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Fix a finitely generated Coxeter system $(W, S)$, and let $W_J$ denote the standard parabolic proper subgroup generated by a subset $J \subset S$. It is well known that the poset of cosets $\{xW_J\}$ ordered by reverse inclusion is face poset of Coxeter complex.

Consider the poset of cosets $\{xH\}$, ordered by reverse inclusion, $H$ run over proper reflection subgroup of $W$(subgroup generated by a collection of reflections in $W$). This poset may not be face poset of a complex. Do we know any topological property of this poset, is this poset spherical ?

Edit 2021/9/18

This question comes from our consideration about orbit configuration space $$F_G(X,n)= \{(x_1,x_2,...,x_n)\in X^n | Gx_i \cap Gx_j = \emptyset \text{ for } i\neq j\}$$ for a space $X$ with $G$-action, we have calculated the cohomology ring of orbit configuration space of standard $\mathbb{Z}_k^m$-action (a standard $\mathbb{Z}_k^m$-action of $\mathbb{C}^m$ is a map $\varphi : \mathbb{Z}_k^m \times \mathbb{C}^m \rightarrow \mathbb{C}^m$ that $\varphi((z_1,z_2,...), (x_1,x_2,...))=(e^{2\pi iz_1/k}x_1, e^{2\pi iz_2/k}x_2,...)$). The result is beautiful and I will put the paper on arxiv recently.

Furthermore, it seems that I need deep knowledge about reflection group if I want to consider orbit configuration space of general reflection group on $\mathbb{R}^m$ or $\mathbb{C}^m$. The poset in the question appears naturally for two point orbit configuration space, i.e., $n=2$ and observe the intersection lattice (all possible intersection) of these subspaces $x_1=gx_2$, where $g$ runs over elements of a reflection group.

I am not an expert of reflection group, so I wonder may be there is some known results about this question and I haven't found it.

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  • $\begingroup$ What examples have you tried? Does it have nice properties for rank 2 for example? Other small finite examples? $\endgroup$ Commented Sep 17, 2021 at 19:06
  • $\begingroup$ @NathanReading I edited this question again and explain the motivation $\endgroup$
    – J.D.Chern
    Commented Sep 18, 2021 at 2:12
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    $\begingroup$ Thanks for adding this motivation. I'm more asking about which small examples you have tried. There are lots of small examples where you can draw this poset and see what its topology is. For example, does the finite reflection group of type B_2 (the symmetry group of a square) have topological properties that you find interesting? More generally, symmetry groups of other regular polygons, or the symmetric groups, etc. $\endgroup$ Commented Sep 18, 2021 at 15:16

2 Answers 2

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I don't have enough reputation to comment, but this paper of Bux and Welsch seems like it would be relevant (although I don't believe that it completely answers your question): Bux and Welsch - Coset Posets of Infinite Groups.

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I assume that $W$ is finit (not just $S$) and I take parabolic subgroups rather than reflection subgroups. Then the coset poset is indeed Cohen-Macaulay.

In the recent work Cluster Parking Functions, we consider the coset poset associated to the family of noncrossing parabolic subgroups. (A coset with respect to a noncrossing parabolic subgroup is, combinatorially, a parking function.) We prove that this is Cohen-Macaulay, and the same method is good in the present case where we take all parabolic subgroups.

I just prove sphericality. Cohen-Macaulayness follows by 'self-similarity': if you take a strict open interval in your poset, its topology is known by induction on the rank.

1)

Recall that the partial order on cosets is reversed inclusion, so that maximal elements in the coset poset are singletons (cosets for the trivial subgroup). Take a total order on your group: $W = \{w_1,w_2,\dots\}$ (see below for what is the needed technical assumption). Let $Y_i$ be the order complex of the ideal generated by the maximal element $\{w_i\}$, and let $X_i = Y_1 \cup \dots \cup Y_i$. The idea is to prove by induction on $i$ that each $X_i$ is $d$-spherical where $d = \#S - 1$. Note that each $Y_i$ (and in particular $X_1=Y_1$) is topologically trivial because there is a cone point. We use the following:

  • If $X_i$ is $d$-spherical and $X_i \cap Y_{i+1}$ is $(d-1)$-spherical, then $X_{i+1}$ is $d$-spherical.

Indeed, the Mayer-Vietoris long exact sequence gives the homology of $X_{i+1} = X_i \cup Y_{i+1}$, and a pure wedge of spheres is characterized by its homology. The nontrivial hypothesis is in the next point and we conclude by induction.

2)

I claim that $X_i \cap Y_{i+1}$ is $(d-1)$-spherical when $w_1,w_2,\dots$ is a linear extension of the Bruhat-Chevalley order. Note that the order ideal of elements below $\{w_i\}$ in the coset poset identifies with the lattice of strict parabolic subgroups of $W$, by $w_i P \mapsto P$. Recall that the lattice of parabolic subgroup is the dual of a geometric lattice.

Consider a coset $ w_{i+1} P \in X_i \cap Y_{i+1}$ (I keep the same notation either for the order complex or the poset). The condition $ w_{i+1} P \in X_i$ means that $w_{i+1}$ is not a minimal length representative, ie., some reflection $t$ of $P$ is a right inversion of $w_{i+1}$. So $w_{i+1} P \leq \{ w_{i+1} , w_{i+1}t \}$ and $\{ w_{i+1} , w_{i+1}t \} \in X_i \cap Y_{i+1}$. This proves that the maximal elements of $ X_i \cap Y_{i+1}$ are covered by $\{ w_{i+1} \}$. There is a property of geometric lattices that permits to complete this point (see next point).

3)

A technicality that arises and gives the last step of the proof is the following. Let $L$ be a geometric lattice, and $L'$ be its proper part (with minimum and maximum removed). Take any nonempty subset $X$ of the atoms of $L$ (minima of $L'$). Then the order filter generated by $X$ in $L'$ is $d$-spherical, where $d = \operatorname{rank}(L)-2$.

This follows from a construction in poset topology, as in Poset topology: tools and applications by Michelle Wachs. The section about geometric lattices gives the construction of an EL-labelling on $L$, starting from an arbitrary total order on the atoms. Choose the total order so that elements of $X$ come first, and consider the lexicographic shelling coming from this EL-labelling. By construction, the maximal chains going through an element of $X$ come first in the shelling order. So keeping only these chains gives a shelling of the order filter generated by $X$.

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  • $\begingroup$ Doesn't the sphericity of (3) follow from the following easier observation? A set of rank(L) elements in a geometric lattice generates a Boolean lattice, so (passing to the order complex) a subdivision of the boundary of a simplex. $\endgroup$ Commented Aug 25 at 9:49

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