Timeline for Cohomology of lattice subgroups
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2018 at 21:20 | comment | added | Tom | @AndyPutman What about $H_2(Sp_4(\mathbb{Z}))$? | |
| May 12, 2011 at 18:47 | comment | added | Igor Rivin | MR0387496 (52 #8338) Borel, Armand Stable real cohomology of arithmetic groups. Ann. Sci. École Norm. Sup. (4) 7 (1974), 235–272 (1975). 22E40 (20G10) | |
| May 12, 2011 at 17:49 | comment | converted from answer | Kamlesh Parwani | Andy. You mention Borel's Theorem a lot. Can you give a reference or please state the theorem. I have been informed that for n bigger than 8, the second cohomology of a lattice in Sl(n, Z) has trivial rank. I am looking for a reference. | |
| May 12, 2011 at 16:43 | vote | accept | Igor Rivin | ||
| May 12, 2011 at 16:43 | comment | added | Igor Rivin | @Andy: thanks! @Jim: thanks also, I will try to internalize Soule (might take a while...) | |
| May 11, 2011 at 22:19 | comment | added | Jim Humphreys | @Igor: Concerning Soule's paper and some of the other literature, the strategy seems to depend mainly on the action of an arithmetic group on a well-chosen space (inspired here by the work of Borel and Serre). Soule does get precise information about each degree of cohomology in the rank 2 case, but he also has some results in the general case on how to treat the group as an amalgamated product. It's unfortunate that no high-level survey of the whole subject seems to exist. It gets technical but has many concrete aspects. | |
| May 11, 2011 at 19:50 | comment | added | Andy Putman | To determine $H^3$ of the congruence subgroups, you would need to know $H_2$ of them, and I'm pretty sure that this is not known except for small levels and dimensions. | |
| May 11, 2011 at 19:49 | comment | added | Andy Putman | Here's an example of what you do. Let's say you want to know $H^3(SL_n(\mathbb{Z}))$ for $n$ large. Borel's theorem tells you that this has rank $0$ over $\mathbb{Q}$, and I gave a reference above for the fact that $H_2(SL_n(\mathbb{Z})) = \mathbb{Z}/2$. It follows then from the universal coefficients theorem that $H^3(SL_n(\mathbb{Z})) \cong \mathbb{Z}/2$. Using this procedure, I gave you enough to determine $H^i$ for $1 \leq i \leq 3$ for $SL_n(\mathbb{Z})$ and $\Sp_{2g}(\Z)$ and enough to determine $H^i$ for $1 \leq i \leq 2$ for the congruence subgroups. | |
| May 11, 2011 at 19:42 | comment | added | Andy Putman | @Igor : The homology groups determine the cohomology groups completely -- the universal coefficients exact sequence splits (though not in a natural way). They don't tell you anything about the ring structure on cohomology, but since you indicated that you only care about low degrees this probably isn't relevant to you. | |
| May 11, 2011 at 19:13 | comment | added | Igor Rivin | @Jim: yes, I looked a bit at the Soule paper (OK, I lie, at the math review). That is certainly a good start, but I wonder how those results generalize, since (as it says in the math review), SL(3, Z) has very special structure (of amalgamation), and it is conceivable that the results are special to $n=3$ | |
| May 11, 2011 at 18:49 | comment | added | Jim Humphreys | @Igor: Soule's paper describes the cohomology in all degrees for this particular group. There are by now many other related papers, but I'm not sure what is most relevant to your question. Wilberd van der Kallen should comment further, if he is tuned in. | |
| May 11, 2011 at 18:01 | comment | added | Igor Rivin | @Andy: thanks! You speak mostly of homology, not cohomology -- this gives you information about cohomology from the universal coefficient theorem, but does this tell you everything? (I am a bit of a peasant, homologically, so this is probably a really stupid question...) | |
| May 11, 2011 at 16:25 | history | edited | Andy Putman | CC BY-SA 3.0 |
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| May 11, 2011 at 3:11 | history | edited | Andy Putman | CC BY-SA 3.0 |
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| May 11, 2011 at 2:56 | history | edited | Andy Putman | CC BY-SA 3.0 |
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| May 11, 2011 at 1:58 | history | answered | Andy Putman | CC BY-SA 3.0 |