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Given a Coxeter system $(W, S)$, we can form its Coxeter graph, and say that the system is irreducible if the graph is connected. Now, irreducibility is not solely a function of $W$; it depends also on $S$. E.g., for $W=D_{12}$ we could have $|S|=2$ or $3$, giving the systems $G_2$ or $A_1 \times A_2$.

Question: For which (other) irreducible $(W,S)$ is there an $S'$ with $(W,S')$ reducible? When this phenomena occurs, does it always change $|S|$ as above? In particular, is it obvious one way or the other for $W=S_n$?

I think this should at least be answerable for finite Coxeter groups since these are completely classified, but I'd also be happy to hear examples in the infinite case (e.g. affine Weyl groups or hyperbolic Coxeter groups). The systems for which it cannot be done would seem to be in some sense "more irreducible," or at least the irreducibility is more purely algebraic than geometric.

Related: Coxeter group generators.

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It's good to add a tag coxeter-groups, since by now this part of group theory has acquired its own identity. Aside from that, I'd recommend looking carefully at the substantial recent literature on the isomorphism problem for Coxeter groups, where this kind of question inevitably comes into play. –  Jim Humphreys Jan 19 '13 at 0:21
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2 Answers

For infinite Coxeter groups, there is a paper of Luis Paris, "Irreducible Coxeter groups", which proves that every infinite Coxeter group cannot be decomposed as a non-trivial direct product.In particular they must be irreducible.

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I had forgotten about this paper, but it's clearly most helpful for the questions raised (and has been used by people working on the isomorphism problem for Coxeter groups). Note that it's posted on arXiv: front.math.ucdavis.edu/0412.5214 but was formally published in Internat. J. Algebra Comput. 17 (2007). I'm not sure which version is actually later, but the arXiv version includes also the list of exceptions for finite Coxeter groups. –  Jim Humphreys Jan 19 '13 at 16:15
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Yes, for $S_n$ the answer is obvious: None of the $S_n$ is decomposable hence there cannot be a reducible Coxeter system for $S_n$.

In general one can do the following construction: $W/W' = C_2^k$ where $k$ is the number of connected components in the graph that is obtained from the Coxeter graph of $(W,S)$ by deleting all edges $s-t$ with $ord(st)\in 2\mathbb{N}\cup\lbrace\infty\rbrace$. In particular $W/W' = C_2$ for all finite Coxeter groups other than $B_n$ and $I_2(m)$ with even $m$. If there was a decomposition $W \cong W(X_1 \times X_2)$ for two nonempty Coxeter diagrams $X_1$ and $X_2$, then we would have $\dim_{\mathbb{F}_2} W/W' \geq 2$. This is a contradiction. Therefore we only need to care about the types $B_n$ and $I_2(m)$ with $m$ even.

If $m$ is even, then $I_2(m)$ is isomorphic to the Coxeter group of $A_1\times I_2(\frac{m}{2})$.

For $W(B_n) = (\mathbb{Z}/2)^n \rtimes S_n$ with uneven $n\geq 3$ the answer is also positive because the natural $\mathbb{F}_2[S_n]$-module $\mathbb{F}_2^n$ is decomposable: $C_2^n \rtimes S_n = X \times (Y \rtimes S_n)$ with $X=\langle(1,1,...)\rangle$ and $Y:=\lbrace x | \sum_i^n x_i = 0\rbrace$. This decomposition is a Coxeter group of type $A_1\times D_n$.

I think the answer should be negative for $W(B_n)$ with even $n\geq 4$ but I cannot think of a short argument at the moment. Here is what I got so far:

If $W(B_n) = N_1\times N_2$ then the projections of $N_i$ must be commuting, normal subgroups of $S_n$ that generate $S_n$ together. Ergo one of them projects onto 1 the other on $S_n$. Therefore we may assume wlog $N_1\leq (\mathbb{Z}/2)^n$ and for every $\sigma\in S_n$ there is a $(v_\sigma, \sigma)\in N_2$. Since $N_1$ is abelian, it acts trivially on itself. $N_2$ acts trivially on $N_1$ by assumption, hence $N_1\subseteq Z(W)=\langle(1,1,...)\rangle=X$.

The length-function of $B_n$ maps $(1,1,...)$ to zero since $n$ is even. The length-function is a non-zero homomorphism $N_1/N_1' \times N_2/N_2' = W/W' \to \mathbb{Z}/2$ and it maps $N_1$ to zero. Because we know $W/W' \cong (\mathbb{Z}/2)^2$ we get $\ker(l)/W' = N_1/N_1'$ and therefore $\ker(l) = N_1 \times N_2'$. Hence $v:=(1,1,0,...)$ lies in $N_1\times N_2'$. Therefore either $v$ itself or $v+(1,1,1,...) = (0,0,1,...)$ lies in $N_2$. The conjugates of both vectors generate $Y:=\lbrace x | \sum_i v_i = 0\rbrace$. Hence $N_1=X\leq Y\leq N_2$ and we finally have a contradiction: $W(B_n)$ is indecomposable if $n$ is even.

EDIT: I totally forgot $F_4$. Well one can show that $W(D_4)$ is indecomposable as a group similar to the above proof: $W:=W(F_4) = W(D_4) \rtimes S_3$ where $S_3$ acts as the group of diagram automorphisms. In particular $s_1, s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4$ are a set of simple reflections for the $D_4$ subgroup and $s_3, s_4$ generate the $S_3$ subgroup.

If $N_1, N_2$ are direct factors of $W$, then wlog $N_2 \twoheadrightarrow S_3$ and $N_1\subseteq C_{N_2}(W) = \langle s_2, s_3 s_2 s_3, s_4 s_3 s_2 s_3 s_4 \rangle \cong C_2^3$ is abelian and hence central. But now we have $N_1\subseteq Z(W) \cap W(D_4) \subseteq Z(W(D_4)) = 1$.

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