# If $A$ is a non-unital $C^*$-algebra then $\widehat {A}$ can never be compact with respect to the Gelfand topology.

Let $$A$$ be a non-unital $$C^*$$-algebra. Let $$\widehat {A}$$ be the space of all non-zero multiplicative linear functionals on $$A$$ endowed with the Gelfand topology. Then $$\widehat {A}$$ is locally compact but not compact.

I can show that $$\widehat {A}$$ is locally compact since we can take a closed ball around it which is weak* -compact by Banach-Alaoglu theorem. Also since $$A^{\ast}$$ is Hausdorff with respect to weak* topology it follows that $$\widehat {A}$$ is also Hausdorff with respect to the Gelfand topology. I have also shown that $$\widehat {A^+}$$ is homeomorphic to $$\widehat {A} \cup \{\phi_{0}\}$$ where $$A^+$$ is the unitization of $$A$$ and $$\phi_{0}$$ is the zero linear functional on $$A.$$ Therefore we can conclude that $$\widehat {A} \cup \{\phi_{0}\}$$ is compact with respect the induced weak* topology. But I can't able to show that $$\widehat {A}$$ can never be compact unless $$A$$ is unital. In case $$A$$ is unital, $$\widehat {A}$$ is compact being a closed subset of the unit ball which is weak*-compact. But how to deal with the situation if $$A$$ is non-unital? Any help or suggestion in this regard would be warmly appreciated.

Thanks a bunch.

For the commutative case, your claim is true: $$\widehat{C_0(X)}$$ is homeomorphic to $$X$$ and $$C_0(X)$$ is unital iff $$X$$ is compact.

For the non-commutative case, there are counter-examples. Take any non-unital, simple $$C^*$$-algebra with dimension at least 2, for example the compact operators over an infinite dimensional Hilbert space $$K(H)$$. If $$\phi:K(H)\to\mathbb{C}$$ is a *-homomorphism, then $$\ker(\phi)$$ is a closed, two sided ideal. Since $$K(H)$$ is simple, $$\phi=0$$ or $$\ker(\phi)=0$$. If $$\ker(\phi)=0$$, then $$\phi$$ is injective and thus $$K(H)$$ is one dimensional, which is false. So there exist no non-zero *-homomorphisms $$K(H)\to\mathbb{C}$$, i.e. $$\widehat{K(H)}=\emptyset$$.

Do you consider the empty set as a non-compact space? No problem, here is another one:

consider $$A=K(H)\oplus \mathbb{C}$$ (coordinate wise operations and supremum norm on coordinates). This is a non-unital $$C^*$$-algebra and we have natural inlcusions (i.e. injective $$*$$-homomorphisms) $$K(H)\subset K(H)\oplus \mathbb{C}$$ and $$\mathbb{C}\subset K(H)\oplus\mathbb{C}$$. Now if a non-zero $$*$$-homomorphism $$\phi:K(H)\oplus\mathbb{C}\to\mathbb{C}$$ exists, then $$\phi\vert_{K(H)}=0$$, by the preceding paragraph. Also, $$\phi\vert_{\mathbb{C}}$$ is a $$*$$-homomorphism on $$\mathbb{C}$$ so $$\phi\vert_\mathbb{C}=0$$ or $$\phi\vert_{\mathbb{C}}=\text{id}$$, and since we want $$\phi$$ to be non-zero, we must have $$\phi\vert_{\mathbb{C}}=\text{id}$$. In other words, there exists a unique non-zero $$*$$-homomorphism $$K(H)\oplus\mathbb{C}\to\mathbb{C}$$, namely $$(x,\lambda)\mapsto\lambda$$ and thus $$\widehat{A}$$ is a singleton. Whatever topology you put on a singleton, it has to be compact and not just locally compact. You can also take more copies of $$\mathbb{C}$$ in your direct sum and make non-unital $$C^*$$-algebras with maximal ideal space being finite sets (thus compact) and have larger cardinality.

To conclude: you claim that a $$C^*$$-algebra is unital if and only if its maximal ideal space is compact. For the commutative case this is true, for the non-commutative case this is false.

• I think everybody considers the empty set as a compact space. Every open cover has a finite subcover is trivially true :) Commented Oct 10, 2021 at 14:03
• @QuantumSpace It's just that some people (sometimes including me) are not convinced when the counterexample is the emptyset :P Commented Oct 10, 2021 at 15:33
• Can you suggest some proof of the facts used in commutative case? Thanks.
– RKC
Commented Oct 10, 2021 at 16:53
• @RabinKumarChakraborty For the commutative case, it is a classic thing that you can find in the first chapters of Murphy that $\hat{C_0(X)}$ is homeomorphic to $X$. Modulo this part, $C_0(X)$ is unital iff the constant function $f(x)=1$ is contained in $C_0(X)$, i.e. it vanishes at infinity, i.e. for any $\epsilon>0$ the sets $\{x\in X:f(x)\geq\varepsilon\}$ are compact. For $\epsilon=1/2$, you get that $X$ is compact. Commented Oct 10, 2021 at 17:29
• Very nice answer. Many many thanks for your kind help.
– RKC
Commented Oct 11, 2021 at 5:28