If $G$ is nonabelian, the group of linear characters is too small to be an orthonormal basis of $L^2 (G)$. Still the problem of giving an orthonormal basis of $L^2 (G)$ has a satisfactory solution.

If $V$ is a complex vector space, the ring $\operatorname{End} (V)$ of vector space endomorphisms of $V$ is a ring. This ring has as its multiplicative group the group $\operatorname{GL} (V)$ of invertible linear transformations. Assuming that $V$ has finite dimension $d$, $\operatorname{End} (V) \cong \operatorname{Mat}_d (\mathbb{C})$ and $\operatorname{GL} (V) \cong \operatorname{GL} (d, \mathbb{C})$.

By a (complex) *representation* of $G$ we mean an ordered pair $(\pi,
V)$, where $V$ is a complex vector space and $\pi : G \longrightarrow
\operatorname{GL} (V)$ is a homomorphism; or we say that $\pi : G \longrightarrow
\operatorname{GL} (V)$ is a representation. For finite groups, we may limit
ourselves to the case where $V$ is finite-dimensional, and indeed when we say
that $(\pi, V)$ is a representation, we will always assume that $V$ is
finite-dimensional.

If $V$ is a vector space over a field $F$, a *linear functional* is a
linear map $V \longrightarrow F$. The linear functionals themselves form a
vector space $V^{\ast}$, the *dual space* of $V$. If $V$ is finite
dimensional, then the dual space has the same complex dimension; indeed, if
$x_1, \cdots, x_D$ are a basis of $V$ there is a *dual basis*
$\lambda_1, \cdots, \lambda_D$ of $V^{\ast}$ such that $\lambda_i (x_j) = 1$
if $i = j$, and $0$ if $i \neq j$. The situation is quite analogous to that of
the dual group, already discussed: $V \cong V^{\ast}$ but not canonically.
However $V \cong (V^{\ast})^{\ast}$ and this isomorphism is natural.

The dual space is a contravariant functor: if $\phi : V \longrightarrow W$ is any linear transformation, then composition with $\phi$ gives a linear transformation $\phi^{\ast} : W^{\ast} \longrightarrow V^{\ast}$. Thus if $\Lambda : W \longrightarrow F$ is any linear functional, then $\phi^{\ast} (\Lambda) \in V^{\ast}$ is the linear functional $\Lambda \circ \phi : V \longrightarrow F$. We will mainly be concerned with the case where $F =\mathbb{C}$.

**Exercise 2.3.1:***
Show that if $T : V \longrightarrow V$ is a linear
transformation, and if $M$ is the matrix of $T$ with respect to some basis
$v_1, \cdots, v_D$, then the matrix of $T^{\ast} : V^{\ast} \longrightarrow
V^{\ast}$ with respect to the dual basis $\lambda_1, \cdots, \lambda_D$ is
the transpose of $M$. Thus if $M = (m_{i j})$ we have $T (v_i) = \sum_j m_{j
i} v_j$, and you must prove that $T^{\ast} (\lambda_i) = \sum_j m_{i j}
\lambda_j$.
*

Now let $(\pi, V)$ be a representation. We then have another representation $( \hat{\pi}, V^{\ast})$ of $G$ on the dual space, defined by $\hat{\pi} (g) = \pi (g^{- 1})^{\ast}$. Note that we need the inverse to compensate for the fact that the dual is a contravariant functor, so that $\hat{\pi} (g h) = \hat{\pi} (g) \hat{\pi} (h)$. In concrete terms if $\Lambda \in V^{\ast}$ then $\hat{\pi} (g) \Lambda$ is the linear functional

\[ ( \hat{\pi} (g) \Lambda) (v) = \Lambda (\pi (g)^{- 1} v) . \] | (2.3.1) |

**Theorem 2.3.1:***
If $\pi : G \longrightarrow \operatorname{GL} (V)$ is a
representation, then there exists an inner product on $V$ such that
$\left\langle \pi (g) v, \pi (g) w \right\rangle = \left\langle v, w
\right\rangle$ for all $g \in G$ and $v, w \in V$.
*

We can express the identity $\left\langle \pi (g) v, \pi (g) w \right\rangle =
\left\langle v, w \right\rangle$ by saying that $\left\langle \;, \;
\right\rangle$ is *invariant* under the action of $G$. Or we may say
that $\pi (g)$ is *unitary*. Indeed, if $T : V \longrightarrow V$ is a
linear transformation of a finite-dimensional Hilbert space $V$ such that
$\left\langle T v, T w \right\rangle = \left\langle v, w \right\rangle$ for
all $v, w \in V$, we say that $T$ is *unitary*. The unitary linear
transformations form a group, called the *unitary group* $U (V)$.
Unlike $\operatorname{GL} (V)$, it has the important topological property of being
compact. The content of the theorem is that given any representation, an inner
product can be chosen so that $\pi (G)$ is contained in the unitary group.

**Proof. **(Click to Expand/Collapse)

Choose *any* inner product $\left\langle \left\langle \;, \;
\right\rangle \right\rangle$ on $V$. Modify it by averaging:
\[ \left\langle v, w \right\rangle = \frac{1}{| G|} \sum_{g \in G}
\left\langle \left\langle \pi (g) v, \pi (g) w \right\rangle
\right\rangle . \]
It is easy to check that $\left\langle v, w \right\rangle$ is still a
positive definite Hermitian form, and is invariant under the action of $G$.

Now if $(\pi, V)$ is a representation and $U$ is a subspace of $V$, we say
that $U$ is *invariant* if $\pi (g) U \subseteq U$ for all $g \in G$.
The representation is called *irreducible* if $V \neq 0$, and if the
only invariant subspaces are $0$ and $U$. Clearly every nonzero representation
contains an irreducible subspace. If $V$ is not irreducible, and is nonzero,
then we say $V$ is *reducible*, and this means it has a proper nonzero
invariant subspace.

**Theorem 2.3.2:***
( Maschke.) Let $(\pi, V)$ be a representation of the finite group $G$.
Then $V$ is a direct sum of irreducible subspaces.
*

**Proof. **(Click to Expand/Collapse)

Let $(\pi, V)$ be a minimal counterexample. Thus $V$ is nonzero (otherwise
$V$ is a direct sum of zero invariant subspaces) and not irreducible (since
otherwise $V$ is a direct sum of one invariant subspace). So it is
reducible, and has a proper nonzero invariant subspace $U$. Given a
$G$-invariant subspace $V$, let $U'$ be the orthogonal complement of $U$.
Thus
\[
U'=\{u'\in V|\text{$\left\langle u',u\right\rangle = 0$ for all $u\in U$}\}.
\]
We claim that $U'$ is invariant. Indeed, if $u' \in U'$ and
$u \in U$ then
\[ \left\langle \pi (g) u', u \right\rangle = \left\langle u', \pi (g)^{- 1}
u \right\rangle = 0 \]
since $\pi (g)^{- 1} u \in U$, and so $u' \in U'$. Now $V = U \oplus U'$
(Exercise 2.1.2). The invariant subspaces $U$ and $U'$ both
afford representations of $G$, so by induction they are both direct sums of
irreducible subspaces. Combining these gives a direct sum decomposition of
$V$ itself.

We recall that a matrix $A$ is called *diagonalizable* $M^{- 1} A M$ is
diagonal for some invertible matrix $M$. We also say that a linear
transformation is diagonalizable if and only if its matrix is diagonalizable
with respect to any basis.

**Lemma 2.3.1:***
If $A$ represents the linear transformation $T$ of the vector space $V$,
then $A$ is diagonalizable if and only if $V$ has a basis of eigenvectors
of $A$.
*

**Proof. **(Click to Expand/Collapse)

Indeed, if we identify $V$ with $\mathbb{C}^d$, and if $v_1, \cdots, v_d$
are these column vectors, so $A v_i = \lambda_i v_i$, let $M$ be the matrix
with columns $v_1, \cdots, v_d$. Then
\[ A M = A (v_1, \cdots, v_d) = (A v_1, \cdots, A v_d) = (\lambda_1 v_1,
\cdots, \lambda_d v_d) . \]
This equals
\[ (v_1, \cdots, v_d) \left(\begin{array}{ccc}
\lambda_1 & & \\
& \ddots & \\
& & \lambda_d
\end{array}\right) = M D, \hspace{2em} D = \left(\begin{array}{ccc}
\lambda_1 & & \\
& \ddots & \\
& & \lambda_d
\end{array}\right) . \]
Since $A M = M D$, we see that $M^{- 1} A M = D$.

**Proposition 2.3.1:***
Let $V$ be a finite-dimensional vector space and let $T
: V \longrightarrow V$ be a linear transformation of finite order, so $T^n =
I$ (the identity map) for some $n$. Then $V$ has a basis of eigenvectors for
$T$. (So $T$ is diagonalizable.)
*

This can be proved using the Jordan canonical form, but we will prove it as a consequence of Maschke's theorem (Theorem 2.3.2).

**Proof. **(Click to Expand/Collapse)

Let $\Gamma$ be a cyclic group of order $n$ with generator $\gamma$. There
is a representation $(\pi, V)$ of $\Gamma$ in which $\pi (\gamma^k) = T^k$.
By Maschke's theorem, $V$ is a direct sum of irreducible invariant
subspaces, so we will be done if we check that these are one-dimensional. If
$U$ is an irreducible subspace of $V$, then $T$ induces a linear
transformation of $U$, and since $\mathbb{C}$ is algebraically closed, $T$
has an eigenvector $u \in U$. Now $\mathbb{C}u$ is a nonzero invariant
subspace of $U$ and since $U$ is irreducible, $U =\mathbb{C}u$ is
one-dimensional.

We now introduce a ring called the *group algebra*, denoted
$\mathbb{C}[G]$. As a vector space over $\mathbb{C}$ it is just the space of
all functions on the group. It has a multiplication called
*convolution* and denoted $\ast$ in which if $f_1, f_2 \in
\mathbb{C}[G]$, then $f_1 \ast f_2$ is the function

\[ (f_1 \ast f_2) (x) = \sum_{y \in G} f_1 (y) f_2 (y^{- 1} x) = \sum_{\begin{array}{c} y, z \in G\\ y z = x \end{array}} f_1 (y) f_2 (z) = \sum_{z \in G} f_1 (x z^{- 1}) f_2 (z) . \] | (2.3.2) |

**Exercise 2.3.2:***
If $g \in G$ let $\delta_g$ be the function defined by $\delta_g (x) = 1$ if
$x = g$, 0 otherwise. Show that $\delta_{g h} = \delta_g \ast \delta_h$ and
conclude that $g \longmapsto \delta_g$ is an injective homomorphism of $G$
into $\mathbb{C}[G]^{\times}$.
*

If one makes use of this exercise, then one may identify $G$ with its image under the homomorphism $\delta$. This works for finite groups, or more generally discrete topological groups, but not in general. For example, there is no analog of $\delta_g$ in $C_c^{\infty} (\mathbb{R})$.

If $(\pi, V)$ is a representation, then we may define a multiplication $\mathbb{C}[G] \times V \longrightarrow V$ by \[ f \cdot v = \sum_{g \in G} f (g) \pi (g) v. \] Let us check that $V$ becomes a $\mathbb{C}[G]$-module with this multiplication. If $f_1, f_2$ are functions on $G$ then \[ f_1 \cdot (f_2 \cdot v) = f_1 \cdot \left( \sum_{z \in G} f_2 (z) \pi (z) v \right) = \sum_{y, z \in G} f_1 (y) f_2 (z) \pi (y z) v. \] Collecting the terms with $y z = g$, this may be written \[ \sum_{g \in G} \left( \sum_{y, z = g} f_1 (y) f_2 (z) \right) \pi (g) v = (f_1 \ast f_2) \cdot v. \]

**Proposition 2.3.2:***
Let $V$ be a $ \mathbb{C}[G]$-module. Since $\mathbb{C} \subset
\mathbb{C}[G]$, $V$ is a vector space over $\mathbb{C}$, and we assume it
to be finite-dimensional. Then there exists a representation $(\pi, V)$ such
that the $ \mathbb{C}[G]$-module structure on $V$ is the one determined by
this representation.
*

Thus representations and modules over the group algebra are really the same
thing. We will sometimes use the expression *$G$-module* to mean
$\mathbb{C}[G]$-module.

**Proof. **(Click to Expand/Collapse)

Given the $\mathbb{C}[G]$-module $V$, define $\pi : G \longrightarrow
\operatorname{GL} (V)$ by $\pi (g) v = \delta_g \cdot v$. We leave it to the reader
to check that this is a representation, and that it determines the given
$\mathbb{C}[G]$-module structure on $V$.

Now let $(\pi, V)$ and $(\sigma, W)$ be two representations. Then we call a
linear map $\phi : V \longrightarrow W$ an *intertwining operator* if
for all $g \in G$ the following diagram is commutative:

**Exercise 2.3.3:***
Prove that $\phi : V \longrightarrow W$ is an intertwining operator if and
only if it is a $\mathbb{C}[G]$-module homomorphism.
*

We will sometimes say that $\phi : V \longrightarrow W$ is a *$G$-module
homomorphism* to mean that it is a $\mathbb{C}[G]$-module homomorphism; we
see that this is the same as saying that it is an intertwining operator. We
may also express the same thing by saying that the map is
*$G$-equivariant*. If $R$ is a ring and $V$, $W$ are $R$-modules, the
space of $R$-module homomorphisms $M \longrightarrow N$ will be denoted
$\operatorname{Hom}_R (V, W)$; if $R =\mathbb{C}[G]$ we will also use the notation
$\operatorname{Hom}_G (V, W)$ to mean the same thing.

The notion of irreducibility extends to an arbitrary ring $R$. If $R$ is a
ring and $M$ is an $R$-module, then $M$ is called *simple* if it has no
proper, nontrivial submodules. If $R =\mathbb{C}[G]$, then simple and
irreducible are synonymous terms.

**Proposition 2.3.3:***
( Schur's Lemma.) (i) Let $R$ be a ring and let $M,
N$ be simple $R$-modules. Let $\phi \in \operatorname{Hom}_R (M, N)$. Then $\phi$ is
either zero or an isomorphism.
*

*
(ii) Let $R =\mathbb{C}[G]$ and let $V$ be an irreducible
$G$-module that is finite-dimensional as a complex vector space. Let $\phi
\in \operatorname{End}_R (V)$. Then there exists a complex number $\lambda \in
\mathbb{C}$ such that $\phi (v) = \lambda v$ for all $v \in V$.
*

**Proof. **(Click to Expand/Collapse)

For (i), we observe that $\ker (\phi)$ is an $R$-submodule of $M$, and since
$M$ is simple, the only $R$-submodules are $M$ and $0$. Thus either $\ker
(\phi) = M$ (so $\phi$ is zero) or $\ker (\phi) = 0$ (so $\phi$ is
injective). Similarly $\operatorname{im} (\phi)$ is an $R$-submodule of $N$, which
is also simple, so either $\operatorname{im} (\phi) = N$ (so $\phi$ is surjective)
or $\operatorname{im} (\phi) = 0$ (so $\phi$ is zero). We see that either $\phi$is
zero or bijective, proving (i).

For (ii), since $\phi$ is a linear transformation of a complex vector space it has an eigenvalue $\lambda$. Thus $U =\{v \in V| \phi (v) = \lambda v\}$ is nonzero. It is a submodule since if $f \in R$ and $v \in U$ then $\phi (f v) = f \phi (v) = \lambda f v$, so $f v \in U$. Since it is a nonzero submodule of a simple module, $U = V$.

**Proposition 2.3.4:***
Let $R =\mathbb{C}[G]$ and let $V, W$ be irreducible
$G$-modules that are finite-dimensional as complex vector spaces. Then
\[ \dim_{\mathbb{C}} \; \operatorname{Hom}_{\mathbb{C}[G]} (V, W) = \left\{
\begin{array}{ll}
1 & \text{if $V \cong W$ as $G$-modules;}\\
0 & \text{otherwise.}
\end{array} \right. \]
*

**Proof. **(Click to Expand/Collapse)

If $V \ncong W$ then by Schur's Lemma (i) there are no $G$-module
homomorphisms and $\operatorname{Hom}_{\mathbb{C}[G]} (V, W) = 0$. On the other
hand if $V \cong W$ there is no loss of generality in assuming that $V = W$,
in which case by Schur's Lemma (ii), every $G$-module homorphism is a scalar
map, so $\operatorname{Hom}_{\mathbb{C}[G]} (V, W) \cong \mathbb{C}$.

We may also use Schur's Lemma to complement Theorem 2.3.1
with a uniqueness result. If $V$ is a finite-dimensional complex vector space,
we may find a real vector space $V_0 \subset V$ such that $V = V_0 \oplus i
V_0$. Indeed, choose a basis $x_1, \cdots, x_d$ of $V$, and let
\[ V_0 =\mathbb{R}x_1 \oplus \cdots \oplus \mathbb{R}x_d . \]
Now if $v \in V$ we may write (uniquely) $v = v_0 + i v_0'$ with $v_0$ and
$v_0'$ in $V_0$. Define $c (v) = v_0 - i v_0'$. We call $V_0$ a *real
structure* on $V$, and $c$ the *conjugation* relative to this real
structure.

**Proposition 2.3.5:***
Let $V$ be an irreducible, finite-dimensional $G$-module. Then any
$G$-invariant two inner products on $V$ are proportional.
*

Since $\left\langle v, v \right\rangle > 0$ for $v \in V$, the constant of proportionality is positive.

**Proof. **(Click to Expand/Collapse)

Pick a real structure $V_0$ on $V$, and let $c : V \longrightarrow V$ be the
corresponding conjugation. The main properties that we need is that it is
conjugate linear, that is, $c (\alpha v) = \bar{\alpha} c (v)$ for $\alpha
\in \mathbb{C}$, and an $\mathbb{R}$-linear isomorphism. Define another
representation $\pi' : G \longrightarrow \operatorname{GL} (V)$ by $\pi' (g) = c^{-
1} \pi (g) c$. We note that $\pi' (g)$ is $\mathbb{C}$-linear, since if
$\alpha \in \mathbb{C}$ we have
\[ c^{- 1} \pi (g) c (\alpha v) = c^{- 1} \pi (g) \bar{\alpha} c (v) = c^{-
1} \bar{\alpha} \pi (g) c (v) = \alpha c^{- 1} \pi (g) c (v) . \]

To define $\theta$, we use the inner product as follows. If $v \in V$, then
$\theta (v)$ is to be the linear functional defined by $\theta (v) x =
\left\langle x, c (v) \right\rangle$. Note that since $c$ is conjugate
linear, and since the inner product is conjugate linear in the second
variable, $\theta (v)$ is actually linear.

We will show that given an inner product on $V$ we can construct an intertwining operator $\theta : V \longrightarrow V^{\ast}$ that intertwines the representations $\pi'$ on $V$ and $\hat{\pi}$ on $V^{\ast}$. That is, we have a commutative diagram:

To check the commutativity of this diagram, consider \begin{eqnarray*} ( \hat{\pi} (g) \theta (v)) (x) & = & \\ \theta (v) \pi (g)^{- 1} x = \left\langle \pi (g)^{- 1} x, c (v) \right\rangle & = & \\ \left\langle x, \pi (g) c (v) \right\rangle = \left\langle x, c \pi' (g) v \right\rangle & = & \theta (\pi' (g) v) (x) . \end{eqnarray*} Since this is true for all $x \in V$, we have $\hat{\pi} (g) \theta (v) = \theta \pi' (g) (v)$, and since this is true for all $v \in V$, the diagram commutes.

One may check that $\pi'$ and $\hat{\pi}$ are both irreducible representations (Exercise 2.3.4), and so by Schur's Lemma, $\theta$ is determined up to constant multiple. Hence the inner product is determined up to constant multiple.

**Exercise 2.3.4:***
Complete the last proof by showing that $\pi'$ and
$\hat{\pi}$ are both irreducible.
*

The irreducible representations will play the same role in the representation theory of nonabelian groups as the linear characters in the case of abelian groups.

Let $(\pi, V)$ be an irreducible representation. We choose a $G$-invariant inner product on $V$ (Theorem 2.3.1). Let $\mathcal{R}_V$ be the vector space spanned by functions on $G$ of the form $f_{v, w} : V \longrightarrow \mathbb{C}$, where for $v, w \in V$ we define

\[ f_{v, w} (g) = \left\langle \pi (g) v, w \right\rangle . \] | (2.3.3) |

The reason for the term "matrix coefficient'' can be understood as follows. Using Proposition 2.3.1 we may find an orthonormal basis $v_1, \cdots, v_D$ of $V$. Now let us represent $\pi (g)$ as a matrix \[ \pi (g) = \left(\begin{array}{ccc} \pi_{11} (g) & \cdots & \pi_{1 d} (g)\\ \vdots & & \vdots\\ \pi_{d 1} (g) & \cdots & \pi_{d d} (g) \end{array}\right) \] with respect to this basis. This means that $\pi (g) v_j = \sum_i \pi_{i j} (g) v_i$. We will show that

\[ \left\langle \pi (g) v_j, v_i \right\rangle = \pi_{i j} (g) . \] | (2.3.4) |

**Spoiler Warning!** We will eventually show that
the functions $\sqrt{d} \pi_{i j} (g)$ are orthonormal. and that taking the
union of these over all the isomorphism classes of irreducible representations
of $G$ gives an orthonormal basis of $L^2 (G)$.

We will find an orthonormal basis of $L^2 (G)$ from among the matrix coefficients, where as before $L^2 (G)$ is the space of functions on $G$, with the inner product

\[ \left\langle \alpha, \beta \right\rangle_2 = \frac{1}{|G|} \sum_{x \in G} \alpha (x) \overline{\beta (x)} . \] | (2.3.5) |

**Proposition 2.3.6:***
( Schur Orthogonality I.) Let $V$ and $W$ be
simple $G$-modules, and let $\alpha \in \mathcal{R}_V$, $\beta \in
\mathcal{R}_W$. If $V \ncong W$ then $\alpha, \beta$ are orthogonal in $L^2
(G)$.
*

This is the first of several results that go under the umbrella called
*Schur orthogonality*.

**Proof. **(Click to Expand/Collapse)

We assume that $\alpha$ and $\beta$ are not orthogonal. We may assume that
$\alpha = f_{v_1, v_2}$ and $\beta = f_{w_1, w_2}$ where $v_1, v_2 \in V$
and $w_1, w_2 \in W$. Indeed, $\alpha$ is a linear combination of matrix
coefficients, and so is $\beta$, and so if all the matrix coefficients that
appear in $\phi$ are orthogonal to all the matrix coefficients that appear
in $\psi$ we would have $\left\langle \alpha, \beta \right\rangle = 0$, and
we are assuming this is not the case.

Therefore assume $\alpha = f_{v_1, v_2}$ and $\beta = f_{w_1, w_2}$. We assume that $\left\langle \alpha, \beta \right\rangle \neq 0$, and we will construct a $G$-module isomorphism $\phi : V \longrightarrow W$. Specifically, define \[ \phi (x) = \frac{1}{|G|} \sum_{g \in G} \left\langle \pi (g) x, v_2 \right\rangle \sigma (g)^{- 1} w_2, \hspace{2em} x \in V. \]

Let us show that this is $G$-equivariant. We have, for $h \in G$ \[ \phi (\pi (h) x) = \frac{1}{|G|} \sum_{g \in G} \left\langle \pi (g h) x, v_2 \right\rangle \sigma (g)^{- 1} w_2 . \] We make the variable change $g \longmapsto g h^{- 1}$, so $\sigma (g)^{- 1} \longmapsto \sigma (h) \sigma (g)^{- 1}$, and we see that this equals \[ \frac{1}{|G|} \sum_{g \in G} \left\langle \pi (g h) x, v_2 \right\rangle \sigma (h) \sigma (g)^{- 1} w_2 = \sigma (h) \phi (x) . \] Thus $\phi$ is equivariant, so by Schur's Lemma it is either zero or an isomorphism. To see that it is nonzero, we note that \begin{eqnarray*} \left\langle \phi (v_1), w_1 \right\rangle = \frac{1}{|G|} \sum \left\langle \pi (g) v_1, v_2 \right\rangle \left\langle \sigma (g)^{- 1} w_2, w_1 \right\rangle & = & \\ \frac{1}{|G|} \sum \left\langle \pi (g) v_1, v_2 \right\rangle \left\langle w_2, \sigma (g) w_1 \right\rangle & = & \\ \frac{1}{|G|} \sum \left\langle \pi (g) v_1, v_2 \right\rangle \overline{\left\langle \sigma (g) w_1, w_2 \right\rangle} & = & \left\langle f_{v_1, v_2}, f_{w_1, w_2} \right\rangle \end{eqnarray*} which is nonzero; so $\phi$ cannot be the zero map.

Now we know that representative functions for *nonisomorphic*
irreducible $G$-modules are orthogonal. We need a complementary result for
$G$-modules that *are* isomorphic. But if the modules are isomorphic,
we may as well assume that they are the same.

**Proposition 2.3.7:***
( Schur Orthogonality II.) Let $V$ be an irreducible
$G$-module. Then there exists a positive constant $d (V)$ such that for
$v_1, v_2, w_1, w_2 \in V$ we have
\[ \left\langle f_{v_1, v_2}, f_{w_1, w_2} \right\rangle_2 = \frac{1}{d (V)}
\left\langle v_1, w_1 \right\rangle \overline{\left\langle v_2, w_2
\right\rangle} . \]
*

Note that on the left-hand side, the inner product is the one on $L^2 (G)$, and on the right-hand side, the inner product is the one in $V$. We will eventually prove that $d (V) = \dim (V)$.

**Proof. **(Click to Expand/Collapse)

We proceed as in the proof of Proposition 2.3.6 and define
$\phi : V \longrightarrow V$ by
\[ \phi (x) = \frac{1}{|G|} \sum_{g \in G} \left\langle \pi (g) x, v_2
\right\rangle \pi (g)^{- 1} w_2, \hspace{2em} x \in V. \]
As in Proposition 2.3.6 this is $G$-equivariant. By part
(ii) of Schur's Lemma, it is a scalar map, that is, there exists a constant
$\lambda$ such that $\phi (x) = \lambda x$ for all $x \in V$. The constant
$\lambda$ depends on $v_2$ and $w_2$, and to keep track of this dependence
we write $\lambda = \lambda (v_2, w_2)$. Now the same calculation as in \
Proposition 2.3.6 shows that
\[ \left\langle \phi (v_1), w_1 \right\rangle_2 = \left\langle f_{v_1, v_2},
f_{v_2, w_2} \right\rangle, \]
that is
\[ \left\langle f_{v_1, v_2}, f_{w_1, w_2} \right\rangle_2 = \lambda (v_2,
w_2) \left\langle v_1, w_1 \right\rangle, \]
and this identity is true for all $v_1$ and $w_1$, with $v_2$ and $w_2$ held
constant. Now interchanging $v_1 \longleftrightarrow w_1$ and $v_2
\longleftrightarrow w_2$ and taking complex conjugates, remembering that
$\left\langle f, f' \right\rangle = \overline{\left\langle f', f
\right\rangle}$, we obtain the identity
\[ \left\langle f_{v_1, v_2}, f_{w_1, w_2} \right\rangle_2 =
\overline{\lambda (v_1, w_1)} \overline{\left\langle v_2, w_2
\right\rangle} . \]
Now since
\[ \lambda (v_2, w_2) \left\langle v_1, w_1 \right\rangle =
\overline{\lambda (v_1, w_1)} \overline{\left\langle v_2, w_2
\right\rangle} \]
for all $v_1, v_2, w_1, w_2$, we have
\[ \overline{\lambda (v_1, w_1}) / \left\langle v_1, w_1 \right\rangle =
\lambda (v_2, w_2) / \overline{\left\langle v_2, w_2 \right\rangle} \]
for all $v_1, v_2, w_1, w_2$ such that the denominators are nonzero. The
left-hand side is independent of $v_2$ and $w_2$, and the right-hand side is
independent of $\lambda (v_1, w_1)$, so this must be a constant, and we
obtain
\[ \left\langle f_{v_1, v_2}, f_{w_1, w_2} \right\rangle_2 = \operatorname{constant}
\times \left\langle v_1, w_1 \right\rangle \left\langle v_2, w_2
\right\rangle . \]
To see that the constant is positive, taking $v_1 = w_1$ and $v_2 = w_2$
makes all three inner products positive. Since the constant is positive, we
are justified in calling it $1 / d (V)$ for some $d (V) > 0$.

**Lemma 2.3.2:***
Suppose that $V$ is a finite-dimensional Hilbert
space, and let $\Lambda : V \longrightarrow \mathbb{C}$ be any linear
functional. Then there exists a unique vector $v \in V$ such that $\Lambda
(x) = \left\langle x, v \right\rangle$.
*

**Proof. **(Click to Expand/Collapse)

Define $\lambda_v : V \longrightarrow \mathbb{C}$ by $\lambda_v (x) =
\left\langle x, v \right\rangle$. Then $\lambda_v \in V^{\ast}$. The map $v
\longmapsto \lambda_v$ is a map of complex vector spaces $V \longrightarrow
V^{\ast}$. It is not linear, but antilinear; that is, if $\alpha$ is a
complex number, $\lambda_{\alpha v} = \bar{\alpha} \lambda_v$.

Still, the map $v \longmapsto \lambda_v$ is $\mathbb{R}$-linear, and $\dim_{\mathbb{R}} (V) = 2 \cdot \dim_{\mathbb{C}} (V) = 2 \cdot \dim_{\mathbb{C}} (V^{\ast}) = \dim_{\mathbb{R}} (V^{\ast})$. As a linear transformation of real vector spaces of the same dimension, it will be surjective if and only if it is injective. To see that it is injective, we observe that $\lambda_v (v) = \left\langle v, v \right\rangle > 0$ if $v \neq 0$, so $\lambda_v$ is not the zero map unless $v = 0$.

There exists a representation $\rho : G \longrightarrow \operatorname{GL} (L^2 (G))$,
which is the action by *right translation*. Specifically, if $f \in L^2
(G)$ and $g \in G$, then

\[ (\rho (g) f) (x) = f (x g) . \] | (2.3.6) |

**Exercise 2.3.5:***
Check that the inner product (2.3.5) is invariant with respect to the
regular representation. We recall that this means $\left\langle \rho (g)
f_1, \rho (g) f_2 \right\rangle_2 = \left\langle f_1, f_2 \right\rangle_2$.
*

Let us consider how the right regular representation affects matrix coefficients.

**Lemma 2.3.3:***
Let $\pi : G \longrightarrow \operatorname{GL} (V)$ be a
representation, where $V$ is given a $G$-invariant inner product. If $v, w
\in V$, then $\rho (g) f_{v, w} = f_{\pi (g) v, w}$.
*

**Proof. **(Click to Expand/Collapse)

If $x \in G$, we have
\[ \rho (g) f_{v, w} (x) = f_{v, w} (x g) = \left\langle \rho (x g) v, w
\right\rangle = \left\langle \rho (x) \rho (g) v, w \right\rangle =
f_{\rho (g) v, w} (x) . \]

**Lemma 2.3.4:***
Suppose that $f \in L^2 (G)$ is orthogonal to the matrix
coefficients of all irreducible representations. Then $f = 0$.
*

**Proof. **(Click to Expand/Collapse)

Let $W$ be the vector space of functions that are orthogonal to all matrix
coefficients of irreducible representations. Then we claim that $W$ is
invariant under $\rho$. Indeed, by Lemma 2.3.3, the set of matrix
coefficients is invariant under $\rho$, and since the inner product
(2.3.5) is invariant with respect to $\rho$, it follows that the
orthogonal complement $W$ is $\rho$ invariant – that is, $\rho (g) W
\subset W$. Thus $W$ is a $G$-submodule of $L^2 (G)$. If it is nonzero, it
contains an irreducible submodule $W_0$. Let $f$ be a nonzero element of
$W_0$. We will actually show that $f$ is itself a matrix coefficient.

Evaluation at the identity is a linear functional $\phi \longmapsto \phi (1)$ on $W_0$. By Lemma 2.3.2, there exists a vector $\xi \in W_0$ such that $\phi (1) = \left\langle \phi, \xi \right\rangle_2$ for all $\phi \in W_0$. Now consider that \[ \left\langle \rho (g) f, \xi \right\rangle_2 = \rho (g) f (1) = f (1 \cdot g) = f (g) . \] We see that $f$ itself is a matrix coefficient of the irreducible representation of $G$ on $W_0$. Since $f$ is nonzero, it is not orthogonal to itself, which is a contradiction since it was assumed to be orthogonal to all matrix coefficients.

The following theorem gives a nearly complete description of an orthonormal basis of $L^2 (G)$.

**Theorem 2.3.3:***
The finite group $G$ has only finitely many isomorphism classes of
irreducible representations. If $V_1, \cdots, V_h$ are representatives of
these isomorphism classes (so that any irreducible $G$-module is isomorphic
to $V_i$ for some unique $i$), then $L^2 (G) = \bigoplus_i
\mathcal{R}_{V_i}$. The spaces $\mathcal{R}_{V_i}$ are mutually orthogonal,
so if we exhibit an orthonormal basis of $\mathcal{R}_{V_i}$ for each $i$,
the union of these is an orthonormal basis of $L^2 (V)$. The dimension of
$\mathcal{R}_{V_i}$ is $D_i^2$, where $D_i = \dim (V_i)$. Let $v_1, \cdots,
v_{D_i}$ be an orthonormal basis of $V_i$ with respect to a $G$-invariant
inner product on $V_i$. Then with $d_i = d (V_i)$ as in
Proposition 2.3.7, the $D_i^2$ quantities $\sqrt{d_i} f_{v_k,
v_l}$ are an orthonormal basis of $\mathcal{R}_{V_i}$.
*

Eventually we will prove that $d_i = D_i$ (Proposition 2.4.5).

**Proof. **(Click to Expand/Collapse)

Everything asserted here has already been proved. The fact that the
$\mathcal{R}_{V_i}$ are mutually orthogonal is
Proposition 2.3.6. To see that $L^2 (G)$ is a direct product
of these, it is therefore enough to check that they span $L^2 (G)$, and if
they do not, then the orthogonal complement of their span is nonzero,
contradicting Lemma 2.3.4. Finally, to see that the functions
$\sqrt{d_i} f_{v_k, v_l}$ are orthonormal, by Proposition 2.3.7,
we have
\[ \left\langle \sqrt{d_i} f_{v_k, v_l}, \sqrt{d_i} f_{v_m, v_n}
\right\rangle_2 = \left\langle v_k, v_m \right\rangle \left\langle v_l,
v_n \right\rangle, \]
which is $1$ if $k = m$ and $l = n$, and zero otherwise.

**Proposition 2.3.8:***
Let $G$ act on $\mathcal{R}_{V_i}$ by the right regular representation. Then
as $G$-modules
\[ \mathcal{R}_{V_i} \cong V_i \oplus \cdots \oplus V_i \hspace{2em}
\text{($\dim (V_i)$ copies.)} \]
*

**Proof. **(Click to Expand/Collapse)

Let $v_1, \cdots, v_d$ be an orthonormal basis of $V_i$, and let $U_j
=\{f_{v, v_j} |v \in V_i \}$. Then since the $f_{v_i, v_j}$ are an
orthonormal basis of $\mathcal{R}_{V_i}$, it is clear that
\[ \mathcal{R}_{V_i} = U_1 \oplus \cdots \oplus U_{D_i} \]
as an internal direct sum; for the basis elements are divided up evenly
amongst the $U_j$, and those in $U_j$ span it. Now it follows from
Lemma 2.3.3 that $v \longmapsto f_{v, v_j}$ is an isomorphism of
$V_i$ onto $U_j$ for every $j$, and the statement follows.

**Proposition 2.3.9:***
If $V_1, \cdots, V_h$ are representatives of the irreducible isomorphism
classes of $G$ and $D_i = \dim (V_i)$, then
\[ \sum_i D_i^2 = |G|. \]
*

**Proof. **(Click to Expand/Collapse)

This follows from the decomposition of $L^2 (G)$ as $\bigoplus_i
\mathcal{R}_i$ since $\dim (\mathcal{R}_i) = D_i^2$.

We recall that the group algebra $\mathbb{C}[G]$ is, like $L^2 (G)$, the vector space of complex valued functions on $G$. But whereas $L^2 (G)$ is an inner product space, $\mathbb{C}[G]$ is a ring. In view of these different algebraic structures, we will use different notations to denote them. We can rewrite the identity $L^2 (G) = \bigoplus_i \mathcal{R}_{V_i}$ in the form \[ \mathbb{C}[G] = \prod_i \mathcal{R}_{V_i}, \] and this is appropriate since we will see that this is actually a direct product of rings. After all, we discussed in Section 1.3 that the Cartesian product of rings has the universal property of the product, but not the universal property of the coproduct, so it is now more appropriate to use $\prod$ instead of $\bigoplus$. In the Exercises, you will show that $\mathcal{R}_{V_i}$ is a two-sided ideal, and that $\mathcal{R}_{V_i}$ is a ring isomorphic to $\operatorname{Mat}_{D_i} (\mathbb{C})$, so \[ \mathbb{C}[G] \cong \prod_i \operatorname{Mat}_{D_i} (\mathbb{C}) . \]

**Exercise 2.3.6:***
Let $V$ be an irreducible $G$-module. Show that the convolution of two
matrix coefficients is computed by the formula
\[ f_{t, u} \ast f_{v, w} = \frac{\left\langle t, w \right\rangle}{d (V)}
f_{v, u} . \]
*

**Exercise 2.3.7:***
Show that if $f_1$ and $f_2$ are matrix coefficients for two nonisomorphic
irreducible $G$-modules, then $f_1 \ast f_2 = 0$.
*

**Exercise 2.3.8:***
Prove that $\mathcal{R}_V$ is a two-sided ideal in $L^2 (G)$, and that it
has the structure of a ring isomorphic to $\operatorname{Mat}_D (\mathbb{C})$.
Hint: choose an orthonormal basis $v_1, \cdots, v_d$ of $V$.
Every element of $\mathcal{R}_V$ can be written uniquely in the form
\[ d (V) \sum_{i, j} a_{i j} f_{v_j, v_i} . \]
Map this to the matrix $(a_{i j})$.
*