16
$\begingroup$

Let $S$ be a compact connected orientable surface, and let $G$ be a nontrivial finite group acting freely on $S$ and preserving orientation (note the the action being free is a strong condition, since automorphisms usually have fixed points). Then $H^1(S)$ also has an action of $G$. I know how to prove, using Riemann-Hurwitz and Artin's induction theorem, that $$ H^1(S,\mathbb{Q}) \cong \mathbb{Q}^2 \oplus \mathbb{Q}[G]^{2k} $$ for some integer $k\ge 0$, where $\mathbb{Q}$ denotes the trivial $\mathbb{Q}[G]$-module.

What is the $\mathbb{Z}[G]$-module $H^1(S,\mathbb{Z})$ ?

I could not figure this out, although it feels like it should be well-known. In particular, when do we have $H^1(S,\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ ?


EDIT: In case it helps, here is a proof over $\mathbb{Q}$:

Let $k = g(M/G)-1$ and $V = \mathbb{Q}^2\oplus\mathbb{Q}[G]^{2k}$, where $g(\cdot)$ denotes the genus. For every subgroup $H\le G$, the covering $M\to M/H$ is unramified, so we have $g(M)-1 = |H|\cdot(g(M/H)-1)$, so that $$ \dim H^1(M,\mathbb{Q})^H = \dim H^1(M/H,\mathbb{Q}) = 2+2\frac{\dim H^1(M,\mathbb{Q})-2}{|H|}\cdot $$ On the other hand, we have $$ \dim V^H = 2+2k[G:H] = 2+2\frac{\dim V-2}{|H|}\cdot $$ By Artin's induction theorem, $\mathbb{Q}[G]$-modules are characterized by the dimensions of the spaces fixed by the subgroups of $G$, so $H^1(M,\mathbb{Q})\cong V$.

$\endgroup$
  • 2
    $\begingroup$ I would start with a model for the quotient surface (either as simplicial complex, delta-complex, CW-complex, simplicial set...). Then the inverse image of each "simplex" is a disjoint union of $|G|$ simplices permuted simply transitively by $G$. This gives you a "triangulation" of $S$, hence the complex of ${\mathbb{Z}}G$-modules giving the (co)homology of $S$ (with $G$-action). I think it is better to work with the complex rather than the cohomology. It is a complex of free ${\mathbb{Z}}G$-modules of respective ranks $1$, $2g(S/G)$, $1$. $\endgroup$ – Daniel Juteau Mar 22 '16 at 9:43
  • $\begingroup$ @DanielJuteau Thanks for the suggestion! In fact, since we know $H_0$ and $H_2$, maybe we should work with the Euler-Poincaré characteristic of the complex as an element of the representation ring of $G$: we can see it in terms of the complex instead of the homology as you suggest, it is used in the proof over $\mathbb{Q}$, and it sees the difference with 3-manifolds (which is good since I know they have many more possible representations). $\endgroup$ – Aurel Mar 22 '16 at 10:19
  • $\begingroup$ @DanielJuteau Also, why are the ranks only $1, 2g(S/G), 1$ and not larger? That seems to say that you can represent $S/G$ with a simplicial complex with only $2g(S/G)$ simplices. $\endgroup$ – Aurel Mar 22 '16 at 10:39
  • $\begingroup$ Hmmm, actually the Euler characteristic of $\mathbb{Z}[G]$-modules is not going to work well... It might not be preserved when passing to the homology. $\endgroup$ – Aurel Mar 22 '16 at 11:01
  • $\begingroup$ The minimal numbers of cells depends on the framework you work with. I think what I said is OK for CW-complexes. $\endgroup$ – Daniel Juteau Mar 22 '16 at 13:21
3
$\begingroup$

I don't see how to answer this is general, but the following partial result might be of interest to you.

Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.

Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence $$S \to S/P \to BP$$ in $\mathbb{F}_p$-cohomology, which has the form $E_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}_p)$.

The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence $$0 \to H^1(P;\mathbb{F}_p) \to H^1(S/P;\mathbb{F}_p) \to H^0(P; H^1(S;\mathbb{F}_p)) \overset{d_2}\to H^2(P;\mathbb{F}_p) \to X \to 0$$ and $X$ injects into $H^2(S/G;\mathbb{F}_p)$. This exact sequence is $$0 \to \mathbb{F}_p^3 \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \overset{d_2}\to \mathbb{F}_p^{6} \to X \to 0$$ so $X \cong \mathbb{F}_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}_p) = \mathbb{F}_p$. QED

The point of this argument is that if $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then (i) the same is true of any subgroup, and (ii) it follows from the spectral sequence argument that there are classes $a,b,c,d \in H^2(G;\mathbb{Z})$ such that $$0 \to H^i(G;\mathbb{Z}) \overset{(a,b)}\to H^{i+2}(G;\mathbb{Z}) \oplus H^{i+2}(G;\mathbb{Z}) \overset{c+d}\to H^{i+4}(G;\mathbb{Z}) \to 0$$ is exact (in fact, with any ring coefficients not just $\mathbb{Z}$). This ought to give many non-examples.

EDIT: In fact, the argument obstructs being $\text{projective} \oplus \mathbb{Z}^2$, not just $\text{free} \oplus \mathbb{Z}^2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, very nice! In your exact sequence, should the $\mathbb{F}_p^6$ be $\mathbb{F}_p$ and $X\cong \mathbb{F}_p^2$ ? Or maybe I made a mistake when recomputing your example. Also, this is quite sharp since for $G = \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ there are examples with $H_1 \cong \mathbb{Z}^2\oplus\mathbb{Z}[G]^{2k}$. $\endgroup$ – Aurel Apr 4 '16 at 9:21
  • $\begingroup$ No, I don't think so. By the Kunneth formula $H^2(P;\mathbb{F}_p)$ should have rank 6. $\endgroup$ – Oscar Randal-Williams Apr 4 '16 at 9:26
  • $\begingroup$ Right, my mistake. But I still fail to see how this breaks with $(\mathbb{Z}/p\mathbb{Z})^2$ instead of $(\mathbb{Z}/p\mathbb{Z})^3$. $\endgroup$ – Aurel Apr 4 '16 at 9:28
  • $\begingroup$ In that case the sequence becomes $0 \to \mathbb{F}_p^2\to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{3} \to X \to 0$ so $X=\mathbb{F}_p$ which is not a contradiction. $\endgroup$ – Oscar Randal-Williams Apr 4 '16 at 9:33
  • $\begingroup$ I see, thanks! Any idea what module $H_1$ could be in this case? Your arguments show that it cannot even be $\mathbb{Z}^2\oplus P$ with $P$ projective since $P$ would be free over $\mathbb{Z}_p[G]$, right? [edit: Ok, this is the content of your last edit] $\endgroup$ – Aurel Apr 4 '16 at 9:41
2
$\begingroup$

This answer complements Oscar Randal-Williams's. First of all, if $G=\mathbb{Z}/p$ acts freely and orientation-preservingly on a closed orientable surface $S$ then $H_{1}(S) =\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$. One way to see this is to put the action into a sort of "normal form". The covering $S\to S/G$ is determined by a classifying map $c:S/G\to BG$, or, equivalently, the corresponding surjective homomorphism $c_{\#}:\pi_{1}(S/G)\to G$. Such a homomorphism can be completely described by listing its values on a standard set of simple closed curves representing a symplectic basis for $H_{1}(S/G)$. One can argue that a system of such curves can be chosen so that all but one curve maps trivially. This result goes back to P.A. Smith, if not further. For details see my old paper Allan L. Edmonds, MR 654478 Surface symmetry. I, Michigan Math. J. 29 (1982), no. 2, 171--183. From this normal form, the main claim is now immediate, since the covering is trivial over all except a torus.

On the other hand if $G=(\mathbb{Z}/p)^{2}$ it can happen that the representation has the desired form and also happen that the representation is different. Again one can understand the covering $S\to S/G$ by putting the classifying homomorphism $c_{\#}:\pi_{1}(S/G)\to G$ into standard form. It turns out in this case that up to equivariant homeomorphism and automorphisms of $G$ there are exactly two such forms with respect to a suitable system of simple closed curves representing a symplectic basis for $H_{1}(S/G)$. If $G$ has generators $x,y$, then the normal forms are

(1) $(x,y;1,1;\dots ;1,1)$

(2) $(x,1; y,1; 1,1;\dots ;1,1)$

Again see my old paper for more details. For Case (1) the corresponding representation on $H_{1}(S)$ again has the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$, since the covering $S\to S/G$ is trivial over all but the core torus.

In Case (2), however, things are different. The two cases are distinguished by $c_{*}[S/G]$ in $H_{2}(BG)=\mathbb{Z}/p$. In Case (1) $c_{*}[S/G]\neq 0$. But in Case (2) $c_{*}[S/G]= 0$, as one can visibly see, by surgering curves to create a cobordism from the map $c$ to a map $S^{2}\to BG$, which is necessarily null-homotopic. In this situation we can adapt Oscar's argument to see that $H_{1}(S)$ cannot be of the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$.

The spectral sequence of the fibration $S\to S/G \to BG$ leads to a five-term exact sequence of integral homology groups $$ H_{2}(S/G)\to H_{2}(BG)\to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to H_{1}(BG)\to 0. $$ Because we are in Case (2), the left hand homomorphism is trivial, and the sequence becomes $$ 0\to \mathbb{Z}/p \to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to (\mathbb{Z}/p)^{2}\to 0. $$ Now if $H_{1}(S)=\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$ we must have $H_{1}(S/G)=\mathbb{Z}^{n+2}$ by consideration of Euler characteristics. Then the exact sequence becomes $$ 0\to \mathbb{Z}/p \to \mathbb{Z}^{n+2}\to \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0, $$ which is impossible. One could also work with coefficients $\mathbb{F}_{p}$ and reach a similar contradiction. One can describe the $\mathbb{Z}[G]$ module $H_{1}(S)$ more or less explicitly in this case, presumably involving augmentation ideals. Note also that in Case (1) this sequence becomes $$ 0\to \mathbb{Z}^{n+2}\to \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0, $$ which does occur.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much for your answer and the reference to your paper! I will need some time to digest all of it, but it looks very interesting. $\endgroup$ – Aurel Apr 29 '16 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.