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Taken from C*-algebras and Operator Theory by Gerald Murphy.

A useful way of thinking of the theory of C*-algebras is as “non-commutative topology”. This is justified by the correspondence between abelian C*-algebras and locally compact Hausdorff spaces given by the Gelfand representation. The algebras studied in this chapter, von Neumann algebras, are a class of C*-algebras whose study can be thought of as “non-commutative measure theory”. The reason for the analogy in this case is that the abelian von Neumann algebras are (up to isomorphism) of the form L^\infty(\Omega,\mu), where (\Omega,\mu) is a measure space.

The theory of von Neumann algebras is a vast and very well-developed area of the theory of operator algebras. We shall be able only to cover some of the basics. The main results of this chapter are the von Neumann double commutant theorem and the Kaplansky density theorem.

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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

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In this section we introduce the important GNS construction and prove that every C*-algebra can be regarded as a C*-subalgebra of B(H) for some Hilbert space H. It is partly due to this concrete realisation of the C*-algebras that their theory is so accessible in comparison with more general Banach spaces.

A representation of a C*-algebra A is a pair (H,\varphi) where H is a Hilbert space and \varphi:A\rightarrow B(H) is a *-homomorphism. We say (H,\varphi) is faithful if \varphi is injective.

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For abelian C*-algebras we were able to completely determine the structure of the algebra in terms of the character space, that is, in terms of the one-dimensional representations. For the non-abelian case this is quite inadequate, and we have to look at representations of arbitrary dimension. There is a deep inter-relationship between the representations and the positive linear functionals of  a C*-algebra. Representations will be defined and some aspects of this inter-relationship investigated in the next section. In this section we establish the basis properties of positive linear functionals.

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I’m getting the impression that the bra-ket notation is more useful for linear ON THE LEFT!

An approximate unit for a C*-algebra is an increasing net \{u_\lambda\}_{\lambda\in\Lambda} of positive elements in the closed unit ball of A such that a= \lim_{\lambda }au_\lambda=\lim_\lambda u_\lambda a for all a\in A.

Example

Let H be a Hilbert space with infinite orthonormal basis \{e_n\}. The C*-algebra K(H) is now non-unital. If P_n is the projection onto \langle e_1,\dots,e_n\rangle, then the increasing sequence \{P_n\}\subset K(H) is an approximate unit for K(H). It will suffice to show that T=\lim_np_nT if T\in F(H), since F(H) is dense in K(H). Now if  T\in F(H), there exist x_1,\dots,x_my_1,\dots,y_m\in H such that:

T=\sum_{k=1}^m|x_k\rangle\langle y_k|.

Hence,

P_nT=\sum_{k=1}^m|P_nx_k\rangle\langle y_k|.

Since \lim_n P_nx=x for all x\in H, therefore for each $k$:

\lim_{n\rightarrow \infty}\||P_nx_k\rangle\langle y_k-|x_k\rangle\langle y_k|\|=\lim_{n\rightarrow \infty} \|P_nx_k-x_k\|\|y_k\|=0.

Hence, \lim_{n}P_nT=T.

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Question 1

Let A be a Banach algebra such that for all a\in A the implication

Aa=0 or aA=0 \Rightarrow a=0

holds. Let LR be linear mappings from A to itself such that for all a,b\in A,

L(ab)=L(a)bR(ab)=aR(b), and R(a)b=aL(b).

Show that L and R are necessarily continuous.

Question 2

Let A be a unital C*-algebra.

(a)

If a,b are positive elements of A, show that \sigma(ab)\subset \mathbb{R}^+.

Solution (Wills)

For elements a,b of a unital algebra A:

\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}

If a\in A^+ then a^{1/2}\in A^+ so that

\sigma(ab)\cup\{0\}=\sigma(a^{1/2}(a^{1/2}b))\cup\{0\}=\sigma(a^{1/2}ba^{1/2})\cup\{0\}

Now if b\in A^+, for any c\in Ac^*bc\in A^+. Hence \sigma(a^{1/2}ba^{1/2})\subset \mathbb{R^+} and the result follows (note that ab need not be hermitian) \bullet

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Let X be a compact Hausdorff space and H a Hilbert Space. A spectral measure E relative to (X,H) is a map from the \sigma-algebra of all Borel sets of X to the set of projections in B(H) such that

  1. E(\emptyset)=0E(X)=1;
  2. E(S_1\cap S_2)=E(S_1)E(S_2) for all Borel sets S_1,\,S_2 of X;
  3. for all x,y\in H, the function E_{x,y}:S\mapsto \langle E(S)x,y\rangle, is a regular Borel complex measure on X.

A Borel measure \mu is a measure defined on Borel sets. If every Borel set in X is both outer and inner regular, then \mu is called regular. A measurable A\subset X is inner and outer regular if

\mu(A)=\sup\left\{\mu(F):\text{ closed }F\subset A\right\}, and

\mu(A)=\inf\left\{\mu(G):A\subset G\text{ open }\right\}

Denote by M(X) the Banach space of all regular Borel complex measures on X, and by B_\infty(X) the C*-algebra of all bounded Borel-measurable complex-valued functions on X (I assume with respect to the Borel \sigma-algebra on \mathbb{C}).

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This starts on P.58 rather than p.53. I underline my own extra explanations, calculations, etc. The notation varies on occasion from Murphy to notation which I prefer.

H is always a Hilbert space. H^1 is the set of unit vectors.

If P is a finite-rank projection on H, then the C*-algebra A=PB(H)P is finite dimensional. To see this, write P=\sum_{j=1}^ne_j\otimes e_j, where e_1,\dots,e_n\in H. If T\in B(H), then

PTP=\sum_{j,k=1}^n(e_j\otimes e_j)T(e_k\otimes e_k)=\sum_{j,k=1}^n\langle Te_k,e_j\rangle (e_j\otimes e_k)

Hence, A\subset <e_j\otimes e_k>, (j,k=1,\dots,n) (*), and therefore is finite dimensional.

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