1
$\begingroup$

I am trying to prove the equivalence of categories between compact Hausdorff spaces and unital $C^*$ algebras.

The maps \begin{align*} f: X \mapsto C(X) \\ g: A \mapsto \hat{A} \end{align*} is a contravariant category equivalence between the category of compact Hausdorff spaces and continuous maps and the category of commutative unital $C^*$ algebras and unital $*$-homomoprhisms.

I know one direction $fg \simeq id:A \rightarrow C(\hat{A})$ is Gelfand Naimark.

It then suffices to prove $gf \simeq id$, i.e. $$ X \mapsto \widehat{C(X)}, \quad x \mapsto \phi_x:f \mapsto f(x)$$ is a homeomorphism?

$\endgroup$
1
$\begingroup$

Yes. And the only nontrivial part is that the map $x\longmapsto \phi_x$ is surjective. Below is the argument that I know.

Assume that $\varphi:C(X)\to\mathbb C$ is a character. We prove a few things

  • if $\varphi(f)=0$, then there exists $x_f\in X$ such that $f(x_f)=0$. Indeed, if $f(x)\ne0$ for all $x$, then by compactness there exists $c>0$ with $|f(x)|\geq c$ for all $x$. Then $1/f\in C(X)$ and $1=\varphi(f/f)=\varphi(f)\varphi(1/f)$ and $\varphi(f)\ne0$.

  • for each $f\in C(X)$, there exists $x_f\in X$ with $\varphi(f)=f(t_x)$. Apply the above to the function $f-\varphi(f)$.

  • For each $f\in C(X)$, consider the set $T_f=\{x:\ f(x)=\varphi(f)\}$. By the above, $T_f\ne\varnothing$. As $T_f=f^{-1}(\{\varphi(f)\})$, it is compact.

  • Given $f_1,\ldots,f_n\in C(X)$, we have $\bigcap_{j=1}^n T_{f_j}\ne\varnothing$. Indeed, let $$ g=\sum_{j=1}^n|f_j-\varphi(f_j)|^2. $$ As $\varphi$ is linear and multiplicative, $$ \varphi(g)=\sum_{j=1}^n \varphi[(f_j-\varphi(f_j))\overline{(f_j-\varphi(f_j))}] =\sum_{j=1}^n \varphi[f_j-\varphi(f_j)]\varphi[\overline{f_j-\varphi(f_j)}]=0 $$ As we showed above, there exists $x\in X$ with $g(x)=0$. It follows that $f_j(x)=\varphi(f_j)$ for $j=1,\ldots,n$ so $x\in\bigcap_{j=1}^n T_{f_j}$.

  • The family $\{T_f\}_{f\in C(X)}$ has the finite intersection property. As $X$ is compact, it follows that $\bigcap_{f\in C(X)}T_f\ne\varnothing$. That is, there exists $x\in X$ such that $\varphi(f)=f(x)$ for all $f\in C(X)$.

$\endgroup$
  • $\begingroup$ Thanks a lot Martin, I will take some time to digest this, mean while, if you have time, I hope you could have a look at the problems I have regarding Higson's works. $\endgroup$ – Cy L Shih Apr 6 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.