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 $\mathbb{Q}\dots$
1 Answer
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
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$\begingroup$ @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...) $\endgroup$ Commented May 11, 2011 at 18:01
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$\begingroup$ @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. $\endgroup$ Commented May 11, 2011 at 18:49
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$\begingroup$ @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 $n=3$ $\endgroup$ Commented May 11, 2011 at 19:13
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$\begingroup$ @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. $\endgroup$ Commented May 11, 2011 at 19:42
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2$\begingroup$ 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. $\endgroup$ Commented May 11, 2011 at 19:49
$SL(4,\mathbb{Z})$
and related number theory, partly in collaboration with Mark McConnell and my colleague Paul Gunnells. Also, there is an older but possibly helpful ICM 1978 address by K.S. Brown: MR562617 (81e:20058) Cohomology of infinite groups. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 285–290, Acad. Sci. Fennica, Helsinki, 1980, online at mathunion.org/ICM $\endgroup$