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I have been reading Gelfand theory for a while and it just occurs to me that the whole theory is an analogy to what we did for Banach spaces.

For a Banach space $X$, we investigate its dual $X'$ and double dual $X''$. Some times these spaces give us information about $X$, eg. the embedding of $X$ in $X''$ gives notions such as reflexivity.

In Gelfand theory, we impose more algebraic structure on the space, namely, the object is algebras $A$. Now the dual space become the spectrum $\sigma(A)$ and the embedding becomes the Gelfand transformation $\Gamma:A\to C(\sigma(A))$.

Thus I wonder whether there is similar notion like reflexivity, that is, $\Gamma(A)=C(\sigma(A))$ and whether this gives some information about the algebra.

I have not gotten time to look into this problem myself. But intuitively such case should be rare for each step in $A\to\sigma(A)\to\Gamma(A)$ we lose some thing more than in the case for Banach spaces. But if $\Gamma(A)=C(\sigma(A))$ for some special $A$, it seems to tell a lot.

Thanks!

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    For commutative $C^\ast$-algebras the Gelfand transformation is an isometric-$^\ast$-isomorphism.2011-12-31
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    @AlexE can you point to a reference for that? Thanks!2011-12-31
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    A good reference is the book _Murphy: $C^\ast$-Algebras and Operator Theory_.2011-12-31
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    See also: http://mathoverflow.net/questions/82871/reference-request-for-translating-from-top-to-c-alg2011-12-31
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    Another place to find a detailed exposition: Chapter 11 of Rudin's Functional Analysis (2nd edition). See also Simmons's book Introduction to Topology and Modern Analysis, where the result mentioned by Alex E (the so-called commutative Gelfand-Neumark theorem) is proved somewhere towards the end of the book.2012-01-08
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    I should also mention that as far as I know, no one uses the terminology used in the title of the question. My guess is that almost every functional analyst would interpret the phrase "reflexive Banach algebra" to mean "a Banach algebra whose underlying Banach space is reflexive"2012-01-08

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Here is a more detailed list of properties of Gelfand's transformation, which shows us necessary conditions for algebra $A$ to be isometrically isomorphic to $C_0(\Omega(A))$

Theorem (A. Ya. Helemskii, Banach and locally convex algebras)

Let $\Gamma:A\to C_0(\Omega(A))$ be Gelfand's transformation of commutative Banach algebra $A$, then

  • $\Gamma$ is continuous homomorphism of Banach algebras, and if norm of algebra $A$ is submultiplicative then $\Vert \Gamma\Vert\leq 1$
  • $\text{Ker}(\Gamma)=\text{Rad}(A)$, as the consequence for semisimple Banach algebras Gelfand's transformation is injective.
  • $\text{Im}(\Gamma)$ separates points in $\Omega(A)$
  • If $A$ is unital, then $\Gamma(1_{A})=1_{C_0(\Omega(A))}$

Theorem (D. P. Blecher, C. Le Merdy, Operator algebras and their modules. An operator space approach)

Let $A$ be a $C^*$-algebra, $B$ a Banach algebra and $\pi: A\to B$ a contractive homomorphism. Then $\pi(A)$ is a norm closed and it possesses an involution with respect to which it is a $C^*$-algebra. Moreover $\pi$ is then a $*$-homomorphism into $C^*$-algebra. If $\pi$ is one-to-one then $\pi$ is an isometry.

Thus if we assume that $\Gamma:A\to C_0(\Omega(A))$ is an isometric isomorphism we see that $A$ can be endowed with involution which makes it a $C^*$-algebra. Moreover $\Gamma$ becomes an isometric $*$-isomorphism.

The main result here that if $A$ is a commutative Banach algebra and $\Gamma$ is an isometric isomorphism then $A$ can be made a $C^*$ algebra and, moreover now $\Gamma$ preserves additional structure - involution.

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    Does "multiplicative" mean $\|ab\|\leq\|a\|\|b\|$? (This is sometimes called "submultiplicative.")2011-12-31
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    I don't see how your last sentence is supposed to be related to the theorem you cited. If $\Gamma$ is an isometric isomorphism, then yes, $A$ is a $C^*$-algebra, because it is isometrically isomorphic to the $C^*$-algebra $C_0(\Omega(A))$. You have not given reasons why the converse also holds.2011-12-31
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    I am confused by the last sentence too.2012-01-01
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    I have made some corrections to my answer. Hope this will clarify2012-01-01
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    Norbert, I still find the chain of logic in your presentation a bit confusing. Isn't it just simpler to state the theorem as Alex E did above: if $A$ is a commutative unital C*-algebra then the Gelfand transform $A\to C(\Omega(A))$ is bijective. (When I learned the subject this was called the Gelfand-Neumark theorem, but perhaps other people would debate the terminology.)2012-01-08
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    See edits of my answer2012-01-08