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Is it true that if I take two surfaces that are topologically equivalent, I can find a conformal mapping between them?

Kieran Cooney
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  • No. Consider $\mathbb{C}$ and the unit disk. – Daniel Fischer Oct 08 '13 at 23:19
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    Ok, so is there some other way to characterize an equivalence class of conformally equivalent surfaces? For example, in the topological case we can use the genus. – Kieran Cooney Oct 08 '13 at 23:22
  • According to the wiki article, genus is not sufficient for the topological case. $;$ –  Oct 08 '13 at 23:29
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    Riemann was aware of this problem. In general for a closed surface of genus g there are 6g - 6 parameters for a closed surface and 6g - 6 + n when the surface is open and we have n boundary components. – Alan Oct 08 '13 at 23:35

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The simplest nontrivial case is an open cylinder. Each of these is conformally equivalent to a circular annulus $\{z: r<|z|<R\}$, and two such annuli are conformally equivalent iff they have the same ratio $R/r$. In other words, annuli are classified by one real parameter, conformal modulus.

The case of torus was discussed in conformally equivalent flat tori.

In general, the key term is the moduli space of a surface. As Alan noted in comments, its dimension can be expressed in terms of the genus and the number of boundary components. The formidable [to me] Wikipedia article suggests that the subject is not an easy one to enter, while the MathOverflow discussion Intuition behind moduli space of curves attempts to give some low-tech insights.

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user98130
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