Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to find a reference for lower cohomology groups $H^i(G, \mathbb{Z}),$ for $i=1, 2, 3$ for lattices in higher rank (for example, $SL(n, \mathbb{Z}), Sp(2n, \mathbb{Z}),$ and possibly congruence subgroups thereof. There is the Borel periodicity formula, but that's over $\mathbb{Q}\dots$

share|improve this question
2  
Closely related: mathoverflow.net/questions/57183/… –  Theo Buehler May 11 '11 at 0:07
    
@Theo: Thanks! That's very useful! –  Igor Rivin May 11 '11 at 1:16
1  
You doubtless know this, but it may be worthwhile pointing out that Matsushima formula gives a general approach to calculating cohomology over R and that the first integral cohomology group is finite by Property (T). Of course, torsion is not covered by either of these results, but the second cohomology group is known for G(Z) because of the relation to central extensions. This should be in Milnor, Algebraic K-theory. –  Victor Protsak May 11 '11 at 1:18
    
Thanks! I know $H^1$ is finite, as you say (actually, I think both the groups I mean are easily shown to be perfect without prop. T, but that's another matter...) I will look in Milnor... –  Igor Rivin May 11 '11 at 1:25
1  
Igor, as Andy explains below, congruence subgroups need not be perfect. The point of Property (T) is that it gives somewhat crude information about all lattices, not even assumed arithmetic a priori. (By Margulis arithmeticity, they all are in the higher rank case, but the proof uses Property (T).) Likewise, Matsushima formula works for more general congruence subgroups. –  Victor Protsak May 13 '11 at 3:03

2 Answers 2

up vote 8 down vote accepted

I don't think this is known in complete generality. Here are some results. Everything is easy in rank 1, so I'll restrict myself to $SL_n(\mathbb{Z})$ for $n \geq 3$ and $Sp_{2g}(\mathbb{Z})$ for $g \geq 2$. Denote the level $L$ congruence subgroups of $SL_n(\mathbb{Z})$ and $Sp_{2g}(\mathbb{Z})$ by $SL_n(\mathbb{Z},L)$ and $Sp_{2g}(\mathbb{Z},L)$.

First homology

Of course, both $SL_n(\mathbb{Z})$ and $Sp_{2g}(\mathbb{Z})$ are perfect, so $H_1$ of them are trivial. This can be found in many places -- a suitable textbook reference (that definitely includes the symplectic group, which many sources omit) is Hahn-O'Meara's book "The classical groups and K-theory".

For the congruence subgroups, it is proven in

R. Lee and R. H. Szczarba, On the homology and cohomology of congruence subgroups, Invent. Math. 33 (1976), no. 1, 15–53.

that $H_1(SL_n(\mathbb{Z},L)) \cong \mathfrak{sl}_n(\mathbb{Z}/L)$, the abelian group of $n \times n$ matrices over $\mathbb{Z}/L$ with trace $0$. The associated abelian quotient is easy to construct -- an arbitrary element of $SL_n(\mathbb{Z},L)$ is of the form $1+L M$ for some matrix $M$, and the associated homomorphism $\phi : SL_n(\mathbb{Z},L) \rightarrow \mathfrak{sl}_n(\mathbb{Z})$ is defined by $\phi(1+LM)=M$ modulo $L$. This is a homomorphism since $$(1+LM)(1+LN) = 1+L(M+N)+L^2 MN.$$ This was extended to prove that for $L$ odd, $H_1(Sp_{2g}(\mathbb{Z},L)) \cong \mathfrak{sp}_{2g}(\mathbb{Z}/L)$ independently in

B. Perron, Filtration de Johnson et groupe de Torelli modulo p, p premier, C. R. Math. Acad. Sci. Paris 346 (2008), no. 11-12, 667–670.

and

A. Putman, The abelianization of the level L mapping class group, arXiv:0803.0539

and

M. Sato, The abelianization of the level d mapping class group, J Topology (2010) 3 (4): 847-882.

For $L$ even, Sato in the last paper above proves that the abelianization of $Sp_{2g}(\mathbb{Z},L)$ is an extension of $\mathfrak{sp}_{2g}(\mathbb{Z}/L)$ by $H_1(\Sigma_g;\mathbb{Z}/2)$, where $\Sigma_g$ is a genus $g$ surface.

Second homology

Let me now turn to the second homology group. For the special linear group, it follows from Corollary 10.2 and the remark after Theorem 5.10 of Milnor's book on algebraic k-theory that $H_2(SL_n(\mathbb{Z})) \cong \mathbb{Z}/2$ for $n \geq 5$. For $n=3,4$, it is proven in

W. van der Kallen, The Schur multipliers of SL(3,Z) and SL(4,Z), Math. Ann. 212 (1974/75), 47-49.

that $H_2(SL_n(\mathbb{Z})) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2$.

It is also known that for $g$ at least $4$, we have $H_2(Sp_{2g}(\mathbb{Z})) \cong \mathbb{Z}$. This is an old result, but I don't know a great reference for it. Because of this, I gave a (somewhat lame) proof of it in my paper "The Picard group of the moduli space of curves with level structures". See the remark after the proof of Lemma 7.5 of that paper. For $g = 3$, it is proven in

M. Stein, The Schur multipliers of Sp6(Z), Spin8(Z), Spin7(Z), and F4(Z), Math Ann., Volume 215, Number 2, 165-172.

that $H_2(Sp_{2g}(\mathbb{Z})) \cong \mathbb{Z} \oplus \mathbb{Z}/2$.

I don't know any general results for the second homology groups of congruence subgroups. The only calculation I am aware of is in

R. Lee and R. H. Szczarba, On the homology and cohomology of congruence subgroups, Invent. Math. 33 (1976), no. 1, 15–53.

which calculates $H_2(SL_{3}(\mathbb{Z},3))$.

Of course, for $n$ large (I think that $n \geq 4$ should work) Borel's theorem tells you that $H^2(SL_n(\mathbb{Z},L))$ has rank $0$, so you can determine it integrally using the above calculation of $H_1$. A similar remark applies to the symplectic group. However, I'm pretty sure that $H_2$ is not known intergrally for congruence subgroups except in sporadic cases.

Third homology

The only results I am aware of for $H_3$ are the rational calculations that follow from Borel's work and Soule's complete calculation of the cohomology ring of $SL_3(\mathbb{Z})$ in

C. Soule, The cohomology of SL3(Z), Topology 17 (1978) 1–22

share|improve this answer
    
@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...) –  Igor Rivin May 11 '11 at 18:01
    
@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. –  Jim Humphreys May 11 '11 at 18:49
    
@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 $n=3$ –  Igor Rivin May 11 '11 at 19:13
    
@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. –  Andy Putman May 11 '11 at 19:42
1  
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. –  Andy Putman May 11 '11 at 19:49

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.

share|improve this answer
1  
MR0387496 (52 #8338) Borel, Armand Stable real cohomology of arithmetic groups. Ann. Sci. École Norm. Sup. (4) 7 (1974), 235–272 (1975). 22E40 (20G10) –  Igor Rivin May 12 '11 at 18:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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