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The Tits building for a vector space $V$ denoted by $T(V)$ is defined as a simplicial complex whose vertices are non-zero proper sub-vector spaces and edges are inclusion of subspaces and $i$-simplices are flags of non-zero proper subspaces of length $i$. The homotopy type of $T(V)$ is wedge of $n-2$-spheres. Each $n-2$-sphere corresponds to something called an apartment which are choice of $n$ linearly independent lines in $V$. Now consider an integral domain $A$ and the free module $A^n$. We can define something similar to $T(A^n)$ which vertices are non-zero proper free split sub-modules of $A^n$ like $F$ such that $A^n/F$ is also free. More generally $i$-simplices corresponds to flags like $F_1\subset F_2 \cdots \subset F_i$. Such that each of them are vertices and also each inclusion $F_j\subset F_{j+1}$ for $1\leq j \leq i-1$ is a split injection and has the proeprty that $F_{j+1}/F_j$ is also a free $A$-module.

Is the homotopy type of $T(A^n)$ known? Or is it possible to describe the generators of its homology groups combinatorially.

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  • $\begingroup$ I don't know if its true or not. But Quillen in "Finite generation of the groups $K_i$ of rings of algebraic integers" gives an argument for the vector space case which might well generalize easily to your case. $\endgroup$ – Charles Rezk Apr 29 '19 at 15:00
  • $\begingroup$ His argument doesn't work here. The claim that Y is contractible doesn't work. More precisely the condition ii of proposition 3.1.2 in arxiv.org/pdf/1108.2441.pdf. Fails. But there might be another way of proving its contractibility. $\endgroup$ – user127776 Apr 29 '19 at 15:50
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If $A = \mathbb{Z}/p^m$ (not an integral domain for $m\ge2$), then $T(A^n)$ has the homotopy type of a wedge sum of $(n-2)$-spheres. See Proposition 3 in my unpublished note http://folk.uio.no/rognes/papers/Tits.pdf from 1991. The proof uses a "Graph Lemma" which, if I recall correctly, replaced one of Quillen's arguments.

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  • $\begingroup$ Thanks for the reference. I was reading your paper folk.uio.no/rognes/papers/rank_filtration.pdf your definition on the first page (Def 14.5') looks very similar to $T(A^n)$. I think it is the suspension of $T(A^n)$ . Is that correct? or what is its relation with $T(A^n)$? I'm confused as for Euclidean and local $R$ you conjecture that its homologies are in degree $2n-2$. Which I don't get it since $T(A^n)$ is of dimension $n-2$. $\endgroup$ – user127776 May 1 '19 at 23:04
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    $\begingroup$ @user127776 The stable building $D(A^n)$ (denoted $D(R^k)$ in Def. 14.5') receives a map from the suspension of the Tits building $T(A^n)$ (denoted $B(R^k)$ in that paper, but they are not the same for $n\ge2$. $\endgroup$ – John Rognes May 2 '19 at 1:26
  • $\begingroup$ @user127776 The stable building $D(A^n)$ is a spectrum with $GL_n(A)$-action, related to a filtration $\{F_n K(A)\}_n$ of the algebraic $K$-theory spectrum $K(A)$. The homotopy cofiber of $F_{n-1} K(A) \to F_n K(A)$ is equivalent to the homotopy orbit spectrum $D(A^n)_{hGL_n(A)}$. On the other hand, the Tits building is related to a filtration $\{ F_n BK(A)\}_n$ of the first delooping of the $K$-theory space. Quillen's $Q$-construction $QP(A) = BK(A)$ gives one model for that delooping. The homotopy cofiber of $F_{n-1} BK(A) \to F_n BK(A)$ is equivalent to $(\Sigma^2 T(A^n))_{hGL_n(A)}$. $\endgroup$ – John Rognes May 2 '19 at 1:33
  • $\begingroup$ @user127776 For any spectrum $X = \{X_m\}_m$ there is a stable map $\Sigma^{-1} X_1 \to X$. You can interpret $\Sigma^2 T(A^n)$ as the level $m=1$ space in $D(A^n)$. This gives the map from $\Sigma T(A^n) \to D(A^n)$. There is a chain complex concentrated in degrees $n-1 \le * \le 2n-2$ with homology calculating the homology of $D(A^n)$. The lowest (degree $n-1$) part of that chain complex is realized by the homology of $\Sigma T(A)$. Viewed as a $GL_n(A)$-representation it is usually called the Steinberg representation. $\endgroup$ – John Rognes May 2 '19 at 1:38
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    $\begingroup$ @user127776 Maybe the later exposition and references in folk.uio.no/rognes/papers/bergen10.pdf can be useful. $\endgroup$ – John Rognes May 2 '19 at 1:39

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