Connection between two analytic notions of "spectrum"

Question

In Spectral Theory and Analytic Geometry over Non-Archimedean Fields Berkovich defines the spectrum of a commutative, unital Banach ring $A$ to be the set of all bounded multiplicative seminorms on $A$ with the weak topology with respect to the family of functions $$\phi_f: || \mapsto |f|$$ In another class, I learned the definition of spectrum $\sigma(T)$ of an element of a Banach algebra to be the set of all the $\lambda$ such that $T - \lambda I$ is non-invertible. Many times I try to look something up related to Berkovich's text (such as the Gel'fand transform, characters), and I find a concept by the same or similar name that seems similar but is definitionally different. So are the two definitions connected somehow? This section of Wikipeda seems promising, as well as the page on the Gel'fand representation, but I could use some help connecting the dots.

Answer 1

Make note of the following theorem:

Theorem (Mazur)

If $A$ is a normed algebra over $\Bbb R$ together with a multiplicative norm (ie $\|a\cdot b\|=\|a\|\cdot\|b\|$ for all $a,b\in A$) then $A$ is isometrically isomorphic (as an $\Bbb R$-algebra) to one of the following:

1. $\Bbb R$
2. $\Bbb C$
3. $\mathcal H$ (the quaternions)

Now if $A$ is Banach-algebra and $\|\cdot\|_s$ is a multiplicative semi-norm, let $N=\{a\in A\mid \|a\|_s=0\}$, then $N$ is a (closed) two-sided ideal and $\|\cdot\|_s$ descends to a multiplicative norm on $A/N$. So by the theorem you've got that $A/N$ is either $\Bbb R, \Bbb C$ or $\mathcal H$. Since we are asking $A$ to be commutative $A/N$ must also be commutative and $\mathcal H$ is out of the picture. Additionally I am assuming that when you say "Banach algebra" you are talking about a complex Banach algebra, since thats the common convention, so $\Bbb R$ is also out.

Hence every multiplicative semi-norm $\|\cdot\|_s$ is of the form:

$$A\to \Bbb C \to \Bbb R_{≥0}, \qquad a\mapsto \psi(a) \mapsto |\psi(a)|$$ for some unital algebra morphism $\psi: A\to \Bbb C$, these morphisms are also called "characters".

This is the connection of the first definition to the Gel'fand transform story, this story is very well expositioned and you can find it explained in almost every introductory textbook on operator algebras. For convenience I'll summarise the connection with the spectrum $\sigma(a)$ of elements of $A$.

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Denote with $\Delta$ the space of characters and give it the weak* topology, ie the topology induced by the semi-norms $|\psi|_a= |\psi(a)|$, since $\Delta$ is weak* closed in the unit ball of the dual $A'$ you get that $\Delta$ is compact. Further there is a continuous algebra morphism $$A\to C(\Delta), \qquad a\mapsto [\psi\mapsto \psi(a)]$$ called the Gel'fand transform. You have that:

1. Each element $\psi\in\Delta$ is uniquely determined by the semi-norm $a\mapsto |\psi(a)|$ and the inclusion into the space of multiplicative semi-norms is a homoemorphism if you give the semi-norms the topology you have indicated, ie $\Delta$ is the spectrum in the sense of Berkovich.
2. $a(\Delta) \subseteq \sigma(a)$, ie the image of the spectrum under $a$ is contained in the spectrum of $a$.
3. The map $\Delta\mapsto \mathrm{Spec}_m(A)$, $\psi\mapsto \ker(\psi)$ is well defined and bijective, ie $\Delta$ is equal to the maximal spectrum (in the sense of algebraic geometry) - but this map won't generally be a homeomorphism since in the Banach algebra setting you have a very nice analytic topology on $\Delta$ thats not available in the general algebraic setting.
4. The map $A\to C(\Delta)$ is an isomorphism iff $A$ is a $C^*$-algebra.
5. However there are commutative Banach-algebras for which $\Delta$ is just a point.

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The proofs look roughly like this:

1. Note that for any $\psi\in\Delta$ the sub-space $\ker(\psi)$ is an ideal and it has co-dimension $1$, the quotient $A/\ker(\psi)$ is then isomorphic to $\Bbb C$ as a one-dimensional unital $\Bbb C$ algebra, $\psi$ necessarily factors over the quotient to $A/\ker(\psi)$, but there is only one non-zero morphism from this algebra to $\Bbb C$, so $\psi$ is uniquely determined by $\ker(\psi)$. Hence for each semi-norm $\|\cdot\|_s$ there is at most one character $\psi$ with $\|a\|_s=|\psi(a)|$ for all $a$.
2. Note that $a-\psi(a)\Bbb1$ cannot be invertible because under $\psi$ its image is $0$, if it were invertible there would be a $b$ with $1=\psi(\Bbb1)=\psi(b(a-\psi(a)\Bbb1))=\psi(b)(\psi(a)-\psi(a))=0$. As such $\psi(a)\in\sigma(a)$ for all $a$.
3. $\ker(\psi)$ is a co-dimension $1$ ideal and hence maximal. Further if $\mathfrak m$ is maximal then every non-zero element of $A/\mathfrak m$ must be invertible (else if $a+\mathfrak m$ were not invertible the ideal generated by $a$ and $\mathfrak m$ would be proper) - by a general theorem you get that $A/\mathfrak m$ is isomorphic to $\Bbb C$ and then the quotient map $A\to A/\mathfrak m$ is a character with kernel $\mathfrak m$. Since characters are uniquely determined by their kernels (see point 1) this correspondence is then bijective.
4. Consult a book on C* algebras for this one.
5. Look at a Banach space, choose some element to be the $\Bbb1$ and let any multiplication $a\cdot b=0$ when neither $a$ or $b$ are proportional to $\Bbb 1$.