Math 202B: Lecture 20

Let $G$ be a finite group. recall that a unitary representation of $G$ is a pair $(V,\rho)$ consisting of a nonzero Hilbert space $V$ together with a group homomorphism $\varphi \colon G \to U(V),$ where $U(V)$ is the group of unitary operators on $V.$

We are going to spend the rest of the course studying the category $\mathbf{Rep}(G)$ whose objects are finite-dimensional unitary representations of $G.$ Morphisms in $\mathbf{Rep}(G)$ are defined as follows.

Definition 20.1. Let $(V,\rho)$ and $(W,\psi)$ be unitary representations of $G.$ A linear transformation $T \in \mathrm{Hom}(V,W)$ is said to intertwine the actions $\varphi$ and $\psi$ if

$T \circ \varphi(g) = \psi(g) \circ T, \quad g \in G.$

Declare the set of $\mathrm{Hom}_G(V,W)$ of morphisms from $(V,\rho)$ to $(W,\psi)$ to be the set of intertwining maps as above.

We have now defined the category $\mathbf{Rep}(G)$ of unitary representations of $G.$ This category encodes all the ways in which $G$ can act on finite-dimensional Hilbert spaces. It has more objects than the category $\mathbf{FHil}$ of finite-dimensional Hilbert spaces, but its Hom-spaces are smaller.

Problem 20.1. Prove that $\mathrm{Hom}_G(V,W)$ is a subspace of $\mathrm{Hom}(V,W).$ Moreover, prove that if $T \in \mathrm{Hom}_G(V,W)$ then $T^* \in \mathrm{Hom}_G(W,V).$

Now let us discuss a few ways in which the definition of $\mathrm{Hom}_G(V,W)$ is well-chosen.

First consider $\mathrm{End}_G(V)=\mathrm{Hom}_G(V,V),$ the space of endomorphisms of a unitary representation $(V,\varphi)$ of $G.$ This is the subspace of $\mathrm{End}(V)=\mathrm{Hom}(V,V)$ consisting of linear operators $A$ on $V$ which commute with each of the unitary operators $\varphi(g),$

$T \varphi(g) = \varphi(g)T, \quad g \in G.$

Equivalently, if we let $(V,\Phi)$ be the linear representation of $\mathcal{C}(G)$ corresponding to the unitary representation $(V,\phi)$ via $\Phi(E_g)=\varphi(g)$ , then $\mathrm{End}_G(V)$ is the centralizer of the image of $\mathcal{C}(G)$ in $\mathrm{End}(V)$ under $\Phi.$ Thus, since the centralizer of any subalgebra is again a subalgebra, we have that $\mathrm{End}_G(V)$ is a subalgebra of $\mathrm{End}(V).$

Isomorphisms in $\mathbf{FHIl}$ are easy to understand: $\mathrm{Iso}(V,W)$ is nonempty if and only if $\dim V=\dim W,$ and it consists of all linear bijections $T \colon V \to W.$ We know from Math 202A that if $\mathrm{Iso}(V,W)$ is nonempty then there exists an isometry $U \in \mathrm{Iso}(V,W).$ The set $\mathrm{Iso}_G(V,W)$ of isomorphisms between unitary representations $(V,\varphi)$ and $(W,\psi)$ is more constrained, since it consists of linear bijections $T \colon V \to W$ which intertwine the actions $\varphi$ and $\psi,$ and it may be empty even if a linear bijection $V \to W$ exists.

Problem 20.2. Prove that $\mathrm{Iso}_G(V,W)$ is nonempty if and only if it contains an isometry $U \colon V \to W.$ (This can be done using either singular value decomposition or polar decomposition).

Problem 20.2 gives a matrix interpretation of what it means for two unitary representations $(V,\varphi)$ and $(W,\psi)$ to be isomorphic. Namely, let $U \in \mathrm{Iso}_G(V,W)$ be an isometry. Choose any orthonormal basis $X \subset V,$ and let $Y = \{Ux \colon x \in V\}$ be the corresponding orthonormal basis of $W.$ Then, we have

$\langle y',\psi(g)y\rangle_W = \langle Ux',\psi(g)Ux\rangle_W = \langle Ux',U\varphi(g)x\rangle_W=\langle x',\varphi(g)x\rangle_V$ .

In other words, for any ordering $x_1,\dots,x_n$ , taking the corresponding ordering $Ux_1,\dots,Ux_n$ of $Y$ we have that the $n \times n$ unitary matrices $[\varphi(g)]_X$ and $[\psi(g)]_Y$ coincide for all $g \in G.$ So, isomorphic representations of $G$ represent the elements of $G$ as exactly the same unitary matrices. An obvious but important consequence of this is the following.

Theorem 20.1. Isomorphic objects in $\mathbf{Rep}(G)$ have the same character.

One of the many miracles of group representation theory is that the converse of Theorem 20.1 holds: the character of a unitary representation characterizes it up to isomorphism (hence the name). As a first step towards establishing this result, we consider another important aspect of $\mathbf{Rep}(G)$ , namely that given any two objects we can build a third as follow.

Definition 20.2. Let $(V,\rho)$ and $(W,\psi)$ be unitary representations of $G.$ The corresponding adjoint representation $(\mathrm{Hom}(V,W),\omega)$ is defined by

$\omega(g)T=\psi(g)T\varphi(g^{-1}), \quad g \in G.$

Let us place this construction in a more general framework. Given two finite-dimensional Hilbert spaces $V$ and $W,$ equip $\mathrm{Hom}(V,W)$ with the Hilbert-Schmidt scalar product.

Definition 20.3. The adjoint action of $U(V) \times U(W)$ on $\mathrm{Hom}(V,W)$ is the group homomorphism

$\omega \colon U(V) \times U(W) \longrightarrow U(\mathrm{Hom}(V,W))$

defined by

$\omega(A,B)T=BTA^*.$

Problem 20.3. Check that $\omega(A,B)$ is indeed a unitary operator on $\mathrm{Hom}(V,W)$ , and that $\omega$ is indeed a group homomorphism.

Now let us calculate the matrix elements of $\omega(A,B) \in U(\mathrm{Hom}(V,W))$ in terms of the matrix elements of $A \in U(V)$ and $B\in U(W).$ To do so, let $X \subset V$ and $Y \subset W$ be orthonormal bases, and let $\{E_{yx} \colon x \in X, y \in Y\}$ be the corresponding orthonormal basis of $\mathrm{Hom}(V,W).$

Proposition 20.3. We have

$\langle E_{y'x'},\omega(A,B)E_{yx}\rangle = \overline{\langle x',Ax\rangle}{\langle y',By\rangle}.$

Proof: We have

$\langle E_{y'x'},\omega(A,B)E_{yx}\rangle =\mathrm{Tr}\ E_{y'x'}^*BE_{yx}A^*=\mathrm{Tr}\ BE_{yx}A^*E_{x'y'},$

which is the trace of an operator in $\mathrm{End}(W).$ We compute this trace as

$\mathrm{Tr}\ BE_{yx}A^*E_{x'y'}=\sum\limits_{w \in Y} \langle w,BE_{yx}A^*E_{x'y'}w\rangle = \langle y',BE_{yx}A^*x'\rangle.$

Now expand

$A^*x' = \sum\limits_{x'' \in X} \langle x'',A^*x'\rangle x''$

to get

$\langle y',BE_{yx}A^*x'\rangle=\sum\limits_{x'' \in X} \langle y',BE_{yx}x''\rangle \langle x'',A^*x'\rangle=\langle y',By\rangle \langle x,A^*x'\rangle = \overline{\langle x',Ax \rangle}\langle y',By\rangle,$

which completes the proof. $\square$

Corollary 20.4. We have

$\mathrm{Tr}\omega(A,B) = \overline{\mathrm{Tr} A}\ \mathrm{Tr}B.$

Proof: Proposition 20.3 gives

$\mathrm{Tr}\omega(A,B) = \sum\limits_{x \in X}\sum\limits_{y \in Y}\langle E_{yx},\omega(A,B)E_{yx}\rangle =\sum\limits_{x \in X}\sum\limits_{y \in Y} \overline{\langle x,Ax \rangle}\langle y,By\rangle=\overline{\mathrm{Tr} A}\ \mathrm{Tr}B,$

as claimed. $\square$

From the above, we get the following relationship between the characters of two given unitary representations $(V,\varphi)$ and $(W,\psi),$ and the character of the corresponding adjoint representation $(\mathrm{Hom}(V,W),\omega).$