Math 202B: Lecture 6

The main objects of study in Math 202B are finite-dimensional algebras. Unlike Hilbert spaces, which were the primary focus of Math 202A, the product in an algebra is vector-valued and not scalar-valued. The question arises as to whether we can unify the two by introducing a scalar product on a given algebra \mathcal{A}, so that it is also a Hilbert space.

Certainly, the answer is yes: since \mathcal{A} is in particular a finite-dimensional vector space, we may simply choose a vector space basis of \mathcal{A} and equip \mathcal{A} with the scalar product in which this basis is orthonormal. However, this scalar product has nothing to do with the algebra structure on \mathcal{A}. We would prefer a Hilbert space structure on \mathcal{A} which interfaces meaningfully with the algebra structure.

For example, we might want to find a scalar product on \mathcal{A} such that the multiplicative identity I \in \mathcal{A} is a unit vector in the corresponding norm. This is easy: choose a basis of \mathcal{A} which contains \mathcal{A} and apply the above construction. But our notion of a scalar product on \mathcal{A} which is compatible with the algebra structure will be much more demanding than this.

Definition 6.1. A Frobenius scalar product on \mathcal{A} is a scalar product which satisfies

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

Definition 6.1 describes a scalar product on \mathcal{A} which satisfies two identities, called the left Frobenius identity and the right Frobenius identity. If \mathcal{A} is the endomorphism algebra of a finite-dimensional Hilbert space, we know that such a scalar product exists from Math 202A, where we constructed the Frobenius scalar product on \mathrm{End}(V) using the scalar product on the underlying Hilbert space V. The question we address now is whether such a scalar product can be obtained more generally, when \mathcal{A} is not necessarily the endomorphism algebra of a Hilbert space.

To explore this question, our first step is to choose a linear functional on \mathcal{A} rather than a linear basis in \mathcal{A}. Indeed, associated to every linear functional

\tau \colon \mathcal{A} \longrightarrow \mathbb{C}

is a sesquilinear form

\langle \cdot,\cdot \rangle_\tau \colon \mathcal{A} \times \mathcal{A} \longrightarrow \mathbb{C}

defined by

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

Here is the computation verifying sesquilinearity. First,

\langle \alpha_1A_1+\alpha_2A_2,\beta_1B_1+\beta_2B_2\rangle_\tau = \tau((\alpha_1A_1+\alpha_2A_2)^*(\beta_1B_1+\beta_2B_2))=\tau(\overline{\alpha}_1\beta_1A_1^*B_1+\overline{\alpha}_1\beta_2A_1^*B_2+\overline{\alpha}_2\beta_1A_2^*B_1+\overline{\alpha}_2\beta_2A_2^*B_2),

which uses both antillinearity of conjugation and bilinearity of multiplication in \mathcal{A}. Second, linearity of \tau gives

\tau(\overline{\alpha}_1\beta_1A_1^*B_1+\overline{\alpha}_1\beta_2A_1^*B_2+\overline{\alpha}_2\beta_1A_2^*B_1+\overline{\alpha}_2\beta_2A_2^*B_2)=\overline{\alpha}_1\beta_1\tau(A_1^*B_1)+\overline{\alpha}_1\beta_2\tau(A_1^*B_2)+\overline{\alpha}_2\beta_1\tau(A_2^*B_1+\overline{\alpha}_2\beta_2\tau(A_2^*B_2).

Third, remembering the definition of \langle \cdot,\cdot \rangle_\tau gives

\langle \alpha_1A_1+\alpha_2A_2,\beta_1B_1+\beta_2B_2\rangle_\tau =\overline{\alpha}_1\beta_1\langle A_1,B_1\rangle_\tau+\overline{\alpha}_1\beta_2\langle A_1,B_2\rangle_\tau+\overline{\alpha}_2\beta_1\langle A_2,B_1\rangle_\tau+\overline{\alpha}_2\beta_2\langle A_2,B_2\rangle_\tau,

which is sesquilinearity.

Since a scalar product is a Hermitian sesquilinear form, so we want it to be the case that

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

coincides with

\overline{\langle B,A \rangle_\tau}=\overline{\tau(B^*A)}.

Since conjugation is antimultiplicative, we have

\overline{\tau(B^*A)}=\overline{\tau((A^*B)^*)},

and we see that the property we really need from \tau is

\tau(A^*)=\overline{\tau(A)}.

So a linear functional \tau which yields a Hermitian form on \mathcal{A} via the recipe \langle A,B \rangle_\tau=\tau(A^*B) must have this special homomorphism-like feature. There is no guarantee that such a functional exists.

Problem 6.1. Show that \tau(A^*)=\overline{\tau(A)} if and only if \tau(X) \in \mathbb{R} for selfadjoint X.

We have now shown how to construct a Hermitian form \langle \cdot,\cdot \rangle_\tau on \mathcal{A} using a linear functional on \langle \cdot,\cdot \rangle_\tau which has the extra feature \tau(A^*)=\overline{\tau(A)}. We also want this form to be nonnegative, meaning that

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

is a nonnegative real number. This is in fact a stronger assumption than \tau(A^*)=\overline{\tau(A)}, as Evan pointed out in lecture.

Problem 6.2. Show that \tau(A^*A) \geq 0 for all A \in \mathcal{A} implies \tau(A^*)=\overline{\tau(A)} for all A \in \mathcal{A}.

Linear functionals on an algebra which are normalized and nonnegative have a special name.

Definition 6.1. A linear functional \tau \colon \mathcal{A} \to \mathbb{C} is called a state if it satisfies \tau(I_\mathcal{A})=1 and \tau(A^*A) \geq 0 for all A \in \mathcal{A}. If moreover \tau(A^*A) =0 implies A=0_\mathcal{A}, then \tau is called a faithful state.

Problem 6.3. Finish the proof that if \tau is a faithful state on \mathcal{A} then \langle \cdot,\cdot \rangle_\tau is a scalar product on \mathcal{A}, and I_\mathcal{A} is a unit vector in the corresponding norm.

Now comes the question of whether the scalar product \langle \cdot,\cdot \rangle_\tau on \mathcal{A} induced by a faithful state \tau \colon \mathcal{A} \to \mathbb{C} is a Frobenius scalar product, as per Definition 6.1. Let us see: we have

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

so we get the left Frobenius identity for free. For the right Frobenius identity, we need it to be the case that

\tau(B^*(A^*C))=\tau((A^*C)B^*),

so we require yet more from \tau.

Definition 6.2. A linear functional \tau \colon \mathcal{A} \to \mathbb{C} is called a trace if it satisfies \tau(AB)=\tau(BA) for all A,B \in \mathcal{A}.

Of course, if \mathcal{A} is a commutative algebra then every linear functional is a trace. If not, there is no reason why a trace need exist.

Definition 6.3. A von Neumann algebra is a pair (\mathcal{A},\tau) consisting of an algebra \mathcal{A} together with a faithful tracial state \tau.

Recall that in Math 202B all algebras are assumed finite-dimensional unless stated otherwise; the same convention applies to von Neumann algebras. Thus, while infinite-dimensional Von Neumann algebras are very interesting objects which have been and continue to be much-studied, they are not on our menu.

We have one example of a von Neumann algebra from Math 202A: the algebra \mathcal{E}(X)=\mathrm{End} \mathcal{F}(X) of linear operators on the function algebra of a finite set X (or equivalently, the endomorphism algebra of any finite-dimensional Hilbert space V, since V contains an orthonormal basis X). In Math 202B, we will soon see a whole new class of von Neumann algebras, namely convolution algebras of finite groups. Abstractly, we can characterize Von Neumann algebras as follows: \mathcal{A} is a von Neumann algebra (i.e. admits a faithful tracial state) if and only if it is isomorphic to a subalgebra of \mathrm{End}(V) for some Hilbert space V. We will prove this next lecture, and this characterization will motivate our quest to classify the subalgebras of \mathrm{End}(V).

To end this lecture, let us consider the existence question for faithful states for our tamest example algebra, namely the function algebra \mathcal{F}(X) of a finite set X. As we have seen, this algebra is very easy to analyze and we can classify faithful states on \mathcal{F}(X) without much difficulty.

Problem 6.4. Show that states on \mathcal{F}(X) are in bijection with probability measures on X. (Hint: think about expected value).

For an abstract, possibly noncommutative algebra \mathcal{A} we cannot make any concrete statements about the existence of states and traces without assuming that \mathcal{A} has additional attributes. However, assuming such functionals exist we can make an important statement about the region of the space of linear functionals on \mathcal{A} which they occupy.

Problem 6.5. Let \mathcal{A} be an algebra such that the sets

\{\text{states on }\mathcal{A}\} \supseteq \{\text{faithful states on }\mathcal{A}\} \supseteq \{\text{faithful tracial states on }\mathcal{A}\}

are nonempty. Show that they are convex subsets of the linear dual of \mathcal{A}.

Assuming the set of states on \mathcal{A} is nonempty, it is a convex set whose extreme points are called pure states.

Problem 6.6. Classify the pure states on \mathcal{F}(X), and show that they are precisely the algebra homomorphisms \mathcal{F}(X) \to \mathbb{C}. (Hint: this will help you to understand the general principle that if expectation is a multiplicative functional, then the underlying distribution must be a delta measure).

Published by Jonathan Novak

Mathematician at UC San Diego.

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