# Mackey Theory

Mackey theory asks the following question: if $H_1$ and $H_2$ are subgroups of $G$ and $\psi_1$ and $\psi_2$ are characters, then what is $\left\langle \psi_1^G, \psi_2^G \right\rangle$? This may seem like a technical question, and indeed many accounts of Mackey theory may not do much to dispel this impression. However Mackey's theorem is extremely important and useful, and properly understood it has a conceptual basis, which we hope to convey.

For simplicity, we will limit ourselves to the special case where $\psi_1$ and $\psi_2$ are linear characters, which makes for a minor simplification, and is already enough for some important examples.

We go back to the viewpoint that $\mathbb{C}[G]$ is the ring (under convolution) of complex valued functions on $G$. We recall the right regular representation $\rho : G \longrightarrow \operatorname{End} (\mathbb{C}[G])$, which is the action $(\rho (g) f) (x) = f (x g)$.

Lemma 4.3.1: Let $T : \mathbb{C}[G] \longrightarrow \mathbb{C}[G]$ be a linear transformation that commutes with $\rho (g)$; that is, $T (\rho (g) f) = \rho (g) T (f)$. Then there exists a unique $\lambda \in \mathbb{C}[G]$ such that $T (f) = \lambda \ast f$.

Proof. (Click to Expand/Collapse)

Define $\delta_0 (g) = 1$ if $g = 1$, and $0$ if $g \neq 1$. Then $\delta_0$ is the unit in the convolution ring $\mathbb{C}[G]$, that is, $\delta_0 \ast f = f \ast \delta_0 = f$ for all $f \in \mathbb{C}[G]$. If $\lambda$ exists such that $T (f) = \lambda \ast f$ for all $f$, then $\lambda = \lambda \ast \delta_0 = T (\delta_0)$. Hence it is unique, and it remains to be shown that $\lambda = T (\delta_0)$ works. We claim that any $f \in \mathbb{C}[G]$ can be written as
 $f = \sum_{g \in G} f (g) \rho (g^{- 1}) \delta_0 .$ (4.3.1)

Indeed, applying the right-hand side to $x \in G$ gives $\sum_{g \in G} f (g) (\rho (g^{- 1}) \delta_0) (x) = \sum_{g \in G} f (g) \delta_0 (x g^{- 1}) .$ Only one term contributes, which is $g = x$, and that term equals $f (x)$. This proves (4.3.1).

Now applying $T$ to (4.3.1) gives $T f = \sum_{g \in G} f (g) T \left( \rho (g^{- 1}) \delta_0 \right) = \sum_{g \in G} f (g) \rho (g^{- 1}) T \left( \delta_0 \right) = \sum_{g \in G} f (g) \rho (g^{- 1}) \lambda .$ Thus $T f (x) = \sum_g (\rho (g^{- 1}) \lambda) (x) f (g) = \sum_g \lambda (x g^{- 1}) f (g) = (\lambda \ast f) (x) .$

We may regard $\psi_i$ as the character of $H_i$ acting on $V_i =\mathbb{C}$ (since $V_i$ is one-dimensional) with the representation $\pi_i (g) v = \psi_i (g) v$ when $g \in G$ and $v \in V_i =\mathbb{C}$. Then $\psi_i^G$ acts on the space $V_i^G$ of all functions $f_i : G \longrightarrow \mathbb{C}$ $(= V_i$) such that $f_i (h_i g) = \psi_i (h_i) f (g)$ when $h_i \in H_i$. The subspaces $V_i^G$ are thus invariant subspaces of $\mathbb{C}[G]$ under the action $\rho$, and the action of $G$ on $V_i^G$ is given by $\rho$.

Theorem 4.3.1: (Geometric form of Mackey's Theorem.) Let $\Lambda \in \operatorname{Hom}_G (V_1^G, V_2^G)$. Then there exists a function $\Delta \in \mathbb{C}[G]$ such that
 $\Delta (h_2 g h_1) = \psi_2 (h_2) \Delta (g) \psi_1 (h_1), \hspace{2em} h_i \in H_i,$ (4.3.2)

and $\Lambda f = \Delta \ast f$ for all $f \in \psi_1^G$. The map $T \longmapsto \Delta$ is a vector space isomorphism of $\operatorname{Hom}_G (\psi_1^G, \psi_2^G)$ with the space of all functions satisfying (4.3.2).

Proof. (Click to Expand/Collapse)

If $f \in \mathbb{C}[G]$ define $p (f)$ to be the function $p (f) (g) = \frac{1}{|H_1 |} \sum_{h \in H_1} \psi_1 (h) f (h^{- 1} g) .$ The following facts are easily verified (Exercise 4.3.1). First $p$ is a projection operator with image $V_1^G$; this means that $p (f) \in V_1^G$, and that $p (f) = f$ if $f \in V_1^G$. Moreover, $p (\rho (g) f) = \rho (g) p (f)$. Now we define $T : \mathbb{C}[G] \longrightarrow \mathbb{C}[G]$ to be $\Delta \circ p$. Then since $\Lambda$ is a $G$-module homomorphism, we have $\Lambda \circ \rho (g) = \rho (g) \circ \Lambda$, and so $T$ satisfies $T \circ \rho (g) = \rho (g) \circ T$. Therefore by the Lemma we have $T f = \Delta \ast f$ for some unique $\Delta$. Let us check that $\Delta$ has the property (4.3.2). This can be separated into two statements,
 $\Delta (g h_1) = \Delta (g) \psi_1 (h_1), \hspace{2em} h_1 \in H_1,$ (4.3.3)

and
 $\Delta (h_2 g) = \psi_2 (h_2) \Delta (g), \hspace{2em} h_2 \in H_2 .$ (4.3.4)

We will prove (4.3.3) and leave (4.3.4) to the reader, with a hint. For (4.3.3) we note that $\Delta \ast p (f) = \Lambda \circ p \circ p (f) = \Lambda (p (f)) = \Delta \ast f.$ This means that \begin{eqnarray*} \sum_{y \in G} \Delta (y) f (y^{- 1} g) = (\Delta \ast f) (g) = (\Delta \ast p (f)) (g) & = & \\ \sum_{y \in G} \Delta (y) \frac{1}{|H_1 |} \sum_{h \in H_1} \psi_1 (h_1) f (h^{- 1} y^{- 1} g) & & \end{eqnarray*} Interchanging the order of summation and (for fixed $h_1$) making the variable change $y \longmapsto y h_1^{- 1}$ gives $\sum_{y \in G} \Delta (y) f (y^{- 1} g) = \frac{1}{|H_1 |} \sum_{h \in H_1} \sum_{y \in G} \Delta (y h^{- 1}) \psi_1 (h) f (y^{- 1} g) .$ Since this is true for all $f$, we can deduce that $\Delta (y) = \frac{1}{|H_1 |} \sum_{h \in H_1} \Delta (y h^{- 1}) \psi_1 (h) .$ From this identity (4.3.3) is easily deduced: we have $\Delta (y h_1) = \frac{1}{|H_1 |} \sum_{h \in H_1} \Delta (y h_1 h^{- 1}) \psi_1 (h) .$ On making the variable change $h \longmapsto h h_1$ this turns into $\left( \frac{1}{|H_1 |} \sum_{h \in H_1} \Delta (y h^{- 1}) \psi_1 (h) \right) \psi_1 (h_1) = \Delta (y) \psi_1 (h_1) .$ We leave (4.3.4) to the reader, with the hint that it follows from the fact that the image of $T$ is contained in $V_2^G$.

Exercise 4.3.1: Fill out the details in the proof of Theorem 4.3.1.

Corollary 4.3.1: The inner product $\left\langle \psi_1^G, \psi_2^G \right\rangle$ equals the dimension of the vector space of functions $\Delta$ on $G$ that satisfy (4.3.2).

Proof. (Click to Expand/Collapse)

This follows from Proposition 2.4.4.

In every application, the first step is to compute the double cosets $H_2 \backslash G / H_1$. A double coset is a subset $H_2 x H_1$, and by definition the set of these is $H_2 \backslash G / H_1$. One can think of this as follows: the group $H_2 \times H_1$ acts on $G$ by $(h_2, h_1) : x \longmapsto h_2 x h_1^{- 1},$ and the double cosets are the orbits. In many cases, there aren't very many. Once the double cosets are known, $\left\langle \psi_1^G, \psi_2^G \right\rangle$ can be computed by Mackey's theorem, and conclusions can be deduced from this information. For example if $H_1 = H_2$ and $\psi_1 = \psi_2$ we learn whether the induced representation is irreducible.

We will give typical applications in the following sections.