Re: Symmetrized complex space

On Feb 28, 10:58 pm, p...@xxxxxxxx (Graven Water) wrote:
I wrote:
So C^3/A_3 looks like
 X = { (x,s,t,u) in C^4 : s t = u^3 }
which is nonsingular except at the origin.  Even if you delete the
origin the surface is connected.  I'm not sure what kind of Riemann
domain it is with the origin deleted, whether it's homeomorphic to
C^3.  

I doubt it's a Riemann domain at all, that is no global map to C^3 that's
locally a homeomorphism.  

Laura

Why would there exist such a map? It is a rather unlikely hope...

I don't know what you mean by "Riemann domain". But since
C^3 is simply connected, it has no coverings and, in
consequence, there is no (connected) space Z such that
there exists a map Z --> C^3 which is locally a homeomorphism.
This has nothing to do with anything discussed in this thread.

-- m
.

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