Gelfand Representation

The main goal is to represent a commutative Banach algebra as an algebra of continuous functions on a topological space. Specifically, we want to find a locally compact Hausdorff space such that is isomorphic to , the algebra of continuous functions on vanishing at infinity.

Topological Interlude: The Weak* Topology

Weak* Topology

Let be a Banach space and its dual space of all bounded linear functionals on .

• We say a net converges in the weak* topology to if for all .
• The weak* topology is the topology generated by the subbase of open sets of the form for , , and . Any open set can be written as an intersection of finitely many such sets.

Theorem

The weak* topology is Hausdorff.

Proof This is a corollary of the Hahn-Banach theorem.

Banach-Alaoglu Theorem

Let be a Banach space. Then the closed unit ball in (with respect to the norm of ) is weak*-compact.

The Character Space

Character

Let be a commutative Banach algebra. A character on is a non-zero algebra homomorphism ; that is, a non-zero linear functional such that for all . The set of all characters on is denoted by and is called the character space or spectrum of .

e.g. Let be a compact Hausdorff space and let . For any fixed , the evaluation map defined by is a character.

Character Space is boring for non-commutative algebras

Characters can be defined on any Banach algebra . However, for non-commutative algebras, the character space may not be very interesting. For example, if for , then is empty. Let us consider the case. For the sake of contradiction, assume is a character of , then we have At the same time, This shows that . Similarly, we can show that all of the are zeroes as well. Therefore is the zero map, yielding a contradiction.

Lemma

Any character on a Banach algebra is bounded. More precisely, . If is unital, then .

Since is a subset of the dual space , we can equip with the weak* topology.

Theorem

The character space of a commutative Banach algebra is a locally compact Hausdorff space when equipped with the weak* topology. If is unital, then is compact.

Proof is Hausdorff because the weak* topology is Hausdorff. By the lemma, any character has , so is a subset of the closed unit ball in . By the Banach-Alaoglu theorem, this unit ball is weak*-compact. Case 1: is a unital algebra. We will show that is a weak*-closed subset of the unit ball (ref. theorem). For any , define the map by , and by . The maps and are continuous in the weak* topology. A functional is a character if and only if for all and (which implies ). Therefore, we can write as: Since and are weak*-continuous, the sets and are weak*-closed. The intersection of closed sets is closed. Thus, is a weak*-closed subset of the compact unit ball, and is therefore compact. Case 2: is any commutative Banach algebra. We claim that the set is weak*-compact. A functional is in if and only if it is a homomorphism (we allow the zero homomorphism here). As before, this shows is a weak*-closed subset of the unit ball (since any homomorphism has ), so is weak*-compact. The character space is . Since is a compact Hausdorff space, and is a closed point, is an open subset of . An open subset of a compact Hausdorff space is locally compact. Hence, is locally compact.

There is a deep connection between characters and maximal ideals in a commutative Banach algebra.

3

Let be a unital commutative Banach algebra.

1. If , then is a maximal ideal in .
2. (Gelfand-Mazur, part of) If is a maximal ideal in , then the quotient algebra is isometrically isomorphic to via the map where .

Proof of (i) Let . Since is a homomorphism, its kernel is an ideal. Since , is a proper ideal. Suppose is an ideal such that . Let . Then . Consider the element , which is in . We have . Since is unital, , so . This means . Since both and are in , their sum must also be in . Thus, , which proves that is a maximal ideal.

Theorem

Let be a unital commutative Banach algebra. The mapping is a bijection from onto the set of all maximal ideals of . In particular, since every unital commutative Banach algebra has a maximal ideal, .

Proof By Lemma 3(i), the map is well-defined.

• Injectivity: Let with . For any , the element has . So, . Applying gives , which means . Since , we get . As this holds for all , .
• Surjectivity: Let be a maximal ideal in . Let be the quotient homomorphism, with . By Lemma 3(ii) (Gelfand-Mazur), there is an isomorphism . We can define . This is a homomorphism, and its kernel is . Since is a proper ideal, is not the zero homomorphism. Thus, and .

e.g. For where is a compact Hausdorff space, we saw that the evaluation maps are characters. The kernel of is , which is a maximal ideal. One can prove that every maximal ideal of is of the form for some unique . This establishes a bijection between and the set of maximal ideals. The map is a homeomorphism from onto .

5

Let be a unital commutative Banach algebra and let . The following are equivalent:

1. (i.e., is invertible).
2. does not belong to any maximal ideal of .
3. for all .

Note: The whiteboard contains a slightly different phrasing for TFAE conditions, which have been corrected here to the standard theorem.

The Gelfand Transform

Gelfand Transform

Let be a commutative Banach algebra. The Gelfand transform is the map defined by where is the function given by for all . The set is the image of the Gelfand transform.

The function is continuous on because the topology on is the weak* topology, which is precisely the weakest topology making all such evaluation maps continuous. If is non-unital, one can show vanishes at infinity, so . If is unital, is compact, so .

The Spectrum Formula

Let be a unital commutative Banach algebra. For any , the spectrum of is precisely the range of its Gelfand transform: If is non-unital, the formula is .

Proof (Unital Case) is not invertible. By Lemma 5, this is equivalent to there existing a character such that . This means , and since , we get . This is equivalent to saying is in the range of .

Gelfand Representation Theorem

For a commutative Banach algebra , the Gelfand transform is a norm-decreasing algebra homomorphism from to .

• The kernel of the Gelfand transform is the Jacobson radical of , .
• The spectral radius of is given by the supremum norm of its transform: .