Math 202B: Lecture 13

***All problems in this lecture due Feb. 15 at 23:59***

Let G be a finite group. We have been comparing and contrasting the function algebra \mathcal{F}(G) and the convolution algebra \mathcal{C}(G). The function algebra \mathcal{F}(G) sees G as a finite set, and does not interact with its group structure. The convolution algebra does interact with the group structure of G, and in particular is commutative if and only if G is abelian.

We understand subalgebras of \mathcal{F}(G) very well: they are in bijection with partitions of G. Namely, if \mathfrak{p} is a partition of G, then we have a corresponding subalgebra of \mathcal{F}(G), the set of functions on G which are constant on the blocks of \mathfrak{p}.

Our understanding of subalgebras of \mathcal{C}(G) is less complete. We know that every subgroup H of G gives rise to a corresponding subalgebra \mathcal{A}(H) of \mathcal{C}(G), namely the set of functions on G which vanish outside H. We also used the averaging trick to construct a two-dimensional subalgebra of \mathcal{C}(G) of a different kind.

Since pointwise multiplication of functions and convolution of functions are very different operations, there is no reason to expect that \mathcal{A}(\mathfrak{p}) should be closed under convolution. But perhaps this could be the case for some “special” partition of G which arises from its group structure.

Partitions of G are the same thing as equivalence relations on G. In the context of group theory, there is a particularly important equivalence relation: conjugacy. By definition, two points g,h \in G are conjugate to one another if and only if g = khk^{-1} for some k \in G. This equivalence relation defines a partition of G whose blocks are called conjugacy classes. In group theory, the significance of this concept is that a subgroup of G is normal if and only if it is a union of conjugacy classes. We will now consider the partition of G into conjugacy classes from the algebra perspective. Functions on G which are constant on conjugacy classes are know as class functions. Could it be the case that the set of class functions on G is a subalgbra of G?

Let \{C_\alpha \colon \alpha \in \Lambda\} be the partition of G into conjugacy classes. Here \Lambda is a finite set of labels which parameterize the blocks of this partition.

Problem 13.1. Prove that |\Lambda|=|G| if and only if G is abelian, and that |\Lambda|=1 if and only if |G|=1. Next, prove that the cardinality of the set \{(g,h) \in G \times G \colon gh=hg\} of commuting pairs of elements from G is |G||\Lambda|.

We have defined class functions to be those functions on G which are constant on its conjugacy classes. A useful alternative characterization of such functions is that they are precisely those functions which are insensitive to any noncommutativity in G.

Problem 13.2. A function A \in \mathcal{C}(G) is a class function if and only if A(gh)=A(hg).

Now we return to our original question: is the set of class functions on G a subalgebra of \mathcal{C}(G)? The answer is yes, as the following theorem shows.

Theorem 13.1. The set of class functions on G is precisely the center of \mathcal{C}(G).

Proof: Suppose first that A \in \mathcal{C}(G) is a class function, and let B \in \mathcal{C}(G) be any function. We want to prove that A commutes with B. According to the definition of convolution, we have

[AB](g) = \sum\limits_{h \in G} A(gh^{-1})B(h) = \sum\limits_{h \in G} A(g(gh^{-1})^{-1})B(gh^{-1})=\sum\limits_{h \in G}B(gh^{-1})A(ghg^{-1}),

where to get the second equality we used the substation h \rightsquigarrow gh^{-1} and to get the third inequality we used the commutativity of \mathbb{C}. Since A is a class function, we conclude that

[AB](g) =\sum\limits_{h \in G}B(gh^{-1})A(ghg^{-1})=\sum\limits_{h \in G}B(gh^{-1})A(h)=[BA](g).

Conversely, suppose that A \in \mathcal{C}(g) is a central function on G. Then, it commutes with each of the elementary functions E_g, which form a basis of \mathcal{C}(G) as g ranges over G. Now observe that

[AE_g](h) = \sum\limits_{k \in G} A(hk^{-1})E_g(k)=A(hg^{-1}),

whereas

[E_gA](h)=\sum\limits_{k \in G} E_g(hk^{-1})A(k)=A(g^{-1}h).

Since these are equal due to the centrality of A, Problem 13.2 shows that A is a class function. \square

The fact that the set of class functions is a subalgebra of \mathcal{C}(G) now follows from the fact that the center of any algebra is a subalgebra thereof. Let us write \mathcal{K}(G):=Z(\mathcal{C}(G)) for brevity.

Theorem 13.2. The dimension of \mathcal{K}(G) is equal to |\Lambda|, the number of conjugacy classes in G.

Proof: Let

K_\alpha = \sum\limits_{g \in C_\alpha} E_g

be the indicator function of the conjugacy class C_\alpha. Then, the set \{K_\alpha \colon \alpha \in \Lambda\} spans \mathcal{K}(G). Moreover, for the $L^2$-scalar product on \mathcal{C}(G) we have

\langle K_\alpha,K_\beta\rangle = \sum\limits_{g \in C_\alpha, h \in C_\beta} \langle E_g,E_h\rangle = |C_\alpha \colon C_\beta|.

Since \{C_\alpha \colon \alpha \in \Lambda\} is a partition of G, this gives

\langle K_\alpha,K_\beta\rangle = \delta_{\alpha\beta}|C_\alpha|,

hence \{K_\alpha \colon \alpha \in \Lambda\} is an orthogonal set of nonzero functions. \square

It is useful to think of \mathcal{K}(G) and \mathcal{C}(G) on the same footing, as two algebras associated to any finite group. The class algebra \mathcal{K}(G) is always commutative, while the convolution algebra \mathcal{C}(G) is commutative precisely when G is abelian, and coincides with \mathcal{K}(G) in this case. Any group G with at least two elements has at least two conjugacy classes, hence the dimension of \mathcal{K}(G) is at least two – the convolution algebra \mathcal{C}(G) of a nontrivial group can be noncommutative, but not a noncommutative as the endomorphism algebra of a Hilbert space, whose center is one-dimensional.

An interesting aspect of the class algebra \mathcal{K}(G) is that the connection coefficients of the class basis \{K_\alpha \colon \alpha \in \Lambda\} count solutions to equations in G. More precisely, consider the |\Lambda|^3 structure constants defined by

K_\alpha K_\beta = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}K_\gamma, \quad \alpha,\beta \in \Lambda.

Problem 13.3. Let k \in C_\gamma. Show that

c_{\alpha\beta\gamma}=|\{(x,y) \in C_\alpha \times C_\beta \colon xy=k\}|.

Math 202B: Lecture 12

Let G be a group with group law written multiplicatively: for any elements g,h \in G, their product is gh \in G. Since G is a finite set, we can form the algebra \mathcal{F}(G) of functions A \colon G \to \mathbb{C}. As we know, this function algebra has a natural basis consisting of the elementary functions

E_g(h) = \delta_{gh},\quad g,h \in G

which are selfadjoint orthogonal idempotents in \mathcal{F}(G). However, the function algebra \mathcal{F}(G) has nothing to do with the group structure of G, so from the point of view of this construction the fact that G is a group and not just a set is irrelevant.

We can use the underlying group structure to define a different algebra \mathcal{C}(G) whose elements are functions A \colon G \to \mathbb{C}, but whose operations differ from those of \mathcal{F}(G) and are defined using the group structure of G. In \mathcal{C}(G), multiplication of functions not defined pointwise, but rather by convolution: we declare

E_gE_h = E_{gh}, \quad g,h \in G.

Thus, the multiplication tensor of \mathcal{C}(G) is not the three-dimensional identity matrix, but rather a two-dimensional object: the multiplication table of G. Thus for any two functions

A = \sum\limits_{g \in G} \alpha_g E_g \quad\text{and}\quad B=\sum\limits_{g \in G}\beta_gE_g,

their convolution product is given by

AB=\sum\limits_{g,h \in G} \alpha_g\beta_h E_{gh},

which can equivalently be written

AB = \sum\limits_{k \in G} \left(\sum\limits_{gh=k}\alpha_g\beta_h \right)E_k,

or

AB= \sum\limits_{g \in G} \left(\sum\limits_{h \in G}\alpha_{gh^{-1}}\beta_h \right)E_g

or

AB= \sum\limits_{g \in G} \left(\sum\limits_{h \in G}A(gh^{-1})B(h) \right)E_g.

This is very different from the product of A and B viewed as elements of \mathcal{F}(G), which is

AB=\sum\limits_{g \in G} A(g)B(g)E_g.

Similarly, instead of defining conjugation in \mathcal{C}(G) using pointwise using conjugation in \mathbb{C}, we define it using the the natural involution in G, which is taking inverses,

E_g^* = E_{g^{-1}}, \quad g \in G,

and extend this antilinearly,

A^* = \left(\sum\limits_{g \in G} \alpha_g E_g\right)^*=\sum\limits_{g \in G}\bar{\alpha}_g E_{g^{-1}} = \sum\limits_{g \in G} \bar{\alpha}_{g^{-1}}E_g.

This group-theoretic conjugation is indeed antimultiplicative with respect to convolution,

(E_g*E_h)^* = E_{gh}^*=E_{(gh)^{-1}}=E_{h^{-1}g^{-1}} = E_{h^{-1}}*E_{g^{-1}}=E_h^**E_g^*,

as well as involutive,

(E_g^*)^* = E_{g^{-1}}^*=E_{(g^{-1})^{-1}}= E_g.

Finally, the multiplicative unit in \mathcal{C}(G) is I= E_e, where e \in G is the group unit, as opposed to being the constant function equal to 1 on every g \in G, which is the multiplicative unit in \mathcal{F}(G).

Let us record the above as an official definition.

Definition 12.1. The convolution algebra (aka a group algebra) of a group G is the vector space

\mathcal{C}(G) = \{A \colon G \to \mathbb{C}\}

of complex-valued functions on G with multiplication defined by

[AB](g) = \sum\limits_{h \in G}A(gh^{-1})B(h)

and conjugation defined by

[A^*](g) = \overline{A(g^{-1})}.

In summary, \mathcal{F}(G) and \mathcal{C}(G) are exactly the same as vector spaces, but they are different as algebras. In particular, the elementary functions \{E_g \colon g \in G\} are a basis of \mathcal{F}(G) consisting of orthogonal projections, but in \mathcal{C}(G) they form a basis consisting of unitary elements.

Problem 12.1. Prove that if G,H are isomorphic groups then their convolution algebras \mathcal{C}(G) and \mathcal{C}(H) are isomorphic algebras (the converse is false, bonus points if you give a counterexample). Also show that \mathcal{C}(G) is commutative if and only if G is abelian.

To further highlight the difference between \mathcal{C}(G) and \mathcal{F}(G), let us consider their selfadjoint elements. To say that a function A \colon G \to \mathbb{C} is selfadjoint as an element of \mathcal{F}(G) simply means that it is real-valued:

A^*(g) = A(g) \iff \overline{A(g)} = A(g).

However, A being selfadjoint in \mathcal{C}(G) means something else:

A^*(g)=A(g) \iff \overline{A(g^{-1}})=A(g).

To further compare and contrast the function algebra \mathcal{F}(G) and the convolution algebra \mathcal{C}(G), let us consider their subalgebras. From Lecture 3 we know that the lattice of subalgebras of \mathcal{F}(G) is isomorphic to the lattice of partitions of G, which has nothing to do with the group structure on G. On the other hand, we have a subalgebra of \mathcal{C}(G) associated to every subgroup of H of G, defined by

\mathcal{A}(H) = \{A \in \mathcal{C}(g) \colon A(g) = 0 \text{ unless }g \in H\}.

In lecture, we also gave a characterization of unitary elements in \mathcal{C}(G).

Problem 12.2. Prove that \mathcal{A}(H) is a subalgebra of \mathcal{C}(G), and moreover that \mathcal{A}(H) is isomorphic to \mathcal{C}(H).

This raises the question of whether subalgebras of \mathcal{C}(G) are in bijection with subgroups of G, i.e. whether the lattice of subalgebras of \mathcal{C}(G) is isomorphic to the lattice of subgroups of G, which would be analogous to the situation with function algebras. This is not true in general. Consider the following element of \mathcal{C}(G),

P = \frac{1}{|G|} \sum_{g \in G} E_g,

which is the average of the elementary functions. Equivalently, P\colon G \to \mathbb{C} is the constant function

P(g) = \frac{1}{|G|}, \quad g \in G.

Then P is selfadjoint (make sure you understand why), and also

P^2 = \frac{1}{|G|^2} \sum_{h \in G} \sum_{g \in G} E_gE_h=\frac{1}{|G|^2} \sum_{h \in G} \sum_{g \in G} E_{gh}.

Now, the internal sum is just

\sum_{g \in G} E_{gh} = \sum_{g \in G} E_g,

since multiplying each element g \in G by a fixed group element h \in G just permutes the points of the group (make sure you understand why). Therefore, P^2=P and we conclude that the average of all elementary functions E_g is a projection in \mathcal{C}(G). Now consider the subalgebra of \mathcal{C}(G) generated by the the multiplicative identity I=E_e and P,

\mathcal{A}=\mathrm{alg}\{I,P\} = \{\alpha I + \beta P \colon \alpha,\beta \in \mathbb{C}\}.

Since I,P are linearly independent (make sure you understand why), we have located a two-dimensional subalgebra inside \mathcal{C}(G), for any group G. If G is a group of odd order, Lagrange’s theorem tells us that any subgroup H \leq G must also have odd order, and consequently the subalgebra \mathcal{A}=\mathrm{alt}\{I,P\} cannot be the algebra of functions vanishing outside some subgroup H of G, for then \mathcal{A} would be isomorphic to \mathcal{C}(H) and its dimensions would be the cardinality of H, which is impossible since |H| cannot be even.

Math 202B: Lecture 11

Let \mathcal{A} be an abelian subalgebra of the endomorphism algebra \mathrm{End}(V) of a finite dimensional Hilbert space V. We are going to use Math 202A linear algebra to show that \mathcal{A} is isomorphic to a function algebra. Combining this with the work we did last week, this result allows us to conclude the following: if \mathcal{A} is a commutative algebra which supports a scalar product satisfying the left-Frobenius identity (or, equivalently, admits a faithful state), then \mathcal{A} is isomorphic to a function algebra.

Problem 11.1. Prove that a one-dimensional subalgebra of \mathrm{End}(V) is isomorphic to the function algebra of a point.

Now let \mathcal{A} be an arbitrary m-dimensional abelian subalgebra of \mathrm{End}(V), and let \{A_1,\dots,A_m\} be a basis of \mathcal{A}. Since \mathcal{A} is commutative, all its elements are normal, as we proved in Week 1. Thus, A_1,\dots,A_m are commuting normal operators on V and we can apply the spectral theorem.

Theorem 11.1. There exists an orthonormal basis X \subset V such that

A_1x = \widehat{A}_1(x),\dots,A_mx=\widehat{A}_m(x)x, \quad x \in X,

where \widehat{A}_1(x), \dots, \widehat{A}_m(x) are scalars.

In possibly more familiar terms, Theorem 11.1 says that a finite family of commuting normal operators is simultaneously diagonalizable. For our purposes, we want to interpret this result as defining a function \mathcal{A} \to \mathcal{F}(X). Since \{A_1,\dots,A_m\} is a basis of \mathcal{A}, every A \in \mathcal{A} can be uniquely represented as a linear combination

A=\alpha_1A_1,\dots,\alpha_mA_m,

so we have a well-defined function

\mathcal{A} \longrightarrow \mathcal{F}(X)

which sends A \in \mathcal{A} to the function \widehat{A} \in \mathcal{F}(X) defined by

\widehat{A}(x) = \alpha_1\widehat{A}_1(x) + \dots + \alpha_m\widehat{A}_m(x), \quad x \in X.

We call this mapping \mathcal{A} \to \mathcal{F}(X) the spectral transform on \mathcal{A} relative to the orthonormal basis X \subset V.

Theorem 11.2. The spectral transform is an injective algebra homomorphism.

Proof: We carefully checked this in class, and if you were not there you should do the same. \square

Theorem 11.2 proves that every abelian subalgebra \mathcal{A} of \mathrm{End}(V) is isomorphic to a subalgebra of \mathcal{F}(X) for a finite set X. Since we have already shown in Math 202B that every subalgebra of a function algebra is isomorphic to a function algebra, this completes the proof that every commutative subalgebra of an endomorphism algebra is isomorphic to a function algebra.

However, we can be more precise than this: we can say which subalgebra of \mathcal{F}(X) the abelian subalgebra \mathcal{A} \leq \mathrm{End}(V) is transformed into. Namely, each of the operators A_i in our chosen basis \{A_1,\dots,A_m\} of \mathcal{A} induces a partition \mathfrak{p}_i of X obtained by partition the points of X into distinct eigenspaces. That is, two points x,y \in X are in the same block of \mathfrak{p}_i if and only if \widehat{A}_i(x)=\widehat{A}_i(y).

Problem 11.2. Prove that the image of \mathcal{A} under the spectral transform is the subalgebra of \mathcal{F}(X) consisting of all functions constant on the blocks of \mathfrak{q}, where \mathfrak{q} is the coarsest partition of X finer than each of the partitions \mathfrak{p}_1,\dots,\mathfrak{p}_m. (Hint: Show that \mathcal{A} maps into the stated subalgebra is straightforward, and Stone-Weierstrass finishes the job).

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})

by

\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 9

Let V be a finite-dimensional Hilbert space and let \mathcal{A}=\mathrm{End}(V) be the algebra of linear operators on V. We have shown that the unique faithful tracial state on \mathcal{A} is

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

the average of the diagonal matrix elements of A \in \mathcal{A} with respect to any orthonormal basis X \subset V. Equivalently, the unique Frobenius scalar product on \mathcal{A}=\mathrm{End}(V) is

\langle A,B \rangle_F = \frac{1}{\dim V}\sum\limits_{x \in X} \langle Ax,Bs \rangle.

We also showed that \tau is not an algebra homomorphism, so that algebra homomorphisms \mathrm{End}(V) \to \mathbb{C} simply do not exist.

We can also go the other way and consider states which are not necessarily faithful or tracial. That is, we consider linear functionals \sigma \colon \mathcal{A} \to \mathbb{C} which are required to satisfy

\sigma(I) = 1 \quad\text{and}\quad \sigma(A^*A) \geq 0,

but nothing more. The corresponding sesquilinear form

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

is Hermitian, satisfies \langle I,I \rangle_\sigma=\sigma(I)=1, and the left Frobenius identity holds,

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

In the case where \mathcal{A}=\mathcal{F}(X) is the function algebra of a finite set X, states on \mathcal{A} are in bijection with probability measures on X. Faithful states correspond to probability measures whose support is all of X, i.e. \mathbb{P}(x) >0 for all x \in X. We will now obtain an analogous classification of states on \mathcal{A}=\mathrm{End}(V). In this noncommutative setting, the role of the probability measure \mathbb{P} on X is played by a special kind of operator on P on V called a density operator.

Let \sigma be an arbitrary linear functional on \mathcal{A}=\mathrm{End}(V). Equip \mathcal{A} with the Hilbert-Schmidt scalar product. By the Riesz representation theorem, there is R \in \mathcal{A} such that

\sigma(A) = \langle R,A \rangle_{HS} = \mathrm{Tr}(R^*A), \quad A \in \mathcal{A}.

Equivalently, setting P=R^* we have

\sigma(A)=\mathrm{Tr}(PA), \quad A \in \mathcal{A}.

Proposition 9.1. We have \sigma(I)=1 if and only if \mathrm{Tr}(P)=1.

Proof: \sigma(I)=\mathrm{Tr}(P). \square

Proposition 9.2. We have \sigma(A^*A) \geq 0 for all A \in \mathcal{A} if and only if P=Q^*Q for some Q \in \mathcal{A}.

Proof: If P=Q^*Q, then

\sigma(A^*A)=\mathrm{Tr}(PA^*A)=\mathrm{Tr}(Q^*QA^*A)=\mathrm{Tr}(QA^*AQ^*)=\langle AQ,AQ\rangle_{HS} \geq 0.

Conversely, suppose \sigma(A^*A) \geq 0 for all A \in \mathcal{A}. Then, choosing an orthonormal basis X\subset V we have that

\mathrm{Tr}(PA^*A) =\sum\limits_{x \in X} \langle x,PA^*Ax \rangle \geq 0, \quad A \in \mathcal{A}.

Taking A=E_{yx} to be an elementary operator, we have

\mathrm{Tr}(PA^*A)=\mathrm{Tr}(PE_{xy}E_{yx}) = \sum\limits_{z \in X} \langle z,PE_{xx}z\rangle = \langle x,Px \rangle,

so that

\langle x,Px \rangle \geq 0 \quad\text{ for all } x \in X.

Thus, nonnegativity of \sigma means that all diagonal matrix elements of P, in any orthonormal basis X \subset V, are nonnegative. Writing P=H+iK with H,K \in \mathcal{A} selfadjoint, we have

\langle x,Px \rangle = \langle x,Hx \rangle + i\langle x,Kx \rangle.

Thus, the fact that x,Px \rangle is real for all x \in X is enough to force P to be selfadjoint.

What remains now is to show that a selfadjoint operator P which satisfies \langle v,Pv \rangle \geq 0 for all v \in V can be factored as P=Q^*Q for some Q \in \mathcal{A}. To do this we use the spectral theorem: since P is selfadjoint we have

P=\sum\limits_{\lambda \in \Lambda} \alpha_\lambda P_\lambda,

where |\Lambda| is a finite nonempty set of cardinality at most |X|=\dim V, and $P_\lambda,$ \lambda \in \Lambda are selfadjoint orthogonal idempotents: for all \lambda,\mu \in \Lambda we have

P_\lambda^*=P_\lambda, \quad P_\lambda P_\mu =\delta_{\lambda\mu}.

Indeed, this is the algebraic form of the spectral theorem for selfadjoint operators in \mathrm{End}(V). The geometric form of the theorem is that there exists an orthogonal decomposition

V=\bigoplus_{\lambda \in \Lambda} V_\lambda

of V into nonzero subspaces such that P restricted to V_\lambda is a scalar multiple \alpha_\lambda of the identity operator. The connection between the algebraic statement and the geometric one is that P_\lambda is the orthogonal projection of V onto V_\lambda.

Now if e_\lambda is a unit vector in the image V_\lambda of P_\lambda then for any \lambda \in \Lambda we have

e_\lambda,Pe_\lambda \rangle = \langle e_\lambda,\alpha_\lambda e_\lambda\rangle \geq 0,

so \alpha_\lambda is a nonzero nonnegative number, aka a positive number. Thus we can define

Q=\sum\limits_{\lambda \in \Lambda}\sqrt{\alpha_\lambda}P_\lambda

and we have P=Q^*Q. \square

What we have proved is that every state on the algebra \mathcal{A}=\mathrm{End}(V) has the form

\tau(A)=\mathrm{Tr}(PA), \quad A \in \mathcal{A}

where \mathrm{Tr}(P)=1 and P=Q^*Q for some Q \in \mathcal{A}. Any P \in \mathcal{A}=\mathrm{End}(V) with these two properties is called a density operator.

Thus, we have shown that states on the endomorphism algebra \mathrm{End}(V) of a finite-dimensional Hilbert space V are in bijection with density operators on V. This is reminiscent of our earlier theorem that states on the function algebra \mathcal{F}(X) of a finite set X are in bijection with probability measures on X. In fact, density operators are the noncommutative analogue of probability measures in a very precise way. In the proof above, we saw that the condition that there exists a factorization P=Q^*Q is equivalent to the condition that

P=\sum\limits_{\lambda \in \Lambda} \alpha_\lambda P_\lambda

where \Lambda is a nonempty set of cardinality bounded by \dim V and P_\lambda are pairwise orthogonal selfadjoint idempotents weighted by positive numbers \alpha_\lambda. This gives

\mathrm{Tr}(P) = \sum\limits_{\lambda \in \Lambda} \alpha_\lambda (\dim V_\lambda),

where V_\lambda is the eigenspace of P which P_\lambda projects onto. Together with the condition \mathrm{Tr}(P)=1, this means the density operator P induces a probability measure \mathbb{P} on \Lambda defined by

\mathbb{P}(\lambda) = \alpha_\lambda (\dim V_\lambda).

Problem 9.1. Let P \in \mathrm{End}(V) be a density operator. Show that the corresponding state \tau_P on \mathrm{End}(V) is a faithful state if and only if \mathrm{rank}(P)=\mathrm{dim}(V).

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.

Problem 8.2. Show that \tau(A) = \frac{1}{\dim V}\mathrm{Tr}(A) is not a homomorphism when \dim V >1.

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 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).

Math 202B: Lecture 5

***All problems in this lecture are due 01/20 at 23:59***

***No lecture on 01/16***

By now you should have a reasonably good feeling for subalgebras. Unfortunately, this is only half the battle. Let \mathcal{A} and \mathcal{B} be algebras, and let \Phi \colon \mathcal{A} \to \mathcal{B} be an algebra homomorphism. Since \mathcal{A} and \mathcal{B} are vector spaces, and since \Phi is a linear transformation, \mathrm{Ker}(\Phi) is a subspace of \mathcal{A} and \mathrm{Im}(\Phi) is a subspace of \mathcal{B}. Since \Phi is not just a linear map but an algebra homomorphism, one hopes that its kernel and image will have additional structure.

Problem 5.1. Prove that \mathrm{Im}(\Phi) is a subalgebra of \mathcal{B}.

It is not necessarily true that \mathrm{Ker}(\Phi) is a subalgebra of \mathcal{A}, because it is not necessarily true that \Phi(I_\mathcal{A})=0_\mathcal{B}. However, the other subalgebra properties, namely closure under conjugation and closure under multiplication, do hold for \mathrm{Ker}(\Phi). In fact, \mathrm{Ker}(\Phi) is not just closed under multiplication – it has the stronger property that AK \in \mathrm{Ker}(\Phi) for any A \in \mathcal{A} and any K \in \mathrm{Ker}(\Phi). This observation leads to the consideration of subspaces of \mathcal{A} which have this absorption property.

Definition 5.1. A subspace \mathcal{J} of \mathcal{A} is said to be an ideal if it is closed under conjugation and absorbs multiplication, in the sense that AJ and JA lie in \mathcal{J} for any J \in \mathcal{J} and all A \in \mathcal{A}.

Subalgebras of \mathcal{A} are smaller algebras embedded in \mathcal{A}. They inherit the algebra structure of \mathcal{A} under restriction. Ideals on the other hand are special subspaces of \mathcal{A} which can be used to produce smaller algebras by collapsing rather than restricting. More precisely, since an ideal \mathcal{J} in \mathcal{A} is a subspace we can form the quotient vector space \mathcal{A}/\mathcal{J} whose points are translations of \mathcal{J},

[A]=\{A+J \colon J \in \mathcal{J}\}.

The zero vector in the quotient space is [0_\mathcal{A}]=\mathcal{J}, and linear combinations are defined by

\alpha_1[A_1]+\alpha_2[A_2]=[\alpha_1 A_1 + \alpha_2 A_2].

The linear transformation

\Pi \colon \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{J}

defined by \Pi(A) = [A] has kernel \mathcal{J}, so by the rank-nullity theorem

\dim \mathcal{A}/\mathcal{J}=\dim \mathcal{A}-\dim \mathcal{J}.

The fact that \mathcal{J} is not just a subspace but an ideal allows us to go further and put an algebra structure on \mathcal{A}/\mathcal{J} by defining conjugation as [A]^*=[A^*] and multiplication as [A][B]=[AB].

Problem 5.2. Check that \mathcal{A}/\mathcal{J} really is an algebra.

We conclude that in our quest to understand the structure of a given algebra \mathcal{A}, we have to understand not just its subalgebras but also its ideals. Thankfully, in many ways ideals are simpler and easier to understand than subalgebras. For example, recall that we have decided to view the set of all subalgebras of \mathcal{A} as a poset under inclusion, making it an induced subposet of the lattice of subspaces of \mathcal{A}, but not an induced sublattice: the max of two subalgebras of \mathcal{A} is not just the span of their union, but the algebra generated by their union. Concerning the poset of ideals of \mathcal{A}, this is an induced sublattice of the lattice of all subspaces without needing any additional constructions.

Problem 5.3. Given two ideals \mathcal{J},\mathcal{K} in \mathcal{A}, show that

\mathrm{Span}(\mathcal{J} \cup \mathcal{K})= \mathcal{J}+\mathcal{K}=\{J+K \colon J \in \mathcal{J},\ K \in \mathcal{K}\}.

As we saw in Lecture 4, we have a correspondence between partitions of X and subalgebras of \mathcal{F}(X). The combinatorial objects which parameterize ideals of \mathcal{F}(X) are simpler — they are just subsets of \mathcal{X}. Given a subset S \subseteq X, we define a corresponding ideal in \mathcal{F}(X) by

\mathcal{I}(S) = \{A \in \mathcal{F}(X) \colon A(x)=0 \text{ for all }x \in S\}.

Thus \mathcal{I}(S) is the set of all functions in \mathcal{F}(X) which vanish on every point of S. Hence, we call \mathcal{I}(S) the vanishing ideal of \mathcal{S}.

Problem 5.4. Prove that \mathcal{I}(S) really is an ideal in \mathcal{F}(X). Moreover, show that for subsets S \leq T of X we have \mathcal{I}(T) \leq \mathcal{I}(S).

A nice feature of vanishing ideals in \mathcal{F}(X) is that they are coordinate spaces relative to the elementary basis \{E_x \colon x \in X\}.

Problem 5.5. Show that elementary functions \{E_x \colon x \in X\backslash S\} form a basis of \mathcal{I}(S). Moreover, show that the quotient algebra \mathcal{F}(X)/\mathcal{I}(S) is isomorphic to the function algebra \mathcal{F}(S).

We can now view S \mapsto \mathcal{I}(S) as an order-preserving function from the lattice of subsets of X to the lattice of ideals of \mathcal{F}(X).

Problem 5.6. Show that the mapping S \mapsto \mathcal{I}(X) is injective.

With the above problems solved, we will have an isomorphism between the lattice of subsets of X and the lattice of ideals of \mathcal{F}(X) once we show surjectivity.

Theorem 5.2. The mapping S \mapsto \mathcal{J}(S) is surjective.

Proof: Given an arbitrary ideal \mathcal{J} of \mathcal{F}(X), we need to construct a corresponding point set S \subseteq X such that \mathcal{I}(S) = \mathcal{J}. A natural candidate is the variety defined by \mathcal{J}, which by definition is the set

V(\mathcal{J}) = \{x \in X \colon A(x)=0 \text{ for all }A \in \mathcal{J}\}

all points in X at which every function A \in \mathcal{J} vanishes. By construction, we have \mathcal{J} \leq \mathcal{I}(V(\mathcal{J})) and we need to show that the reverse inclusion also holds

Let S = V(\mathcal{J}). In order to show \mathcal{I}(S) \leq \mathcal{J}, it is sufficient to show that E_x \in \mathcal{J} for all x \in X\backslash S. This is because we know that the set \{E_x \colon x \in X \backslash S\} is a basis of \mathcal{I}(S). So, fix a point x \in X \backslash S. Since \mathcal{S} is the variety cut out by \mathcal{J}, there must exist a function F_x \in \mathcal{J} which does not vanish at x, else x would be a point of S. Furthermore, we can assume WLOG that F_x(x)=1. We do not have any information about the values of the function F_x at any other points. However, multiplying F_x by the elementary function E_x produces something very simple, namely E_xF_x=E_x. Since F_x \in \mathcal{J} and \mathcal{J} is an ideal, we have E_x \in \mathcal{J}. \square

The above proof showed that for any subset S \subseteq X, we have

\mathcal{I}(V(\mathcal{I}(S)))=\mathcal{I}(S).

This is a very simple finite set version of Hilbert’s Nullstellensatz, a basic theorem of algebraic geometry on vanishing sets of polynomial ideals which will be discussed in Math 202C.

Math 202B: Lecture 4

***All problems assigned in this lecture are due January 20th at 23:59***

Let us begin this lecture with a physical way of thinking about the function algebra \mathcal{F}(X) of a finite set X. Imagine that X is the set of all possible outcomes of an experiment, for example a chemical reaction whose yield depends on environmental factors, the proportions in which the constituents are mixed, and so forth. After the experiment has been performed, we want to determine what substance has been produced. From this point of view, \mathcal{F}(X) represents the collection of all numerical measurements or “observables” of the yield that can theoretically be computed. For example, T \in \mathcal{F}(X) assigns to each x \in X its temperature T(x), and H \in \mathcal{F}(X) which assigns to each x \in X its specific heat H(x), etc.

In practice (as opposed to in theory), our experiment is performed in a laboratory with limited resources. For example, we may have access to a thermometer but not to a Bunsen burner, so we can compute T(x) but not H(x). The set of all measurements which we can perform in our laboratory is then a proper subalgebra \mathcal{A} of \mathcal{F}(X) which contains T but not H. A natural question now is: given a proper subalgebra \mathcal{A} <\mathcal{F}(X) of accessible observables, can we compute enough measurements to distinguish between any two outcomes? In operational terms, \mathcal{A} separates the points of X if our laboratory is advanced enough that we can perform at least one measurement which distinguishes any two distinct outcomes.

Definition 4.1. We say that \mathcal{A} separates the points of X if, for any distinct x,y \in X, there is A \in \mathcal{A} such that A(x) \neq A(y).

Unfortunately, as soon as our laboratory is limited in any way operationally indistinguishable outcomes exist.

Theorem 4.2. A subalgebra \mathcal{A} \leq \mathcal{F}(X) separates the points of X if and only if \mathcal{A}=\mathcal{F}(X).

Proof: One direction is clear: if \mathcal{A}=\mathcal{F}(X) then we have access to the set \{E_x \colon x \in X\} of all elementary functions, and for any distinct points x,y \in X we have E_x(x)=1 and E_x(y)=0.

Conversely, suppose that \mathcal{A} \leq \mathcal{F}(X) is a subalgebra which separates X. Pick an arbitrary point x \in X. Then, for each y \in X\backslash \{x\} there exists a function F_y \in \mathcal{A} such that

\alpha = F_y(x) \quad\text{and}\quad \beta = F_y(y)

are distinct numbers. The centered and scaled function

\tilde{F}_y(z) = \frac{F_y(z)-\beta}{\alpha-\beta}

then satisfies

\tilde{F}_y(x) =1 \quad\text{and}\quad \tilde{F}_y(y)=0.

We thus have the factorization

E_x =\prod\limits_{y \in X\backslash \{x\}} \tilde{F}_y.

Since \mathcal{A} is closed under products, we have shown that \mathcal{A} contains the elementary function E_x. Since x \in X was arbitrary, we have shown that \mathcal{A} contains the elementary basis \{E_x \colon x \in X\}. \square

The category of finite sets is a full subcategory of the category whose objects are compact Hausdorff spaces and whose morphisms are continuous functions — give a finite sets the discrete topology, in which every singleton set is open. The fact that the algebra of all continuous functions on any Hausdorff topological space can separate any two distinct closed sets holds in this larger topological category (this is Urysohn’s Lemma). The opposite direction, that a separating subalgebra \mathcal{A} is equal to the algebra of continuous functions, has to be weakened to the statement that \mathcal{A} is uniformly dense in the algebra of all continuous functions (this is the Stone-Weierstrass theorem).

Let us now generalize Theorem 4.2 within the category of finite sets.

Definition 4.2. A partition of X is a set \mathfrak{p} of disjoint nonempty subsets of X whose union is X. The elements of \mathfrak{p} are referred to as its “blocks.”

Let \mathfrak{P}(X) denote the set of all partitions of X. Note that \mathfrak{P}(X) is in bijection with the set of equivalence relations on X — a partition \mathfrak{p} defines an equivalence relation on X in which points are equivalent if they are elements of the same block, and conversely an equivalence relation on X defines a partition whose blocks are equivalence classes.

We make \mathfrak{P}(X) into a poset as follows: for each \mathfrak{p},\mathfrak{q} \in \mathfrak{P}(X), we declare \mathfrak{p} \leq \mathfrak{q} if and only if \mathfrak{q} can be obtained by partitioning blocks of \mathfrak{p}. Equivalently, every block of \mathfrak{q} is contained in a block of \mathfrak{p}. This is called the refinement order on \mathfrak{P}(X). If \mathfrak{p} \leq \mathfrak{q}, we say that \mathfrak{q} is finer than \mathfrak{p}, or equivalently that \mathfrak{p} is coarser than \mathfrak{q}. Going one step further, we can make \mathfrak{P}(X) into a lattice, where \max(\mathfrak{p},\mathfrak{q}) is the coarsest partition (weakly) finer than both \mathfrak{p} and \mathfrak{q}, and \min(\mathfrak{p},\mathfrak{q}) is the finest partition (weakly) coarser than both \mathfrak{p} and \mathfrak{q}.

There is a natural mapping from the lattice of partitions of X to the lattice of subalgebras of \mathcal{F}(X) — for each \mathfrak{p} \in \mathfrak{P}(X), declare \mathcal{A}(\mathfrak{p}) to be the set of all functions in \mathcal{F}(X) which are constant on the blocks of \mathfrak{p}.

Problem 4.1. Prove that \mathcal{A}(\mathfrak{p}) really is a subalgebra of \mathcal{A}(X), and moreover that the mapping \mathfrak{p} \to \mathcal{A}(\mathfrak{p}) is injective.

Problem 4.2. Prove that \mathfrak{p} \leq \mathfrak{q} implies \mathcal{A}(\mathfrak{p}) \leq \mathcal{A}(\mathfrak{q}), meaning that our mapping \mathfrak{P}(X) \to \mathcal{F}(X) is an order homomorphism.

Problem 4.3. Work a bit harder and show that the mapping \mathfrak{p} \mapsto \mathcal{A}(\mathfrak{p}) is a lattice homomorphism.

Problem 4.4. Prove that \mathcal{A}(\mathfrak{p}) is isomorphic to \mathcal{F}(\mathfrak{p}).

We can make use of Problem 4.4 as follows. Let us say that a function A \in \mathcal{A}(\mathfrak{p}) separates the blocks of \mathfrak{p} if, for any two distinct blocks P and Q of \mathfrak{p}, the constant functions A|_P and A|_Q are distinct. This generalizes Definition 4.1, which is the case where \mathfrak{p} is the partition of X with |X| blocks. The corresponding generalization of Theorem 4.2 is the following.

Theorem 4.3. If \mathcal{A} is a subalgebra of \mathcal{A}(\mathfrak{p}) which separates the blocks of \mathfrak{p}, then \mathcal{A}=\mathcal{A}(\mathfrak{p}).

Proof: By Problem 4.4, this reduces to Theorem 4.2. (Make sure you understand this). \square

At this point the following classification theorem is essentially complete.

Theorem 4.4. The lattice of subalgebras of \mathcal{F}(X) is isomorphic to the lattice of partitions of X.

Proof: It remains only to show that our method of assigning subalgebras to partitions is surjective. Let \mathcal{A} be an arbitrary subalgebra of \mathcal{F}(X). Define an equivalence relation on X by

x \sim y \quad \iff \quad A(x)=A(y) \text{ for all }A \in \mathcal{A},

and let \mathfrak{p} be the partition of X whose blocks are the corresponding equivalence classes. Then, by definition of \mathfrak{p} we have \mathcal{A} \leq \mathcal{A}(\mathbf{p}). Furthermore, by definition of \mathfrak{p} we have that A separates the blocks of \mathfrak{p}. By Theorem 4.3, we have \mathcal{A}=\mathcal{A}(\mathfrak{p}). \square