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 

Advertisements