# Irreducible representations of hyperoctahedral group

Consider the hyperoctahedral group $H_n=\mathbb{Z}_2\wr S_n$. This group can be represented by matrices of the form $$\pmatrix{0&0&-1\cr 1&0&0\cr 0&-1&0}$$ that is permutation matrices, where instead of each one, there can be plus or minus one. Formally, $$H_n\simeq\{A\in{\rm GL}(n)\mid Ae_i=\varepsilon_i e_{\sigma(i)}, \sigma\in S_n, \varepsilon_1,\dots,\varepsilon_n\in\{-1,1\}\}.$$

This representation is faithful and irreducible. (Proof of irreducibility: Consider $S_n\subset H_n$. The representation restricted to this subgroup is the representation by permutation matrices, which splits to a direct sum of the trivial representation and the standard representation. It is easy to check that the invariant submodule corresponding to the trivial representation is not invariant with respect to the action of the whole $H_n$.)

I was trying to find out, how to construct a general representation of this group. Following another thread I studied the book of Kerber. He provides a construction of all irreducible representations of $G\wr H$ for general groups $G$ and $H\subset S_n$. If I apply his procedure to our case it means that every irreducible representation is constructed as follows (cf. Section 5):

Denote $\phi_0$ the trivial rep. of $\mathbb{Z}_2$ and $\phi_1$ the non-trivial one. (Since there are just those two, am I right?) Consider $\phi^*:=\phi_{i_1}\boxtimes\dots\boxtimes\phi_{i_n}$ a general irreducible representation of $\mathbb{Z}_2^n$, where $i_1,\dots,i_n\in\{0,1\}$. Now, one should construct a representation $\tilde\phi^*$ of $\mathbb{Z}_2\wr S_n=\mathbb{Z}_2^n\rtimes S_n$ from $\phi^*$. Since $\phi^*$ is one-dimensional, this construction just means that $\tilde\phi^*$ acts identically in the second argument.

Denote $S_{(n)}:=\{\sigma\in S_n\mid i_{\sigma(j)}=i_j\;\hbox{for all$j$}\}$. Choose an irreducible representation $\psi$ of $S_{(n)}\subset S_n\subset H_n$. Then $\phi:=\tilde\phi^*\otimes\psi$ extended to the whole group $H_n$ forms its irreducible representation. Every irreducible representation is of this form.

My problem is the following. Where is the representation described in the beginning in this construction? It seems to me that every representation constructed in this way is unfaithful.

First, a corrected description of the irreducible representations.

You cannot just extend $\phi^*$ to obtain a $1$-dimensional representation of $H_n$.

Exercise: check that the map you are describing is not a group homomorphism.

One way to think of this is that you are trying to "lift $\phi^*$ from a quotient", but $(\mathbb{Z}/2\mathbb{Z})^n$ is not a quotient of $H_n$, since the subgroup $S_n$ is not normal. Having said that, what you can do is extend $\phi^*$ in the way you have described to the subgroup $(\mathbb{Z}/2\mathbb{Z})^n\rtimes S_{\phi^*}$, where $S_{\phi^*}$ is what you called $S_{(n)}$ (I do not like that notation, because the notation should depend on $\phi^*$, it is not just a function of $n$). Why can you do that? Because $S_n$ acts on the set of irreducible representations of $(\mathbb{Z}/2\mathbb{Z})^n$ by $(\sigma\cdot \phi)(g) = \phi(\sigma^{-1}g)$ for $\sigma\in S_n$, $\phi\in {\rm Irr}\,(\mathbb{Z}/2\mathbb{Z})^n$ and $g\in (\mathbb{Z}/2\mathbb{Z})^n$; and the subgroup $S_{\phi^*}$ is precisely the stabiliser of $\phi^*$ under this action. So, vaguely speaking, "from the point of view of $\phi^*$, $S_{\phi^*}$ looks like a normal subgroup" and you can "lift $\phi^*$ from the quotient by that subgroup". Waffle aside, if you have done the above exercise, you will see that stabilising $\phi^*$ under the above action is exactly what you need in order to be able to extend the character to a group homomorphism on this larger group by defining it to be the identity on $S_{\phi^*}$.

Now, what you wrote actually makes sense: if $\psi$ is an irreducible representation of $S_{\phi^*}$, then you can lift it from the quotient $((\mathbb{Z}/2\mathbb{Z})^n\rtimes S_{\phi^*})\big/(\mathbb{Z}/2\mathbb{Z})^n\cong S_{\phi^*}$ to an irreducible representation $\psi$ of $(\mathbb{Z}/2\mathbb{Z})^n\rtimes S_{\phi^*}$, and you can take the tensor product of $\phi^*$ and $\psi$, which is now an irreducible representation of $(\mathbb{Z}/2\mathbb{Z})^n\rtimes S_{\phi^*}$, not of $H_n$.

As a final step, you induce, rather than "extend" that representation to $H_n$.

Exercise: The faithful representation you have described at the beginning is obtained via this construction if you take $i_1=1$ and $i_k=0$ for all other $k$, i.e. $\phi^*$ is non-trivial on the first entry, and trivial on all the others; check that the stabiliser of $\phi^*$ inside $S_n$ is the point stabiliser of $1$, which is isomrphic to $S_{n-1}$; and you take $\psi$ to be the trivial representation of $S_{n-1}$ (recall that the usual representation of $S_n$ in terms of permutation matrices is obtained by inducing the trivial representation of $S_{n-1}$ to $S_n$).

The above description of irreducible representations is a special case of the general description of irreducible representations of semi-direct products by abelian groups, and I believe it is easier to understand in that generality.