202B – Page 2 – Jonathan Novak

Math 202B: Lecture 17

In this lecture and the next, we will develop the Discrete Fourier Transform (DFT), which is probably the most widely used algebra isomorphism. Abstractly, we already know about this isomorphism from our first characterization of function algebras long ago: a given algebra $\mathcal{A}$ is isomorphic to a function algebra if and only if it admits a basis $\{F^\lambda \colon \lambda \in \Lambda\}$ of selfadjoint and pairwise orthogonal idempotents – a “Fourier basis.” Whenever we have such a basis, we can expand each $A \in \mathcal{A}$ as

$A=\sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)F^\lambda,$

and in doing so we associate a function $\widehat{A}(\lambda) \in \mathcal{F}(\Lambda)$ to each $A \in \mathcal{A}.$ This association is an algebra isomorphism which we call the Fourier transform on $\mathcal{A}.$

Thus, returning to the setting of convolution algebras, the basic question is how to concretely construct a Fourier basis in $\mathcal{C}(G)$ . It is clear that this construction must somehow leverage the group structure of $G,$ and the basic ingredient is the following.

Definition 17.1. Given a finite group $G,$ a group homomorphism $\chi \colon G \to \mathbb{U}$ from $G$ to the unit circle $\mathbb{U} = \{u \in \mathbb{C} \colon |u|=1\}$ is called a multiplicative character of $G.$

Every group $G$ has at least one multiplicative character, namely the “trivial character” which sends every group element to the number $1.$ In last week’s homework you constructed an example of a group for which the trivial character is the the only multiplicative character. For abelian groups, the opposite extreme holds: there are as many multiplicative characters as there are group elements. This fact is the key to constructing a Fourier basis of the convolution algebra of a finite abelian group.

To begin, recall that every finite abelian group is isomorphic to a product of cyclic groups. We thus fix a positive integer $r \in \mathbb{N},$ and $r$ positive integers $n_1,\dots,n_r \in \mathbb{N},$ and consider the group

$G = G_1 \times \dots \times G_r,$

where $G_i$ is a cyclic group of order $n_i$ with generator $g_i.$ Define the dual group of $G$ to be

$\Lambda = \{\alpha=(\alpha_1,\dots,\alpha_r) \colon \alpha_k \in \{0,1,\dots,n_k-1\},\ 1 \leq k \leq r\}.$

That is,

$\Lambda = \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_r},$

where $\mathbb{Z}_n=$ is the additive group of integers modulo $n$ . We can parameterize $G$ by the points of $\Lambda,$ writing

$g_\alpha=(g_1^{\alpha_1},\dots,g_r^{\alpha_r}), \quad \alpha \in \Lambda.$

Problem 17.1. Prove that the parameterization $\alpha \mapsto g_\alpha$ is a group isomorphism $\Lambda \to G$ (Note that because $|\Lambda|=|G|$ it is sufficient to show the parameterization is an injective group homomorphism).

Theorem 17.1. For every $\lambda \in \Lambda,$ the function $\chi^\lambda \colon G \to U(\mathbb{C})$ defined by

$\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda^r},$

where $\omega_k=\exp\left(\frac{2\pi i}{n_k}\right)$ is a principal $n_k$ th root of unity, is a multiplicative character of $G.$

Proof: For any $\lambda \in \Lambda,$ it is clear that $\chi^\lambda(e)=1,$ because the identity element $e \in G$ has parameters $\alpha=(0,0,\dots,0).$ Moreover, for any $\alpha,\beta \in \Lambda$ we have

$\chi^\lambda(g_\alpha g_\beta) = \chi^\lambda(g_{\alpha+\beta})=(\omega_1)^{(\alpha_1+\beta_1)\lambda_1} \dots (\omega_r)^{(\alpha_r+\beta_r)\lambda_r)} = (\omega_1)^{\alpha_1\lambda_1} \dots (\omega_r)^{\alpha_r\lambda_r}(\omega_1)^{\beta_1\lambda_1} \dots (\omega_r)^{\beta_r\lambda_r}=\chi^\lambda(g_\alpha)\chi^\lambda(g_\beta),$

so $\chi^\lambda$ is indeed a group homomorphism $G \to U(\mathbb{C}).$

-QED

Problem 17.2. Enhance Theorem 17.1 by proving that there are no other multiplicative characters of $G.$

We now have a special subset $\{\chi^\lambda \colon \lambda \in \Lambda\}$ of the convolution algebra $\mathcal{C}(G)$ of the finite abelian group $G$ , the set of all its multiplicative characters. We now claim that this set forms a basis of $\mathcal{C}(G).$ Since the number of characters is $|\Lambda|=|G|,$ which is the dimension of $\mathcal{C}(G)$ , it is sufficient to show that $\{\chi^\lambda \colon \lambda \in \Lambda\}$ is a linearly independent set in $\mathcal{C}(G).$

Theorem 17.2. The set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ is orthogonal with respect to the $\ell^2$ -scalar product on $G$ – we have

$\langle \chi^\lambda,\chi^\mu\rangle = \sum\limits_{g \in G} \overline{\chi^\lambda(g)} \chi^\mu(g) = \delta_{\lambda\mu}|G|.$

Proof: For any $\lambda,\mu \in \Lambda,$ we have

$\langle \chi^\lambda,\chi^\mu \rangle = \sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda(g_\alpha)}\chi^\mu(g_\alpha) = \sum\limits_{\alpha \in \Lambda}(\omega_1)^{\alpha_1(\mu_1-\lambda_1)} \dots (\omega_r)^{\alpha_r(\mu_r-\lambda_r)}=\left(\sum\limits_{\alpha_1=0}^{n_1-1} \zeta_1^{\alpha_1}\right) \dots \left(\sum\limits_{\alpha_r=0}^{n_r-1} \zeta_r^{\alpha_r}\right),$

where

$\zeta_1 = \omega_1^{\mu_1-\lambda_1},\ \dots,\ \zeta_r=\omega_r^{\mu_r-\lambda_r}.$

Thus if $\lambda=\mu$ we have

$\langle \chi^\lambda,\chi^\mu \rangle = n_1 \dots n_r = |\Lambda|=|G|,$

and if $\lambda \neq \mu$ we have

$\langle \chi^\lambda,\chi^\mu \rangle =\frac{1-\zeta_1^{n_1}}{1-\zeta_1} \dots \frac{1-\zeta_r^{n_r}}{1-\zeta_r},$

where the denominator of each factor is nonzero and the numerator of some factor is zero, because $\zeta_k=\omega_k^{\mu_k-\lambda_k}$ is an $n_k$ th root of unity and $\mu_k \neq \lambda_k$ for some $k$ .

-QED

The orthogonal basis $\{\chi^\lambda \colon \lambda \in \Lambda\}$ of $\mathcal{C}(G)$ is called its character basis. In fact, the rescaled character basis

$F^\lambda = \frac{1}{|G|}F^\lambda, \quad \lambda \in \Lambda,$

is even better, for the following reason.

Theorem 17.3. The set $\{F^\lambda \colon \lambda \in \Lambda\}$ is a Fourier basis of $\mathcal{C}(G).$

Proof: For any $\lambda \in \Lambda,$ we have

$(F^\lambda)^* = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)^*=\frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)} E_g^* =\frac{1}{|G|} \sum\limits_{g \in G} \chi^\lambda(g^{-1}) E_{g^{-1}} = F^\lambda.$

For any $\lambda,\mu \in \Lambda,$ we have

$F^\lambda F^\mu = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)\left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\mu(g) E_g\right)=\frac{1}{|G|^2}\sum\limits_{g \in G} \left( \sum\limits_{h \in G}\chi^\lambda(gh^{-1})\chi^\mu(h)\right)E_g = \frac{1}{|G|^2}\sum\limits_{g \in G} \chi^\lambda(g)\left( \sum\limits_{h \in G}\chi^\lambda(h^{-1})\chi^\mu(h)\right)E_g.$

The internal sum is

$\sum\limits_{h \in G}\chi^\lambda({h^{-1}})\chi^\mu(h)=\sum\limits_{h \in G}\overline{\chi^\lambda(h)}\chi^\mu(h)=\delta_{\lambda\mu}|G|,$

where the final equality is Theorem 8.3. Thus

$F^\lambda F^\mu = \frac{1}{|G|^2}\delta_{\lambda\mu}|G|\sum\limits_{g \in G} \chi^\lambda(g)E_g=\delta_{\lambda\mu}F^\lambda.$

-QED

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 15

In this lecture we will review the abstract theory of algebras.

Algebras. We began with the general definition of an algebra: a finite-dimensional complex vector space $\mathcal{A}$ together with a multiplication $(A,B) \mapsto AB$ and a conjugation $A \mapsto A^*$ satisfying certain natural axioms. These axioms are to a large extent inspired by the key example $\mathcal{A}=\mathrm{End}(V),$ the endomorphism algebra of a finite-dimensional Hilbert space $V$ , which was the sole subject of Math 202A. We use much of the same terminology for general algebras:

Normal: $A$ and $A^*$ commute.
Unitary: $U$ is invertible with inverse $U^*$ .
Selfadjoint: Important subtypes:
- Idempotent: satisfies $E^2=E.$
- Positive: factors as $P=A^*A$ .

We obtained a simple classification of commutative algebras.

Theorem: $\mathcal{A}$ is commutative if and only if all its elements are normal.

Subalgebras. We next had a general discussion of subalgebras: subspaces $\mathcal{B}$ of an algebra $\mathcal{A}$ which are themselves algebras. More precisely, $\mathcal{B}$ is a vector subspace of $\mathcal{A}$ which is closed under multiplication and conjugation, and contains the multiplicative identity $I$ . In particular, the zero space is not a subalgebra, and $\mathbb{C}I$ is the smallest subalgebra of $\mathcal{A}$ in the sense that it is contained in every subalgebra. We discussed the lattice of subalgebras and for any set $S \subseteq \mathcal{A}$ defined the algebra $\mathrm{Alg}(S)$ it generates to be the intersection of all subalgebras containing $S.$

A particularly important subalgebra is the center $Z(\mathcal{A}),$ the set of elements in $\mathcal{A}$ which commute with every other element. The dimension of $Z(\mathcal{A}$ )$ is a measure of how commutative $\mathcal{A}$ is. Whenever we encounter a particular algebra we wish to understand, our first step should be to understand its center and compute its dimension.

Subalgebras always come in pairs: if $\mathcal{B}$ is a subalgebra of $\mathcal{A},$ there is an affiliated subalgebra $Z(\mathcal{B}),$ the centralizer of $\mathcal{B},$ the set of elements in $\mathcal{A}$ commuting with every element of $B.$ This concept also helps us to identity maximal abelian subalgebras in $\mathcal{A}$ .

Theorem: An abelian subalgebra $\mathcal{B} \leq \mathcal{A}$ is a MASA if and only if $Z(\mathcal{B})=\mathcal{B}.$

Ideals. We did not spend much time on ideals because we don’t really need to know much about them for our purposes. However, this concept arises naturally from the fact that, given an algebra homomorphism $\Phi \colon \mathcal{A} \to \mathcal{B},$ the image of $\Phi$ is a subalgebra of $\mathcal{B}$ but the kernel of $\Phi$ is not generally a subalgebra of $\mathcal{A}.$ . However, $\mathrm{Ker}(\Phi)$ is a subspace of $\mathcal{A}$ closed under conjugation with a special absorption property: if $J \in \mathrm{Ker}(\Phi),$ then both $AJ$ and $JA$ are contained in $\mathrm{Ker}(\Phi)$ for all $A \in \mathcal{A}.$ Any $*$ -closed subspace $\mathcal{J}$ of $\mathcal{A}$ with this property is called an ideal and the quotient vector space $\mathcal{A}/\mathcal{J}$ whose points are sets of the form

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

has a natural algebra structure given by

$[A]_\mathcal{J}[B]_\mathcal{J}=[AB]_\mathcal{J} \quad\text{and}\quad [A]_\mathcal{J}^*= [A^*]_\mathcal{J},$

with the multiplicative identity being $[I]_\mathcal{J}.$ The most basic theorem here is the First Isomorphism Theorem for the category of algebras.

Theorem: For any algebra homomorphism $\Phi \colon \mathcal{A} \to \mathcal{B}$ the algebras $\mathcal{A}/\mathrm{Ker}(\Phi)$ and $\Phi(\mathcal{A})$ are isomorphic via $[A]_{\mathrm{Ker}(\Phi)}\mapsto \Phi(A).$

von Neumann algebras. We arrived at this special class of algebras by wondering whether we can equip a given algebra $\mathcal{A}$ with a scalar product that is compatible with its algebra structure in the sense that the Frobenius identities

$\langle AB,C \rangle = \langle B,A^*C\rangle \quad\text{and}\quad \langle AB,C\rangle = A,CB^*\rangle$

hold for all $A,B,C \in \mathcal{A}$ . When such a scalar product exists and further satisfies $\langle I,I \rangle =1,$ we refer to it as a Frobenius scalar product. We defined a von Neumann algebra to be a pair $(\mathcal{A},\langle \cdot,\cdot\rangle)$ consisting of an algebra equipped with a Frobenius scalar product.

We reduced the search for Frobenius scalar products on $\mathcal{A}$ to the search for linear functionals with certain special properties. First, we established that the left Frobenius identity holds for a given scalar product if and only if it is of the form

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

with $\tau \colon \mathcal{A} \to \mathbb{C}$ a linear functional satisfying $\tau(A^*A) \geq 0$ for all $A \in \mathcal{A},$ with equality if and only if $A=0_\mathcal{A}.$ If such a $\tau$ does exist then it is unique, and the condition $\langle I,I \rangle =1$ is equivalent to $\tau(I)=1$ . We thus defined states to be linear functionals on $\mathcal{A}$ satisfying

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

and resolved to call a state faithful if $\tau(A^*A) \geq 0$ forces $A=0_\mathcal{A}.$ We thus conclude that scalar products on $\mathcal{A}$ satisfying the left Frobenius identity and the normalization condition $\langle I,I \rangle =1$ are in bijection with faithful states on $\mathcal{A}.$ Moreover, we the scalar product $\langle A,B\rangle=\tau(A^*B)$ defined by a faithful state $\tau$ satisfies the right Frobenius identity if and only if $\tau$ is a trace, meaning that $\tau(AB)=\tau(BA)$ for all $A,B \in \mathcal{A}.$ Hence we may equivalently define a von Neumann algebra to be a pair $(A,\tau)$ consisting of an algebra together with a faithful tracial state.

Summary. The above is essentially the abstract theory of algebras that we have developed so far. The rest of the course content concerns particular algebras: function algebras, endomorphism algebras, convolution algebras, and class algebras. A review of these concrete algebras is up next.

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 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

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

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

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

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 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 $\dim V >1.$