# Symmetrized complex space

1. Feb 26, 2009

### lark

A polynomial $p(z)=z^n+a_{n-1}z^{n-1}+...+a_0$ has n roots $\lambda_1,...,\lambda_n$, and there's a map from the coefficients $(a_0,...,a_{n-1})\in C^n$ to $(\lambda_1,...,\lambda_n)\in C^n/S_n$, where $S_n$ is the symmetry group on n elements, and $C^n/S_n$ is complex n-space quotiented by permutations on the elements (since it doesn't matter what order the roots are in). $C^n/S_n$ has the quotient topology. This map $C^n\rightarrow C^n/S_n$ is injective because of unique factorization, surjective, and continuous, and it has a continuous inverse.
Does that mean that $C^n$ is homeomorphic to $C^n/S_n?$ That seems remarkable.
Does anybody recognize this space $C^n/S_n$, or know how to find out more about it?
Laura

2. Feb 27, 2009

### yyat

$$\mathbb{C}^2/S_2$$ is an example of an http://en.wikipedia.org/wiki/Orbifold" [Broken].

Here is a geometric explanation (sort of) for why $$\mathbb{C}^2/S_2$$ is homemorphic to $$\mathbb{C}^2$$. Under the linear transformation that maps (z,w) to (z-w,z+w), the map (z,w)->(w,z) becomes (z,w)->(-z,w), hence $$\mathbb{C}^2/S_2$$ is homeomorphic to $$\mathbb{C}/\{1,-1\}\times\mathbb{C}$$. So the problem is reduced to showing that $$\mathbb{C}/\{1,-1\}$$ is homeomorphic to $$\mathbb{C}$$.

I will denote the real projective line by $$\mathbb{R}P_1$$.
If you map an element $$z\in\mathbb{C}/\{1,-1\}$$ to the pair
(the line spanned by z, |z|)
you get a homeomorphism to $$(\mathbb{R}P_1\times [0,\infty))/(\mathbb{R}P_1\times \{0\})$$. This is basically the half open cylinder $$S^1\times[0,\infty)$$ with the boundary circle shrunk to a point, i.e. an infinite cone, which is homeomorphic to $$\mathbb{C}$$.

Last edited by a moderator: May 4, 2017
3. Feb 27, 2009

### lark

Apparently it's true in general that $C^n/S_n$ is homeomorphic to $C^n$! I found a paper online with a long, detailed proof, that seems ok as far as I've read, about why the roots of a polynomial are continuous functions of the coeffs.. It is using the quotient topology on $C^n/S_n$.

The map $f:z\rightarrow z^2$ is a homeomorphism from $C^2/\{1,-1\}$ to $C^2.$

Laura

Last edited: Feb 27, 2009
4. Feb 27, 2009

### lark

Here is the link to the paper about $C^n/S_n$ being homeomorphic to $C^n$.
http://arxiv.org/PS_cache/math/pdf/0502/0502037v1.pdf" [Broken]
Laura

Last edited by a moderator: May 4, 2017