# Commutative unital Banach algebra not isomorphic to $C(X)$ for some compact Hausdorff space $X$

According to some lecture notes I am reading, "it is not so difficult to find an example of a commutative unital Banach algebra which is not isomorphic to $$C(X)$$ for some compact Hausdorff space $$X$$"...

Well I was thinking about using the disk algebra $$\mathcal{A}$$ of continuous functions on the disk $$D = \{z \mid |z| \leq 1\}$$ which are holomorphic on the interior of $$D$$, with the sup norm.

This should be a commutative unital Banach algebra. However, since it is the beginning of the chapter about the Gelfand transformation, I would like to prove that $$\mathcal{A}$$ is not isomorphic to $$C(X)$$ without using Gelfand's result.

What other tools could I use to prove this fact?

• Fourier series? }:) – logarithm May 9 at 17:50
• @logarithm Could you elaborate? I would be interested to see a proof of this that uses Fourier series! – Laurent Hayez May 10 at 9:20

Here is an argument. Assume that $$\pi:\mathcal A\to C(X)$$ is a unital Banach algebra isomorphism. Let $$f\in \mathcal A$$. Put $$g=\pi^{-1}(\overline{\pi(f)})$$ (that is, map $$f$$ to $$C(X)$$, conjugate it, and come back). Now $$\pi(gf)=\pi(g)\pi(f)=|\pi(f)|^2\geq0.$$ Then, for any $$r>0$$, $$\pi(gf+r+is)=|\pi(f)|^2+r+is$$ takes values at distance $$r$$ or more from $$0$$, so $$gf+r+is$$ is invertible. This says that $$-r+is$$ (we can write a plus since $$s$$ was arbitrary) is not in the image of $$gf$$. In other words, $$\operatorname{Re}(gf)\geq0$$. You can see proof here (applied to $$-gf$$, and it may require a rotation if $$gf(0)$$ is not real) that then $$gf$$ is constant. In other words, there exists $$c\in\mathbb R$$ such that $$gf=c1$$. Then $$|\pi(f)|^2=\pi(gf)=\pi(c1)=c1.$$ Now $$f$$ was arbitrary and $$\pi$$ is onto, so every $$h\in C(X)$$ has $$|h|^2$$ constant. This can only happen if $$X$$ consists of a single point, and in that case $$\mathcal A$$ would be one-dimensional, a contradiction.
• Nice proof, thanks! Just a thing in your post, you might have mixed $fg$ with $gf$ when you say "$\pi(fg + r1)$ is invertible and so $fg + r1$ is invertible". – Laurent Hayez May 10 at 9:22
• Indeed, edited. Note that $\mathcal A$ is commutative, though. – Martin Argerami May 10 at 14:25