Model characters for wreath products with symmetric groups

Let $G$ be a finite group. The model character for $G$ is $\sum_{\chi \in \mathrm{Irr}(G)} \chi$. A nice short paper by Inglis, Richardson and Saxl gives a self-contained inductive proof that if $\pi_{2r}$ is the permutation character of $\mathrm{Sym}_{2r}$ acting by conjugacy on its set of fixed-point-free involutions then

$\sum_r (\pi_{2r} \times \mathrm{sgn}_{n-2r}) \bigl\uparrow^{\mathrm{Sym}_n}$

is the model character for $\mathrm{Sym}_n$. Assuming Pieri’s rule, that if $\alpha$ is a partition of $r$ then $\chi^\alpha \times \mathrm{sgn}_{\mathrm{Sym}_t} \uparrow^{\mathrm{Sym}_{r+t}} = \sum \chi^\lambda$, where the sum is over all partitions $\lambda$ obtained from $\alpha$ by adding $t$ boxes, no two in the same row, this follows from the well-known fact (proved inductively in the paper) that $\pi_{2r} = \sum_{\gamma \in \mathrm{Par}(r)} \chi^{2\gamma}$.

Note that $\pi_{2r} \times \mathrm{sgn}_{n-2r}\bigl\uparrow^{\mathrm{Sym}_n}$ is the induction of a linear character from the centralizer of the involution $(1,2)\ldots (2r-1,2r) \in \mathrm{Sym}_n$. (When $r=0$ we count the identity as an involution as an honorary involution.) Up to conjugacy, each involution is used exactly once to define the model character.

In an interesting paper, Baddeley generalizes the Inglis–Richardson–Saxl result to a larger class of groups. He makes the following definition.

Definition. An involution model for a finite group $G$ is a collection $\{ (\tau_1, \rho_1), \ldots, (\tau_c, \rho_c) \}$ such that $\{\tau_1,\ldots, \tau_c\}$ is a set of conjugacy-class representatives for the involutions of $G$ and $\rho_i : C_G(u_i) \rightarrow \mathbb{C}$ is a linear character for each $i \in \{1,\ldots, c\}$, chosen so that

$\sum_{\chi \in \mathrm{Irr}(G)} \chi = \sum_{i=1}^c \rho_i \bigl\uparrow_{\mathrm{Cent}_G(\tau_i)}^G$

For example, if an abelian group $A$ has an involution model then, since each centralizer is $A$ itself, comparing degrees shows that $c = |A|$, and so $A$ is an elementary abelian 2-group. Conversely, any such group clearly has an involution model. By the Frobenius–Schur count of involutions, a necessary condition for a group to have an involution model is that all its irreducible representations are defined over the reals.

Baddeley’s main theorem is as follows.

Theorem. [Baddeley] If a finite group $H$ has an involution model then so does $H \wr \mathrm{Sym}_n$.

The aim of this post is to sketch my version of Baddeley’s proof of his theorem in the special case when $H = C_2$. Some familiarity with the theory of conjugacy classes and representations of wreath products in Chapter 4 of James & Kerber, Representation theory of the symmetric group is assumed. The characters $\rho_i$ defined below differ from Baddeley’s by a factor of $\mathrm{Inf}_{\mathrm{Sym}_n}^{H \wr \mathrm{Sym}_n} \mathrm{sgn}_{\mathrm{Sym}_n}$; this is done to make $\phi_r$ (defined below) a permutation character, in analogy with the Inglis–Richardson–Saxl character $\pi_r$.

Aside: the Hyperoctahedral group

The group of all $n \times n$ matrices with entries $\pm 1$ that become permutation matrices when all $-1$ entries are changed to $1$ is isomorphic to $H \wr S_n$. Thus $C_2 \wr \mathrm{Sym}_n$ is the hyperoctahedral group of symmetries of the $n$-hypercube. It is a nice exercise to identify $\mathrm{Sym}_4$, the rotational symmetry group of the cube, as an explicit index $2$ subgroup of $H \wr \mathrm{Sym}_3$.

Preliminaries

From now on let $H = \langle h \rangle \cong C_2$. The group $\mathrm{Sym}_n$ acts on $H^{\times n}$ by

$(b_1,\ldots,b_n)^\sigma = (b_{1\sigma^{-1}},\ldots,b_{n\sigma^{-1}}).$

This is a place permutation: the element $b_i$, in position $i$ on the left-hand side, occupies position $b_{i\sigma}$ on the right-hand side. Let $G = H^{\times n} \rtimes \mathrm{Sym}_n \cong H \wr \mathrm{Sym}_n.$

We write elements of $G$ as $(b_1,\ldots,b_n;t)$ where each $b_i \in \{1,h\}$.

Imprimitive action of $G$

For each $i \in \{1,\ldots, n\}$ introduce a formal symbol $\overline{i}$. (This could be thought of as $-i$, but I find that bar makes for a more convenient notation.) Let $\Omega = \{1, \ldots, n, \overline{1}, \ldots, \overline{n} \}$. Given $\sigma \in \mathrm{Sym}_{\{1,\ldots,n\}}$, we define $\overline{\sigma} \in \mathrm{Sym}_{\{\overline{1},\ldots,\overline{n}\}}$ by $i \overline{\sigma} = i$ for all $i \in \{1,\ldots, n\}$ and $\overline{i} \overline{\sigma} = \overline{i\sigma}$ for all $\overline{i} \in \{\overline{1},\ldots,\overline{n} \}$. Then $G$ is isomorphic to the subgroup $G_n \le \mathrm{Sym}_\Omega$ defined by $G_n = B_n \rtimes T_n$ where

$B_n = \langle (1, \overline{1}), \ldots, (n, \overline{n}) \rangle$

and

$T_n = \langle \sigma \overline{\sigma} : \sigma \in \mathrm{Sym}_n \rangle.$

So $G_n$ acts imprimitively on $\Omega$ with blocks $\{1, \overline{1}\}$, $\ldots$, $\{n,\overline{n}\}$.

Irreducible representations of $G$

Let $\epsilon : \{1,h\} \rightarrow \{\pm 1\}$ be the faithful character of $H \cong C_2$ and let $\widetilde{\epsilon}^{\times s}$ denote the linear character of $H \wr \mathrm{Sym}_s$ on which each of the $s$ factors of $H$ in the base group factor acts as $\epsilon$. Given a bipartition $(\lambda|\mu) \in \mathrm{BPar}(n)$ with $\lambda \in \mathrm{Par}(t)$ and $\mu \in \mathrm{Par}(n-t)$, we define

$\chi^{(\lambda|\mu)} = \bigl( \mathrm{Inf}_{\mathrm{Sym}_t}^{H \wr \mathrm{Sym}_t} \chi^\lambda \times \widetilde{\epsilon}^{\times (n-t)} \mathrm{Inf}_{\mathrm{Sym}_{n-t}}^{H \wr \mathrm{Sym}_{n-t}} \chi^\mu \bigr) \bigl\uparrow^{H \wr \mathrm{Sym}_n}.$

Basic Clifford theory shows that the characters $\chi^{(\lambda|\mu)}$ for $(\lambda|\mu) \in \mathrm{BPar}(n)$ form a complete irredundant set of irreducible characters of $G$. For example, the $n$-dimensional representation of $G$ as the symmetry group of the $n$-hypercube has character labelled by the bipartition $\bigl((n-1)|(1)\bigr)$.

Conjugacy classes of involutions in $G$

Since $G_n/B_n \cong T_n \cong \mathrm{Sym}_n$, any involution in $G$ is of the form $(b_1,\ldots,b_n ; \sigma)$ where $\sigma \in \mathrm{Sym}_n$ is an involution. Moreover, as the calculation $(h,1;(1,2))^2 = (h,h)$ suggests, the place permutation action of $\sigma$ on $(b_1,\ldots,b_n)$ permutes amongst themselves the indices $i$ such that $b_i = h$. By applying a suitable place permutation we may assume that $b_1,\ldots b_{n-s} = 1$ and $b_{n-s+1} = \ldots = b_n = h$, for some $s \in \{0,\ldots, n\}$. Now using that $(h,h ; (1,2))$ is conjugate, by $(h,1)$, to $(1,1;(1,2))$, we see that a set of conjugacy class representatives for the involutions in $G$ is

$\{ (\stackrel{n-s}{\overbrace{1, \ldots, 1}} , \stackrel{s}{\overbrace{h, \ldots, h}} ; (1,2) \ldots (2r-1,2r) \}$

for $r$ and $s$ such that $2r+s \le n$. The generalized cycle-type invariant defined in James–Kerber can be used to show no two of these representatives are conjugate.

Centralizers of involutions in $G$

As an element of $G_n$, the involution defined above is

$\tau_s^{(r)} = (n-s+1,\overline{n-s+1}) \ldots (n,\overline{n}) \theta_r\overline{\theta_r}$

where, by definition, $\theta_r = (1,2)\ldots (2r-1,2r) \in T_{2r} \le \mathrm{Sym}_r$. If $(b_1,\ldots,b_n;\sigma)$ commutes with $\tau_s^{(r)}$ then, passing to the quotient, $\sigma$ commutes with $\theta_r$. Therefore the non-singleton orbits

$\{1,2\}, \{\overline{1},\overline{2}\}, \ldots, \{2r-1,2r\}, \{\overline{2r-1},\overline{2r}\}$

of $\tau_s^{(r)}$ are permuted by $(b_1,\ldots,b_n;\sigma)$, as are the remaining non-singleton orbits $\{n-s+1, \overline{n-s+1}\}, \ldots, \{n,\overline{n} \}$.

Therefore

$\mathrm{Cent}_{G_n}(\tau_s^{(r)}) = \mathrm{Cent}_{G_{2r}}(\theta_r\overline{\theta_r}) \times G_{n-(2r+s)} \times G_s$

where $G_{n-(2r+s)}$ acts on $\{2r+1, \ldots, n-s, \overline{2r+1}, \ldots, \overline{n-s}\}$ and $G_s$ acts on $\{n-s+1,\ldots,n,\overline{n-s+1},\ldots,\overline{n}\}$. Clearly $(b_1,b_2,\ldots, b_{2r-1},b_{2r}) \in G_r$ commutes with

$(1,2)\ldots (2r-1,2r) (\overline{1},\overline{2})\ldots(\overline{2r-1},\overline{2r})$

if and only if $b_1 = b_2, \ldots, b_{2r-1} = b_{2r}$. Therefore the first factor is permutation isomorphic to

$D_r \rtimes \mathrm{Cent}_{T_{2r}}(\theta_r\overline{\theta_r})$

where

$D_r = \{(b_1,b_1,\ldots,b_r,b_r) : b_1, \ldots, b_r \in H \}$.

Set $E_r = \mathrm{Cent}_{T_{2r}}(\theta_r\overline{\theta_r})$. Note that $E_r$ is permutation isomorphic to $C_2 \wr \mathrm{Sym}_r$, acting with one orbit on $\{1,\ldots,r\}$ and another on $\{\overline{1},\ldots,\overline{r}\}$. (One has to get used to the two different ways in which the group $C_2 \wr \mathrm{Sym}_r$ arises; in this post I’ve used $H$ when the $C_2$ comes from the base group.)

For example, if $n=7$ then

$\tau_2^{(2)} = (1,2)(\overline{1},\overline{2})(3,4)(\overline{3},\overline{4})(6,\overline{6})(7,\overline{7}) \in G_7$

and the centralizer of $\tau_2^{(2)}$ is generated by $(1, \overline{1})(2, \overline{2})$, $(3,\overline{3})(4,\overline{4})$, $(5, \overline{5})$, $(6,\overline{6})$, $(7,\overline{7})$ in the base group $B_7$ and $(1,2)(\overline{1},\overline{2})$, $(1,3)(2,4)(\overline{1},\overline{3})(\overline{2},\overline{4})$ and $(6,7)(\overline{6},\overline{7})$ in the top group $T_7$. The first two top group generators generate $E_2$.

Definition of the linear representations and reduction

The second and third factors of the centralizer are both complete wreath product, so, by analogy with the Inglis–Richardson–Saxl paper, it is natural guess to define $\rho_s^{(r)}: \mathrm{Cent}_{G_n}(\tau_s^{(r)}) \rightarrow \mathbb{C}$ so that $\rho_s^{(r)}$ restricts to:

• The trivial character on $D_r \rtimes \mathrm{Cent}_{T_{2r}}(\theta_r\overline{\theta_r}) = D_r \rtimes E_r$;
• $\mathrm{Inf}_{\mathrm{Sym}_{n-(2r+s)}}^{H \wr \mathrm{Sym}_{n-(2r+s)}} \mathrm{sgn}_{\mathrm{Sym}_{n-(2r+s)}}$ on $G_{n-(2r+s)}$;
• $\widetilde{\epsilon}^{\times s} \mathrm{Inf}_{\mathrm{Sym}_s}^{H \wr \mathrm{Sym}_s} \mathrm{sgn}_{\mathrm{Sym}_s}$ on $G_s$.

That is (omitting the details of the inflations for brevity),

$\rho_s^{(r)} = 1_{D_r \rtimes E_r} \times \mathrm{Inf} \;\mathrm{sgn}_{\mathrm{Sym}_{n-(2r+s)}} \times \widetilde{\epsilon}^{\times s} \mathrm{Inf}\; \mathrm{sgn}_{\mathrm{Sym}_s}.$

Define

$\phi_r = 1_{D_r \rtimes E_r}\bigl\uparrow^{G_{2r}}.$

Since $\widetilde{\epsilon}^{\times 2r}$ restricts to the trivial character of $D_r \rtimes E_r$, we have $\phi_r = \widetilde{\epsilon}^{\times r}\phi_r$. The definition of $\rho_s^{(r)}$ above is therefore symmetric with respect to $1_H$ and $\epsilon$. Moreover, if

$\phi_r = \sum_{(\alpha|\beta) \in \mathrm{BPar}(2r)} m_{\alpha\beta} \chi^{(\alpha|\beta)}$

then, by Pieri’s rule for the hyperoctahedral group (this follows from Pieri’s rule for the symmetric group in the same way as the hyperoctahedral branching rule follows from the branching rule for the symmetric group — for the latter see Lemma 4.2 in this paper),

$\rho_s^{(r)} \bigl\uparrow^{G_n} = \sum_{(\alpha|\beta) \in \mathrm{BPar}(2r)} \sum_{(\lambda|\mu)} m_{\alpha\beta} \chi^{(\lambda|\mu)}$

where the second sum is over all bipartitions $(\lambda|\mu)$ of $n$ such that $\lambda$ is obtained by adding $n-(2r+s)$ boxes, no two in the same row, to $\alpha$ and $\mu$ is obtained by adding $s$ boxes, again no two in the same row, to $\beta$. Therefore Baddeley’s theorem holds if and only if $\phi_r$ is multiplicity-free, with precisely the right constituents for the Pieri inductions as $s$ varies to give us every character of $H \wr \mathrm{Sym}_n$ exactly once. This is the content of the following proposition.

Proposition. $\phi_r = \sum_{(\gamma|\delta) \in \mathrm{BPar}(r)} \chi^{(2\gamma|2\delta)}$

Proof of the proposition

To avoid some messy notation I offer a ‘proof by example’. I believe it shows all the essential ideas of the general case.

Proof by example. Take $r=3$. We have

$D_3 = \langle (1,\overline{1})(2, \overline{2}), (3,\overline{3})(4,\overline{4}), (5, \overline{5})(6, \overline{6}) \rangle \le B$

and so $\phi_3$ is induced from the trivial character of

$\mathrm{Cent}_{G_3}(\tau_3^{(0)}) = D_3 \rtimes E_3$

(Recall that $\tau_3^{(0)} = \theta_3 \overline{\theta_3}$ where $\theta_3 = (1,2)(3,4)(5,6)$ and $E_3 = \mathrm{Cent}_{T_6}(\theta_3\overline{\theta_3})$.) To apply Clifford theory, it would be much more convenient if we induced from a subgroup of $G_6 = B_6 \rtimes T_6$ containing the full base group $B_6$. We arrange this by first inducing up to $B_6 \rtimes E_3$. (For the action of $E_3$, it is best to think of $B_6$ as $(H \times H)^{\times 3}$.) The calculation

$1_{D_1}\bigl\uparrow^{G_2} = \widetilde{1_H}^{\times 2} + \widetilde{\epsilon}^{\times 2}$

shows that, on restriction to $B_6$, the induced character

$1_{D_3 \rtimes E_3} \bigl\uparrow^{B_6 \rtimes E_3}$

is the sum of all products $\psi_1 \times \psi_2 \times \psi_3$ where each $\psi_i$ is one of the irreducible characters $1_{H \wr C_2} = \widetilde{1_H}^{\times 2}$ or $\eta_{H \wr C_2} = \widetilde{\epsilon}^{\times 2}$ on the right-hand side above. The centralizer $\mathrm{Cent}_{T_6}(\theta_3 \overline{\theta_r})$ acts transitively on the 3 factors: glueing together the products in the same orbits into induced characters we get that $1_{D_3 \rtimes E_3} \uparrow^{B_6 \rtimes E_3}$ has the following irreducible constituents:

• $\widetilde{1_{H \wr C_2}}^{\times 3} 1_{S_3}$
• $\bigl( \widetilde{1_{H \wr C_2}}^{\times 2} 1_{S_2} \times \eta_{H \wr C_2} \bigr) \bigl\uparrow_{(B_4 \rtimes E_2) \times (B_2 \rtimes E_1)}^{B_6 \rtimes E_3}$
• $1_{H \wr C_2} \times \widetilde{\eta_{H \wr C_2}}^{\times 2} 1_{S_2} \bigl\uparrow_{(B_2 \rtimes E_1) \times (B_4 \rtimes E_2)}^{B_6 \rtimes E_3}$
• $\widetilde{\eta_{H \wr C_2}}^{\times 3} 1_{S_3}$.

Note that the ’tilde-construction’ enters in two ways: once to combine characters of each two $H$-factors in the same orbit of $\tau_r$, and then again to combine the characters obtained in this way. As a small check, observe that the sum of degrees is $1 + 3 + 3 + 1 = 8$, which is the index of $D_3 \rtimes E_3$ in $B_6 \rtimes E_3$.

Reflecting the isomorphisms

$(H \wr C_2) \wr S_3 \cong H \wr (C_2 \wr S_3) \cong H \wr E_3 \cong B_6 \rtimes E_3$

we rewrite these characters as follows:

• $\widetilde{1_H}^{\times 6} 1_{E_3}$
• $\widetilde{1_H}^{\times 4} 1_{E_2} \times \widetilde{\epsilon}^{\times 2} 1_{E_1} \uparrow^{B_6 \rtimes E_3}$
• $\widetilde{1_H}^{\times 2} 1_{E_1} \times \widetilde{\epsilon}^{\times 4} 1_{E_2} \uparrow^{B_6 \rtimes E_3}$
• $\widetilde{\epsilon}^{\times 6} 1_{E_3}$.

It is now routine to induce ‘in the top group’ up to

$G_6 = C_2 \wr S_6 = B_6 \rtimes T_6$

using the decomposition of $\pi_r$ into characters labelled by even partitions. For the second summand we use transitivity of induction, starting at the subgroup $(B_4 \rtimes E_2) \times (B_2 \rtimes E_1)$ and going via $(B_4 \rtimes T_4) \times (B_2 \rtimes T_2)$. The third summand is dealt with similarly. Thus $\phi_3$ is the sum of the $\chi^{(\lambda|\mu)}$ for the following bipartitions $(\lambda|\mu)$:

• $\bigl((6)|\varnothing\bigr), \bigl((4,2)|\varnothing\bigr), \bigl((2,2,2)|\varnothing\bigr)$
• $\bigl((4)|(2)\bigr), \bigl((2,2)|(2)\bigr)$
• $\bigl((2)|(4)\bigr), \bigl((2)|(2,2)\bigr)$
• $\bigl(\varnothing|(6)\bigr), \bigl(\varnothing|(4,2)\bigr), \bigl(\varnothing|(2,2,2)\bigr)$.

as required. $\Box$