1. May 8, 2008

### neworder1

Gelfand-Naimark theorem states that every commutative C-* algebra is isometric to $C(M)$, the ring of continuous functions over its spectrum. Is the theorem true for ordinary commutative Banach semisimple algebras, i.e. without *? Every proof that the Gelfand transform is an isometry uses the fact that in C-* algebra $|xx^{*}| = |x|^{2}$, so I wonder whether it is true when we don't have the * and that identity. If it is not isometric, is it isomorphic?

2. May 8, 2008

### Hurkyl

Staff Emeritus
Well, my first thought is just to think of interesting algebras. The simplest 'interesting' one I can think of is the ring of polynomials with complex coefficients. I'd expect that you could make this into a normed algebra -- how do things work out for this example?

3. May 8, 2008

### gel

No, it won't work for non-C* algebras.

It can't be isometric for anything else, because C(M) is always a C* algebra.
For example, consider the set of integrable functions f:R->C with convolution as the 'multiplication' operator.
This is a Banach algebra with the L1 norm, $\Vert f\Vert = \int |f(x)|\,dx$. However, ||a* a|| <= ||a||2, and equality doesn't always hold, so it isn't a C*-algebra.

Last edited: May 8, 2008
4. May 8, 2008

### gel

Also, in my example, you can see that the Gelfand transform is a Fourier transform, so C(M) is isomorphic to C(R) and the original algebra maps to those elements of C(R) which can be written as a Fourier transform of an integrable function.

Last edited: May 8, 2008
5. May 8, 2008

### gel

I think the way you phrased your question, you were only talking about unitial alegebras (otherwise you should replace C(M) by the continuous functions vanishing at infinity).
In that case, you could let A be the alegebra of functions f:Z->C such that $\sum_n |f(n)| < \infty$, again with convolution as the multiplication operator.

Using fourier series, C(M) is (isomorphic to) the set of continuous functions f:R->C of period 1 or, equivalently, C(S1). However, the Gelfand transform maps A only to those elements whose fourier series is absolutely convergent.

Interesting to note that in both my examples, C(M) is isomorphic to the C*-algebra generated by the Banach alegebra. Wouldn't be surprised if that is always the case.

Last edited: May 8, 2008
6. May 9, 2008

### neworder1

Thanks. What do you mean by "C* algebra generated by a Banach algebra"?

If B is semisimple, then its Gelfand transform f(B) is dense in C(M) - so if Gelfand transform isn't "onto", f(B) isn't closed in C(M)?

7. May 9, 2008

### gel

I'm not sure if this is standard terminology. You could try taking the completion w.r.t. the largest continuous C*-norm, which should give a closed subspace of C(M), and maybe is isometric to C(M).

It isn't closed in my examples above. It is closed (and complete) under the original Banach norm, but not the C*-norm on C(M).
Do you need the semisimple condition?

8. May 10, 2008

### neworder1

What do you mean by C-* norm? $C(M)$ is equipped naturally with standard supremum norm.

I need semisimplicity only to assure that the translation is injective.

9. May 12, 2008

### gel

Given any Banach *-algebra A, you could consider the C* seminorms on A. That is, F is a seminorm satisfying F(a* a) = F(a*) F(a). If F is continuous then it follows that F(a) <= ||a||. So, the maximum of all such seminorms exists and gives a unique maximum C* seminorm. Then, A can be completed with respect to this, to give a C* algebra.
That's what I meant by the C* algebra generated by A.
If this C* algebra is B, then the map A -> C(M) extends uniquely to a continuous homomorphism B->C(M), which I was suggesting gave an isometry.

However, you've already stated that semisimple => map is injective, and that the map is dense, so it seems that you know quite a bit already.