The Gelfand Theorem for Commutative C*-Algebras

Spectral Theory

Spectral Permanence

Let be a unital C*-algebra and let be a C*-subalgebra of containing the unit of . Then for any element , its spectrum in is the same as its spectrum in .

Proof The inclusion is always true. To prove the other direction, one shows that if is invertible in , it must also be invertible in . This is first proven for self-adjoint elements, leveraging the fact that for such elements , , which implies that is connected. Since the spectra must agree for self-adjoint elements, one can extend the argument to arbitrary elements by considering the self-adjoint element .

To state the main theorem, we first need the (complex version) Stone-Weierstrass theorem:

The Stone-Weierstrass Theorem

Stone-Weierstrass Theorem (Real Version)

Let be a locally compact Hausdorff space. Let be the algebra of all continuous real-valued functions on vanishing at infinity, equipped with the sup-norm. Assume that is a real subalgebra of such that:

1. separates the points of (i.e., for any with , there exists an such that ).
2. is non-vanishing (i.e., for any , there exists an such that ).

Then, is dense in .

Proof The core of the proof is to show that the closure of , denoted , is a sublattice. That is, if , then and are also in . Step 1: For any distinct , there exists such that and . This extends the initial conditions. By the separation and non-vanishing properties, we can construct a function that is non-zero at both and . Step 2: For any distinct and any , there exists an such that and . Using Step 1, we can find functions that separate and . We can then solve a system of linear equations to find coefficients such that satisfies the conditions. Since is an algebra, .

The determinant is , which is non-zero, guaranteeing a unique solution. Step 3: If , then . The function can be uniformly approximated by a sequence of polynomials on any closed interval . We can ensure . Since is bounded on the compact space , its range is contained in some . The function is in for any polynomial (since is an algebra). As uniformly on , the sequence of functions converges uniformly to . Since is closed, the limit must be in . From this, it follows that is a sublattice, because: Step 4: . Let and . We want to construct an such that . For any pair of points , by Step 2, we can find a function such that and .

1. Fix . For each , define the open set . Since is compact, the open cover has a finite subcover . Define . By Step 3, . This function satisfies for all and .
2. Now, for each , define the open set . Again, this forms an open cover of , so we can find a finite subcover . Define . This is in .

By construction, near . The final function satisfies for all . Also, since each , their minimum also satisfies . Therefore, we have for all , which means .

Stone-Weierstrass Theorem (Complex Version)

Let be a locally compact Hausdorff space, and let be the algebra of complex-valued continuous functions on vanishing at infinity. Let be a complex subalgebra of satisfying:

1. separates the points of .
2. is non-vanishing.
3. is a -subalgebra (i.e., if , then its complex conjugate is also in ).

Then, is dense in with respect to the supremum norm.

Proof Let be the set of real-valued functions in . is a real subalgebra of . For any , we can write and . Since is a -subalgebra, both and are in . separates points and is non-vanishing. Thus, by the real version of the Stone-Weierstrass theorem, . Since any function can be written as , and both and can be approximated by functions in , it follows that is dense in . Specifically, .

Remark

The condition of being a *-subalgebra is essential. For example, consider the disk algebra , the set of functions holomorphic on the open unit disk and continuous on its closure. Viewed as a subalgebra of , it separates points and is non-vanishing, but it is not a *-subalgebra and is not dense in .

The Gelfand Theorem

This fundamental theorem provides a concrete representation for any abstract abelian C*-algebra.

Existence of Characters

Any non-zero abelian C*-algebra has a character.

Proof If is unital, this is a standard result from Banach algebra theory. If is non-unital, since , there exists with . For an abelian C*-algebra, . The spectral radius is given by , where is the character space of the unitization. Since , there must be a character on such that . The restriction of this character to is the required non-zero character on .

Gelfand Theorem for Commutative C*-Algebras

If is an abelian C*-algebra, let be its character space (the set of all non-zero characters on ) equipped with the weak-* topology. The Gelfand representation defined by is an isometric *-isomorphism.

Proof

• -Homomorphism: Since characters are *-homomorphisms, for any , we have . Thus , making a *-homomorphism.
• Isometry: For any , the element is self-adjoint. So , which means is an isometry and thus injective.
• Surjectivity: The image is a *-subalgebra of . It can be shown to separate points and be non-vanishing. By the Stone-Weierstrass theorem, is dense in . Since is an isometry and is a Banach space (complete), its image is also complete and therefore closed in . A dense and closed subspace must be the whole space, so . Thus is surjective.

Continuous Functional Calculus in C*-Algebras

The Gelfand theorem allows us to “do calculus” with elements of a C*-algebra by applying continuous functions to them.

Lemma

Suppose is a unital commutative C*-algebra. Let be normal, , and be its Gelfand representation. Then the map , is a homeomorphism onto the spectrum of .

Proof Assume . Because is generated by and , has to agree with on the whole , thus the mapping is injective. It is continuous by the definition of the weak-* topology on . Moreover, is compact Hausdorff and hence it is a homeomorphism.

Lemma

Suppose and are compact Hausdorff spaces, and is a homeomorphism. Then the pullback , is an isometric -isomorphism.

Continuous Functional Calculus

Let be a unital C*-algebra and let be a normal element (). Let be the spectrum of . There exists a unique unital -homomorphism such that , where is the identity function on . Moreover, is an isometry from onto the C*-subalgebra generated by and , denoted . Therefore, we are allowed write for any continuous function .

Proof We establish a chain of isomorphisms to show existence. Let be the spectrum of the commutative C*-algebra generated by and . We have the following facts:

• The Gelfand representation is an isometric -isomorphism.

• There is a homeomorphism given by .
• This homeomorphism induces an isometric -isomorphism where . We define . This is an isometric -isomorphism from onto , and one can check that .

We now prove the uniqueness. Suppose and are two such unital -homomorphisms. They both map the constant function to the identity element , and they both map the identity function to . The set generates the algebra of all polynomials in . By the Stone-Weierstrass theorem, the algebra of polynomials in and is dense in . Since and are continuous and agree on a dense subset, they must be equal on all of .

Spectral Mapping Theorem

Let be a normal element in a unital C*-algebra . For any , we have: Furthermore, is normal, and for any , we have that .

Proof The spectrum of within the full algebra is the same as its spectrum within the subalgebra . Using the Gelfand representation for , the spectrum of an element is the range of its Gelfand transform: The -homomorphism property means . So, . Since the map is a homeomorphism from onto , the set of values is precisely . Therefore, .