Math 202B: Lecture 7

*** Problems in this lecture due Feb. 1 ***

Let $\mathcal{A}$ be an algebra. In Lecture 6, we introduced the notion of a Frobenius scalar product on $\mathcal{A}.$ This is by definition a scalar product on $\mathcal{A}$ which is compatible with its algebra structure in the sense that

$\langle B,A^*C\rangle =\langle AB,C\rangle = \langle A,CB^*\rangle$

holds for all $A,B,C \in \mathcal{A}.$ The first equality above is called the left Frobenius identity, and the second is called the right Frobenius identity. By scaling, we may assume that the multiplicative unit $I \in \mathcal{A}$ is a unit vector with respect to the corresponding norm, and we build this normalization condition into the definition of a Frobenius scalar product.

The upshot of Lecture 6 is that the existence of a Frobenius scalar product on $\mathcal{A}$ is equivalent to the existence of a special kind of linear functional on $\mathcal{A},$ namely a faithful tracial state.

Theorem 7.1. An algebra $\mathcal{A}$ admits a Frobenius scalar product if and only if it admits a faithful tracial state.

Proof: In Lecture 6, we showed that if $\tau$ is a faithful tracial state on $\mathcal{A}$ then

$\langle A,B \rangle_\tau = \tau(A^*B)$

defines a Frobenius scalar product on $\mathcal{A}.$ Conversely, suppose we have a Frobenius scalar product on $\mathcal{A}$ and define a corresponding linear functional by

$\tau(A) = \langle I,A \rangle.$

Applying the left Frobenius identity, we have

$\tau(A^*A) = \langle I,A^*A\rangle = \langle A,A \rangle \geq 0,$

with equality if and only if $A = 0_\mathcal{A}.$ This shows that $\tau$ is a faithful state on $\mathcal{A}$ . Furthermore, the left Frobenius identity gives

$\tau(AB) = \langle I,AB\rangle = \langle A^*I,B \rangle=\langle A^*,B\rangle$

and the right Frobenius identity gives

$\tau(BA) = \langle I,BA \rangle = \langle IA^*,B\rangle =\langle A^*,B\rangle,$

which shows that $\tau$ is a trace. $\square$

Note that in the above argument shows that existence of a scalar product on $\mathcal{A}$ which need only verify the left Frobenius scalar product is equivalent to existence of a faithful but not necessarily tracial state on $\mathcal{A}.$

Definition 7.2. A von Neumann algebra is an algebra $\mathcal{A}$ equipped with a Frobenius scalar product. Equivalently, a von Neumann algebra is an algebra $\mathcal{A}$ equipped with a faithful tracial state.

In Lecture 6, we classified states on the function algebra $\mathcal{F}(X)$ of a finite set $X,$ showing that they are in bijection with probability measures on $X.$ Under this bijection, faithful states correspond to probability measures whose support is all of $X.$ The trace notion is irrelevant because $\mathcal{F}(X)$ is commutative.

Problem 7.1. Show that the normalized $L^2$ -scalar product

$\langle A,B \rangle = \frac{1}{|X|}\sum\limits_{x \in X} \overline{A(x)}B(x),$

is a Frobenius scalar product on $\mathcal{F}(X)$ . Which probability measure on $X$ does it correspond to?

As we have stressed from the beginning of Math 202B, $\mathcal{F}(X)$ is the fundamental example of a commutative algebra. The fundamental example of a noncommutative algebra is $\mathcal{E}(X)$ = \mathrm{End}\mathcal{F}(X)$, the algebra of linear operators on the Hilbert space $\mathcal{F}(X)=L^2(X).$ In this lecture, we will classify states, faithful states, and faithful tracial states on $\mathcal{E}(X).$

For notational purposes, it is convenient to view $\mathcal{F}(X)$ as a Hilbert space $V$ containing the finite set $X$ as an orthonormal basis – this is the algebraist’s notation, where we identify the elementary function $E_x$ with $latex,$ so that the decomposition

$A = \sum\limits_{x \in X} A(x)E_x$

of a function $A$ on $X$ is identified with a formal linear combination

$A \sum\limits_{x \in X} \alpha_x x$

of the points of $X$ . Then, $\mathcal{E}(X) = \mathrm{End}(V)$ is the vector space of all linear operators on the Hilbert space $V.$ We are now considering not just the vector space structure of $\mathrm{End}(V)=\mathcal{E}(X)$ , but its algebra structure, where multiplication is composition and conjugation is adjoint.

Let us briefly review the basic aspects of $\mathrm{End}(V)$ familiar from Math 202A, where we analyzed its vector space structure. In particular, a basis of $\{E_{yx} \colon x,y \in X\}$ is given by the elementary operators

$E_{yx}v = y \langle x,v\rangle, \quad v \in V,$

and the expansion of any $A \in \mathrm{End}(V)$ in the elementary basis is

$A=\sum\limits_{x,y \in X} \langle y,Ax \rangle E_{yx},$

where the scalar product is that in the underlying Hilbert space $V.$ This is nothing more or less than saying that the matrix of the elementary operator $E_{yx}$ with respect to the orthonormal basis $X \subset V$ is the elementary matrix with a single $1$ into row $y$ and column $x$ and all other entries equal to $0,$ and that every matrix can be written as a linear combination of elementary matrices. The advantage to doing things our way is that we don’t need to choose an ordering of the basis $X$ and keep track of indices.

For the purposes of Math 202B, we also want to know how the elementary operators behave with respect to conjugation and multiplication.

Proposition 7.3. We have

$E_{yx}^*=E_{xy} \quad\text{and}\quad E_{zy}E_{xw} = \langle y,x\rangle E_{zw}.$

Proof: Compare two calculations: first

$\langle w,E_{yx}v\rangle = \langle w,y \langle x,v\rangle\rangle= \langle x,v \rangle \langle w,y\rangle,$

and second

$\langle E_{xy}w,v \rangle = \langle x \langle y,w\rangle,v\rangle=\langle x,v \rangle \overline{\langle y,w\rangle}=\langle x,v \rangle \langle w,y\rangle.$

The fact that these two computations produce the same result proves that $E_{yx}^*=E_{xy}.$ For the multiplication rule, we have

$E_{zy}E_{xw}v=E_{zy}x\langle w,v \rangle = z\langle y,x\rangle\langle w,v\rangle,$

and also

$\langle y,x\rangle E_{zw}v=\langle y,x\rangle z \langle w,v\rangle,$

which coincide. $\square$

From Proposition 7.3, we get that $\{E_{xx} \colon x \in X\}$ is a set of orthogonal selfadjoint idempotents which span the space of operators acting diagonally on the basis $X.$ So, we have associated to every finite set $X$ three algebras,

$\mathcal{F}(X) = \mathrm{Span}\{E_x \colon x \in X\},$

$\mathcal{E}(X)=\mathrm{Span}\{E_{yx} \colon x,y \in X\},$

$\mathcal{D}(X) = \mathrm{Span}\{E_{xx} \colon x \in X\},$

related as follows.

Problem 7.2. Prove that $\mathcal{D}(X)$ is isomorphic to $\mathcal{F}(X)$ , and that $\mathcal{D}(X)$ a maximal abelian subalgebra of $\mathcal{E}(X) = \mathrm{End}\, \mathcal{F}(X).$

You may wish to ponder the above, rewrite it in various ways, think of matrices versus operators, etc. At some point I want to be able to make statements like “consider the maximal abelian subalgebra of the symmetric group algebra consisting of all operators acting diagonally in the Young basis,” and I want you to have the muscles required to lift this heavy statement off the board and drop it in your head.

Coming back to Frobenius scalar products, in Math 202A we put a scalar product on $\mathrm{End}(V)=\mathcal{E}(X)$ by declaring the elementary basis to be orthonormal.

Definition 7.4. The Hilbert-Schmidt scalar $\langle \cdot,\cdot\rangle_{HS}$ product on $\mathrm{End}(V)$ is the scalar product in which $\{E_{yx} \colon x,y \in X\}$ is an orthonormal basis,

$\langle E_{zy},E_{xw} \rangle_{HS} = \langle z,x\rangle \langle y,w\rangle.$

As we showed in Math 202A, the above definition leads easily to the following formula for calculating the Hilbert-Schmidt scalar product of any two operators in terms of the scalar product in the underlying Hilbert space.

$\langle A,B \rangle_{HS}= \sum\limits_{x \in X} \langle Ax,Bx\rangle,$

and we used this scalar product for various linear algebraic purposes. Now we want to show that, up to a minor detail, the Hilbert-Schmidt scalar product on $\mathrm{End}(V)$ is a Frobenius scalar product, and in fact it is the only Frobenius scalar product on the full operator algebra $\mathrm{End}(V)$ . The minor detail is that

$\langle I,I \rangle_{HS} = \sum\limits_{x \in X} \langle E_{xx},E_{xx}\rangle_{HS}=\dim V,$

so that the identity operator $I \in \mathrm{End}(V)$ is not a unit vector in the Hilbert-schmidt norm $\|A\|_{HS}=\sqrt{\langle A,A\rangle_{HS}}.$ Therefore, we will normalize and define

$\langle A,B\rangle_F := \frac{1}{\dim V}\langle A,B\rangle_{HS}, \quad A,B \in \mathrm{End}(V).$

We will prove the following on Monday.

Theorem 7.5. The normalized Hilbert-Schmidt scalar product $\langle \cdot,\cdot \rangle_F$ is the unique Frobenius scalar product on $\mathrm{End}(V).$

Math 202B: Lecture 7

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