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:

Theorem 1(Frobenius Reciprocity). Let $G$ be a finite group and $H$ is a subgroup of $G$. $\text{Ind}_H^G$ is a left and right adjoint functor to $\text{Res}^G_H$, i.e., $\begin{equation}\label{eq:left-adjoint} \text{Hom}_G(\text{Ind}_H^G \, \pi, \rho) \cong \text{Hom}_H(\pi, \text{Res}^G_H \, \rho) \end{equation},$ and $\begin{equation}\label{eq:right-adjoint} \text{Hom}_G(\rho, \text{Ind}_H^G \, \pi) \cong \text{Hom}_H(\text{Res}^G_H \, \rho, \pi) \end{equation},$

where $\rho \in \text{Rep}(G)$ and $\pi \in \text{Rep}(H)$.

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 $K$.

Theorem 2(Mackey’s Theorem). As representations of $K$, $\text{Res}^G_K\,\text{Ind}_H^G\, W \cong \bigoplus_{s \in S} \text{Ind}_{H_s}^K \, W_s$

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 $L_\Delta$.

Theorem 2$^\prime$(Mackey’s Theorem). As vector spaces, $\text{Hom}_G(\text{Ind}_H^G\, \pi, \text{Ind}_K^G\,\lambda) \cong \mathcal{D}.$ Moreover, given $\Delta \in \mathcal{D}$, $\Delta \mapsto L_\Delta$ is the corresponding intertwining operator.

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 $\eqref{eq:calculation}$. $\square$