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Suppose $\Sigma$ is a Riemann surface of genus $g$ and with $b$ points removed. Is there any restriction on the possible homotopy type that $\Sigma$ can possess? What about the case when $\Sigma$ has no points removed?

user02138
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    A Riemann surface of genus $0$ with empty boundary is already not homotopy equivalent to a bouquet of circles. – Chris Eagle Mar 06 '13 at 16:59
  • Did you think at all about your new version? No compact boundaryless Riemann surface is equivalent to a bouquet of circles, for the same reason as for genus $0$: the 2nd homology groups don't match. – Chris Eagle Mar 06 '13 at 17:21
  • It's true that a torus with $n$ points removed is homotopic to a wedge sum of $n+1$ circles, no? – user02138 Mar 06 '13 at 17:28
  • No, e.g. deleting a point from a sphere gives you something contractible. – Chris Eagle Mar 06 '13 at 17:31

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This is more or less a summary of the answers you already received in the comments. I am quite confused, because you use the term "Riemann surface" and "boundary component", even though Riemann surfaces - being complex manifolds - are assumed to have no boundary. I assume you mean compact, orientable surface, possibly with boundary. The homotopy type of such surface can be completely described.

Let $\Sigma$ be a compact surface of genus $g$ with $b$ boundary components. Note that boundary components are circles, because this is the only compact 1-manifold with boundary. By looking at what happens if one glues disks along the boundary circles one sees that $\Sigma$ has the homotopy type (actually - is homeomorphic/diffeomorphic) to a compact Riemann surface $\Sigma _{g, b}$ of genus $g$ with $b$ points removed.

If $b = 0$, then $\Sigma_{g} = \Sigma _{g, 0}$ is compact and, thus, by Poincare duality, $H_{2}(\Sigma_{g}, \mathbb{Q}) \simeq \mathbb{Q}$. Moreover, $H_{1}(\Sigma_{g}, \mathbb{Q}) \simeq \mathbb{Q}^{2g}$, so such surfaces are classified up to homotopy by their genus, which is a non-negative integer.

If $b > 0$, then $\Sigma_{g, b}$ is non-compact and $H_{2}(\Sigma _{g, b}, \mathbb{Q}) = 0$, so these are all up to homotopy different than the previous case. By examining the "polygon with opposite sides glued model" for $\Sigma _{g}$ one sees - as was remarked in the comments - that $\Sigma_{g, b}$ is homotopy equivalent to $\bigvee _{2g+b-1} S^{1}$, ie. a wedge of circles. (Note that if $2g+b-1 = 0$, which happens only if $g = 0$ and $b = 1$, then $\bigvee _{0} S^{1} = \{ * \}$ by convention, since $\{ * \}$ is the neutral element for wedge sum). Thus, in this case the homotopy type of $\Sigma _{g, b}$ is classified completely by the number $2g-1+b$.

If you actually wanted to ask about all - not necessarily compact - surfaces, then I believe this classification was done up to some extent. By uniformization theorem a Riemann-surface has a universal covering of either $\mathbb{H} ^{+}, \mathbb{C}$ or $S^{2}$, but by looking at the complement of the Cantor set in $\mathbb{R}^{2} \simeq \mathbb{C}$ one sees that the invidual surfaces might still be very difficult to describe.

Piotr Pstrągowski
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