Timeline for $G$-action on the integral homology of a compact surface

Current License: CC BY-SA 3.0

10 events

when toggle format	what		by	license	comment
May 17, 2016 at 10:24	vote	accept	Aurel
May 17, 2016 at 10:24	comment	added	Aurel		I will accept this answer since it shows that the question is difficult and that this module does not always have a simple form. Thanks!
Apr 4, 2016 at 9:41	comment	added	Aurel		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]
Apr 4, 2016 at 9:34	history	edited	Oscar Randal-Williams	CC BY-SA 3.0	added 132 characters in body
Apr 4, 2016 at 9:33	comment	added	Oscar Randal-Williams		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.
Apr 4, 2016 at 9:28	comment	added	Aurel		Right, my mistake. But I still fail to see how this breaks with $(\mathbb{Z}/p\mathbb{Z})^2$ instead of $(\mathbb{Z}/p\mathbb{Z})^3$.
Apr 4, 2016 at 9:26	comment	added	Oscar Randal-Williams		No, I don't think so. By the Kunneth formula $H^2(P;\mathbb{F}_p)$ should have rank 6.
Apr 4, 2016 at 9:21	comment	added	Aurel		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 $H_1 \cong \mathbb{Z}^2\oplus\mathbb{Z}[G]^{2k}$.
Apr 4, 2016 at 9:18	history	edited	Oscar Randal-Williams	CC BY-SA 3.0	added 525 characters in body
Apr 4, 2016 at 8:57	history	answered	Oscar Randal-Williams	CC BY-SA 3.0