# Why is the Berkovich spectrum of a C*-Algebra the same as the Gelfand spectrum?

Let $A = \mathcal{C}(X)$ be a commutative (unital) C*-Algebra. Let $Spec(A)$ denote its Gelfand spectrum $$Spec(A) = \{A \rightarrow \mathbb{C} : \text{non-zero *-homomorphism} \} \simeq X.$$ Now view $A$ as a normed algebra over $\mathbb{C}$ with respect to the supremum norm $\| a\|_{sup} = \sup_{x \in X} (\|a(x)\|)$. Its Berkovich spectrum $Sp^B(A)$ is then given by $$Sp^B (A) = \{|\cdot|: A \rightarrow \mathbb{R}_+ \text{ multiplicative seminorm bounded by }\| \|_{sup} \}.$$

Why do we have $Spec(A) = Sp^B(A)$? (as for example stated in the wikipedia article on Berkovich spectra https://en.wikipedia.org/wiki/Berkovich_space#Berkovich_spectrum)

It is clear that every homomorphism $A \rightarrow \mathbb{C}$ composed with the absolute value gives a seminorm of the required form. However I struggle to see the converse.

Given a multiplicative seminorm $|\cdot| : A \rightarrow \mathbb{R}_+$ bounded by $\| \|_{sup}$, why must there be an $x \in X$ such that for all $a \in A:$ $|a| = \| a(x) \|$?

Using the boundedness and multiplicativness of $|\cdot|$, I can show that for each $a \in A$ there must be an $x$ such that $|a| = \| a(x) \|$. But a priori these $x$ can be different for different $a$ - I simply fail to see why this cannot happen.

• If you have a multiplicative continuous seminorm, its kernel is a maximal ideal. – user1688 Jul 8 '16 at 13:58
• Is it obvious why the kernel is \emph{maximal} and not just prime? Please forgive if I am ignorant of basic C*-theory, not my area of expertise, any pointers of proofs welcome. – Niki Jul 9 '16 at 10:00
• The kernel is a closed ideal such that the quotient admits a full multiplicative norm. A C* algebra with a full multiplicative norm must be equal to the algebra of complex numbers. – user1688 Jul 9 '16 at 10:10
• Niki, this follows from Urysson lemma. In $C(X)$, $X$ locally compact not a point, you could always find two functions with disjoint supports, ie zero divisors. Apply this to $X$ being the spectrum of $A/p$, $p$ prime. – Uri Bader Jul 10 '16 at 6:15

## 1 Answer

Observe that every commutative C*-algebra $A \not\simeq \mathbb{C}$ has a non-trivial zero-divisor. To see this view $A$ as continuous functions over its spectrum which contains at least two points (and is Hausdorff) and use Urysohn lemma to construct two continuous functions supported at separating open sets of these two points. It follows that every prime ideal is the kernel of a homomorphism to $\mathbb{C}$.

Now, given a multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a prime ideal and it defines a multiplicative norm on $A/p\simeq \mathbb{C}$. This norm must coincide with the usual absolute-value $|\cdot|$. Denoting the map $A\to\mathbb{C}$ by $a\mapsto a(x)$ we conclude that for every $a\in A$, $s(a)=|a(x)|$.

Note that the commutativity assumption was inessential in all of the above.

• Thanks! But why is $A/p \simeq \mathbb{C}$? Edit: Oh I see, use Gelfand-Mazur since $A/p$ is a complex Banach Algebra without zero divisors! Thanks!! – Niki Jul 11 '16 at 9:16
• Niki, the first paragraph up there meant to prove that $A/p\simeq \mathbb{C}$. – Uri Bader Jul 11 '16 at 12:29