Signed Permutations and Coxeter Groups

Question

Context: (most of which is pulled from comments and answers to https://mathoverflow.net/questions/431964/signed-permutations-and-operatornameson)

The diagonal subgroup $ C_2^n $ of $ O_n(\mathbb{Z}) $ is a generated by reflections. The permutation subgroup $ S_n $ of $ O_n(\mathbb{Z}) $ is also generated by reflections. Thus the subgroup of $ O_n(\mathbb{Z}) $ consisting of all signed permutation matrices, also known as the Weyl group $ W(B_n) $, is also generated by reflections.

Viewing these statements abstractly we have that $ C_2^n $ is a Coxeter group, $ S_n $ is a Coxeter group and the group of signed permutation matrices $ W(B_n) $ (which has shape $ 2^{n}:S_n $) is also a Coxeter group.

It is also known that $ W(D_n) $, the index $ 2 $ normal subgroup of $ W(B_n) $ consisting of monomial matrices with an even number of sign changes, is a Coxeter group. It has shape $ 2^{n-1}:S_n $.

What about the subgroup of $ W(B_n) $ of matrices with determinant $ +1 $? We will denote this subgroup of $ SO(n) $ by $ W_n $. This is an a priori different index $ 2 $ normal subgroup of $ W(B_n) $ than $ W(D_n) $.

It is the case that $ W(D_3)=W(A_3)=S_4=W_3 $. Strange that while $ W_2=C_4 $ on the contrary $ W(D_2)=C_2 \times C_2 $. Something similar must happen to $ W_4 $ and $ W(D_4) $ with $ W_4 $ having more elements of order $ 4 $. In general I guess $ W_n $ will have more elements of order $ 4 $ than $ W(D_n) $ (with the exception of $ n=3 $ when the groups coincide).

$ W_n $ has shape $ 2^{n-1}.S_n $. $ W_n $, for $ n \geq 3 $, is generated by involutions using blocks of the form $$ \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$ however such involutions have two $ -1 $ eigenvalues instead of one so they are not reflections in a codimension 1 hyperplane and so it is still hard to say if $ W_n $ is a coxeter group or not.

As previously stated, it is the case that $ C_2^n,S_n,W(B_n),W(D_n) $ are all Coxeter groups. But what about $ W_n $?

Question:

Is it the case that $ W_n \cong W(D_n) $ if and only if $ n $ is odd? And moreover is it the case that $ W_n $ is a Coxeter group if and only if $ n $ is odd?

Answer 1

When $n$ is odd, the permutation module $2^n$ for $S_n$ splits as $2^1 \oplus 2^{n-1}$, with $2^{n-1}$ irreducible. It follows easily from Shapiro's Lemma that $H^k(S_n,2^n) \cong H^k(S_{n-1},2^1)$ for all $k \ge 0$, and so $H^2(S_n,2^{n-1})=0$, so there is no non-split extension of $2^{n-1}$ by $S_n$, and we must have $W(D_n) \cong W_n \cong 2^{n-1}:S_n$.

When $n$ is even, the module $2^n$ is uniserial with factors $2^1$, $2^{n-2}$, $2^1$, and $W(D_n)$ and $W_n$ are both extensions of $2^{n-1}$ by $S_n$, where the $2^{n-1}$ here is uniserial with factors $2^1$ and $2^{n-2}$.

In that case $W_n$ is a nonsplit extension of $2^{n-1}$ by $S_n$, so it is not isomorphic to $W(D_n)$. I think this can be shown by a more complicated cohomological calculation, but I would need to think about it some more.

Answer 2

Just repeating the relevant comments from https://mathoverflow.net/a/450128/20598:

Claim: $W_n$ is a split extension $2^{n-1}:S_n$ (hence isomorphic to $W(D_n)$) if and only if $n$ is odd.

Proof: If $n$ is odd then a splitting map $S_n\to W_n$ is given by sending $\pi$ to $\operatorname{sgn}(\pi) P_\pi$, where $P_\pi$ is the permutation matrix representing $\pi$. Suppose $n$ is even and let $\pi \in S_n$ be an $n$-cycle. The preimages of $\pi$ in $W_n$ are the matrices of the form $g = s P_\pi$, where $s$ is a diagonal matrix with $\pm1$ entries, and we must have $1 = \det g = - \det s$, which means that $s$ must have an odd number of $-1$ entries. But then $g^n = -I$, so $g$ has order $2n$. Since this holds for every preimage of $\pi$, the extension does not split. $\square$

In fact $W_n$ is not a Coxeter group when $n$ is even. There is probably an easier proof of this, but here is a tedious and heavy-handed argument based on the classification of finite Coxeter groups (which I think OP more or less already knows). First observe that if $W_n$ is a Coxeter group then it must be an irreducible one, because $W_n$ is not a nontrivial direct product $A \times B$. Indeed, if $W_n = A \times B$ then one of $A$ or $B$, say $A$, projects trivially onto $S_n$, so $A \le C_2^{n-1}$ and $C_2^{n-1} = A \times (B \cap C_2^{n-1})$. Since $C_2^{n-1}$ is an indecomposable $S_n$-module, it follows that $A$ is trivial or $A = C_2^{n-1}$, and the latter is not possible because $[A, S_n] = 1$. Therefore $W_n$ is a Coxeter group then it appears in the list linked to above. But, it doesn't. The only group in that list with the structure $C_2^{n-1}S_n$ is $W(D_n)$, which is a split extension.