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Take $G$ to be a standard semisimple algebraic $\mathbb{Q}$-group, e.g. $Sp_{2g}$ or $SO(h)$ for $h$ a nondegenerate quadratic form over $\mathbb{Q}$. The arithmetic group $\Gamma=G_{\mathbb{Z}}$ has an obvious simplicial action on the associated rational Tits building $\Delta_{G, \mathbb{Q}}$. It is well known (a theorem of Borel) that $G_{\mathbb{Z}}$ acts with finitely many orbits on the vertices of $\Delta_{G, \mathbb{Q}}$, i.e. finiteness of the double cosets $G_\mathbb{Z} \backslash G_\mathbb{Q} /P_\mathbb{Q}$ for $P_\mathbb{Q}$ a rational parabolic. Hence the naive quotient $\Gamma \backslash \Delta$ is some finite complex.

Question: are the combinatorics of the naive quotient $\Gamma \backslash \Delta$ known?

My true motivation: I would like to know more about $\Gamma$-equivariant maps with target $\Delta$. Tits-Solomon theorem says $\Delta$ has the homotopy-type of a countable wedge of spheres $S^d$ (where $d=d(G)$ is known).

Question: can somebody give me a reference or explanation describing the $\Gamma$-homotopy type of $\Delta$? Is the equivariant homotopy type of the associated adele groups known, i.e. $G_\mathbb{A}(\infty)$-homotopy type of $\Delta_{G, \mathbb{A}}$?

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2 Answers 2

$\newcommand\SL{\text{SL}}\newcommand\O{\mathcal{O}}\newcommand\cl{\text{cl}}$For $\text{SL}_n$ over a number field $K$, the rational Tits building is the classical Tits building $T_n(K)$, namely the poset of proper nonzero $K$-subspaces of $K^n$. In fact, starting here we can assume simply that $\O_K$ is a Dedekind domain with fraction field $K$.

For each $d=1,\ldots,n-1$, the orbits of $d$-dimensional subspaces under $\SL_n\O_K$ can be identified with the ideal class group $\widetilde{K}_0(\O_K)$: for each $d$-dimensional $K$-subspace $V$, the intersection $V\cap \O_K^n$ is a rank-$d$ projective $\O_K$-module, and $V$ and $W$ lie in the same $\SL_n\O_K$-orbit iff $V\cap \O_K^n$ and $W\cap \O_K^n$ are isomorphic.

More than this, using the fact that $\O_K$ is a Dedekind domain, you can show that one chain of subspaces is in the same orbit as another iff the ideal classes match up (i.e. iff the $i$th subspaces lie in the same orbit, for each $i$).

This shows that the quotient $T_n(K)/\SL_n\O_K$ is the $(n-1)$-fold join of the discrete set $\widetilde{K}_0(\O_K)$ with itself. This is an $\ell^{n-1}$-fold wedge of $(n-2)$-spheres, where $\ell=\lvert \widetilde{K}_0(\O_K)\rvert-1$, and indeed a spherical building of type $A_1\times \cdots \times A_1$: a chamber is determined by a sequence $(c_1,\ldots,c_{n-1})$ of ideal classes $c_i\in\widetilde{K}_0(\O_K)$. If $\O_K$ is a PID (i.e. $\lvert\widetilde{K}_0(\O_K)\rvert=1$), this tells you that the quotient $T_n(K)/\SL_n\O_k$ is contractible.

All of this so far is elementary, but the following theorem is not. It is joint with Benson Farb and Andrew Putman and will appear in our forthcoming paper "Modular symbols, class numbers, and the top-dimensional cohomology of $\SL_n\O$" (which we hope to post to the arXiv in the next few weeks).

Theorem: The projection map $T_n(K)\twoheadrightarrow T_n(K)/\SL_n\O_K$ is surjective on homology.

An even stronger theorem is true: for every apartment in the quotient $T_n(K)/\SL_n\O_K$, there is an apartment in $T_n(K)$ whose fundamental class hits its homology class, on the nose.

To get a feeling for the strength of this result, it's a fun game to pick your favorite Dedekind domain with nontrivial class group, pick $2n-2$ ideal classes to determine an apartment in the quotient, and then try to find an apartment in $T_n(K)$ that hits it. (Warning: you probably won't be able to do this, except for $n=2$ or $n=3$! --- at least we weren't able to for a while. But the theorem we prove says that you always can.)

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Very impressive! I think a similar argument could be used for the symplectic group, using projective modules with alternating forms. Does the same argument also show that the quotient is more complicated for the orthogonal group? –  Matthias Wendt Jun 15 at 16:45
    
I look forward to reading your paper (and of course seeing what you et al. have to say about homological stability for $SL_n Z$). While I'm sure your paper will clarify, what homology are you considering on the quotient $T_n / SL_n \mathcal{O}$? Are you considering a cellular homology (with $Z$ coefficients) on the naive quotient, or is your theorem stating surjectivity in rational homology (i.e. finite reducible subgroups in $SL_n \mathcal{O}$ prevent the action on $T_n$ from being free). –  J. Martel Jun 29 at 14:46

This is more an extended comment than an answer.

First, in the simplest case $SL_2$, the group $SL_2\mathbb{Z}$ acts transitively on $\mathbb{P}^1(\mathbb{Q})$, viewed as the boundary of the upper half plane in the Borel-Serre compactification. The stabilizer subgroup is the intersection of the Borel subgroup of $SL_2\mathbb{Q}$ with $SL_2\mathbb{Z}$. Note, however, that this is something non-trivial, it depends on the fact that $\mathbb{Z}$ has trivial class group. There are similar statements for other number fields, where you can identify the number of cusps of the action of $SL_2\mathcal{O}_K$ on the symmetric space with the class group of $\mathcal{O}_K$. I think statements like this go back to Siegel.

Second, I can answer your question in the function field case. If $G$ is a split semisimple algebraic group over $\mathbb{F}_q(T)$, and we let $\Gamma=G(\mathbb{F}_q[T])$ act on the spherical building $\Delta_{G,\mathbb{F}_q(T)}$ associated to $G(\mathbb{F}_q(T))$, then the quotient $\Gamma\backslash\Delta_{G,\mathbb{F}_q(T)}$ is a single chamber. The stabilizer of the whole chamber is the intersection of the Borel of $G(\mathbb{F}_q(T))$ with $G(\mathbb{F}_q[T])$. Similarly, all the other faces of the chamber are stabilized by the various intersections of parabolic subgroups of $G(\mathbb{F}_q(T))$ with $G(\mathbb{F}_q[T])$. This follows from a theorem of Soulé (C. Soulé. Chevalley groups over polynomial rings. in: Homological group theory. London Math. Soc. Lecture Notes 36, Cambridge University Press (1979) 359--367). In this paper, he computes the quotient of the euclidean building associated to the valuation for the infinite place of $\mathbb{F}_q(T)$ modulo the action of $G(\mathbb{F}_q[T])$. But since the spherical building $\Delta_{G,\mathbb{F}_q(T)}$ can be identified with the boundary of the euclidean building, this also produces a computation of the quotient of the spherical building, as described above.

Taking these two observations together, I would expect (but I do not know of a reference or argument) that for $Sp_{2g}$ the quotient you are asking for is again a single chamber in the spherical building. I am not so sure about the orthogonal group case: the classification of anisotropic forms over $\mathbb{Z}$ is fairly complicated, but rather easy over $\mathbb{F}_q(T)$. This might lead to different statements in the orthogonal group case.

[Edit: I thought a bit more about this, and there is a bit more to say about the expectation voiced above. First, we can look at the analogous situation for $SL_n\mathbb{Z}$. First, the upper half-plane retracts onto Serre's tree, on which $SL_2\mathbb{Z}$ acts with fundamental domain an interval. This is another way to see that the action on the boundary, the spherical building, is transitive. In a similar vein, Soulé has computed the quotient of the symmetric space modulo $SL_3\mathbb{Z}$ (C. Soulé. The cohomology of $SL(3,\mathbb{Z})$. Topology 17 (1978), 1-22) by first retracting onto a subspace on which the group acts compactly. More recently, these methods have been used to compute the cohomology of $SL_n\mathbb{Z}$ in some range, cf. the paper of Elbaz-Vincent, Gangl and Soulé, arXiv:1001.0789. The method generally is to retract onto a subcomplex determined by perfect forms, on which the group acts with compact quotient. Looking at the concrete quotient computations in these low degrees, it is probably possible to actually compute the simplicial structure of the quotient of the spherical building.

Ok, but this is only for $SL_n\mathbb{Z}$ and certainly does not answer your question about symplectic or orthogonal groups. The case of $Sp_4\mathbb{Z}$ acting on the associated symmetric space is discussed in (Brownstein, Lee: Cohomology of the symplectic group $Sp_4\mathbb{Z}$. Part I: the odd torsion case. Transactions Amer. Math. Soc. 334 (1992), 575-596). They discuss the action, and in particular show that the symmetric space equivariantly decomposes into two components. I think this can be worked out to show that the quotient of the spherical building modulo $Sp_4\mathbb{Z}$ is a single chamber. I am however unaware of any literature dealing with higher rank symplectic groups.]

As a final comment (sorry for asking new questions instead of answering the old ones first) it would be very interesting to know if there are statements like this not just for $\mathbb{Z}$ but for more general number rings. The answer most likely depends on a lot of number-theoretic data, starting (but not ending) with the class groups of the number ring.

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I find your comment informative, thanks. However I do not see (and don't believe) the existence of a $G_Z$-equivariant cocompact retract forcing equivariant contractibility at-infinity. Not even for a finite-index torsion free (and possibly neat) $\Gamma < G_Z$. What argument do you have in mind for this point? –  J. Martel Jun 15 at 12:50
    
@J.Martel: I wanted to do something with the relative homology long exact sequence, but looking at this argument again I realized that it did not work. So consider this more wishful thinking than an actual fact. I removed the doubtful sentence from the answer. –  Matthias Wendt Jun 15 at 14:54
    
@J.Martel, ctd: In the case of $SL_2\mathbb{Z}$ and $SL_3\mathbb{Z}$, the compact quotient is contractible, and the retraction done before is something like the compact quotient times infinite half-space. So in those cases, the quotient of the boundary should indeed be contractible. However, I do not know where the rational cohomology in the higher rank cases comes from. Maybe it is possible to use the Borel-Moore homology exact sequence to see if some of the rational cohomology actually has to be supported on the boundary? –  Matthias Wendt Jun 15 at 14:59

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