In the book "A course in commutative Banach Algebras," Kaniuth shows that the Gelfand representation of $C_0(X),$ where $X$ is locally compact is the identity map. The argument is as follows:

The map $$x \to M_x=\{f \in C_0(X):f(x)=0\}$$ sets up a one-to-one correspondence between the points of $X$ and the maximal modular ideals of $C_0(X).$ On the other hand, we have a bijection $$\Delta(C_0(X))\to \operatorname{Max}(C_0(X)), \phi\mapsto \ker \phi$$ This gives us a bijection $$X \to \Delta(C_0(X)),x\mapsto \phi_x$$ where $\phi_x$ is defined as $\phi_x(f)=f(x).$ This map is then shown to be a homeomorphism.

Everything is clear up to this point. It is then concluded that

After identifying $X$ with $\Delta(C_0(X))$, the Gelfand representation of $C_0(X)$ is the identity mapping.

This last line is unclear to me. I know identifying $X$ with $\Delta(C_0(X))$ makes the Gelfand representation a map from $C_0(X)$ to $C_0(X).$ But how does it make it the identity map?


1 Answer 1


The Gelfand representation is the algebra homomorphism $F: C_0(X) \rightarrow C_0(\Delta C_0(X))$ defined by $$Ff(\phi)=\phi(f)$$ for $\phi \in \Delta C_0(X)=\{\psi: C_0X \rightarrow \mathbb{C}\vert~ \psi \text{ is a nonzero algebra homomorphism}\}$. The homeomorphism $h: X \rightarrow \Delta C_0(X), x \mapsto \phi_x$ induces an algebra isomorphism $h^* : C_0(\Delta C_0(X)) \rightarrow C_0(X) $ given by $$h^*(G)(x)=G(\phi_x).$$ You now want to show that the composition $$ C_0(X) \xrightarrow{F} C_0(\Delta C_0(X))\xrightarrow{h^*} C_0(X) $$ equals the identity map. So take $f\in C_0(X)$ and $x\in X$. Then $$h^* F f(x)=Ff(\phi_x)=\phi_x(f)=f(x),$$ i.e. $h^* F f = f.$


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