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.

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    $\begingroup$ Well A is by definition contained in a commutative Cstar algebra B, so the enveloping Cstar algebra must be contained in B $\endgroup$ – Yemon Choi Jun 29 '16 at 18:55
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    $\begingroup$ 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 $\endgroup$ – Yemon Choi Jun 29 '16 at 18:56
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    $\begingroup$ You may or may not wish to update the "location" field in your profile, btw. $\endgroup$ – Yemon Choi Jun 29 '16 at 18:57
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    $\begingroup$ 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 $\endgroup$ – Yemon Choi Jun 29 '16 at 19:00
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    $\begingroup$ 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". $\endgroup$ – Yemon Choi Jun 29 '16 at 19:42

What they are aiming at is the following result:

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.

  • $\begingroup$ Thanks - I had a vague memory that this is the result, but couldn't remember it precisely off the top of my head $\endgroup$ – Yemon Choi Jun 29 '16 at 20:02

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