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Symmetrized complex space

  1. Feb 26, 2009 #1
    A polynomial [itex]p(z)=z^n+a_{n-1}z^{n-1}+...+a_0[/itex] has n roots [itex]\lambda_1,...,\lambda_n[/itex], and there's a map from the coefficients [itex](a_0,...,a_{n-1})\in C^n[/itex] to [itex](\lambda_1,...,\lambda_n)\in C^n/S_n[/itex], where [itex]S_n[/itex] is the symmetry group on n elements, and [itex]C^n/S_n[/itex] is complex n-space quotiented by permutations on the elements (since it doesn't matter what order the roots are in). [itex]C^n/S_n[/itex] has the quotient topology. This map [itex]C^n\rightarrow C^n/S_n[/itex] is injective because of unique factorization, surjective, and continuous, and it has a continuous inverse.
    Does that mean that [itex]C^n[/itex] is homeomorphic to [itex]C^n/S_n?[/itex] That seems remarkable.
    Does anybody recognize this space [itex]C^n/S_n[/itex], or know how to find out more about it?
    Laura
     
  2. jcsd
  3. Feb 27, 2009 #2
    [tex]\mathbb{C}^2/S_2[/tex] is an example of an http://en.wikipedia.org/wiki/Orbifold" [Broken].

    Here is a geometric explanation (sort of) for why [tex]\mathbb{C}^2/S_2[/tex] is homemorphic to [tex]\mathbb{C}^2[/tex]. 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 [tex]\mathbb{C}^2/S_2[/tex] is homeomorphic to [tex]\mathbb{C}/\{1,-1\}\times\mathbb{C}[/tex]. So the problem is reduced to showing that [tex]\mathbb{C}/\{1,-1\}[/tex] is homeomorphic to [tex]\mathbb{C}[/tex].

    I will denote the real projective line by [tex]\mathbb{R}P_1[/tex].
    If you map an element [tex]z\in\mathbb{C}/\{1,-1\}[/tex] to the pair
    (the line spanned by z, |z|)
    you get a homeomorphism to [tex](\mathbb{R}P_1\times [0,\infty))/(\mathbb{R}P_1\times \{0\})[/tex]. This is basically the half open cylinder [tex]S^1\times[0,\infty)[/tex] with the boundary circle shrunk to a point, i.e. an infinite cone, which is homeomorphic to [tex]\mathbb{C}[/tex].
     
    Last edited by a moderator: May 4, 2017
  4. Feb 27, 2009 #3
    Apparently it's true in general that [itex]C^n/S_n[/itex] is homeomorphic to [itex]C^n[/itex]! 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 [itex]C^n/S_n[/itex].

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

    Laura
     
    Last edited: Feb 27, 2009
  5. Feb 27, 2009 #4
    Here is the link to the paper about [itex]C^n/S_n[/itex] being homeomorphic to [itex]C^n[/itex].
    http://arxiv.org/PS_cache/math/pdf/0502/0502037v1.pdf" [Broken]
    Laura
     
    Last edited by a moderator: May 4, 2017
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