# The C*-envelope of the algebra of continuous functions on a compact topological space is commutative

In my research in operator theory, specifically in C* algebras and enveloping, I came across this strange footnote in a text (locally published in non English where I study) which states the following:

Suppose we have X a compact topological space, now suppose we have A, a sub-algebra of the algebra of continuous functions $C(X)$ and we know A contains 1 (the identity) and that A separates points, in this case we know A is an operator algebra, and we also know the C*-enveloping $C_{e}^{*}(A)$ is commutative and that there exists a topological space F such that $C_{e}^{*}(A) \simeq C(F)$.

OK, I'll admit I am not an expert on enveloping of C* algebras, but this seems really strange to me non the less as it is said as if it is an elementary result but still it is beyond my reach, why is $C_{e}^{*}(A)$ necessarily commutative and why do we must have $C_{e}^{*}(A) \simeq C(F)$, I do not even know what the topological space F is nor do I have any idea how to begin to construct it. Can someone please help me by explaining this strange remark I came across? It seems like valuable information to know but I cannot find it anywhere or understand it. I thank all helpers.

• Well A is by definition contained in a commutative Cstar algebra B, so the enveloping Cstar algebra must be contained in B – Yemon Choi Jun 29 '16 at 18:55
• And then once you define E to be the envelope, E is a commutative unital Cstar algebra, hence it's of the form C(K) -- in fact K is the Gelfand spectrum of E – Yemon Choi Jun 29 '16 at 18:56
• You may or may not wish to update the "location" field in your profile, btw. – Yemon Choi Jun 29 '16 at 18:57
• With a lot of these constructions, any "construction" is usually indirect and not very good at giving a concrete picture. On the other hand, for particular examples it is often possible to guess what F should be, and then check using the definition that this works – Yemon Choi Jun 29 '16 at 19:00
• No I am not saying that. I am saying that if you had a particular example of an A which you "saw in the wild", you could guess what F should be. For related concepts in classical function theory, look up the words "Shilov boundary". – Yemon Choi Jun 29 '16 at 19:42

Let $A \subset C(X)$ be a uniform algebra. Then there exists a unique compact set $F \subset X$, known as the Shilov boundary w.r.t. to $A$, such that every function in $A$ achieves its maximum modulus on $F$. Moreover, $$C_e^*(A) \cong C(F).$$ See chapter 16 in Completely Bounded Maps and Operator Algebras by Vern Paulsen for a proof of this result.