# Gelfand representation theory

1. Jun 20, 2008

### Pere Callahan

Hi,

I'm reading about Gelfand's representation theory for Banach algebras and I have a small problem with one of the proofs.

Let A be a commutative Banach algebra with unit, define

$\Gamma_A=\left\{\varphi:A\to\mathbb{C}:\varphi\neq0, \varphi \text{ an algebra homomorphism }\right\}$

to be the Gelfand space and let

$$\hat{}:A\to C(\Gamma_A),\quad x\mapsto\hat x= \{\varphi\mapsto\varphi(x)\}$$
be the Gelfand transform.

Then the statement is that ^ is a (continuous) algebra homomorphism between the Banach algebra A and $C(\Gamma_A)$, the space of continuous functions on the Gelfand space.

My problem is: Why is ^ surjective? I agree that for any x in A, $\hat x$ is a continuous function on the Gelfand space but why should ANY function in the Gelfand space come as an evaluation functional at some point x...is this not equivalent to assuming that the A as an Banach space is reflexive? If so, does this follow from the fact that A is commutative and has a unit?

Thanks
Pere

oooh, I just see, it is not claimed that ^ is surjective. So is it right that A is isomorphic to $\hat{A}$ which is a closed subspace of $C(\Gamma_A)$ or is ^ not even necessarily injective...?

Last edited: Jun 20, 2008
2. Jun 20, 2008

### morphism

No, the Gelfand transform doesn't have to be injective. (It's only injective when A is semisimple.)

3. Jun 21, 2008

### Pere Callahan

Thanks morphism, seems like I should just have kept on reading,semisimple algerbas are mentioned on the next page and also that for those the Gelfand transform is injective. Next would be C* algebras for which it is stated that the Gelfand transform is indeed an isomorphism (well at least if the algebra is commutative and has a unit).

Am I right that a C* algebra is usually "smaller" than an ordinary commutative Banach algebra (both with units) because ot its additional structure...so for the Gelfand transform to be an isomorphism must not the Gelfand space for a C* algebra be "smaller" as well..is this because the homomorphisms $varphi$ are required to be involutive....? (I can't find that this requirement is mentioned in the text but it seems natural to me

I'm just trying got get a better understanding, the proofs themselves from a technical point of view are now clear.

4. Jun 21, 2008

### morphism

I don't know about "smaller" (they can be arbitrarily "large", e.g. B(H)), but they are more restrictive because of the extra structure involved. And no, the phi's don't need to be involutive.

Maybe the following remarks will be of some use to you. The Gelfand transform (on a commutative, unital Banach algebra A) is isometric iff ||a^2||=||a||^2 for all a in A. Do you see how this immediately implies that the transform is isometric on commutative, unital C*-algebras? This of course doesn't tell us when the transform is surjective -- for this we need "self-adjointness". Once we have an involution this is natural to state, but we can also give some sort of equivalent statement without the presence of an involution. Namely, we say that A is self-adjoint if for each a in A there is a b in A such that $$\hat{b}(M) = \overline{\hat{a}(M)}$$ for every maximal ideal M of A. Then by assuming that A is self-adjoint, we get that $$\hat{A}$$ is dense in $C(\Gamma_A)$.

5. Jun 22, 2008

### DeaconJohn

Personally, I think of Banach algebras as being naturally "smaller." In other words, take a Banach algebra, A. Why is it not a C* algebra? Because it is missing its adjoints, that's why. Add in the adjoints, take the completion again if necessary, and you get your C* algebra back.

Nothing rigouous there, just a viewpoint. Certainly, all of morphism's comments are correct.

Nice comments, I thought, morphism. Well, put!

6. Jun 23, 2008

### philipp_w

surjective?

I would like to know why the Gelfand transform in the case of a commutative, unital Banach Algebra is isomorphic to a SUBalgebra on C(Omega).

So why is the gelfand transform a function "into" C(Omega)? Why does the Gelfand transform not reach all continuous functions which can be defined on Omega?

Thats why we are heading to C*-Algebras where because of the Stone-Weierstrass Theorem the Gelfand transform is a function "onto".

Thanks for help.

-Philipp

7. Jun 23, 2008

### DeaconJohn

I did not mean that as a general theorem, just as the way I look at it.

So, I take you question in the sense of giving examples. After all, that's a mojor palce to get "feelings," "intuitions" that are helpful.

So let L^infinity be the C* algebra of all essentially bounded complex valued functions on the unit interval. No, that doesn't work, any significant Banach subalgebra is also a C* sub algebra.

O.K., I guess a better try is to look at commutative Banach algebras of linear bounded operators on l^p where 1 < p < infinity and p != 2. If those are subalgebras of l^infinity then my intuition has some value in the commutative case, but, the part about the adjoint rounding things out doesn't apply. If those are not equal to l^infinity, and are not subalgebras of l^infinity, then I would like to know about it. Unfortunately, I don't know enough about those Banach algebras to even guess what happens in those cases right off the top of my head.

Deacon John

8. Jun 23, 2008

### DeaconJohn

(continued)

L^p and l^p are standard notations. L is for Lebesque measure on the real line (or some such similar thing). l (little "l") is for the counting measure on the natural numbers. "Green Rudin," Rudin's "Real and COmplex Analysis," is my favorite reference. His beautiful proofs in that book are mostly due to von Nuemann, by the way.

In the non comutative case, take the upper triangular two by two matricies over the complex numbers. Each such matrix has only thre non-zero entries. I believe that the form a Banach algebra over C. I'm not an expert in this area, so please check me on this. C = the complkex numbers, of course.

O.K., not make 'em a C* algebra by throwing in the adjoints. You get the full C* algebra of 2x2 matrices over C.

Now, that's a good example, seems to me! I hope you like it.

Deacon John