Commuting group actions

Wildon's Weblog 2023-04-29

In Schur–Weyl duality one has commuting actions of two algebras $A$ and $B$ on a vector space $V$ such that each is the others centralizer. That is $\mathrm{End}_A(V) = B$ and $A = \mathrm{End}_B(V)$ . In particular the actions of $A$ and $B$ on $V$ commute. Using a left-action for $A$ and a right-action for $B$ , this says that

$(av)b = a(vb)$

for all $a \in A$ , $b \in B$ and $v \in V$ , a fact one would surely expect from the notation. The purpose of this post is to record a lemma about commuting group actions that arose in this context.

Let $G$ and $H$ be finite groups with $G$ acting on the left on $\Omega$ and $H$ acting on the right on $\Omega$ , such the actions commute. For a simple example where $G$ and $H$ may be non-abelian, fix a group $G$ and take $\Omega = H = G$ . Then, for each $x \in \Omega$ , we have

$(gx)h = g(xh)$

by the associativity of group multiplication. The slogan ‘action on places commutes with action on values’ gives another family of examples, typified by $\Omega = \{1,\ldots, b\}^a$ with $G = S_a$ acting on the left on $\Omega$ by position permutation, and $H = S_b$ acting on the right, diagonally on each of the $a$ factors, by permuting values. For instance, if $a = 4$ and $b = 2$ then one joint orbit is

$\{(1,\!1,\!2,\!2),\! (1,\!2,\!1,\!2),\! (2,\!1,\!1,\!2),\! (1,\!2,\!2,\!1), (2,\!1,\!2,\!1), (2,\!2,\!1,\!1 \}.$

The orbit has size $6$ so we expect the stabiliser of points in the orbit ot have order $4! 2! / 6 = 8$ ; a little thought shows that the stabiliser of $(1,1,2,2)$ is

$\langle (1,2)_G, (3,4)_G, (1,3)(2,4)_G (1,2)_H \rangle \cong C_2 \wr C_2.$

(Here group elements are tagged by the group they live in.) Note the ‘diagonally embedded’ element $(1,3)(2,4)_G (1,2)_H$ : even though we are working in $G \times H$ , it is not the case that the stabiliser factors as $\mathrm{Stab}_G(\omega)\mathrm{Stab}_H(\omega)$ .

Since there is no established notation for bimodule induction in the sense I need it, I’m going to have to switch to two commuting right-actions, thought of as a right action of the Cartesian product $G \times H$ , in which, by definition, the two factors commute. Since letters will make it clear which group contains each element, I’ll write $gh$ rather than the formally correct $(g, h)$

Lemma.

Let $G$ and $H$ be finite groups. Suppose that $G \times H$ acts transitively on a set $\Omega$ . Fix $\omega \in \Omega$ .

The orbits of $G$ on $\Omega$ are blocks for the action of $H$ .
Every point in the $G$ -orbit of $\omega$ has the same $H$ -stabiliser.
We have $G\mathrm{Stab}_{G \times H}(\omega) = G\mathrm{Stab}_H(\omega G)$ where $\mathrm{Stab}_H(\omega G)$ is the subgroup of those $h \in H$ fixing setwise the orbit $\omega G$ .
There is an isomorphism of right $\mathbb{C}G$ -modules $\mathbb{C}(\omega G) \cong \mathbb{C}\bigl\uparrow_{\mathrm{Stab}_G(\omega)}^G$ .
There are isomorphisms of right $\mathbb{C}(G \times H)$ -modules $\begin{aligned}\mathbb{C}\Omega &\cong \mathbb{C}\bigl\uparrow_{\mathrm{Stab}_{G \times H} (\omega)}^{G \times H} \\ &\cong \mathbb{C}\bigl\uparrow_{\mathrm{Stab}_G(\omega)}^G \bigl\uparrow_{G\mathrm{Stab}_{G \times H}(\omega)}^{G \times H} \\ &\cong \mathbb{C}\bigl\uparrow_{\mathrm{Stab}_G(\omega)}^G \bigl\uparrow_{G\mathrm{Stab}_{H}(\omega G)}^{G \times H}\end{aligned}$

where we regard $\mathbb{C}\bigl\uparrow_{\mathrm{Stab}_G(\omega)}^G \cong \mathbb{C}(\omega G)$ as a $G \mathrm{Stab}_{G \times H}(\omega)$ -module and as a $G \mathrm{Stab}_H(\omega G)$ -module using (3).

Proof.

More generally, if $N \le K$ is a normal subgroup of a group $K$ then the orbits of $N$ are blocks for the action of $K$ , and for the action of any subgroup of $K$ .
Points in the same orbit have conjugate stabilisers, but here the conjugation action is trivial. Or, to see it very concretely, suppose that $\omega h = \omega$ . Then $(\omega g)h = (\omega h)g = \omega g$ . This shows that $\mathrm{Stab}_H(\omega) \le \mathrm{Stab}_H(\omega g)$ and by symmetry equality holds.
The left-hand side is the set of cosets $Gh$ such that $\omega g h = \omega$ for some $g \in G$ . The right-hand side is the set of cosets $Gh$ such that $\omega h = \omega g'$ for some $g' \in G$ . These are equivalent conditions: take $g' = g^{-1}$ and use that the actions commute.
This is very basic. A nice proof uses the characterisation of induced modules that $V \cong U\uparrow_L^K$ if and only if $V\downarrow_L$ has a submodule isomorphic to $U$ and $\dim V = [K : L]\dim U$ . In this case we take $U$ to to be the right $\mathbb{C}\mathrm{Stab}_G(\omega)$ -module $\langle \omega \rangle \cong \mathbb{C}$ .
The first isomorphism follows as in (4) using that $G \times H$ acts transitively on $\Omega$ . The second follows from the third by (3). Finally the third isomorphism can be proved similarly using the characterisation of induced modules.

To finish, here is one of a family of interesting examples that come from taking two wreath products whose top groups are equal (as abstract groups) but whose base groups are different. To be finished.