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Tits building for an $n$-dimensional vector space $V$ is defined to be the simplicial complex corresponding to the poset of proper and non-zero subspaces of $V$. It is denoted by $T(V)$. This is known to have the homotopy type of $n-2$-spheres. Any choice of linearly independent lines in $V$, define a sphere in $V$. The Steinberg module is defined as $St=H_{n-2}(T(V))$. Often for the choice of linearly independent vectors $v_1,\ldots v_n$ in $V$, we denote the corresponding sphere in the Steinberg module by $[v_1,\ldots ,v_n]$. These symbols satisfy certain properties for example multiplying each vector by a scalar doesn't change it, or switching two vectors multiplies the symbol by $-1$. Especially these symbols generate the Steinberg module. There are some non-trivial relations between these symbols. (We can get rid of them if we choose symbols that correspond to upper-triangular matrices, after a choice of a basis for $V$.)

We choose a subset $S$ of vertices of $T(V)$ i.e. non-zero and proper subspaces of $V$. Let's color the vertices in $S$ by red color. We call an $n-2$ cycle in $T(V)$ to be red iff all of its vertices lie in $S$. We call a homology class of the form $[v_1,\ldots , v_n]$ a red sphere iff all vertices corresponding to it i.e. all subspaces spanned by a subset of $v_i$'s are in $S$. My question is the following:

Is every red cycle a linear combination of red spheres?

The question is equivalent to the following one: Is there a representation for every $n-2$ homology class of $T(V)$ in terms of $[v_1,\ldots , v_n]$, such that none of the vertices of the symbols involved in this representation get cancelled and have an image in the final homology class.

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