Huber spaces-1

One of the motivating examples for Huber spaces comes from some application of valuation theory in Riemann surfaces.

Suppose that $k$ is a algebraically closed field and $k\subset K$ is a field extension of transcendental degree $1$ . Then there is a very famous result in the theory of Riemann surfaces which says that each such extension $K$ corresponds (in some sense uniquely) to a Riemann surface over $k$ . One fascinating proof to this result is to utilize valuation theory.

We say a function $v:K\rightarrow \mathbb{Z}\bigcup \{\infty\}$ is a discret valuation if $v(0)=\infty,v(1)=1$ , and $v|_{K^{*}}$ is a group homomorphism to the group $\mathbb{Z}$ . What is more, $v(a+b)\geq min(v(a),v(b))$ (here the infinity should be understood as the positive infinity). Then we write $V(K/k)$ as the set of all such discret valuations. And for any $k-$ algebra contained in $K$ , we define $U(B)=\{v\in V(K)|v(B)\subset \mathbb{N}\bigcup \{\infty\}\}$ . Then we let $B$ run over all the $k-$ algebras contained in $K$ and take all these $U(B)$ as a basis of topology, so that the topology thus generated is verified to give a Riemann surface structure to $V(K/k)$ . This is the essential part of the proof.

So here we want to generalize the pair $(K,B)$ , and that is where Huber spaces arise.

First we say something on topological rings. A commutative ring $(A,+,*)$ (with unit) with some topology said to be a topological ring if the $A\times A\rightarrow A, (a,a')\mapsto a-a'; A\times A\rightarrow A,(a,a')\mapsto a*a'$ are both continuous. Some typical examples are the $p-$ adic integers $\mathbb{Z}_p$ .

In non-archimedean analysis, we often require that the topology on $A$ is given by some ‘nice’ sets, in other words, one fundamental system of open sets of $0$ can be a system of subgroups of $(A,+)$ . Note that, this is not at all a trivial requirement since that the most common spaces like the real numbers $\mathbb{R}$ do not have this property(and that is why there is non-archimedean analysis). Yet this requirement is on the other hand rather natural for the non-archimedean analysis.

Here we consider when such a system of subgroups of $(A,+)$ can give a topology to $A$ such that it becomes a topological ring. In the following, we will write $S\cdot T$ as the subgroup generated by elements of the form $st(s\in S,t\in T)$ in $A$ .

Suppose that $\mathfrak{G}$ is a set consisting of subgroups of $(A,+)$ , then $\mathfrak{G}$ defines a fundamental system of neighborhood of $0$ such that $A$ becomes a topological ring if and only if $\mathfrak{G}$ satisfies the following three points:(1)for any $G,G'\in\mathfrak{G}$ , there is a $H\in\mathfrak{G}$ such that $H\subset G\bigcap G'$ ;(2)for any $x\in A$ and any $G\in\mathfrak{G}$ , there exists a $H\in\mathfrak{G}$ such that $xH\subset G$ ;(3)for any $G\in\mathfrak{G}$ , there is a $H\in\mathfrak{G}$ such that $H\cdot H\subset G$ .

There is an important remark to make. The notion $S\cdot T$ of two sets is in some sense the same as $ST$ in the setting of non-archimedean analysis, since we can always find a system of subgroups as a fundamental system of neighborhood of $0$ . In the following we shall use frequently this remark.

One important class of subgroups of a ring is the ideals. And this introduces the adic rings.

Definition: A ring $A$ is called an adic ring if there is an ideal $I\subset A$ such that the system $(I^n)_{n\in\mathbb{N}}$ defines a fundamental system of neighborhood of $0$ .

Sometimes for precision, we call a ring an $I-$ adic ring if the ideal in the definition is $I$ . Note that an ideal $J\subset A$ is open in the $I-$ adic topology if there is a positive integer $n$ such that $I^n\subset J$ . So the $I-$ adic topology and $J-$ adic topology are the same if and only if there exist two positive integers $n,m$ such that $I^n\subset J,J^m\subset I$ . This amounts to say that these two ideals are comparable. In particular, $I^i-$ adic topology and $I^j-$ adic topology are always the same( $i,j>0$ ).

Suppose that $A'\subset A$ is an open subring, then the $I-$ adic topology on $A$ gives an induced topology on $A'$ . Since $A'$ is open, then $I^N\subset A'$ for some integer $N$ . Then it is easy to see that the induced subspace topology on $A'$ and the $I^N$ -adic topology on $A'$ are the same. We will also often use this fact.

Next is a very important notion in analysis, the boundedness. For a metric space, this is a rather clear concept, yet if the space has no metric, things becomes not so easy. Here we adopt the definition of boundedness from functional analysis.

$A$ is a topological ring, and $B\subset A$ is a subset. We say that $B$ is bounded in $A$ if for any open neighborhood $U$ of $0$ in $A$ , there exists another open neighborhood of $0$ in $A$ such that $B\cdot V\subset U$ .

Here if $A$ is a non-archimedean topological ring, then we can replace $B\cdot V$ by $BV$ , and the definition remains unchanged. And we call an element $a\in A$ is power-bounded if $\{a^n\}$ is a bounded set. $a$ is called topologically nilpotent if $a^n\rightarrow 0$ . And we write $A^o$ as the set of power-bounded elements of $A$ and $A^{oo}$ as the set of topologically nilpotent elements of $A$ .

A little remark is that any element $a\in A$ as a single-element set $\{a\}$ in a topological ring is bounded due to the very simple fact that $A\times A\rightarrow A,(a,a')\mapsto aa'$ is continuous.

In the next post we will enlarge the notion of adic rings to $f-$ adic rings.

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Huber spaces-1

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