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In Ahlfor's "Complex Analaysis" it is stated (page 147) that:

"Every region (i.e. an open connected subset of $\Bbb C$) with a finite homology basis has finite connectivity and the number of basis elements is one less than the connectivity"

I have two problems with this statement:

  1. I am not sure what is the meaning to assign to homology basis in this context for Ahlfors only defined it on a region that was assumed to have finite connectivity; my best guess, especially in light of the discussione that followed that definition, is that by finite homology basis of a region $\Omega$ here is meant a finite family $\{\gamma_j\}_{j=1,...,n}$ of cycles such that any other cycle $\gamma \subseteq \Omega$ is homologous modulo $\Omega$ to a unique linear combinations of the basis elements with integer coefficients (I am not sure if it is necessary to add that the index of a basis element with respect to any point outside of $\Omega$ is either $0$ or $1$ because this is the property required for the construction of the basis when we already know that the region has finite connectivity);
  2. I have no idea on how to prove it (I tried something by contradiction but it becomes very difficult very fast because it seems to me that you need to do some long construction as in the previous theorems despite the statement being thrown in the book as if it were trivial).

I'd like to hear if anyone else also thought of this statement and if he/she had better judgement than me.

Matteo Menghini
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  • You should always give definitions in the context of such questions. However the idea here is that for any connected component in the complement you can find a cycle in $\Omega$ such that those cycles are not homologous. Observe that a region with trivial homology that is not equal to $\mathbb{C}$ has a connected complement, in particular admits finite connectivity 1. Thus every region with finite homology basis $\gamma_1,\ldots,\gamma_n$ admits finite connectivity $n+1$. – WhenYouHaveNoClue Apr 07 '23 at 19:53
  • You basically obtain the cycles by pushing the boundary of each connected component of the complement a little inwards to $\Omega$ – WhenYouHaveNoClue Apr 07 '23 at 19:55
  • After thinking about the proof of the statement for a bit, I mentioned that the statement actually is wrong in general. You need $\overline{\Omega}$ to be compact, in particular the complement is only allowed to have only 1 non-compact connected component. Otherwise $\Omega:=\left\lbrace z\in\mathbb{C}\ \big|\ 0<\text{Im}(z)<2\right\rbrace$ is a counterexample, as homology is trivial in dimension 1, but $\Omega$ admits finite connectivity $2$. If $\overline{\Omega}$ is compact, then your statement is correct with the above idea. – WhenYouHaveNoClue Apr 07 '23 at 20:10
  • To give a rigourous proof you will probably have to use the classification of compact surfaces to see that the closure of your region (which is homotopic to your region) is homeomorphic to a $g$-holed torus with $b$ disks removed as it will probably be very hard to proof that your constructed cycles are not homologous otherwise. A wonderfully detailed proof in this case can be found here https://math.stackexchange.com/questions/185903/first-homology-of-a-compact-connected-surface-with-boundary – WhenYouHaveNoClue Apr 07 '23 at 20:27
  • You should spell out the definition of finite connectivity used by Ahlfors. – Moishe Kohan Apr 10 '23 at 01:21
  • @MoisheKohan Ahlfors defines an n-connected region as a region whose complement in the extended complex plane (the one homemorphic to the Riemann sphere) has exactly $n$ connected components (he then usually talks about finite regions, that is regions which do not contain the point at infinity). – Matteo Menghini Apr 10 '23 at 13:06
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    You should add this to the question itself instead of the comment. One way to prove the claim is via the Alexander duality. But there is likely an elementary proof as well. – Moishe Kohan Apr 10 '23 at 13:18

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