# Question 1

Let $a,b$ be normal elements of a C*-algebra $A$, and $c$ an element of $A$ such that $ac=cb$. Show that $a^*c=cb^*$, using Fuglede’s theorem and the fact that the element

$d=\left(\begin{array}{cc}a &0\\ 0&b\end{array}\right)$

is normal in $M_2(A)$ and commutes with

$d'=\left(\begin{array}{cc} 0&c\\ 0&0\end{array}\right)$.

This more general result is called the Putnam-Fuglede theorem.

## Solution

Fuglede’s theorem states that if $a$ is a normal element commuting with some $b\in A$, then $b^*$ also commutes with $a$. Now we can show that $d^*d=d^*d$ using the normality of $a$ and $b$. We can also show that $d$ and $d'$ commute. Hence by the theorem $d$ and $d^*$ commute. This yields:

$bc^*=c^*a$.

Taking conjugates:

$cb^*=a^*c$,

as required $\bullet$

# Question 2

Let $\rho$ be a positive linear functional on $A$.

## (a)

If $I$ is a closed ideal in $A$, show that $I\subset \text{ker}\,\rho$ iff $I\subset \text{ker }\varphi_\rho$.

### Solution

Suppose that $I$ is a (non-zero) closed ideal such that $I\subset \text{ker }\rho$. Let $i\in I$ so we have that $\rho(i)=0$. Consider now $\varphi_\rho(i)$. Now

$\varphi_\rho(i)(a+N_\rho)=ia+N_\rho=j+N_\rho$,

as $ia\in I$ as $I$ is an ideal. But $j\equiv 0\text{ mod }N_\rho$ as $\rho(j^*j)=0$ (as $j^*j\in I$). Hence $I\subset \text{ker }\varphi_\rho$.

Now suppose that $I\subset \text{ker }\varphi_\rho$ so that $\varphi_\rho(i)=0$.Then

$\langle\varphi_\rho(i)(u_\lambda+N_\rho),(u_\lambda+N_\rho)\rangle=0$,

$\Rightarrow \rho(u_\lambda^*iu_\lambda)=0$.

Take limits to get $\rho(i)=0$ $\bullet$

## (b)

We say that $\rho$ is faithful if $\rho(a)=0$ $\Rightarrow a=0$ for all $a\in A^+$. Show that if $\rho$ is faithful, then the GNS representation $(H_\rho,\varphi_\rho)$ is faithful.

### Solution

To show that $(H_\rho,\varphi_\rho)$ is faithful we must show that $\varphi_\rho$ is injective. That is we must show that $\varphi_\rho(a)=\varphi_\rho(b)$ implies that $a=b$. Suppose therefore that $\varphi_\rho(a)=\varphi_\rho(b)$ and that $c+N_\rho$ is a non-zero element of $H_\rho$. Then we have that $ac+N_\rho=bc+N_\rho$. If these are equal then we have that $ac-bc=(a-b)c\equiv 0\text{ mod }N_\rho$. That is $(a-b)c\in N_\rho$:

$\rho(c^*(a-b)^*(a-b)c)=0$.

But $\rho$ is faithful and $c^*(a-b)^*(a-b)c$ is positive hence $c^*(a-b)^*(a-b)c=0$. Now by the C*-equation:

$\|c^*(a-b)^*(a-b)c\|=\|(a-b)c\|^2=0$.

We can let an approximate unit $\{u_\lambda\}$ play the role of $c$ so we have that

$\|(a-b)u_\lambda\|=0$

$\Rightarrow \|a-b\|=0$

That is, $a-b=0$ and so $a=b$ as required $\bullet$

## (c)

Suppose that $\alpha$ is an automorphism of $A$ such that $\rho(\alpha(a))=\rho(a)$ for all $a\in A$. Define a unitary on $H_\rho$ by setting $U(a+N_\rho)=\alpha(a)+N_\rho$. Show that $\varphi_\rho(\alpha(a))=U\varphi_\rho(a)U^*$.

### Solution (Wills)

Consider first

$\langle\alpha(a)+N_\rho,\alpha(b)+N_\rho\rangle=\rho(\alpha(b)^*\alpha(a))=\rho(\alpha(b^*a))$

$=\rho(b^*a)=\langle a+N_\rho,b+N_\rho\rangle.$

Therefore $U$ has dense range in $H_\rho$ — because $\alpha$ is an automorphism — and is an isometry so therefore it is a unitary operator.

To find the the adjoint we find a map $U^*:H_\rho\rightarrow H_\rho$ such that $UU^*=I=U^*U$. We can look at $A/N_\rho$ and extend to $H_\rho$ by continuity. Try $U^*(a+N_\rho)=\alpha^{-1}(c)+N_\rho$:

$UU^*(a+N_\rho)=U(\alpha^{-1}(a)+N_\rho)=a+N_\rho$,

and similarly $U^*U=I$.

Now let $b+N_\rho\in H_\rho$. It is now a simple calculation to show that the relation holds. Firstly $\varphi_\rho(\alpha(a))(b+N_\rho)\alpha(a)b+N_\rho$. Now

$U^*(b+N_\rho)=\alpha^{-1}(b)+N_\rho$

$\Rightarrow \varphi_\rho(a)U^*(b+N_\rho)=a\alpha^{-1}(b)+N_\rho$

$\Rightarrow U\varphi_\rho(a)U^*(b+N_\rho)=\alpha(a\alpha^{-1}(b))+N_\rho=\alpha(a)\alpha(\alpha^{-1}(b))+N_\rho$ $\bullet$

# Question 3

If $\varphi:A\rightarrow B$ is a positive linear map between C*-algebras, show that $\varphi$ is necessarily bounded.

## Solution

Just a rehashing of Theorem 3.3.1 $\bullet$

## Question 4

Suppose that $A$ is unital. Let $\alpha$ be an automorphism of $A$ such that $\alpha^2=I_A$. Define $C$ to be the set of matrices

$C=\left\{\left(\begin{array}{cc}a&b\\\alpha(b)&\alpha(a)\end{array}\right):a,b\in A\right\}$.

Show that $C$ is a C*-subalgebra of $M_2(A)$.

## Solution

It is straightforward to show that $C$ is a vector subspace as it contains the zero matrix ($a=0=b$) and is closed under linear combinations ($\alpha$ is linear). We need to show that it is closed under multiplication and under the involution.

Let $c_1,c_2\in C$ and consider $c_1c_2$:

$\left(\begin{array}{cc}a_1&b_1\\\alpha(b_1)&\alpha(a_1)\end{array}\right)\left(\begin{array}{cc}a_2&b_2\\\alpha(b_2)&\alpha(a_2)\end{array}\right)=$

$\left(\begin{array}{cc}a_1a_2+b_1\alpha(b_2)&a_1b_2+b_1\alpha(a_2)\\\alpha(b_1)a_2+\alpha(a_1)\alpha(b_2)&\alpha(b_1)b_2+\alpha(a_1)\alpha(a_2)\end{array}\right)$

Now

$\alpha(a_1a_2+b_1\alpha(b_2))=\alpha(a_1a_2)+\alpha(b_1)\alpha^2(b_2)=\alpha(b_1)b_2+\alpha(a_1)\alpha(a_2)$,

(and vice versa on the other diagonal) as required.

Also

$\left(\begin{array}{cc}a&b\\\alpha(b)&\alpha(a)\end{array}\right)^*=\left(\begin{array}{cc}a^*&\alpha(b)^*\\b^*&\alpha(a)^*\end{array}\right)$

$=\left(\begin{array}{cc}a^* & \alpha(b^*)\\b^*&\alpha(a^*)\end{array}\right)$.

This is an element of $C$ with $a\sim a^*$ and $b\sim \alpha(b^*)$.

That is, $C$ is a C*-subalgebra of $M_2(A)$ $\bullet$

Define  a map $\varphi:A\rightarrow C$ by setting

$\varphi(a)=\left(\begin{array}{cc}a & 0\\ 0&\alpha(a)\end{array}\right)$.

Show that $\varphi$ is an injective *-homomorphism.

## Solution

To do this we must show that $\varphi(a)^*=\varphi(a^*)$$\varphi(ab)=\varphi(a)\varphi(b)$ and if $\varphi(a)=\varphi(b)$, then $a=b$.

Firstly,

$\varphi(a^*)=\left(\begin{array}{cc}a^* & 0\\ 0 & \alpha(a^*)\end{array}\right)=\varphi(a)^*$.

Also

$\varphi(a)\varphi(b)=\left(\begin{array}{cc} a & 0\\ 0&\alpha(a)\end{array}\right)\left(\begin{array}{cc}b&0\\ 0& \alpha(b)\end{array}\right)=\left(\begin{array}{cc}ab &0\\ 0&\alpha(a)\alpha(b)\end{array}\right)=\varphi(ab)$,

once we use the homomorphism property of $\alpha$. Finally the injectivity of $\varphi$ is trivial $\bullet$

We can thus identify $A$ as a C*-subalgebra of $C$. If we set

$U=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$,

then $U$ is a self-adjoint unitary and $C=A+AU$. If $B$ is any unital C*-algebra with a self-adjoint unitary element $V$, and $\psi:A\rightarrow B$ is a *-homorphism such that

$\psi(\alpha(a))=V\psi(a)V^*$

show that there is a unique *-homomorphism $\psi_1:C\rightarrow B$ extending $\psi$ (in the sense that $\psi_1\circ \varphi=\psi$) such that $\psi_1(U)=V$.

### Solution (Wills)

Let $a\in A$. We want to choose $\psi_1$ such that $\psi(a)=\psi_1\circ\varphi(a)$ on $A$. Define a map $\psi_0:B\rightarrow C$ by:

$\psi_0\left(\left[\begin{array}{cc}a&b\\\alpha(a)&\alpha(b)\end{array}\right]\right)=\psi(a)+\psi(b)V$.

Linearity is easy to show and it’s clear that $\psi_0\circ \varphi=\psi$. To show that $\psi_0$ is a *-homomorphism we must show that it is multiplicative and respect the *-operation. Multiplicativity is a straightforward calculation where we use the fact that $\psi(\alpha(a))=V\psi(a)V$ (the fact that $V=V^*$ is very useful here.) Similarly showing that the *-operation is respected it straightforward using this fact.

To show uniqueness, suppose that $\psi_1$ is another map with all these properties. Let $a\in A$. Then

$(\psi_1-\psi_0)\circ\varphi(a)=\psi(a)-\psi(a)=0$.

So that ${\psi_1}_{|\varphi(A)}={\psi_0}_{|\varphi(A)}$. Now let $b\in A$ and examine:

$(\psi_1-\psi_0)(\varphi(b)U)=\psi_1(\varphi(b)U)-\psi(b)V$

$=\psi_1(\varphi(b))\psi_1(U)-\psi(b)V=\psi(b)V-\psi(b)V=0$;

that is $\psi_1=\psi_0$ $\bullet$