ple. If is a compact Hausdorff space then is a commutative C*-algebra with the involution being complex conjugation. It may be regarded as a space of operators acting by multiplication on if is equipped with a suitably nice Borel measure, although I am not sure of the precise hypotheses we need for the operator norm to match the sup norm. At a minimum the measure of every open set must be positive.
We now need the following.
Black Box #2 (Beurling-Gelfand spectral radius formula): Let be an element of a Banach algebra . Then the spectral radius satisfies
This is a surprising result in light of the fact that the LHS depends only on the algebra structure of while the RHS is defined in terms of the norm on . The crux of the proof is the following version of the Cauchy integral formula, which we will not prove here:
for all . If , then it follows that
hence that for all . Since we already know that , the conclusion follows. Again, we defer discussion of the relevant complex analysis to a later post.
Corollary: Let be a self-adjoint element of a C*-algebra. Then .
Proof. The B*-identity gives , and iterating this gives
for all . Taking -th roots of both sides and taking limits, the conclusion follows by the spectral radius formula.
Corollary: Let be an element of a C*-algebra . Then . Consequently, .
Proof. The B*-identity and the above gives since is self-adjoint. Now, , and since is invertible if and only if is invertible, it follows that , hence , and the conclusion follows.
So the norm on a C*-algebra is already uniquely determined by its complex -structure. Moreover:
Corollary: Let be C*-algebras. Any complex -algebra morphism is a weak contraction; consequently, it is already a morphism of C*-algebras. Equivalently, the category of C*-algebras is a full subcategory of the category of complex -algebras.
Proof. If is invertible in then is invertible in . Consequently, , so . By the above, it then follows that .
Before we proceed, we will also need the following.
Lemma: Let be a self-adjoint element of a C*-algebra . Then has real spectrum.
This sum converges absolutely for all and we have . Standard analytic arguments show that
and we know that by the above. It follows that if is any morphism of Banach algebras, it is in particular a weak contraction, so . But and ; combining these two conditions we conclude that has zero imaginary part.
We now have enough to prove the following remarkable result.
Theorem (commutative Gelfand-Naimark): Let be a commutative C*-algebra. Then the Gelfand representation is an isomorphism of C*-algebras.
In particular, commutative C*-algebras are semiprimitive.
Proof. It remains to show that 1) the Gelfand representation respects , 2) the Gelfand representation is an isometry (hence ) and in particular injective, 3) the Gelfand representation is surjective.
1): Let where are self-adjoint (recall that every element of is uniquely of this form). If is a morphism, then are real by the lemma above, hence
2): This follows from and 1), since the Gelfand representation preserves spectral radius. If it follows now that , hence , hence the Gelfand representation is injective.
3): By 2), may be identified with a closed subalgebra of with norm the induced norm which is closed under complex conjugation, and by definition of the maximal spectrum it separates points. By the complex Stone-Weierstrass theorem, it then follows that .
Remember that when we exhibited a contravariant equivalence between compact Hausdorff spaces and certain -algebras we did not specify what algebras arose in this way. The Gelfand-Naimark theorem supplies the answer: they are precisely the self-adjoint subalgebras of commutative C*-algebras. Combined with results we already know about this equivalence, we conclude the following.
Gelfand duality: The functors form a contravariant equivalence between commutative C*-algebras and compact Hausdorff spaces.
Corollary: Let be a topological space. Then is the Stone–Cech compactification of .
Gelfand duality is a very suggestive theorem: it suggests that doing topology with compact Hausdorf