Permutation bases and an unexpected duality

Wildon's Weblog 2018-08-02

Let $G$ be a finite group acting transitively on a set $\Omega$ . An important object in representation theory, much used by Burnside and his successors to study the action of $G$ , is the permutation module $\mathbb{K}\Omega$ , defined to be the vector space of all formal linear combinations of the elements of $\Omega$ , over a chosen field $\mathbb{K}$ . Thus $\mathbb{K} \Omega$ has $\Omega$ as a canonical basis.

It’s easy to fall into the trap of thinking that the canonical basis of a permutation module is in some sense special, or even unique. The latter is certainly wrong in all but the smallest cases. For example, whenever $\mathbb{K}$ has characteristic zero, or the characteristic of $\mathbb{K}$ does not divide $|\Omega|- 1$ , the linear map defined on the canonical basis by

$\omega \mapsto \sum_{\omega' \in \Omega} \omega'$

is an isomorphism of $\mathbb{K} G$ -modules. Generally, if $G$ has rank $r$ in its action on $\Omega$ then $\mathrm{dim}\ \mathrm{Hom}_{\mathbb{K}S_n}(\mathbb{K}\Omega,\mathbb{F}\Omega) = r$ , and a generic linear combination of these $r$ -linearly independent homomorphisms is an automorphism of $\mathbb{K}\Omega$ sending the canonical basis to a basis of $\mathbb{K}\Omega$ permuted by $G$ .

The motivation for this post is an example where $\mathbb{K}\Omega$ has a permutation basis not obtained in this way. Take $\mathrm{GL}_3(\mathbb{F}_2)$ acting on $E = \langle e_1,e_2,e_3 \rangle_{\mathbb{F}_2}$ . Let $\Omega = \bigl\{ \langle e \rangle : e \in E \backslash \{0 \} \bigr\}$ be the set of lines in $E$ . (Each line has a unique non-zero point; but we use the line notation to makes clear the distinction between $e \in E$ and $\langle e \rangle \in \mathbb{K}\Omega$ .) The canonical basis for $\mathbb{K}\Omega$ is therefore

$\bigl\{ \langle e_1 \rangle, \langle e_2 \rangle, \langle e_3 \rangle, \langle e_2 + e_3 \rangle, \langle e_1 + e_3 \rangle, \langle e_1 + e_2 \rangle, \langle e_1 + e_2 + e_3 \rangle \bigr\}.$

For each plane $\Pi$ contained in $E$ , let $v_\Pi = \sum_{e \in \Pi \backslash \{0\}} \langle e \rangle$ . Clearly $v_\Pi g = v_{\Pi g}$ , for $g \in \mathrm{GL}_3(\mathbb{F}_2)$ , so the $v_\Pi$ are permuted by $\mathrm{GL}_3(\mathbb{F}_2)$ . The duality $\langle e \rangle \mapsto \langle e \rangle^\bot$ gives a bijection between the $7$ lines and $7$ planes. Ordering basis vectors as above, the $v_{\langle e \rangle^\bot}$ for non-zero $e \in E$ have the coefficients shown in the matrix $P$ below.

$\left(\begin{matrix} 0&1&1&1&0&0&0 \\ 1&0&1&0&1&0&0 \\ 1&1&0&0&0&1&0 \\ 1&0&0&1&0&0&1 \\ 0&1&0&0&1&0&1 \\ 0&0&1&0&0&1&1 \\ 0&0&0&1&1&1&0 \end{matrix} \right)$

The determinant of $P$ is $-24$ , so whenever $\mathbb{K}$ has characteristic $0$ or $\mathbb{K}$ has prime characteristic $\ge 5$ , the $v_\Pi$ form a permutation basis for $\mathbb{K} \Omega$ . Since $\mathrm{GL}_3(\mathbb{F}_2)$ acts $2$ -transitively on $\Omega$ , the endomorphism algebra of $\mathbb{K}\Omega$ is $2$ -dimensional, spanned by the identity and the ‘complementing’ map defined above. Therefore this alternative basis is not obtained by an endomorphism of $\mathbb{K}\Omega$ .

Now for the surprising duality. When $\mathbb{K}$ has characteristic $2$ , the three linear relations whose first is

$v_{\langle e_1 \rangle^\perp} + v_{\langle e_1 + e_2 \rangle^\perp} + v_{\langle e_1 + e_3 \rangle^\perp} + v_{\langle e_1 + e_2 + e_3 \rangle^\perp} = 0$

show that $\mathrm{rank}\ P \le 4$ . It is clear from the final four rows of $P$ that $v_{\langle e_2 + e_3\rangle^\perp}$ , $v_{\langle e_1 + e_3\rangle^\perp}$ , $v_{\langle e_1 + e_2 \rangle^\perp}$ , $v_{\langle e_1 + e_2 + e_3 \rangle^\perp}$ are linearly independent and span a $4$ -dimensional submodule of $\mathbb{K} \Omega$ . The sum of these vectors, namely $\langle e_1 \rangle + \cdots + \langle e_1 + e_2+e_3\rangle$ , spans the unique trivial submodule of $\mathbb{K}\Omega$ . Since $7$ is odd, it splits off as a direct summand. Observe from rows $4$ and $7$ of the matrix $P$ that

$\begin{aligned} \langle e_1 \rangle + \langle e_1 + e_2 \rangle + \langle e_1 + e_3 \rangle & + \langle e_1 + e_2 + e_3 \rangle \\ &\quad = v_{\langle e_2+e_3\rangle^\perp} + v_{\langle e_1+e_2+e_3 \rangle^\perp}. \end{aligned}$

A convenient basis for a complement to the trivial submodule is therefore

$\begin{aligned} w_1 &= \langle e_1 \rangle + \langle e_1 + e_2 \rangle + \langle e_1 + e_3 \rangle + \langle e_1 + e_2 + e_3 \rangle \\ w_2 &= \langle e_2 \rangle + \langle e_1 + e_2 \rangle + \langle e_2 + e_3 \rangle + \langle e_1 + e_2 + e_3 \rangle \\ w_3 &= \langle e_3 \rangle + \langle e_1 + e_3 \rangle + \langle e_2 + e_3 \rangle + \langle e_1 + e_2 + e_3 \rangle. \end{aligned}$

We have shown that $W = \langle w_1, w_2, w_3 \rangle_\mathbb{K}$ is a $3$ -dimensional $\mathbb{K}\mathrm{GL}_3(\mathbb{F}_2)$ -submodule of $\mathbb{K}\Omega$ . In the special case $K = \mathbb{F}_2$ , the representation homomorphism $\rho : \mathrm{GL}_3(\mathbb{F}_2) \rightarrow \mathrm{GL(W)}$ is even an isomorphism. Looking at the definitions of $w_1, w_2, w_3$ , it might seem almost obvious that $W$ is isomorphic to the natural representation of $\mathrm{GL}_3(\mathbb{F}_2)$ .

This is not the case. There are two $3$ -dimensional irreducible $\mathbb{F}_2\mathrm{GL}_3(\mathbb{F}_2)$ -modules, namely the natural module $E$ and its dual $E^\star$ . These are distinguished by the action of the matrices in $\mathrm{GL}_3(\mathbb{F}_2)$ of order $7$ . Let

$g = \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right)$

be the companion matrix for the irreducible cubic $1 + X + X^3$ . Computing the action of $g$ on $W$ we find that

$\begin{aligned} w_1 g &= \langle e_1 g \rangle + \langle (e_1 + e_2)g \rangle + \langle (e_1 + e_3) g \rangle + \langle (e_1 + e_2 + e_3)g \rangle \\ &= \langle e_2 \rangle + \langle e_2 + e_3 \rangle + \langle e_1 \rangle + \langle e_1 + e_3 \rangle \\ &= w_1 + w_2. \end{aligned}$

Similar calculations show that $w_2 g = w_3$ and $w_3 g = w_1$ . Reordering basis vectors as $w_2, w_3, w_1$ , we see that in its action on $W$ , $g$ is represented by the matrix

$\left( \begin{matrix} 0 &1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{matrix} \right).$

This is the companion matrix of the irreducible cubic $1 + X^2 + X^3$ , which lies in the other conjugacy class of matrices of order $7$ to $g$ . Therefore $W \cong E^\star$ .

Since $\mathrm{dim}\ \mathrm{End}_{\mathbb{F}_2\mathrm{GL}_3(\mathbb{F}_2)}(\mathbb{F}_2\Omega) = 2$ , it follows that $\mathbb{F}_2\Omega \cong \mathbb{F}_2 \oplus U$ where $U$ is an indecomposable $6$ -dimensional $\mathbb{F}_2\mathrm{GL}_3(\mathbb{F}_2)$ -module with $\mathrm{rad}\ U = \mathrm{soc}\ U \cong E^\star$ and $U / \mathrm{rad}\ U \cong E$ . As expected, the summand $U$ is self-dual, but I think it is a mild surprise that it has $E^\star$ and not $E$ in its socle.