202B – Page 2 – Jonathan Novak

Math 202B: Lecture 16

***Problems in this lecture due February 21***

In this lecture we will review the concrete algebras considered so far.

Function algebras. The archetypal commutative algebras: $\mathcal{F}(X)$ of functions on a finite set $X$ with all operations defined pointwise. We understand function algebras completely.

The elementary functions $\{E_x \colon x \in X\}$ are a basis of $\mathcal{F}(X)$ consisting of pairwise orthogonal selfadjoint idempotents: $E_x^*=E_x$ and $E_xE_y = \delta_{xy}E_x.$ In fact, this feature characterizes function algebras up to isomorphism.

Theorem: An algebra $\mathcal{A}$ is isomorphic to a function algebra if and only if it has a basis of pairwise orthogonal selfadjoint idempotents.

We have obtained a complete classification of subalgebras of function algebras.

Theorem: Subalgebras of $\mathcal{F}(X)$ are in bijection with partitions of $X$ . The correspondence is given explicitly by

$\mathcal{A}(\mathfrak{p})=\{A \in \mathcal{F}(X) \colon A \text{ constant on the blocks of }\mathfrak{p}\}.$

Corollary: Every subalgebra of a function algebra is isomorphic to a function algebra.

We have a complete understanding of ideals in function algebras.

Theorem: Ideals in $\mathcal{F}(X)$ are in bijection with subsets of $\mathcal{X}.$ The correspondence is given explicitly by

$\mathcal{J}(S) = \{A \in \mathcal{F}(X) \colon A \text{ vanishes on }S\}.$

We have a complete understanding of states on function algebras.

Theorem: States on $\mathcal{F}(X)$ are in bijection with probability functions on $X.$ The correspondence is given explicitly by

$\tau_P(A) = \sum\limits_{x \in X} A(x)P(x).$

Corollary: Faithful states on $\mathcal{F}(X)$ are in bijection with nonvanishing probability functions on $X.$

Corollary: Homomorphisms $\mathcal{F}(X) \to \mathbb{C}$ are in bijection with points of $X.$ The correspondence is given explicitly by

$\varphi_x(A)=A(x), \quad A \in \mathcal{F}(X).$

Endomorphism algebras. The archetypal noncommutative algebras: $\mathrm{End}(V)$ is the algebra of linear operators on a finite-dimensional Hilbert space $V$ .

If $X \subset V$ is an orthonormal basis, a vector space basis of $\mathrm{End}(V)$ is given by the elementary operators

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

The relations governing the elementary operators are

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

In particular, $\{E_{xx} \colon x \in X\}$ spans a subalgebra $\mathcal{D}(X)$ isomorphic to $\mathcal{F}(X).$

Theorem: Every MASA in $\mathrm{End}(V)$ is of the form $\mathcal{D}(X)$ for some orthonormal basis $X \subset V.$

Corollary: All abelian subalgebras of $\mathrm{End}(V)$ have dimension at most $\dim V.$

Our understand of abelian subalgebras of $\mathrm{End}(V)$ is the following.

Theorem: Every abelian subalgebra of $\mathrm{End}(V)$ is isomorphic to a subalgebra of $\mathcal{F}(X)$ for some orthonormal basis $X \subset V.$ In particular, all abelian subalgebras of $\mathrm{End}(V)$ are isomorphic to function algebras.

At present, this is all we know about subalgebras of $\mathrm{End}(V)$ . A major reason for wanting to know more is the following.

Theorem: Every von Neumann algebra $(\mathcal{A},\tau)$ is isomorphic to a subalgebra of $\mathrm{End}(V)$ for some Hilbert space $V.$

On the bright side, our understanding of states and traces on $\mathrm{End}(V)$ . Let $X \subset V$ be an orthonormal basis and define a corresponding linear functional on $\mathrm{End}(V)$ by

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

Theorem: All traces on $\mathrm{End}(V)$ are scalar multiples of $\mathrm{Tr}.$

Corollary: If $\dim V>1,$ there are no algebra homomorphisms $\mathrm{End}(V) \to \mathbb{C}.$

An operator $P \in \mathrm{End}(V)$ is called nonnegative if $P=Q^*Q$ for some $Q \in \mathrm{End}(V)$ . A nonnegative operator such that $\mathrm{Tr}(P)=1$ is called a density operator.

Theorem: States on $\mathrm{End}(V)$ are in bijection with density operators. The correspondence is given explicitly by $\tau_P(A)=\mathrm{Tr}(AP).$

Convolution algebras. The convolution algebra $\mathcal{C}(G)$ of a finite group $G$ coincides with the function algebra $\mathcal{F}(G)$ as a vector space, but conjugation and multiplication are instead defined by

$E_g^*=E_{g^{-1}} \quad\text{and}\quad E_gE_h=E_{gh}, \quad g,h \in G$ .

In particular, $\{E_g \colon g \in G\}$ is a basis of $\mathcal{C}(G)$ consisting of unitary elements rather than orthogonal idempotents.

Our understanding of convolution algebras is quite basic at present. For example, we know how to associate a subalgebra $\mathcal{C}(H)$ of $\mathcal{C}(G)$ to a subgroup $H \leq G,$ but we do not know much else about subalgebras of $\mathcal{C}(G).$

We did however manage to classify states and traces on $\mathcal{C}(G).$ A function $P \in \mathcal{C}(G)$ is called nonnegative if for any finite choice $\alpha_1,\dots,\alpha_n \in \mathbb{C}$ and $g_1,\dots,g_n \in G$ we have

$\sum\limits_{i,j=1}^n \overline{\alpha_i}\alpha_jP(g_i^{-1}g_j) \geq 0.$

Theorem: States on $\mathcal{C}(G)$ are in bijection with nonnegative functions in $\mathcal{C}(G)$ . The correspondence is given explicitly by

$\tau_P(A) =\sum\limits_{g \in G}A(g)P(g),$

and $\tau_P$ is a trace if and only if $P$ is a central function.

The only function $P \colon G \to \mathcal{C}$ which is both a probability function and a nonnegative function in the above sense is $E_e,$ and it is also a class function. Furthermore, the corresponding tracial state

$\tau_{E_e}(A) =A(e)$

is faithful. Therefore, $(\mathcal{C}(G),\tau_{E_e})$ is a von Neumann algebra and we can conclude the following.

Theorem: The convolution algebras $\mathcal{C}(G)$ is isomorphic to a subalgebra of $\mathrm{End}(V)$ for some Hilbert space $V.$ Furthermore, if $G$ is abelian then $\mathcal{C}(G)$ is isomorphic to a function algebra.

We also have a classification of homomorphisms from $\mathcal{C}(G)$ to $\mathbb{C}.$

Theorem: A state $\tau_P$ on $\mathcal{C}(G)$ is an algebra homomorphism if and only if $P \in \mathcal{C}(G)$ is a group homomorphism from $G$ to the unit circle $\mathbb{U} \subset \mathbb{C}.$

Problem 16.1. Show that there exists at least one algebra homomorphism $\mathcal{C}(G) \to \mathbb{C}.$ Given an example of a group for which this minimum is achieved.

Class algebras. The center of $\mathcal{C}(G)$ is denoted $\mathcal{K}(G)$ and it consists precisely of functions constant on conjugacy classes of $G$ . Therefore, we call $\mathcal{K}(G)$ the class algebra of $G.$ Writing $\{C_\alpha \colon \alpha \in \Lambda\}$ for the set of conjugacy classes in $G$ , the functions

$K_\alpha =\sum\limits_{g \in C_\alpha} E_g, \quad \alpha \in \Lambda,$

form a basis of $\mathcal{K}(G).$ This basis is orthogonal in the $L^2$ -scalar product,

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

Problem 16.2. Show that $\tau_{E_e}(K_\alpha K_\beta)$ is equal to the number of solutions to the equation $xy=e$ in $G$ such that $x \in C_\alpha$ and $y \in C_\beta.$

Four Functors. One way to organize the subject matter so far is to consider that we have defined four functors.

The functor $X \rightsquigarrow \mathcal{F}(X)$ takes us from the category of finite sets to the category of commutative algebras.

The functor $X \rightsquigarrow \mathcal{E}(X)=\mathrm{End}\mathcal{F}(X)$ takes us from the category of finite sets to the category of algebras.

The functor $G \rightsquigarrow \mathcal{C}(G)$ takes us from the category of finite groups to the category of algebras.

The functor $G \rightsquigarrow \mathcal{K}(G)$ takes us from the category of finite groups the the category of commutative algebras.

Math 202B: Lecture 14

We have been comparing and contrasting the function algebra $\mathcal{F}(G)$ of a finite group $G$ with its convolution algebra $\mathcal{C}(G).$ These objects are the same as vector spaces, but very different as algebras. In particular, we understand what is happening inside $\mathcal{F}(G)$ very well, in the sense that we know all of its subalgebras, whereas our understanding of subalgebras of $\mathcal{C}(G)$ is limited to those coming from subgroups $H \leq G$ together with the class algebra $\mathcal{K}(G) =Z(\mathcal{C}(G)).$

Today we will compare and contrast $\mathcal{F}(G)$ and $\mathcal{C}(G)$ through the lens of states and traces. The starting point is the same in both cases: a very simple lemma identifying linear functionals on a function algebra $\mathcal{F}(X)$ with functions in $\mathcal{F}(X).$

Lemma 14.1. For any set $X$ , linear functionals on $\mathcal{F}(X)$ are in linear bijection with functions in $\mathcal{F}(X)$ .

Proof: For any functional $\varphi \colon \mathcal{F}(X) \to \mathbb{C}$ , linear or not, we get a corresponding function $F_\varphi \in \mathcal{F}(X)$ defined by

$F_\varphi(x) = \varphi(E_x), \quad x \in X.$

If $\varphi$ is a linear functional on $\mathcal{F}(X)$ , then it is uniquely determined by its values on the basis $\{E_x \colon x \in X\}$ of elementary functions, hence $\varphi$ is uniquely determined by $F_\varphi.$ Conversely, if $F \in \mathcal{F}(X)$ is any function, we can define a linear functional on $\mathcal{F}(X)$ by

$\varphi_F(E_x) = F(x), \quad x \in X.$

It is clear that the maps $\varphi \mapsto F_\varphi$ and $F \mapsto \varphi_F$ are inverses of one another. $\square$

Recall our classification of states on $\mathcal{F}(X)$ , obtained using Lemma 14.1.

Theorem 14.2. The following are equivalent:

$P$ is a probability function on $X.$
$\tau_P$ is a state on $\mathcal{F}(X)$ .

In fact, we know a few enhancements of Theorem 14.2: first, $\tau_P$ is a faithful state if and only if $P(x)>0$ for all $x \in X$; second, $\tau_P$ is an algebra homomorphism if and only if $P=E_x$ for some $x \in X.$ In particular, all homomorphisms $\mathcal{F}(X) \to \mathbb{C}$ are evaluation at a point, i.e. maps of the form $A \mapsto A(x_0)$ for a particular $x_0 \in X.$

As vector spaces, $\mathcal{F}(G)$ and $\mathcal{C}(G)$ are not just isomorphic they are equal, hence Lemma 14.1 applies verbatim. However, since $\mathcal{F}(G)$ and $\mathcal{C}(G)$ are quite different as algebras, the classification of states on $\mathcal{C}(G)$ is going to be quite different. To prepare yourself for this, solve the following problem.

Problem 14.1. Let $P \in \mathcal{C}(G)$ be a probability function. Prove that the corresponding linear functional $\varphi_P$ is a state on $\mathcal{C}(G)$ if and only if $P=E_e$ is the indicator function of the group identity $e \in G.$ Conclude that in this case $\varphi_P(A) = A(e)$ is evaluation at $e,$ but that this is not an algebra homomorphism $\mathcal{C}(G) \to \mathbb{C}.$

Now let us determine which functions $P \in \mathcal{C}(G)$ correspond to states on $\mathcal{C}(G)$ . The normalization condition is straightforward: since the multiplicative identity $I \in \mathcal{C}(G)$ is $I=E_e,$ we have

$\tau_P(I)=1 \iff \tau_P(E_e)=1 \iff P(e)=1.$

Now let us consider what property $P \in \mathcal{C}(G)$ must have in order for $\tau_P$ to be a nonnegative functional. For an arbitrary function

$A=\sum\limits_{g \in G} \alpha_g E_g,$

we have

$A^*=\sum\limits_{g \in G} \overline{\alpha_g}E_{g^{-1}},$

so that

$\tau_P(A^*A) = \sum\limits_{g,h \in G} \overline{\alpha}_g\alpha_hP(g^{-1}h).$

Hence if we pick an ordering $g_1,\dots,g_n$ of $G$ , and write $\alpha_i=\alpha_{g_i},$ this becomes

$\tau_P(A^*A) = \sum\limits_{i,j=1}^N \overline{\alpha_i}\alpha_j P(g_i^{-1}g_j).$

Definition 14.3. A complex-valued function $P$ on a group $G$ is said to be nonnegative if for any $n \in \mathbb{N},$ any $\alpha_1,\dots,\alpha_n \in \mathbb{C},$ and any $g_1,\dots,g_n \in G$ we have

$\sum\limits_{i,j=1}^n \overline{\alpha}_i\alpha_j P(g_i^{-1}g_j) \geq 0.$

Problem 14.2. Prove $P$ is a nonnegative function on $G$ if and only if, for all $n \in \mathbb{N}$ and $g_1,\dots,g_n \in G$ the matrix $[P(g_i^{-1}g_j)]_{i,j=1}^n$ is Hermitian and has nonnegative eigenvalues. What is this matrix if $P=E_e$ ?

With the above in place, we can state our classification of states on the convolution algebra $\mathcal{C}(G)$ of a finite group $G$ as follows.

Theorem 14.4. The following are equivalent:

$P$ is a nonnegative function on $G$ ;
$\tau_P$ is a state on $\mathcal{C}(G).$

The classification of states on $\mathcal{C}(G)$ is even more straightforward.

Theorem 14.5. The following are equivalent:

$P$ is a central function on $G$ ;
$\tau_P$ is a trace on $\mathcal{C}(G)$ .

Proof: Suppose $P \in \mathcal{C}(G)$ is central. We know from last lecture that this is equivalent to $P(gh)=P(hg).$ Thus,

$\tau_P(E_gE_h)=P(gh)=P(hg)=\tau_P(E_hE_g).$

Conversely, if $\tau_P$ is a trace on $\mathcal{C}(G)$ then

$P(gh)=\tau_P(E_{gh})=\tau_P(E_gE_h)=\tau_P(E_hE_g) = \tau_P(E_{hg})=P(hg),$

whence $P$ is central. $\square$

We now have a complete classification of states and traces on the convolution algebra $\mathcal{C}(G)$ of a finite group $G.$ Namely, every linear functional on $\mathcal{C}(G)$ has the form

$\tau_P(E_g)=P(g), \quad g \in G$

for some function $P \in \mathcal{C}(G)$ . We know that $\tau_P$ is a state if and only if $P(e)=1$ and $[P(g^{-1}h)]_{g,h \in G}$ is a nonnegative Hermitian matrix. We know that $P$ is a trace if and only if $P(gh)=P(hg)$ is a central function on $G.$

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

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

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

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

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

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

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

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 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 0

Welcome to Math 202B at UCSD, Winter quarter 2026. Here is a New Year’s problem you can keep in the back of your mind over the course of the course. Of course, let me know if you solve it.

Problem 0: Prove that 26 is the only positive integer nestled between a square and a cube.

The basic parameters of Math 202B are the same as those in Math 202A: weekly problem sets due on Sundays at 23:59 via GradeScope together with a final exam, with a 70/30 split. The final exam is scheduled for 03/20 at 15:00, and if you cannot sit for the exam at the appointed time you should not enroll in the course.

Math 202A began in the classical computational category $\mathbf{FSet}$ whose objects are finite sets with morphisms being functions. This is in contrast to the quantum computational category $\mathbf{FHil}$ whose objects are finite-dimensional Hilbert spaces with linear transformations as morphisms. We considered the quantization functor $\mathcal{F}$ from $\mathbf{FSet}$ to $\mathbf{FHil}$ which sends a finite set $X$ to the Hilbert space $\mathcal{F}(X)$ of complex-valued functions on $X$ with the pointwise operations and the $L^2$ -scalar product. We then found a miraculous tool, the Singular Value Decomposition, which completely describes all morphism in $\mathrm{Hom}(\mathcal{F}(X),\mathcal{F}(Y))$ for any two finite sets $X$ and $Y.$ In the case $X=Y,$ the SVD gave us the Spectral Theorem for normal operators in $\mathcal{E}(X)=\mathrm{End}\mathcal{F}(X)=\mathrm{Hom}(\mathcal{F}(X),\mathcal{F}(X)).$

In retrospect, we now recognize that Math 202A was actually about two quantization functors departing from the classical computational category, the Schroedinger functor $\mathcal{F}$ and the Heisenberg functor $\mathcal{E},$ both of which land in $\mathbf{FHil}.$ In Math 202B, we recognize something new as well: both quantizations $\mathcal{F}(X)$ and $\mathcal{E}(X)$ come with vector products as well as scalar products. The product of two functions in $\mathcal{F}(X)$ is defined pointwise, and the product of two endomorphisms in $\mathcal{E}(X)$ is defined by composing them.

We now recognize that Schroedinger and Heisenberg are actually telling us to think about a subcategory of $\mathbf{FHil},$ namely the category $\mathbf{FAlg}$ of finite-dimensional algebras. Math 202B is all about the category $\mathbf{FAlg}.$ We will begin by defining algebras precisely and developing their basic theory axiomatically. Given an algebra $\mathcal{A}$ in this category, two natural goals are to classify its subalgebras and measure its commutativity, and we will formulate these goals rigorously.

Our two basic examples, $\mathcal{F}(X)$ and $\mathcal{E}(X)$ , are two opposite extremes: the former is fully commutative and the latter is maximally noncommutative. The classification of subalgebras of $\mathcal{F}(X)$ is elementary, while a complete description of subalgebras of $\mathcal{E}(X)$ is more involved and requires the development of new linear algebraic concepts and methods you may not have seen before.

Once we have said everything there is to say about $\mathcal{F}(X)$ and $\mathcal{E}(X)$ , we will consider the question of what lies between these two extremes. In this range we find a beautiful class of algebras constructed from finite groups. Namely, if $X$ carries a group law, then it becomes possible to associate a third algebra to it, the convolution algebra $\mathcal{C}(X).$ As Hilbert spaces, $\mathcal{C}(X)=\mathcal{F}(X),$ but as algebras the two are very different: multiplication of functions in $\mathcal{C}(X)$ is given by convolution rather than pointwise product. The commutativity index of $\mathcal{C}(X)$ is the number of conjugacy classes $X$ contains, so $\mathcal{C}(X)$ can be as commutative as $\mathcal{F}(X)$ but cannot be as noncommutative as $\mathcal{E}(X).$

When $X$ is an abelian group, $\mathcal{C}(X)$ is isomorphic to $\mathcal{F}(X)$ via an extremely useful map called the Discrete Fourier Transform (DFT), which is perhaps the most widely applied algebra isomorphism there is. When $X$ is nonabelian, $\mathcal{C}(X)$ is isomorphic to a subalgebra of $\mathcal{E}(X)$ via a noncommutative generalization of the DFT whose construction will occupy much of the course and lead us into a new realm where linear algebra and group theory interact in many remarkable ways.

There is no official textbook, but as we move through the course you should regularly consult the following texts:

Algebras of Linear Transformations by Farenick, for the structure theory of finite-dimensional operator algebras.
Linear Representations of Finite Groups by Serre, for the conceptual backbone of representation theory.
Representation Theory of the Symmetric Groups by Ceccherini-Silberstein et al, for a detailed treatment of symmetric groups as the fundamental nonabelian example.
The Symmetric Group by Sagan, for combinatorial viewpoints which make the algebra we encounter more concrete.

DATE	TOPIC	MODALITY
01/05	Algebras I	In Person
01/07	Algebras II	In Person
01/09	Algebras III	In Person
01/12	Function Algebras I	In Person
01/14	Function Algebras II	In Person
01/16	NONE	NO Class
01/19	NONE	NO CLASS
01/21	Operator Algebras I	In Person
01/23	Operator Algebras II	In Person
01/26	Operator Algebras III	In Person
01/28	Operator Algebras IV	In Person
01/30	Operator Algebras V	In Person
02/02	Operator Algebras VI	In Person
02/04	Group Algebras	In Person
02/06	Class Algebras	In Person
02/09	States and traces	In Person
02/11	Review I	Online
02/13	Review II	Online
02/16	NONE	NO CLASS
02/18	Fourier I	In Person
02/20	Fourier II	In Person
02/23	Fourier III	In Person
02/25	Representations I	In Person
02/27	Representations II	In Person
03/02	Representations III	In Person
03/04	Representations IV	In Person
03/06	Symmetric Group I	In Person
03/09	Symmetric Group II	In Person
03/11	Symmetric Group III	In Person
03/13	Symmetric Group IV	In Person

Schedule subject to change, check back regularly.