What does this exercise from Fulton and Harris mean?

Question

The following is an exercise from Representation Theory: A First Course, by Fulton and Harris. Yes, I am assigned this problem as homework. I ask only for help understanding what the claim actually is.

Exercise 2.35: Show that, if the irreducible representations of $G$ are represented by unitary matrices, the matrix entries of these representations form an orthogonal basis for the space of all functions on $G$ ,with inner product given by $$ (\alpha,\beta) = {\frac{1}{|G|}} \sum_{g \in G} \overline{\alpha(g)} \beta(g) $$ Based on context, we should assume $G$ is finite and we're talking about representations over $\mathbb{C}$. My question is, what does this claim mean?

What does it mean that "the irreducible representations of $G$ are represented by unitary matrices"? Does this mean that for each irreducible representation $\rho:G \to \operatorname{GL}_n(\mathbb{C})$, the image is contained in the subgroup $U(n)$ of unitary matrices?

The statement seems to imply that the representations themselves are somehow in correspondence with unitary matrices, not the actions of group elements. In particular, the reference to "matrix entries of these representations" supports that kind of interpretation. I can't make sense of this phrase either.

I assume the phrase "functions on $G$" refers to functions $G \to \mathbb{C}$, but I could be wrong about that.

Any help is much appreciated. Thanks!

Answer 1

Before that exercise, you have exercise 1.14, which states:

Let $V$ be an irreducible representation of the finite group $G$. Show that, up to scalars, there is a unique Hermitian inner product on $V$ preserved by $G$.

So, if $G$ acts on $V$, you can assume that $V$ is endowed with a Hermitian inner product $\langle\cdot,\cdot\rangle$ preserved by $G$, and this means that$$(\forall g\in G)(\forall v,w\in V):\langle gv,gw\rangle=\langle v,w\rangle.$$So, the action of each $g\in G$ is a unitary transformation and therefore, if you fix an orthonormal basis of $V$, you can see the action of $g$ as an unitary matrix (the matrix of the action of $g$ with respect to $b$).

And, yes, here the functions on $G$ are functions from $G$ into $\mathbb C$.

Example: Consider the action $\rho$ of $\mathbb{Z}_3$ on $\mathbb{C}^2$ defined by$$\rho(n)=\begin{pmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{pmatrix}^n$$($n\in\{0,1,2\}$). These matrices are unitary. Let $\alpha\colon\mathbb{Z}_3\longrightarrow\mathbb C$ be the upper left entry of these matrices. Then $\alpha(0)=1$, and $\alpha(1)=\alpha(2)=-\frac12$. And if $\beta\colon\mathbb{Z}_3\longrightarrow\mathbb C$ is the upper right entry of these matrices, then $\beta(0)=0$, $\beta(1)=-\frac{\sqrt3}2$, and $\beta(2)=\frac{\sqrt3}2$.

Answer 2

After I understood what to do (which happened far from immediately), I really enjoyed this exercise.

What the question is asking:

We will assume that the main result of Ex 1.14 holds, i.e. that for every irreducible representation V of a finite group G, up to scalars there is a unique Hermitian inner product on V preserved by G, i.e. $\langle gv, gw \rangle = \langle v, w \rangle$. Later we will also assume Ex 2.34.

Let $V$ be an irreducible representation of $G$ of dimension $n$, and take an orthonormal basis $\{w_1,\ldots,w_{n} \}$. With with respect to this basis, the representation acts as a homomorphism $\rho:G \to \mathcal{U}_{n}$, where $\mathcal{U}_{n}$ is the unitary group of dimension $n$, i.e. the group of $n \times n$ dimensional matrices satisfying $U_{ji} = \bar{U_{i,j}}$.

In particular, for each irreducible representation $V$ of dimension $n_V$, and each pair $1 \leq i,j \leq n_V$, we have a function $U^V_{i,j}:G \to \mathbb{C}$, where $U^V_{i,j}(g)$ is the $(i,j)^{\text{th}}$ slot of the $n_V \times n_V$ matrix representing $g$ with respect to the orthonormal basis of $V$.

Now $\sum_V n_V^2 = |G|$ (sum over distinct irr. reps.), and accordingly $\Gamma := \{ U^V_{i,j} : 1 \leq i,j \leq n, V \text{ irr. rep.} \}$ is a set of $|G|$ functions on $G$. We want to show that these functions span the set $\mathbb{C}[G]$ of complex-valued functions on $G$, which is a $|G|$ dimensional vector space.

The solution:

To prove $\Gamma$ spans $\mathbb{C}[G]$, since $\Gamma$ has cardinality $|G|$, it is sufficient to show $\Gamma$ is linearly independent. Even better, if we define an inner product $(\phi,\psi) := \frac{1}{|G|} \sum_{ g \in G} \overline{\phi(g)} \psi(g)$ on $\mathbb{C}[G]$, it is sufficient to show the distinct functions in $\Gamma$ are orthogonal.

Enter the previous exercise, 2.34. Let $V,W$ be irreducible representations of $G$. Take any linear mapping $T:V \to W$ and define a new linear mapping $L:V \to W$ by $Lv := \frac{1}{|G|} \sum_{g \in G} g^{-1}Tgv$. Then according to Ex 2.34, $L = \frac{ \mathrm{Trace}(T)}{\mathrm{dim}(V)} \mathrm{1}_{V \cong W}$.

Let $V,W$ be irr. reps. of $G$, and with respect to their orthonormal bases above, let $T$ be an $n_W \times n_V$ matrix mapping $V$ to $W$. Let $(v_1,\ldots,v_{n_V})$ be a vector in $V$. Then with respect to these bases, Ex 2.34 reads as the matrix equations \begin{align*} \frac{\mathrm{Trace}(T)}{n_V} \mathbf{v} = \frac{1}{|G|} \sum_{g \in G} U^V(g)^{-1} T U^V(g) \mathbf{v} \qquad \text{(case $V = W$)}, \end{align*} and \begin{align*} 0 = \frac{1}{|G|} \sum_{g \in G} U^W(g)^{-1} T U^V(g) \mathbf{v}. \end{align*} These equations have to hold for all $n_W \times n_V$ matrices $T$, and for all vectors $v$ of length $n_V$. We now show using suitable that this forces $\Gamma$ to be orthogonal. The trick here is to find choices of $v$ and $T$ that pick out the matrix coordinates of the representations.

The $i^{\text{th}}$ slot of the length $n_W$ vector $U^W(g)^{-1} T U^V(g)v$ is given by \begin{align} \label{eq:mat} \left( U^W(g)^{-1} T U^V(g)v \right)_i &= \sum_{ j,k,\ell} U^W(g)^{-1}_{i,j} T_{j,k} U^V(g)_{k,\ell} v_\ell \end{align} where $j,k,\ell$ range over suitable dimensions.

Fix your favourite $a,b,c$ with $1 \leq a \leq n_W$ and $1 \leq b,c \leq n_V$, and take $T_{i,j} := \delta_{i,a}\delta_{j,b}$ and $v_i = \delta_{i,c}$. Then with these choices of $T$ and $v$ the previous equation reads \begin{align*} \left( U^W(g)^{-1} T U^V(g)v \right)_i = U^W(g)^{-1}_{i,a} U^V(g)_{b,c} = \overline{U^W(g)_{a,i}} U^V(g)_{b,c} . \end{align*} Plugging this into the matrix equation with the choice $T^{a,b}_{i,j} := \delta_{i,a}\delta_{j,b}$ and $v_i = \delta_{i,c}$, when $V$ and $W$ are not isomorphic we have \begin{align*} 0 = \sum_{g \in G} \overline{U^W(g)_{a,i}} U^V(g)_{b,c} , \end{align*} i.e. for any $i,a,b,c$ the functions $U^W_{a,i}$ and $U^V_{b,c}$ on $G$ are orthogonal.

As for the case $V = W$, here the trace of the choice of $T$ with $T_{i,j} = \delta_{i,a} \delta_{j,b}$ is given by $\mathrm{Trace}(T) = \delta_{a,b}$. Again taking $v_i = \delta_{i,c}$, the $i^{\text{th}}$ slot of the right-hand-side reads \begin{align*} \left( \frac{\mathrm{Trace}(T)}{n_V} \mathbf{v} \right)_i = \delta_{a,b} \delta_{i,c} / n_V. \end{align*} In this case the matrix equation reads \begin{align*} \frac{ \delta_{a,b}}{ n_V} \delta_{i,c} = \sum_{g \in G} \overline{U^V(g)_{a,i}} U^V(g)_{b,c} , \end{align*} which shows that $U^V(g)_{a,i}$ is orthogonal to $U^V(g)_{b,c}$ unless $a=b$ and $i=c$.

That completes the proof.