# Reflexive Banach algebras?

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. – AlexE Dec 31 '11 at 12:40
@AlexE can you point to a reference for that? Thanks! – Hui Yu Dec 31 '11 at 12:53
A good reference is the book Murphy: $C^\ast$-Algebras and Operator Theory. – AlexE Dec 31 '11 at 13:34
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. – user16299 Jan 8 '12 at 6:36

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.") – Jonas Meyer Dec 31 '11 at 19:11
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. – Jonas Meyer Dec 31 '11 at 19:18
I am confused by the last sentence too. – Hui Yu Jan 1 '12 at 2:37
I have made some corrections to my answer. Hope this will clarify – Norbert Jan 1 '12 at 11:13
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.) – user16299 Jan 8 '12 at 6:34