is $\mathbb{C}^n$, and then we'll talk about the algebra behind this:

Consider the map \begin{alignat*}{1} f:\mathbb{C}^n&\longrightarrow \mathbb{C}^n \\ (x_1,x_2,\cdots,x_n) &\longrightarrow \big( e_1({\bf x}),\cdots,e_n({\bf x})\big) \end{alignat*} where ${\bf x}=(x_1,\cdots,x_n)$ and $e_i$ is the $i^{\text{th}}$ elementary symmetric polynomial in $n$ variables.

Now, $f$ is clearly continuous, so if I convince you that the preimages $f^{-1}({\bf z})$ are exactly the orbits $S_n\cdot {\bf x}$, then you should believe me that $\mathbb{C}^n$ is indeed (homeomorphic to) the topological quotient $S_n\backslash \mathbb{C}^n$.

What is the preimage $f^{-1}({\bf z})$? If it contains an element ${\bf x}$, it contains at least those ${\bf x'}$'s that are in the orbit $S_n\cdot {\bf x}$ (because the polynomials $e_i$ are symmetric, every element of the orbit $S_n\cdot {\bf x}$ should have the same image under $f$...). It is more difficult to show that each $f^{-1}({\bf z})$ contains exactly one orbit though:

Consider two different orbits $S_n\cdot {\bf x}$ and $S_n\cdot {\bf y}$ and pick a polynomial $\theta\in \mathbb{C}[x_1,\cdots,x_n]$ such that $\theta({\bf x})=0$ but $\theta({\bf w})\neq 0$ for all ${\bf w}\in S_n\cdot {\bf y}$. Now consider the polynomial $$\overline{\theta}({\bf w})=\prod_{g\in S_n}\theta(g\cdot{\bf w})$$

By definition, $\overline{\theta}$ is an $S_n$-invariant polynomial that separates ${\bf x}$ and ${\bf y}$ (that is, it takes different values on them). Now, the fundamental theorem of Symmetric polynomials (see also Theorem 1 below) tells us that $\overline{\theta}$ can be written as a polynomial in the $e_i$'s. In particular, not all the $e_i$'s can agree on ${\bf x}$ and ${\bf y}$, or $\overline{\theta}$ should agree too. This means that ${\bf x}$ and ${\bf y}$ cannot be in the same preimage $f^{-1}({\bf z})$ for some ${\bf z}\in \mathbb{C}^n$.

There is one last thing to check; namely, how do we know that for every point ${\bf z}\in \mathbb{C}^n$, the equations $e_i({\bf x})=z_i$ have a common solution? This is because the elementary symmetric functions are exactly $n$-many, and algebraically independent (see also Theorem 2 below). This implies that the system $e_i({\bf x})=z_i$ defines a $(n-n=)\ 0$-dimensional variety, that is, a nonempty, finite set of points. Therefore, every point ${\bf z}\in \mathbb{C}^n$ has a preimage $f^{-1}({\bf z})$ (that has cardinality equal to $n!=|S_n|$ in the generic case).

In this way, the affine space $\mathbb{C}^n$ is realised as the quotient $S_n\backslash \mathbb{C}^n$.

This discussion generalizes to all finite (complex) reflection groups. The identification between preimages $f^{-1}({\bf z})$ and orbits $G\cdot {\bf x}$ (actually this part holds for any finite group $G$ acting linearly on $V$) is given in Chapter 13 of Eisenbud's (very beautiful) book Commutative Algebra with a view toward algebraic geometry. Some of the rest of the arguments are encoded in the two theorems below:

Theorem 1 (Invariant theory) If a finite group $G$ acts on a polynomial algebra $\mathbb{C}[V]:=\mathbb{C}[x_1,\cdots,x_n]$, then the subalgebra $\mathbb{C}[V]^{G}$ is generated by finitely many polynomials $f_1,f_2,\cdots, f_m$ (that may however not be algebraically independent).

Theorem 2 (Chevalley-Shephard-Todd theorem) In the case (and only then) that $G$ acts as a (finite) (complex) reflection group, the invariant subalgebra $\mathbb{C}[V]^{G}$ is actually a polynomial subalgebra. That is, the $f_i$'s are algebraically independent.

The reason that these invariant polynomials appear a lot is algebro-geometric: Those are the best candidates for the coordinate ring of the quotient $G\backslash V$ (if that were to be made into a variety). In the case of a finite group $G$ acting linearly on $V$, this happens and is the introduction to Geometric Invariant Theory.

• Correct me if I am wrong, but it seems like you only show that $f$ is a continuous bijection. How did you show that $f^{-1}$ is continuous? I don't think referring to algebraic geometry resolves the issue. Your answer is enough to show that if the quotient is considered scheme-theoretically, it is isomorphic to $\mathbb{A}^{n}$ over $\mathbb{C}$, but I cannot see whether this shows that $\mathbb{C}^{n}$ is homeomorphic to $\mathbb{C}^{n}/S_{n}$ when one considers the quotient topology coming from the Euclidean topology. – Gil Aug 18 '17 at 4:05
• The interesting thing is that in complex space, topology quotient (consider quotient analytic topology on $\mathbb C^n/S^n$) coincides with variety quotient (spec of invariant ring). But this does not work over real numbers: the variety quotient of $\mathbb R^2$ by $\mathbb Z_2$ is still $\mathbb R^2$ (as fundamental theorem of symmetric polynomials works over any ring), versus the topological quotient $\mathbb R^2/\mathbb Z_2$ is half plane with boundary. – Yilong Zhang Jan 7 at 22:31

I think Theo's answer is valid when you think about the quotient as a scheme. However, if you want to think of the quotient as a topological space whose topology is coming from the Euclidean topology of $\mathbb{C}^{n}$, the situation becomes somewhat different.

The hard part is to show that the inverse of $\mathbb{C}^{n}/S_{n} \rightarrow \mathbb{C}^{n}$ that Theo gives above is continuous with the topology I mentioned above. It seems that in many places, this is stated but not proved, but some kind people gathered nice proofs, which you can find here. They seem to be understandable with a first course in complex analysis.

I kept confusing myself by telling me that having an algebraic map between varieties is enough to give a continuous map in the classical Euclidean topology. This does not seem to be the case. However, feel free to let me know if you find a way to resolve this issue algebraically.