Functional Analysis/C*-algebras

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Functional Analysis
Chapter 5: C^*-algebras
0% developed  as of May 27, 2008 (May 27, 2008). Beware this chapter is less than a draft.

A Banach space over is called a Banach algebra if it is an algebra and satisfies

.

We shall assume that every Banach algebra has the unit unless stated otherwise.

Since as , the map

is continuous.

For , let be the set of all complex numbers such that is not invertible.

5 Theorem For every , is nonempty and closed and

.

Moreover,

( is called the spectral radius of )
Proof: Let be the group of units. Define by . (Throughout the proof is fixed.) If , then, by definition, or . Similarly, we have: . Thus, . Since is clearly continuous, is open and so is closed. Suppose that for . By the geometric series (which is valid by Theorem 2.something), we have:

Thus, is invertible, which is to say, is invertible. Hence, . This complete the proof of the first assertion and gives:

Since is compact, there is a such that . Since (use induction to see this),

Next, we claim that the sequence is bounded for . In view of the uniform boundedness principle, it suffices to show that is bounded for every . But since

,

this is in fact the case. Hence, there is a constant such that for every . It follows:

.

Taking inf over completes the proof of the spectral radius formula. Finally, suppose, on the contrary, that is empty. Then for every , the map

is analytic in . Since , by Liouville's theorem, we must have: . Hence, for every , a contradiction.

5 Corollary (Gelfand-Mazur theorem) If every nonzero element of is invertible, then is isomorphic to .
Proof: Let be a nonzero element. Since is non-empty, we can then find such that is not invertible. But, by hypothesis, is invertible, unless .

Let be a maximal ideal of a Banach algebra. (Such exists by the usual argument involving Zorn's Lemma in abstract algebra). Since the complement of consists of invertible elements, is closed. In particular, is a Banach algebra with the usual quotient norm. By the above corollary, we thus have the isomorphism:

Much more is true, actually. Let be the set of all nonzero homeomorphism . (The members of are called characters.)

5 Theorem is bijective to the set of all maximal ideals of .

5 Lemma Let . Then is invertible if and only if for every

5 Theorem

An involution is an anti-linear map such that . Prototypical examples are the complex conjugation of functions and the operation of taking the adjoint of a linear operator. These examples explain why we require an involution to be anti-linear.

Now, the interest of study in this chapter. A Banach algebra with an involution is called a C*-algebra if it satisfies

(C*-identity)

From the C*-identity follows

,

for and the same for in place of . In particular, (if exists). Furthermore, the -identity is equivalent to the condition: , for this and

implies and so .

For each , let be the linear span of . In other words, is the smallest C*-algebra that contains . The crucial fact is that is commutative. Moreover,

Theorem Let be normal. Then

A state on -algebra is a positive linear functional f such that (or equivalently ). Since is convex and closed, is weak-* closed. (This is Theorem 4.something.) Since is contained in the unit ball of the dual of , is weak-* compact.


5 Theorem Every C^*-algebra is *-isomorphic to where is the spectrum of .

5 Theorem If is isomorphic to , then it follows that and are homeomorphic.

3 Lemma Let be a continuous linear operator on a Hilbert space . Then if and only if for all .

Continuous linear operators with the above equivalent conditions are said to be normal. For example, an orthogonal projection is normal. See w:normal operator for additional examples and the proof of the above lemma.

3 Lemma Let be a normal operator. If and are distinct eigenvalues of , then the respective eigenspaces of and are orthogonal to each other.
Proof: Let be the identity operator, and be arbitrary eigenvectors for , respectively. Since the adjoint of is , we have:

.

That is, , and we thus have:

If is nonzero, we must have .

5 Exercise Let be a Hilbert space with orthogonal basis , and be a sequence with . Prove that there is a subsequence of that converges weakly to some and that . (Hint: Since is bounded, by Cantor's diagonal argument, we can find a sequence such that is convergent for every .)

5 Theorem (Von Neumann double commutant theorem) M is equal to its double commutant if and only if it is closed in either weak-operator topology or strong-operator topology.
Proof: (see w:Von Neumann bicommutant theorem)

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