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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.

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    $\begingroup$ If you have a multiplicative continuous seminorm, its kernel is a maximal ideal. $\endgroup$
    – user1688
    Commented Jul 8, 2016 at 13:58
  • $\begingroup$ 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. $\endgroup$
    – Niki
    Commented Jul 9, 2016 at 10:00
  • $\begingroup$ 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. $\endgroup$
    – user1688
    Commented Jul 9, 2016 at 10:10
  • $\begingroup$ 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. $\endgroup$
    – Uri Bader
    Commented Jul 10, 2016 at 6:15

1 Answer 1

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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 closed$^1$ prime ideal is the kernel of a homomorphism to $\mathbb{C}$.

Now, given a bounded multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a closed 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.

1: Thanks to Robert Furber for correcting my omission here.

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  • $\begingroup$ 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!! $\endgroup$
    – Niki
    Commented Jul 11, 2016 at 9:16
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    $\begingroup$ Niki, the first paragraph up there meant to prove that $A/p\simeq \mathbb{C}$. $\endgroup$
    – Uri Bader
    Commented Jul 11, 2016 at 12:29
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    $\begingroup$ Since this has been linked from another question, a correction is in order: "Prime ideal" should be "maximal ideal" or "closed prime ideal" when written above. There is a huge profusion of non-maximal prime ideals in $C([0,1])$, as is described in Gillman and Jerison's Rings of Continuous Functions, or in the answers to this question: mathoverflow.net/questions/35793/prime-ideals-in-c0-1 $\endgroup$
    – Robert Furber
    Commented Sep 2, 2019 at 5:04
  • $\begingroup$ @RobertFurber thanks for the correction. $\endgroup$
    – Uri Bader
    Commented Sep 2, 2019 at 7:03

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