Timeline for $G$-action on the integral homology of a compact surface
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2016 at 10:24 | vote | accept | Aurel | ||
| Apr 29, 2016 at 14:12 | answer | added | Allan Edmonds | timeline score: 3 | |
| Apr 4, 2016 at 8:57 | answer | added | Oscar Randal-Williams | timeline score: 3 | |
| Apr 2, 2016 at 14:53 | history | edited | Aurel | CC BY-SA 3.0 |
forgot 'connected'
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| Mar 28, 2016 at 17:21 | comment | added | Ben Wieland | Contrary to my prior claim, projective modules over $\mathbb Z_p[G]$ are in fact free. | |
| Mar 27, 2016 at 19:32 | comment | added | Aurel | @BenWieland: thanks for your suggestions. Can you give more details on how you would prove the existence of this projective P? Also, unless I am mistaken, projective over $\mathbb{Z}[G]$ is equivalent to locally free (even though projective over $\mathbb{Z}_p[G]$ is not equivalent to free over $\mathbb{Z}_p[G]$), so in this case there would only be global obstructions left. | |
| Mar 27, 2016 at 19:08 | comment | added | Ben Wieland | This sounds difficult to me. If it is true, I think it should be fairly easy to use homological algebra to prove the weaker statement: $\mathbb Z^2\oplus P$, where $P$ is projective as a $G$-module. There are both local obstructions to projective modules over $\mathbb Z_p[G]$ being free and global obstructions like class groups. A stronger conjecture would be to also take into account the symplectic structure: $(\mathbb Z\oplus\mathbb Z[G]^k)\otimes V$, where $V$ is the standard symplectic structure on $\mathbb Z^2$. | |
| Mar 22, 2016 at 13:21 | comment | added | Daniel Juteau | The minimal numbers of cells depends on the framework you work with. I think what I said is OK for CW-complexes. | |
| Mar 22, 2016 at 11:01 | comment | added | Aurel | 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. | |
| Mar 22, 2016 at 10:39 | comment | added | Aurel | @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. | |
| Mar 22, 2016 at 10:19 | comment | added | Aurel | @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). | |
| Mar 22, 2016 at 9:43 | comment | added | Daniel Juteau | 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 $1$, $2g(S/G)$, $1$. | |
| Mar 21, 2016 at 23:24 | history | edited | Aurel | CC BY-SA 3.0 |
typo in last formula
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| Mar 21, 2016 at 18:07 | history | edited | Aurel | CC BY-SA 3.0 |
added proof over Q
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| Mar 20, 2016 at 22:45 | history | edited | Aurel | CC BY-SA 3.0 |
added orientability hypothesis
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| Mar 20, 2016 at 21:58 | history | asked | Aurel | CC BY-SA 3.0 |