Math 202B: Lecture 16

Let \mathcal{A} be an algebra.

Definition 16.1. A linear representation of \mathcal{A} is a pair (V,\rho) consisting of a Hilbert space V together with an algebra homomorphism \rho \colon \mathcal{A} \to \mathcal{L}(V).

Note that Definition 16.1 does not stipulate that \rho be an algebra isomorphism, so \mathcal{B}=\rho(\mathcal{A}) is just a homomorphic image of \mathcal{A} inside \mathcal{L}(V). In that sense, the term “representation” is a bit misleading. When \rho is an algebra isomorphism, the representation (V,\rho) is said to be faithful.

We saw a special case of Definition 16.1, where we considered the special case of a subalgebra \mathcal{A}_0 of \mathcal{L}(V_0) for some given Hilbert space V_0. Then, Burnside’s theorem says that either \mathcal{A}_0 = \mathcal{L}(V_0) or there is a proper non-trivial subspace V \leq V_0 invariant under all operators A \in \mathcal{A}_0. In this situation, (V,\rho) is a linear representation of \mathcal{A}_0, where \rho \colon \mathcal{A}_0 \to \mathcal{L}(V) is defined by declaring \rho(A) to be the restriction of A to the invariant subspace V \leq V_0. In the case where \mathcal{A}_0=\mathcal{L}(V_0), the pair (V_0,\rho_0) is a linear representation of \mathcal{A}_0 with \rho_0(A)=A the identity homomorphism; this is called the tautological representation or defining representation of \mathcal{A}_0.

Definition 16.2. If (V,\rho) and (W,\sigma) are representations of \mathcal{A}, a linear transformation T \colon V \to W is said to be a homomorphism of representations (or an intertwining transformation) if

T \circ \rho(A) = \sigma(A) \circ T, \quad\text{for all} A \in \mathcal{A}.

The set of all such linear transformations is denoted \mathrm{Hom}_\mathcal{A}(V,W).

Problem 16.1. Prove that \mathrm{Hom}_\mathcal{A}(V,W) is a vector subspace of the space \mathrm{Hom}(V,W) of all linear transformations V \to W.

Definition 16.3. A linear isomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is said to be an isomorphism of representations. When such an intertwining linear isomorphism exists, we say that (V,\rho) and (W,\sigma) are isomorphic representation of \mathcal{A}.

We also encountered Definition 16.3 in a special case in Lecture 15, when we discussed the classification of subalgebras of \mathcal{L}(V).

Problem 16.2. Let (V,\rho) and (W,\sigma) be linear representations of \mathcal{A}. Prove that they are isomorphic representations if and only if there exists a basis \{v_1,\dots,v_n\} of V and a basis \{w_1,\dots,w_n\} of W such that the matrix of \rho(A) in the v-basis of V equals the matrix of \sigma(A) in the w-basis of W, for every algebra element A \in \mathcal{A}.

Let us fix a linear representation (V,\rho) of \mathcal{A} and consider its image \mathcal{B}:=\rho(\mathcal{A})=\{\rho(A) \colon A \in \mathcal{A}\}, which is a subalgebra of \mathcal{L}(V). Recall from Week One (or maybe Week Two) that the centralizer \mathcal{C}=Z(\mathcal{B},\mathcal{L}(V)) is the subalgebra consisting of all operators C \in \mathcal{L}(V) that commute with every operator B \in \mathcal{B}.

Proposition 16.1. We have \mathcal{C} = \mathrm{Hom}_\mathcal{A}(V,V).

Proof: We have C \in \mathcal{C} if and only if C\rho(A)=\rho(A)C, which is exactly the condition for C to be an element of \mathrm{Hom}_\mathcal{A}(V,V).

-QED

Definition 16.4. A linear representation (V,\rho) of \mathcal{A} is said to be irreducible if the only \mathcal{B}-invariant subspaces of V are the zero space and the total space (where once again \mathcal{B} is the image of \mathcal{A} in \mathcal{L}(V) under \rho).

Problem 16.3. Prove that (V,\rho) is irreducible if and only if there does not exist a basis \{v_1,\dots,v_n\} of V such that the matrix of every \rho(A) in the v-basis has block diagonal form, with blocks of positive size.

Proposition 16.2. If (V,\rho) is an irreducible representation of \mathcal{A}, then \mathcal{B}=\mathcal{L}(V).

Proof: This is Burnside’s theorem from Lecture 15: the only subalgebra of \mathcal{L}(V) with exactly two invariant subspaces is \mathcal{B}=\mathcal{L}(V).

-QED

Continuing on with (V,\rho) an irreducible representation of \mathcal{A}, and maintaining the notation \mathcal{C}=Z(\mathcal{B},\mathcal{L}(V))=\mathrm{Hom}_\mathcal{A}(V,V), we have the following.

Corollary 16.3. \mathcal{C}=\mathbb{C}I.

Proof: Since \mathcal{C} is the centralizer of \mathcal{B}=\mathcal{L}(V), the statement is equivalent to the fact that the center of \mathcal{L}(V) consists of scalar multiples of the identity, which we proved in Lecture 15.

-QED

We can refine the above corollary to make the following statement about intertwining maps between two possibly distince representations (V,\rho) and (W,\sigma) of \mathcal{A}, at least one of which is irreducible.

Theorem 16.4. (Schur’s Lemma) If (V,\rho) is irreducible, then every nonzero homomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is injective. If (W,\sigma) is irreducible, then every nonzero homomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is surjective. If both (V,\rho) and (W,\sigma) are irreducible, then every nonzero T \in \mathrm{Hom}_\mathcal{A}(V,W) is an isomorphism of representations.

Proof: Suppose (V,\rho) is irreducible, and consider the kernel of any T \in \mathrm{Hom}_\mathcal{A}(V,W). We claim that \mathrm{Ker}(T) is a \rho(\mathcal{A})-invariant subspace of V. Indeed, if v \in \mathrm{Ker}(T), then for any A \in \mathcal{A} we have

T\rho(A)v=\sigma(A)Tv = \sigma(A)0_W = 0_W,

which shows that \rho(A)v is again in \mathrm{Ker}(T). Since (V,\rho) is irreducible we have either \mathrm{Ker}(T)=\{0_V\} or \mathrm{Ker}(T)=V, and since T is not the zero map we must have T injective.

Now suppose (W,\sigma) is irreducible. Then, we claim that the image of V under T\in \mathrm{Hom}_\mathcal{A}(V,W) is a \sigma(\mathcal{A})-invariant subspace of W. Indeed, suppose w \in \mathrm{Im}(T), i.e. w=Tv for v \in V. Then, for any A \in \mathcal{A} we have that

\sigma(A)w=\sigma(A)Tv=T(\rho(A)v),

showing that \sigma(A)w remains in the image of T. Since T is nonzero and (W,\sigma) is irreducible, we have \mathrm{Im}(T)=W as claimed.

The final statement follows from the two arguments above: any nonzero homomorphism of irreducible representations must be an isomorphism. Equivalently, there does not exist a nonzero homomorphism between two non-isomorphic irreducible representations.

-QED

Published by Jonathan Novak

Mathematician at UC San Diego.

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