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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?

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

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  • $\begingroup$ 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". $\endgroup$ – Laurent Hayez May 10 at 9:22
  • $\begingroup$ Indeed, edited. Note that $\mathcal A$ is commutative, though. $\endgroup$ – Martin Argerami May 10 at 14:25
  • $\begingroup$ Also, the reasoning was not complete as I had written it, so I edited it that, too. $\endgroup$ – Martin Argerami May 10 at 17:25

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