Conformal mapping

  • Thread starter esisk
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  • #1
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Main Question or Discussion Point

how do we describe the biholomorphic self maps of the multiply puncture plane onto itself?
I mean C\{pi,p2,p3..pn}

Plane with n points taken away.

I wanted to generailze the result for the conformal self maps of the punctured plane, but I do feel these are quite different animals.
I thank you for any help/suggestions
 

Answers and Replies

  • #2
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The singularities (and infinity) are removable. So you need to map the sphere to itself (az+b)/(cz+d) in such a way that you permute the points [itex]\{p_1,\dots,p_n,\infty\}[/itex].
 
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  • #3
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I thank you edgar,

I think I see it now. So I suspect we get the full symmetric group on n letters then, as the automorphism group. Thank you again
 

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