Math 202B: Lecture 10

Let $\mathcal{A}$ be an algebra which admits a scalar product

$\langle \cdot,\cdot \rangle \colon \mathcal{A} \times \mathcal{A} \longrightarrow \mathbb{C}$

satisfying the left Frobenius identity,

$\langle AB,C \rangle = \langle B,A^*C \rangle, \quad A,B,C \in \mathcal{A}.$

Equivalently, $\mathcal{A}$ admits a faithful state $\tau \colon \mathcal{A} \to \mathbb{C}.$

Theorem 10.1. There exists a Hilbert space $V$ such that $\mathcal{A}$ is isomorphic to a subalgebra of $\mathrm{End}(V).$

Proof: The proof is constructive: we will present an explicit Hilbert space $V$ together with an explicit injective algebra homomorphism

$\Phi \colon \mathcal{A} \longrightarrow \mathrm{End}(V).$

Pursuing the same strategy as in Cayley’s theorem from group theory, we take our Hilbert space $V$ to be $\mathcal{A}$ itself equipped with a left-Frobenius scalar product, which exists by hypothesis. Define a function

$\Phi \colon \mathcal{A} \longrightarrow \mathrm{End}(\mathcal{A})$

$\Phi(A)B = AB, \quad B \in \mathcal{A}.$

Thus, $\Phi(A)$ is the function $\mathcal{A} \to \mathcal{A}$ defined to be “left-multiply by $A$ .”

Let us first verify that $\Phi$ really does take values in $\mathrm{End}(\mathcal{A}),$ as claimed. Thus for $A \in \mathcal{A}$ we must show that $\Phi(A)$ is a linear operator. We have $\Phi(A)0_\mathcal{A}=A0_\mathcal{A}=0_\mathcal{A}$ , by a problem from Week 1, and also

$\Phi(A)(\beta_1B_1+\beta_2B_2)=A(\beta_1B_1+\beta_2B_2)=\beta_1AB_1+\beta_2AB_2=\beta_1\Phi(A)B_1+\beta_2\Phi(A)B_2,$

by bilinearity of multiplication in $\mathcal{A}.$

Now we check that $\Phi$ itself is a linear transformation from $\mathcal{A}$ to $\mathrm{End}(\mathcal{A}).$ For all $B \in \mathcal{A}$ we have $\Phi(0_\mathcal{A})B =0_\mathcal{A}$ and we also have

$\Phi(\alpha_1A_1+\alpha_2A_2)B=\alpha_1\Phi(A_1)B+\alpha_2\Phi(A_2)B$ ,

so indeed $\Phi$ is a linear transformation. Furthermore, if $\Phi(A)B=0_\mathcal{A}$ for all $B \in \mathcal{A},$ then choosing $B=I$ we have $A=0_\mathcal{A}$ so that

$\mathrm{Ker}(\Phi)=\{0_\mathcal{A}).$

Thus, $\Phi \colon \mathcal{A} \to \mathrm{End}(\mathcal{A})$ is an injective linear transformation, hence a vector space isomorphism of $\mathcal{A}$ onto its image in $\mathrm{End}(\mathcal{A}).$

It remains to show that $\Phi$ is an algebra homoprhism. Since $\Phi(I)B=IB=B$ for all $B \in B,$ we have that $\Phi(I)$ is the identity operator in $\mathrm{End}(\mathcal{A}).$ Next, for any $A,B \in \mathcal{A},$ we have

$\Phi(AB)C=(AB)C=A(BC) = A(\Phi(B)C)=\Phi(A)(\Phi(B)C)$

for all $C \in \mathcal{A},$ so $\Phi(AB)=\Phi(A) \circ \Phi(B)$ and multiplication in $\mathrm{End}(\mathcal{A})$ is indeed composition of functions. Finally, from the left Frobenius identity, for any $A \in \mathcal{A}$ we have

$\langle \Phi(A)B,C\rangle=\langle AB,C\rangle = \langle B,A^*C\rangle = \langle B,\Phi(A^*)C\rangle, \quad B,C \in \mathcal{A},$

which shows that $\Phi(A)^*=\Phi(A^*).$ $\square$

Let us look at a specific case of the above construction.

Take $\mathcal{A}=\mathcal{F}(X)$ to be the algebra of functions on a finite set $X.$ Equip $\mathcal{F}(X)$ with the scalar product

$\langle A,B \rangle = \sum\limits_{x \in X} A(x)\overline{B(x)}.$

This scalar product is left-Frobenius: we have

$\langle AB,C\rangle = \sum\limits_{x\in X} \overline{A(x)B(x)}C(x) = \sum\limits_{x\in X} \overline{B(x)} \overline{A(x)}C(x) = \langle B,A^*C\rangle.$

Theorem 10.1 tells us that $\mathcal{F}(X)$ is isomorphic to a subalgebra of

$\mathcal{E}(X) = \mathrm{End} \mathcal{F}(X),$

and in this very concrete setup we can say precisely which subalgebra of $\mathrm{E}(X)$ it is isomorphic to.

Proposition 10.2. The function algebra $\mathcal{F}(X)$ is isomorphic to the diagonal subalgebra $\mathcal{D}(X)$ of the endomorphism algebra $\mathcal{E}(X).$

Proof: Let $\{E_x \colon x \in X\}$ be the orthonormal basis of elementary functions in $\mathcal{F}(X).$ We have then have a corresponding basis of elementary operators $\{E_{yx} \colon x,y \in X\}$ in $\mathcal{E}(X)=\mathcal{F}(X).$ Let $\Phi \colon \mathcal{F}(X) \to \mathcal{E}(X)$ be the left multiplication map, as in the proof of Theorem 10.1. Then, for any $x,y \in X$ we have

$\Phi(E_x)E_y = E_xE_y = \delta_{xy}E_x,$

where we used the fact that the elementary functions in $\mathcal{F}(X)$ are a basis of orthogonal idempotents. On the other hand, we have

$E_{xx}(E_y) = E_x \langle E_x,E_y \rangle = \delta_{xy}E_x.$

This shows that $\Phi(E_x) = E_{xx}.$ $\square$

One may also consider Proposition 10.2 from a matrix perspective. Let us choose an ordering $x_1,\dots,x_n$ of $X.$ Then, the elementary functions $E_{x_1},\dots,E_{x_n}$ become an ordered orthonormal basis of $\mathcal{F}(X),$ and for any function $A \in \mathcal{F}(X)$ we can represent the operator $\Phi(A) \in \mathcal{E}(X)$ as a matrix $[\Phi(A)]$ . The proposition above says that

$[\Phi(A)]=\begin{bmatrix} A(x_1) & {} & {} \\ {} & \ddots & {} \\ {} & {} & A(x_n) \end{bmatrix}.$

Thus, $\Phi$ is sending a function $A$ on $X$ to the diagonal matrix whose diagonal entries are the values of that function. Make sure you understand this.

Theorem 10.1 is our first result in representation theory.

Definition 10.3. A linear representation of an algebra $\mathcal{A}$ is a pair $(V,\Phi)$ consisting of a Hilbert space $V$ together with an algebra homomorphism $\Phi \colon \mathcal{A} \to \mathrm{End}(V)$ . One says that $V$ carries a representation of $\mathcal{A}$ , and refers to $\Phi$ as an action of $\mathcal{A}$ on $V.$

The representation $(V,\Phi)$ of $\mathcal{A}$ constructed in Theorem 10.1 is called the (left) regular representation of $\mathcal{A}$ . The carrier space in this representation is $V=\mathcal{A}$ , and $\mathcal{A}$ acts on itself by left multiplication.

Problem 10.1 (Due Feb 8). Show that if an algebra $\mathcal{A}$ admits a scalar product satisfying either the left-Frobenius identity or the right-Frobenius identity, then it admits a (possibly different) scalar product satisfying both Frobenius identities. In particular, explain why any algebra that admits a faithful state admits a (possibly different) faithful tracial state.

In this lecture we have seen that every von Neumann algebra $\mathcal{A}$ is isomorphic to a subalgebra of $\mathrm{End}(V)$ for some Hilbert space $V$ . Therefore, it is of interest to classify subalgebras of the endomorphism algebra of a finite-dimensional Hilbert space. We will do this next week, explaining how this situation is in some ways analogous, and in other ways quite different, from the classification of subalgebras of $\mathcal{F}(X)$ achieved earlier in the course.

Math 202B: Lecture 10

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