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Theorem: Suppose $\mathscr{A}$ is a singly generated, commutative, unital $C^\ast$-algebra with $\mathscr{A}=C^\ast(A)$ for some $A$ which is necessarily normal. There is a unique $\ast$-isomorphism of $\mathscr{A}$ onto $C(\sigma(A))$ mapping $A$ to the identity function on $\sigma(A)$.

Claim: If $X$ and $Y$ are homeomorphic compact Hausdorff spaces, then $C(X)$ and $C(Y)$ are $\ast$-isomorphic unital $C^\ast$-algebras.

How can this happen? Is it related to https://math.stackexchange.com/a/268025/593877?

Thanks in advance.

TheCstarLover
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    There is an explicit proof in the link you posted. It's actually an "if and only if" and the proof of the implication you want is simple enough to be stated in one sentence. What are you looking for exactly? – Martin Argerami Nov 20 '22 at 16:19
  • Hi Martin, I was actually looking for a proof of this claim. Most because I would like to see how it is related. But the proof is quite difficult.

    The proof of the claim I give is the first part in the comment, i.e. "=>" direction. For me this is a short written proof which I cannot see how it actually holds. I don't get the details.

     – TheCstarLover Nov 20 '22 at 16:22
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    Still not sure what you are asking. So your question reduces to asking why the map $f\longmapsto f\circ h$ (where $h$ is a homeomorphism) is a $*$-isomorphism? – Martin Argerami Nov 20 '22 at 16:27
  • My actually question is how one can prove the claim as given in OP. Not the theorem I gave. – TheCstarLover Nov 20 '22 at 16:29
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    The proof is that if $h:Y\to X$ is a homeomorphism, then $H:C(X)\to C(Y)$ given by $Hf = f\circ h$ is a $*$-isomorphism. – user469053 Nov 20 '22 at 16:36
  • Hi user! Thank you for your comment. This seems - as the proof in the comment - just like "hand waving". I mean is the proof not bigger than that? – TheCstarLover Nov 20 '22 at 16:39
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    This is what @MartinArgerami was asking. You want to see the details of why $H$ is a $*$-isomorphism? – user469053 Nov 20 '22 at 16:40
  • Yes I want to see it. I've readed the comment in the link I shared but it didn't give me so much yet the other direction in the comment is very detailed however my claim is not "if and only if". – TheCstarLover Nov 20 '22 at 16:43
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    Perhaps you are able to see that the conclusion is obvious on an easy example: $C[0,1]$ is $*$ -isometrically isomorphic to $C[1,2].$ – Ryszard Szwarc Nov 20 '22 at 17:01

1 Answers1

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We know that $C(X)$, $C(Y)$ are unital $C^*$-algebras, which doesn't have anything to do with $X$ and $Y$ being homeomorphic. So I won't elaborate on that part.

Following the proof in your link: Suppose that $h:Y\to X$ is a homeomorphism. Define $H:C(X)\to C(Y)$ by $H(f)=f\circ h$. This is a continuous, scalar valued function on $Y$. It is continuous because it is a composition of continuous functions.

Fix $y\in Y$, $f,g\in C(X)$, and scalars $a,b$. Then $$H(af+bg)(y)=(af+bg)(y)=af(y)+bg(y)=aHf(y)+bHg(y),$$ so $H$ is linear.

We also have that for any $f,g\in C(X)$, $$[H(f)H(g)](y)=f(h(y))g(h(y))=(fg)(h(y))=H(fg)(y),$$ so $H(fg)=H(f)H(g)$.

Moreover, $$H(\overline{f})(y)=\overline{f}(h(y))=\overline{f(h(y))}=\overline{Hf(y)},$$ so $H(\overline{f})=\overline{Hf}$. This shows that $H$ is a $*$-homomorphism.

We note that $H$ is an isomorphism because its inverse $H^{-1}:C(Y)\to C(X)$ is given by $H^{-1}g=g\circ h^{-1}$. To see that this is the inverse of $H$, we note that $HH^{-1}g=g\circ h^{-1}\circ h=g$ and $H^{-1}Hf=f\circ h\circ h^{-1}=f$.

This is also an isometry. Because $h$ is a surjection, $f$ and $f\circ h$ have the same range.

user469053
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    Interesting, I will have a look at your well written proof. Thanks! A question: You say the following: "We know that $C(X), C(Y)$ are unital $C^\ast$-algebras, which doesn't have anything to do with $X$ and $Y$ being homeomorphic." Why? I mean the claim is "If $X$ and $Y$ are homeomorphic compact Hausdorff spaces" THEN $C(X)$ and $C(Y)$ are $\ast$-isomorphic unital $C^\ast$-algebras. I hope it makes sence. :) – TheCstarLover Nov 20 '22 at 16:50
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    If $K$ is any compact, Hausdorff space, then $C(K)$ is a unital $C^$-algebra. The unit is the function $1_K$ given by $1_K(x)=1$ for all $x\in K$. The $$-operation is complex conjugation $f\mapsto \overline{f}$, where $\overline{f}(x)=\overline{f(x)}$. It is straightforward to check that $f\mapsto \overline{f}$ satisfies the requirements of being a $C^*$-algebra, namely that it is a conjugate-linear involution with $\overline{f}f=|f|^2$, so $|\overline{f}f|=|f||\overline{f}|=|f|^2$. – user469053 Nov 20 '22 at 16:53
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    Thanks once again with this great explanation! – TheCstarLover Nov 20 '22 at 17:02