Can anyone tell me what $H_3(SL_n(\mathbb{Z});\mathbb{Z})$ and $H_3(SL_n(\mathbb{F}_p);\mathbb{Z})$ are? It is easy to find references for $H_1$ and $H_2$, but it turns out that I need $H_3$ as well. All I care about are the stable values.
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$\begingroup$ I don't know the F_p case. If memory serves, H_3(SL_n(Z)) = Z/24 for n large. I left this as a comment rather than answer both because I am missing the F_p case and because I don't have a reference handy. (I can dig for one if needed.) $\endgroup$– Theo Johnson-FreydCommented Jan 1, 2017 at 4:58
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$\begingroup$ I happen to remember that H_3(SL_2(3)) = Z/24, in case that helps. This number 24 shows up a lot. $\endgroup$– Theo Johnson-FreydCommented Jan 1, 2017 at 4:59
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3$\begingroup$ Quickest reference is probably 4.1.20 in Weibel's K-book (page 299 there): for integers it gives what @Theo says, for a finite field with $q$ elements --- $\mathbb Z/(q^2-1)\mathbb Z$ (stabilization is from $n=3$ on) $\endgroup$– მამუკა ჯიბლაძეCommented Jan 1, 2017 at 5:27
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$\begingroup$ @მამუკაჯიბლაძე I couldn't find this at the location you suggested, although I don't have the published version available. There is an Exercise 1.20 in Chapter IV on page 16 of math.rutgers.edu/~weibel/Kbook.html, which I think would be page 299 of the book. Probably you mean that these answers can be extracted from Chaper IV Corollary 1.20? Do you mind providing more instruction? I have left a CW answer you could edit, or you could leave a complete answer on your own. $\endgroup$– Theo Johnson-FreydCommented Jan 1, 2017 at 22:44
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$\begingroup$ @Theo Right, I had this in mind. It requires quite a lot of previous material though - calculation of $K_2(\mathbb Z)$ (IV.5.2.2), $K_3(\mathbb Z)$ (VI.10) and the multiplication $\{-1,\_,\_\}$ (1.19) and (1.20) as well as $K$-theory of finite fields (1.13) $\endgroup$– მამუკა ჯიბლაძეCommented Jan 2, 2017 at 6:49
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1 Answer
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Summarizing the comments, the stable ($n\geq 3$) values of $H_3(SL_n;\mathbb Z)$ are
- $H_3(SL_\infty(\mathbb Z);\mathbb Z) = \mathbb Z/24$
- $H_3(SL_\infty(\mathbb F_q);\mathbb Z) = \mathbb Z/(q^2-1)$
and can be found in Weibel's The $K$-book.
Namely, one has $$ K_2(R)\xrightarrow{[-1]}K_3(R)\to H_3(E(R))\to0 $$ for any ring (Corollary IV.1.20 there), with $K_2(R)=H_2(E(R);\mathbb Z)$ and $K_3(R)=H_3(St(R);\mathbb Z)$, where $E(R)\subseteq GL(R)$ is the subgroup generated by elementary matrices and $St(R)\twoheadrightarrow E(R)$ is its universal central extension.
For most decent rings, including integers and fields, $E_n=SL_n$; for $R$ the integers, $K_2(\mathbb Z)=H_2(SL_n(\mathbb Z))=\mathbb Z/2$ (Milnor), $K_3(\mathbb Z)=H_3(St_n(\mathbb Z))=\mathbb Z/48$ (Lee & Szczarba 1976), and the map $[-1]$ is nonzero. For $R=\mathbb F_q$, $K_2$ is zero and $K_3$ is $\mathbb Z/(q^2-1)$ (Quillen).
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$\begingroup$ Thank you for the great answer! Do you know of any results of this type that would help compute higher dimensional homology groups of $SL(n,\mathbb{Z})$, for $n$ in the stable range? $\endgroup$– Noah BCommented Jun 28 at 23:27
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$\begingroup$ @NoahB I don't know. If you had asked about $SL(n,\mathbb{F}_q)$, then Quillen's methods tell you everything, if I recall correctly, by comparing $H_*BSL(n,q)$ to $H_*BSU(n)$. I seem to believe that there is a comparison map between $K_*$ and $H_* BSL$, but this comparison gets worse as the degree increases, something like how the Hurewicz comparison between $H_*$ and $\pi_*$ starts out very accurate, and then gets worse and worse. Maybe someone with actual knowledge will chime in. $\endgroup$ Commented Jun 29 at 15:15