Let $S$ be a closed oriented surface. Consider the (universal) abelian cover $p \colon S_{ab} \rightarrow S$, i.e. the one whose group of deck transformations is the abelianization of $\pi_1(S,\star)$. My professor told me that $H^1(S_{ab},\mathbb{R}) = 0$. Even more so, he said the abelian cover is the smallest cover for which the first cohomology group vanishes.


Here is what I have; I would appreciate any feedback as to correctness of the following arguments.

First of all, $H_1(S,\mathbb{Z}) \cong \mathbb{Z}^{2g}$ ($g$ the genus), so that by the universal coefficients theorem $H_1(S,\mathbb{R}) \cong \mathbb{R}^{2g}$. Thus, the kernel of the Hurewicz map $\pi_1(S) \rightarrow H_1(S,\mathbb{Z})$ is the same as the kernel of the map $\pi_1(S) \rightarrow H_1(S,\mathbb{R})$ and, hence, there is no ambiguity as to which map we use (this refers to one discussion in the comments).

Next, the commutator subgroup of $\pi_1(S)$ is free and for genus $g \geq 2$ is not trivial (let us focus on the higher genus case). Therefore, $$H_1(S_{ab},\mathbb{Z}) \cong \pi_1(S_{ab})^{ab} \cong [\pi_1(S),\pi_1(S)]^{ab}$$ is free-abelian (and not trivial). Let us write this group as $\bigoplus_{j \in J}\mathbb{Z}$, where $J \ne \emptyset$. Then $$H_1(S_{ab},\mathbb{R}) \cong H_1(S_{ab},\mathbb{Z}) \otimes \mathbb{R} \cong \bigoplus_{j \in J} \mathbb{R}.$$ Finally, $$H^1(S_{ab},\mathbb{R}) \cong Hom\big(H_1(S_{ab},\mathbb{Z}),\mathbb{R}\big) \cong Hom\Big(\bigoplus_{j \in J} \mathbb{R},\mathbb{R}\Big)$$ and this group is non-trivial. In conclusion, it seems that the proposed statement in the question is not correct as stated.

  • $\begingroup$ Over $\mathbb{R}$, the universal coefficient theorem tells you precisely that you can compute cohomology in terms of homology, at least the first one (you also know $H_0$) $\endgroup$ – Max Mar 18 at 14:51
  • $\begingroup$ You're right that $\pi_1(S_{ab})$ is the commutator subgroup of $\pi_1(S)$, and $H_1(S_{ab})$ is actually the abelianization of $\pi_1(S_{ab})$. If you can show that the commutator subgroup of $\pi_1(S)$ is perfect then you can conclude that $H_1(S_{ab})$ vanishes, and hence also $H^1(S_{ab})$ by the universal coefficient theorem. $\endgroup$ – William Mar 18 at 15:48
  • $\begingroup$ Thank you for your suggestions thus far. @William $\pi_1(S)$ can be presented with $2g$ generators ($g$ the genus) and 1 relation, so its deficiency is $2g-1$. If its commutator subgroup is perfect, then lemma 1.1 here tells us that the deficiency of $\pi_1(S)$ is at most 1. Hence, only for genus $g=0,1$ can the commutator subgroup be perfect. What can we do in the case $g\geq 2$? $\endgroup$ – Florian R Mar 19 at 12:05
  • $\begingroup$ The more I think about it the less sure I am that your professor's statement about cohomology is true. If $H^1(S_{ab}; \mathbb{R}) = 0$ then $H^1(S_{ab};\mathbb{Z})$ is either trivial or a torsion group, but since $S_{ab}$ is a covering space of an oriented surface it also has to be an oriented surface and the constraint on cohomology means that it has to be a sphere or the plane, and both of these have the wrong fundamental group (should be the commutator subgroup of $\pi_1(S)$ which is not trivial in general). Are you sure you have the right statement? (I might just be confused.) $\endgroup$ – William Mar 19 at 13:17
  • $\begingroup$ For example the answer to this question shows the desired abelian cover for $\Sigma_3$, and it appears that the first cohomology for this space should not be $0$. $\endgroup$ – William Mar 19 at 13:18

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