Is C^n Homeomorphic to C^n/S_n in Polynomial Root Mapping?

[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

 

[yyat]

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

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}.

 

[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.

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}.

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

Laura

 

[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"
Laura