Induced representations and Mackey theory

Let \(G\) be a group. A linear representation of \(G\) is a pair \((\rho, V)\), where \(V\) is a finite dimensional vector space over \(\mathbb{C}\) and \(\rho: G \to GL(V)\) is a group homomorphism. Without ambiguity, we will call \(\rho\) a representation of \(G\). Let \((\rho_1, V_1)\) and \((\rho_2, V_2)\) be two representations of \(G\), an intertwining operator between them is a linear map \(T: V_1 \to V_2\) such that for any \(g \in G\), the following diagram commutes: \[\begin{equation}\label{eq:diagram} \require{AMScd}\begin{CD} V_1 @>T>> V_2 \\ @V{\rho_1(g)}VV @VV{\rho_2(g)}V \\ V_1 @>T>> V_2 \end{CD} \end{equation}\] Let \(\text{Hom}_G(\rho_1, \rho_2)\) be the space of all intertwining operators between \(\rho_1\) and \(\rho_2\).

Let \((\rho, V)\) be a representation of \(G\). Let \(H\) be a subgroup of \(G\), then we can restrict \(\rho\) to \(H\) to get a representation of \(H\). We will use \(\text{Res}^G_H \, \rho\) to denote the restricted representation of \(H\) from \(\rho\). Conversely, if \(\pi\) is a representation of \(H\), then we can construct a representation of \(G\) from \(\pi\), which is known as induced representation of \(G\) from \(\pi\), denoted as \(\text{Ind}_H^G \, \pi\). In this post, I will first talk about the precise description of induced representations and the relations between \(\text{Res}^G_H\) and \(\text{Ind}_H^G\). I will then discuss Mackey’s theorem, which dictates a further relation between restricted and induced representations.

1. Induced Representation

Let \(H\) be a subgroup of a finite group \(G\) and let \((\pi, W)\) be a representation of \(H\). Define \[\text{Ind}_H^G\, W =\{f:G \to W: f(hx)=\pi(h)(f(x)), \forall h \in H\}.\] Then we construct a representation of \(G\) \[\text{Ind}_H^G\, \pi : G \to GL(\text{Ind}_H^G \, W)\] by \(((\text{Ind}_H^G\, \pi)(g)f)(x)=f(xg)\). We call \((\text{Ind}_H^G \, \pi, \text{Ind}_H^G \, W)\) the induced representation of \(G\) from \(\pi\).

Example. Suppose \(H=\{e\}\) is the trivial subgroup of the finite group \(G\). The only irreducible representation of \(H\) is the trivial representation \((\pi, \mathbb{C})\), where \(\pi(e)=Id_\mathbb{C}\). Then \[\text{Ind}_H^G \, \mathbb{C}=\{f:G \to \mathbb{C}\}.\] Since \(G\) is finite, \(\text{Ind}_H^G \, \mathbb{C} \cong \mathbb{C}[G]\), where \(\mathbb{C}[G]\) is the vector space generated by \(\{e_g: g \in G\}\). This is because there is a unique vector \(\sum_{g \in G} f(g)e_g\) attached to \(f \in \text{Ind}_H^G \, \mathbb{C}\). By definition above, G acts on \(\mathbb{C}[G]\) by the obvious right multiplication. Let \(\rho=\text{Ind}_H^G\, \pi\), then \((\rho, \mathbb{C}[G])\) is the right regular representation of \(G\).

To state the relations between \(\text{Ind}_H^G\) and \(\text{Res}^G_H\) better, we first note that they are both functors. Let \(\text{Rep}(G)\) be the category of all finite dimensional linear representations of \(G\), with intertwining operators as morphisms. \(\text{Res}^G_H: \text{Rep}(G) \to \text{Rep}(H)\) is a functor, because the commutative diagram \(\eqref{eq:diagram}\) still holds when restricted to \(H\).

\(\text{Ind}_H^G: \text{Rep}(H) \to \text{Rep}(G)\) is also a functor. This requires some work: we need to know how \(\text{Ind}_H^G\) transfer commutative diagrams in \(\text{Rep}(H)\) to the ones in \(\text{Rep}(G)\), or to understand \(\text{Ind}_H^G T\), where \(T\) is an intertwining operator of representations of \(H\). Let \(T: (\pi_1, V_1) \to (\pi_2, V_2)\) is an intertwining operator of representations of \(H\). Define \(\text{Ind}_H^G\, T: \text{Ind}_H^G\, \pi_1 \to \text{Ind}_H^G\, \pi_2\) by \[(\text{Ind}_H^G\,T\, f)(x)=T(f(x)), \text{for }f \in \text{Ind}_H^G\, V_1.\] It can be checked easily that \(\text{Ind}_H^G\,T\, f \in \text{Ind}_H^G\,V_2\) and \(\text{Ind}_H^G\, T\) is an intertwining operator.

With all above setting up, we can now state the relations between \(\text{Ind}_H^G\) and \(\text{Res}_H^G\) in a nice categorical way:

Instead of giving the proof here, I would rather redirect interested readers to Serre’s book Linear Representations of Finite Groups. He proved the Frobenius Reciprocity(Theorem 13) in Chapter 7 by means of character theory.

2. Mackey Theory

Let \(G\) be a finite group, and let \(H\) and \(K\) be two subgroups of \(G\). Let \((\pi, W)\) be a representation of \(H\). Mackey theory studies how \(\text{Res}^G_K\,\text{Ind}_H^G\, W\) decomposes into irreducible spaces as a representation of \(K\). Suppose \(\lambda\) is an irreducible representation of \(K\), then multiplicity of \(\lambda\) in \(\text{Res}^G_K\,\text{Ind}_H^G\, \pi\) is the same as the dimension of \(\text{Hom}_K(\text{Res}^G_K\,\text{Ind}_H^G\, \pi, \lambda)\) over \(\mathbb{C}\). But by \(\eqref{eq:right-adjoint}\), \[\text{Hom}_K(\text{Res}^G_K\,\text{Ind}_H^G\, \pi, \lambda) \cong \text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda).\] Hence, the multiplicity is also the same as the dimension of \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\).

We then have two different views to work on Mackey thory:

1. Decompose \(\text{Res}^G_K\,\text{Ind}_H^G\, W\) directly.
2. Study the space \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\).

2.1 Decomposition of \(\text{Res}^G_K\,\text{Ind}_H^G\, W\)

Let \(S\) be the set of representatives of \(H\backslash G / K\). For \(s \in S\), let \(H_s = s^{-1}Hs \cap K\). Note that \(H_s\) is a subgroup of \(K\) and \(sH_s s^{-1}\) is a subgroup of \(H\), so we can “restrict” \(\pi\) to \(H_s\) then induce it to \(K\). Define \(\pi^s: H_s \to GL(W)\) by \(\pi^s(x)=\pi(sxs^{-1})\). \(\pi^s\) is then a representation of \(H_s\), let’s call it \((\pi^s, W_s)\). \(\text{Ind}_{H_s}^K\, \pi^s\) is an induced representation of \(K\).

Proof. Let \(f \in \text{Ind}_H^G \, W\). Then \(f=\sum_{s \in S} f_s\), where \(f_s\) is the restriction of \(f\) on \(HsK\) and \(0\) on \(G-HsK\). For each \(s\), we can define \(f^s: K \to W\) by \(f^s(k)=f_s(sk)\). Then for any \(s^{-1}hs \in H_s\), we have \[\begin{align*} f^s(s^{-1}hs\cdot k) &= f_s(hsk) \\ &= \pi(h)f_s(sk) \\ &= \pi^s(s^{-1}hs)f^s(k) \end{align*}\] Hence, \(f^s \in \text{Ind}_{H_s}^K\, W_s\). Thus, \(f \mapsto (f^s)\) is a linear map from \(\text{Ind}_H^G\, W\) to \(\bigoplus_{s \in S} \text{Ind}_{H_s}^K \, W_s\). It’s clear that this map is an injection, because only \(0\) will be sent to \(0\). Since the action of \(K\) is on the right, it’s also easy to see that the map is \(K\)-intertwining. To prove the isomorphism, it then suffices to prove the map is surjective. Given \((f^s)\), we define \(f: G \to W\) by \(f(hsk)=\pi(h)f^s(k)\). Now, an easy calculation can show that \(f\) is well defined and \(f \mapsto (f^s)\). \(\square\)

2.2 Study of \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\)

Let \((\pi, W)\) and \((\lambda, V)\) be representations of \(H\) and \(K\) respectively. Suppose \(\Delta: G \to \text{Hom}_\mathbb{C}(W, V)\) satisfies \[\begin{equation}\label{eq:delta-condition} \Delta(kgh)=\lambda(k)\Delta(g)\pi(h), \text{for }k \in K, g \in G,h \in H. \end{equation}\] Denote \(\mathcal{D}\) the vector space of all these functions satisfying \(\eqref{eq:delta-condition}\).

Let \(f: G \to W\) be a map. We can define the convolution \(\Delta \ast f: G \to V\) by \[(\Delta \ast f)(x)= \frac{1}{|G|}\sum_{g \in G}\Delta(xg^{-1})\left(f(g)\right),\] where \(|G|\) is the cardinality of \(G\). Since \(\Delta\) satisfies \(\eqref{eq:delta-condition}\), if \(f \in \text{Ind}_H^G\, W\), then for \(k \in K\), \[\begin{align*} (\Delta \ast f)(kx) &= \frac{1}{|G|} \sum_{g \in G} \Delta(kxg^{-1})\left(f(g)\right) \\ &= \frac{\lambda(k)}{|G|} \sum_{g \in G} \Delta(xg^{-1})\left(f(g)\right) \\ &= \lambda(k) \circ (\Delta \ast f), \end{align*}\] so \(\Delta \ast f \in \text{Ind}_K^G\, V\). By a change of variable, we can verify that \(g.(\Delta \ast f)=\Delta \ast g.(f)\), where \(g.\) is the right action of \(G\) on the induced representations. This means \(\Delta \ast -\) is an intertwining operator from \(\text{Ind}_H^G\, \pi\) to \(\text{Ind}_K^G\,\lambda\). Let’s denote this intertwining operator \(L_\Delta\).

Proof. We have already seen the map from \(\mathcal{D}\) to \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\). It’s enough to construct the inverse mapping from \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\) to \(\mathcal{D}\).

Let \(T \in \text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\). Suppose \(\Delta \in \mathcal{D}\) and \(T=L_\Delta\). Then \(T\) and \(L_\Delta\) should have the same effects on the basis elements of \(\text{Ind}_H^G\, \pi\). By the definition of induced representation, the basis elements are indexed by cosets of \(H\) in \(G\) and basis of \(W\). Instead of a real basis element, we look at the one indexed by \(g \in G\) and \(w \in W\): \[f_{g, w}(x) = \left\{\begin{array}{ll} \pi(h)w, & \text{if }x=hg \text{ for some }h \in H; \\ 0, & \text{otherwise}. \end{array}\right.\] Then, \[\begin{equation}\begin{aligned} &T(f_{g,w})(x) = L_\Delta(f_{g, w})(x) \\ &= \frac{1}{|G|} \sum_{t \in G} \Delta(xt^{-1})\left(f_{g, w}(t)\right) \\ &= \frac{1}{|G|} \sum_{h \in H} \Delta(xg^{-1}h^{-1})\left(\pi(h)w\right) \\ &= \frac{|H|}{|G|} \Delta(xg^{-1})(w). \end{aligned}\label{eq:calculation}\end{equation}\] From \(\eqref{eq:calculation}\), it’s natural to define \[\Delta(g)(w)=[G:H]T(f_{g^{-1},w})(1).\] Now, using the fact that \(T\) is an intertwining operator, we can easily verify \(\Delta\) satisfies \(\eqref{eq:delta-condition}\), i.e., \(\Delta \in \mathcal{D}\). Hence, we get a map from \(\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda)\) to \(\mathcal{D}\).

It remains to prove that these two maps are inverse to each other. But this is just the repetition of calculation \(\eqref{eq:calculation}\). \(\square\)