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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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    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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    See edits of my answer2012-01-08