Math 202B: Lecture 8

Brief recap. A scalar product on an algebra $\mathcal{A}$ is called a Frobenius scalar product if it is compatible with the algebra operations in the sense that

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

Since all nonnegative scalings of a given Frobenius scalar product are also Frobenius scalar products, we pin the definition down with a normalization condition: we require that the multiplicative unit $I \in \mathcal{A}$ be a unit vector in the norm induced by a Frobenius scalar product.

Given an arbitrary scalar product on $\mathcal{A}$ we have an associate linear functional defined by

$\tau(A) = \langle I,A \rangle, \quad A \in \mathcal{A}.$

We have shown previously that the scalar product in question is Frobenius if and only if this functional is a faithful tracial state. Conversely, a given linear functional $\tau$ on $\mathcal{A}$ is a faithful tracial state if and only if the the sesquilinear form defined by

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

is a Frobenius scalar product. A weaker version of this correspondence is that scalar products on $\mathcal{A}$ which are required to satisfy the left Frobenius identity but not the right one are in bijection with faithful states which need not be tracial.

We have worked out the details of this correspondence in the case where $\mathcal{A}=\mathcal{F}(X)$ is the function algebra of a finite set $X,$ which is our model example of a commutative algebra. We are presently engaged in classifying Frobenius scalar products in the case where $\mathcal{A}=\mathrm{End}(V)$ is the endomorphism algebra of a finite-dimensional Hilbert space $V.$

We have a scalar product on $\mathcal{A}=\mathrm{End}(V)$ which played a major role in Math 202A, the Hilbert-Schmidt scalar product. Let $X \subset V$ be an orthonormal basis. Let $\{E_{yx} \colon x,y \in X\}$ be the elementary operators in $\mathcal{A}$ associated to the orthonormal basis $X \subset V,$

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

These operators form a vector space basis of $\mathcal{A}=\mathrm{End}(V)$ , and for any $A \in \mathcal{A}$ we have

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

The Hilbert-Schmidt scalar product on $\mathcal{A}$ is defined by declaring this basis orthonormal, so that for any $A,B \in \mathcal{A}$ we have

$\langle A,B \rangle_{HS} = \sum\limits_{x,y \in X} \overline{\langle y,Ax\rangle}\langle x,By\rangle.$

We showed in Math 202A that this can also be computed as the single sum

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

Furthermore, we proved in Math 202A that although this construction depends on the chosen orthonormal basis $X,$ any two orthonormal bases of $V$ in fact yield the same scalar product on $\mathcal{A}=\mathrm{End}(V).$

We now want to show that the normalized Hilbert-Schmidt scalar product

$\langle A,B \rangle_F = \frac{1}{\dim V}\langle A,B \rangle_{HS}$

is a Frobenius scalar product. Equivalently, we want to show that the linear functional defined by

$\mathrm{Tr}(A) = \langle I,A \rangle_{HS}, \quad A \in \mathcal{A},$

which is necessarily a faithful state because $\langle \cdot,\cdot\rangle_{HS}$ is a scalar product, is also a trace.

We showed in Lecture 7 that

$\mathrm{Tr}(A) = \sum\limits_{x \in X} \langle x,Ax\rangle$

is the sum of the diagonal matrix elements of $A$ relative to the orthonormal basis $X \subset V.$ Note that, since we know the Frobenius scalar product is basis-independent, we also have that this diagonal matrix elements sum is the same for any two orthonormal bases of $V.$ In order to prove that the (normalized) Hilbert-Schmidt scalar product is a Frobenius scalar product, it remains only to prove the following deliciously subversive statement.

Theorem 8.1. $\mathrm{Tr}$ is a trace.

Proof: For any $A,B \in \mathcal{A}$ , we have

$\mathrm{Tr}(AB) = \sum\limits_{x \in X} \langle x,ABx\rangle =\sum\limits_{x \in X} \left\langle x,A\sum\limits_{y \in X}\langle y,Bx\rangle y\right\rangle=\sum\limits_{x,y \in X} \langle x,Ay\rangle \langle y,Bx\rangle.$

Performing the same computation with $A$ and $B$ swapped, we have

$\mathrm{Tr}(BA) =\sum\limits_{x,y \in X} \langle x,By\rangle \langle y,Ax\rangle=\sum\limits_{x,y \in X} \langle y,Ax\rangle \langle x,By\rangle=\mathrm{Tr}(AB),$

as claimed. $\square$

This shows that the Hilbert-Schmidt scalar product on $\mathcal{A}=\mathrm{End}(V)$ satisfies the right Frobenius identity as well as the left one, so that its normalized version is indeed a Frobenius scalar product on $\mathcal{A}.$ It is a simple but remarkable fact that there is no other Frobenius scalar product on $\mathcal{A}.$

Theorem 8.2. If $\tau \colon \mathcal{A} \to \mathbb{C}$ is a trace, then there exists $\delta \in \mathbb{C}$ such that $\tau(A)=\delta \mathrm{Tr}(A)$ for all $A \in \mathcal{A}.$

Proof: For any $x,y \in X,$ we have

$\mathrm{Tr}(E_{xx}) = \sum\limits_{z \in Y} \langle z,E_{xx}z \rangle =1 = \sum\limits_{z \in Y} \langle z,E_{yy}z \rangle=\mathrm{Tr}(E_{yy}).$

Now let us compute $\tau(E_{xx)})$ and $\tau(E_{yy})$ . Since we do not have a formula for $\tau$ , we must rely on the hypothesis that it is a trace. We have

$\tau(E_{xx})=\tau(E_{xy}E_{yx})=\tau(E_{yx}E_{xy})=\tau(E_{yy}).$

This shows that, the value of $\tau$ on any two diagonal elementary operators $E_{xx},E_{yy}$ is equal to a common value $\delta.$ Thus,

$\tau(E_{xx}) = \delta \mathrm{Tr}(E_{xx}), \quad x \in X.$

Now suppose that $x,y \in X$ are distinct. Then,

$\mathrm{Tr}(E_{yx}) = \sum\limits_{z \in X} \langle z,E_{yx}z\rangle=\sum\limits_{z \in X} \langle z,y\rangle\langle x,z \rangle=\langle x,y\rangle =0.$

Moreover (thanks Amelia),

$\tau(E_{yx}) = \tau(E_{yz}E_{zx}) = \tau(E_{zx}E_{yz})= \delta \langle x,y\rangle=0.$

Thus, $\tau(A) = \delta \mathrm{Tr}(A)$ for all $A \in \mathcal{A}$ . $\square$

Problem 8.1. Prove that the equation $XY-YX=I$ has no solutions in $\mathcal{A}=\mathrm{End}(V),$ where $V$ is a finite-dimensional Hilbert space.

To sum up, we have shown that

$\tau(A) = \frac{1}{\dim V}\mathrm{Tr}(A) = \frac{1}{\dim V}\sum\limits_{x \in X} \langle x,Ax\rangle,$

which is the average of the diagonal matrix elements of $A \in \mathrm{End}(V)$ with respect to any orthonormal basis $X \subset V,$ is the unique faithful tracial state on $\mathcal{A}=\mathrm{End}(V)$ .

This uniqueness gives us another indication that $\mathrm{End}(V)$ is extremely non-commutative: as soon as $\dim V \geq 2$ this algebra does not admit a homomorphism to the complex numbers. Indeed, suppose

$\chi \colon \mathrm{End}(V) \longrightarrow \mathbb{C}$

is an algebra homomorphism. Then $\chi(I)=1$ and $\chi(AB)=\chi(A)\chi(B)=\chi(BA),$ so every algebra homomorphism from $\mathrm{End}(V)$ to $\mathbb{C}$ is a normalized trace. But we know that $\mathrm{End}(V)$ admits only one such functional.