Author: Eiko

Time: 2024-12-18 11:38:47 - 2025-01-20 19:23:43 (UTC)

References:

Spectral Theory and Analytic Geometry over Non-Archimedean Fields by Vladimir Berkovich

Math 731 Topics in Algebraic Geometry I Berkovich Spaces by Mattias Jonsson

Banach Rings

Seminorms

• A seminorm on an abelian group $M$ is a map $\parallel \cdot \parallel :M\to {\mathbb{R}}_{\ge 0}$ such that

  • $\parallel 0\parallel =0$, and is a norm if $\parallel x\parallel =0$ implies $x=0$

  • $\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$

  We also say that it is a non-archimedean seminorm if the strong triangle inequality $\parallel x+y\parallel \le max(\parallel x\parallel ,\parallel y\parallel )$ holds.

• Note that the Hausdorff property of the topology induced by the seminorm corresponds to the seminorm being a norm.

  There exists a group (which is called the separated completion) $\hat{M}$ with a norm $\parallel \cdot {\parallel }_{\hat{M}}$ such that $\hat{M}$ is Hausdorff and complete with respect to $\parallel \cdot \parallel$.

  I think it is probably defined as the limit of a fundamental system of neighborhoods of $0$ given by the balls $\{x\in M:\parallel x{\parallel }_{\hat{M}}\le r\}$.

  $$\hat{M}=limM/M_r=limM/\{x\in M:\parallel x{\parallel }_{\hat{M}}\le r\}$$

  which is going to be separated because the norm zero elements in $\hat{M}$ will be zero.

• Two norms are equivalent if they each differ by at most a bounded constant factor.

• The residue norm can be introduced on any quotient $M/N$ of $M$, defined as the infimum over all pre-images of $\pi :M\to M/N$. (think of element in $M/N$ as a coset $x+N$, and the infimum is the ‘vertical’ distance)

  $$\parallel \bar{x}\parallel :=inf\{\parallel x\parallel :x\in M,x\in {\pi }^{-1}(\bar{x})\}$$

  • The balls in $M/N$ are the direct projections of the balls in $M$

  • residue seminorm is a norm iff $N$ is closed (so $M/N$ is Hausdorff).

  • If $M$ is complete and $N$ is closed, then $M/N$ is complete.

• If $\phi :M\to N$ is an additive map, we say $\phi$ is bounded if it has a finite operator norm with respect to $\parallel \cdot \parallel$. It is admissible if the axiom of abelian category holds

  $$M/\ker (\phi )\cong \mathrm{Im}\phi$$

  as isomorphism of topological groups, where $M/\ker \phi$ is equipped with the quotient norm and topology, and $\mathrm{Im}\phi$ is equipped with the induced norm and topology.

• Question: is the quotient norm induces the same thing as the quotient topology?

Seminormed Rings

Let $A$ be a unital ring, a seminorm on $A$ is an sub-multiplicative seminorm $\parallel \cdot \parallel$ on $A$ such that $\parallel 1\parallel =1$.

• $\parallel \cdot \parallel$ is called power-multiplicative if $\parallel a^n\parallel =\parallel a{\parallel }^n$ for all $n\in \mathbb{N}$ and multiplicative if $\parallel ab\parallel =\parallel a\parallel \parallel b\parallel$, $\parallel 1\parallel =1$.

• A multiplicative norm is called a valuation.

• A Banach ring is a complete sub-multiplicative normed ring.

Example

• Any ring is a Banach ring with trivial norm since trivial norm is always complete.

• $\mathbb{Z}$ is Banach ring with canonical absolute value.

• If $\mathfrak{a}$ is a proper closed bi-ideal of a Banach ring, the quotient $A/\mathfrak{a}$ is complete. By the way all maximal two-sided ideals are closed.

• For a family of Banach rings $\{A_i\}$, consider the product ring but with restrictions: the $(f_i)\in \prod A_i$, with $\parallel f_i\parallel =O(1)$, then this is a Banach ring with sup norm $\parallel f\parallel :=\underset{i}{sup}\parallel f_i\parallel$.

• For a Banach ring $A$ and a positive number $r$, define $A⟨r^{-1}T⟩$ to be the ring of power series with absolute convergence, i.e. the norm is finite

  $$\parallel f\parallel =\sum _{n=1}^{\mathrm{\infty }}\parallel a_n\parallel r^n<\mathrm{\infty }.$$

  This ring $A⟨r^{-1}T⟩$ is Banach with this norm.

• Example: what condition is needed for $1-aT$ be invertible in $A⟨r^{-1}T⟩$?

  $$\frac{1}{1-aT}=1+aT+a^2{T}^2+\cdots$$

  therefore the condition for $1-aT$ to be invertible is

  $$\Vert \frac{1}{1-aT}\Vert =\sum _{n=0}^{\mathrm{\infty }}\parallel a^n\parallel r^n<\mathrm{\infty }.$$

Bounded Seminorm

A seminorm on a Banach ring is bounded if it is bounded on the unit ball, or there exists $C>0$ such that

$$|f|\le C\parallel f\parallel$$

If $|\cdot |$ is power-multiplicative then it suffices to check for $C=1$, since $|f{|}^n\le C\parallel f^n\parallel \le C\parallel f{\parallel }^n$.

Seminormed $A$-modules

For a normed ring $A$, a seminormed $A$-module is an $A$-module $M$ equipped with a seminorm $\parallel \cdot \parallel$ such that

$$\parallel fm\parallel \le C\parallel f\parallel \cdot \parallel m\parallel$$

Complete Tensor Product

If $M$ and $N$ are seminormed $A$-modules, then the tensor product $M{\otimes }_{A}N$ is a seminormed $A$-module with the norm

$$\Vert \sum m_i\otimes n_i\Vert =inf\sum _i\parallel m_i\parallel \cdot \parallel n_i\parallel$$

and for non-archimedean norms, the definition is replaced by

$$\Vert \sum m_i\otimes n_i\Vert =inf\underset{i}{max}\parallel m_i\parallel \cdot \parallel n_i\parallel$$

The completion of this module is denoted by $M{\hat{\otimes }}_{A}N$.

Spectrum

The spectrum $\mathcal{M}(A)$ of a Banach ring $A$ is the set of all bounded multiplicative seminorms on $A$ with the weakest topology (adding inverse images) such that all evaluation maps ${\mathrm{ev}}_f:\parallel \cdot \parallel \mapsto \parallel f\parallel$ are continuous.

• $\mathcal{M}(A)$ is a non-empty compact Hausdorff space.

• Let $x\in \mathcal{M}(A)$ be a point corresponding to $\parallel \cdot {\parallel }_x$, the kernel ${\mathfrak{p}}_x$ is a closed prime ideal of $A$.

  • It is a prime ideal since multiplicativity, $\parallel ab\parallel =0$ implies $\parallel a\parallel =0$ or $\parallel b\parallel =0$.

  • It is closed because of continuity.

  This induces a valuation on the quotient field $A/{\mathfrak{p}}_x$, and that naturally extends to the fraction field $F=\mathrm{Frac}(A/{\mathfrak{p}}_x)$. We denote the closure $\bar{F}$ of $F$ by ${\mathcal{H}}_x$, and the image of an element $f\in A$ in ${\mathcal{H}}_x$ by $f(x)$.

• The morphism

  $$\widehat{(\cdot )}:A\to \prod _{x\in \mathcal{M}(A)}{\mathcal{K}}_x$$

  sending $f\mapsto \hat{f}=(f(x){)}_{x\in \mathcal{M}(A)}$ is called the Gelfand transform.

Interlude On Complex Banach Algebras

Berkovich’s theory is inspired by Gelfand’s theory of complex Banach algebras. Let’s mention some of the ideas here.

• The ring of complex continuous functions on a compact Hausdorff space $X$ denoted by $A=C^0(X,\mathbb{C})$ is a Banach algebra with the sup norm (or called the L-infinity norm).

• The canonical problem is to recover space $X$ from $A$. Consider the spectrum

  $$\sigma (f)=\{\lambda \in \mathbb{C}:\lambda -f\text{ is not invertible in }A\}$$

  It is known that $\sigma (f)$ is non-empty and compact.

• As a corollary, we have the Gelfand-Mazur theorem that any complex Banach field is isomorphic to $\mathbb{C}$, since any $f$ has a spectrum $\lambda$ such that $f-\lambda$ is not invertible, and anything not invertible is zero in a field.

• For any complex Banach algebra $A$, there is an embedding (actually an isomorphism)

  $$\mathrm{mSpec}(A)\stackrel{\sim }{\hookrightarrow }\mathcal{M}(A)$$

  Sending $\mathfrak{m}\mapsto \parallel \cdot {\parallel }_{\mathfrak{m}}=|\cdot {|}_{\mathbb{C}}\circ (A\to A/\mathfrak{m})$.

  The inverse is given by taking a multiplicative seminorm $\parallel \cdot {\parallel }_x$ and find its prime ideal ${\mathfrak{p}}_x$. Since $A/{\mathfrak{p}}_x\hookrightarrow \mathrm{Frac}(A/{\mathfrak{p}}_x)$ is a Banach field, the latter field is $\mathbb{C}$. We have two funny embeddings $\mathbb{C}\hookrightarrow A/{\mathfrak{p}}_x\hookrightarrow \mathbb{C}$, so $A/{\mathfrak{p}}_x\cong \mathbb{C}$ and ${\mathfrak{p}}_x$ is maximal.

• As a summary, in complex Banach algebras, the spectrum is the maximal ideal space. So Berkovich’s theory generalizes the complex theory to the case of non-archimedean fields.

Some Remarks

I feel that we have two layer of structures here, one layer of usual structure of valuations stacked on the possible new structures obtained from generalizing to seminorms. Think about in the classical valuation theory (which is a norm), the maximal ideals are obtained from the set $\{|x|\le 1\}$. Now we have a kernel prime ideal of multiplicative seminorms ${\mathfrak{p}}_x=\{|x|=0\}$, and upon dividing the kernel we are back to the valuation theory, $\mathrm{Frac}(A/{\mathfrak{p}}_x)$ is a field with a valuation, whose completion is ${\mathcal{H}}_x$.

The maximal ideal ${\mathfrak{m}}_x=\{|x|\le 1\}$ contains ${\mathfrak{p}}_x$, effectively give us a 2-d structure.