# Induced Representations

There are a couple of different approaches to induced representations that are commonly found in the literature. The one that we will give here is the one that generalizes to topological groups, such a Lie groups. The other, based on the tensor product, is less versatile.

Let $G$ be a finite group, $H$ a subgroup and $(\pi, V)$ a representation of $H$. We will define $V^G$ to be the vector space of all functions $f : G \longrightarrow V$ such that $f (h x) = \pi (h) f (x)$ when $h \in H$ and $x \in G$. Define, for $g \in G$ $(\pi^G (g) f) (x) = f (x g) .$ Thus $g$ acts on $V^G$ by right translation.

Exercise 4.2.1: Check that if $f \in V^G$ and $g \in G$ then $\pi^G (g) f \in V^G$. Also check that $\pi^G (g_1 g_2) = \pi^G (g_1) \pi^G (g_2),$ so that $(\pi^G, V^G)$ is a representation of $G$.

Theorem 4.2.1: (Frobenius reciprocity.) Let $(\pi, V)$ be a representation of $H$ and let $(\sigma, W)$ be a representation of $G$. We have a vector space isomorphism $\operatorname{Hom}_G (W, V^G) \cong \operatorname{Hom}_H (W, V) .$ In this isomorphism the $G$-module homomorphism $\Phi : W \longrightarrow V^G$ corresponds to the $H$-module homomorphism $\phi : W \longrightarrow V$, where we may express $\Phi$ in terms of $\phi$ and $\phi$ in terms of $\Phi$ by the following formulae. $\phi (w) = \Phi (w) (1), \hspace{2em} \Phi (w) (g) = \phi (\sigma (g) w) .$

Proof. (Click to Expand/Collapse)

We first check that if $\Phi : W \longrightarrow V^G$ is a $G$-module homomorphism, then $\phi (w) = \Phi (w) (1)$ defines an $H$-module homomorphism. Indeed, we have, for $h \in H$ $\phi (\sigma (h) w) = \Phi (\sigma (h) w) (1) = (\pi^G (h) \Phi (w)) (1) = \Phi (w) (1 \cdot h) = \Phi (w) (h \cdot 1) = \pi (h) \Phi (w) (1),$ where we have used the definition of $\phi$, the assumption that $\Phi$ is a $G$-module homomorphism, the definition of $\Phi^G$, the identity $1 \cdot h = h \cdot 1$, and the assumption that $\Phi (w) \in V^G$. This equals $\pi (h) \phi (w)$, so $\phi$ is an $H$-module homomorphism.

We leave the reader to complete the proof (Exercise 4.2.2).

Exercise 4.2.2: Complete the above proof as follows.

(a) Show that if $\phi : W \longrightarrow V$ is an $H$-module homomorphism then $\Phi (w) (g) = \phi (\sigma (g) w)$ defines an element of $V^G$, and that $\Phi : W \longrightarrow V^G$ is a $G$-module homomorphism.

(b) Show that the two constructions $\phi \longmapsto \Phi$ and $\Phi \longmapsto \phi$ are inverse maps between $\operatorname{Hom}_G (W, V^G)$ and $\operatorname{Hom}_H (W, V) .$

Theorem 4.2.2: If $\chi$ is the character of the representation $(\pi, V)$ of $H$, then $\chi^G$ is the character of the representation $(\pi^G, V^G)$ of $G$.

Proof. (Click to Expand/Collapse)

If $\chi'$ is the character of $\pi^G$, and if $\theta$ is the character of the representation $(\sigma, W)$ of $G$, then by Proposition 2.4.4 and Theorem 4.2.1 we have $\left\langle \chi', \theta \right\rangle_G = \dim \; \operatorname{Hom}_G (W, V^G) = \dim \; \operatorname{Hom}_H (W, V) = \left\langle \chi, \theta \right\rangle_H .$ By Frobenius reciprocity for characters, we have $\left\langle \chi', \theta \right\rangle_G = \left\langle \chi^G, \theta \right\rangle_G$ for every character $\theta$ of $G$. In particular, if $\theta$ is irreducible, this mean that $\theta$ has the same multiplicity in the decompositions of $\chi'$ and $\chi^G$ into irreducibles, and so they are equal.

Exercise 4.2.3: Suppose that $(\pi, V)$ is a representation of $H$. Let $x \in G$, and let $v \in V$. Let $f_{x, v} (g) = \left\{ \begin{array}{ll} \pi (h) x & \text{if g = h x for some h \in H;}\\ 0 & \text{if g \notin H x.} \end{array} \right.$ Prove that $f_{x, v} \in V^G$.

A monomial matrix is one that has a unique nonzero element in each row and column. If that entry equals $1$, the matrix is called a permutation matrix. Thus if $A = \left(\begin{array}{cccc} 0 & 3 & 0 & 0\\ 0 & 0 & - 1 & 0\\ 0 & 0 & 0 & 1\\ 4 & 0 & 0 & 0 \end{array}\right), \hspace{2em} P = \left(\begin{array}{cccc} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \end{array}\right)$ both are monomial matrices and $P$ is a permutation matrix.

Exercise 4.2.4: Suppose that $\dim (V) = 1$. Fix a nonzero element $v \in V$. Let $x_1, \cdots, x_m$ be a complete set of coset representatives for $H \backslash G$. Show that $f_{x_i, v}$ are a basis of $V^G$. Show that the endomorphism $\pi (g)$ of $V^G$ is a permutation matrix with respect to this basis.