\documentclass[12pt]{book} \usepackage{amsthm} \theoremstyle{definition} \newtheorem{theorem}{Theorem} \newtheorem{defin}[theorem]{Definition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{sublemma}[theorem]{Sublemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{claim}[theorem]{Claim} \newcommand{\Ex}{\begin{example}} \newcommand{\Eex}{\end{example}} \newcommand{\Z}{{\bf Z}} \newcommand{\ZF}{{\bf Z}_{F}} \newcommand{\Q}{{\bf Q}} \newcommand{\C}{{\bf C}} \newcommand{\R}{{\bf R}} \newcommand{\F}{{\bf F}} \newcommand{\Fp}{{\bf F}_p} \newcommand{\vol}{{\rm vol}} \newcommand{\fa}{{\rm for~all~}} \newcommand{\osc}{{\rm osc}} \newcommand{\defn}[1]{{\bf #1}} \newcommand{\ii}{^{-1}} \begin{document} \section{Topological Groups} \begin{defin} A set $G$ is a {\it topological group} if $G$ is a group, $G$ is a topological space, and the group operations in $G$ are continuous in the topological space $G$, i.e. $(a,b)\mapsto ab : G \times G \to G$ and $a \mapsto a\ii :G\to G$ are continuous maps. \end{defin} \begin{defin} A topological group is {\it locally compact} if there exists a compact neighborhood about each point. Examples are $\Q$, $\R^n$, matrix groups... \end{defin} \begin{defin} A topological field $K$ is a field and a topological space such that addition, multiplication, and inversion on the non-zero element are all continuous maps. \end{defin} I want to classify all locally compact topological fields. I will assume my fields do not have the discrete topology, since any topological group with the discrete topology is locally compact. My goal will be to build a multiplicative homomorphism of the field to the non-negative real number and show that this is in fact, a valuation on the field. \section{Haar measure} On any locally compact group G (written additively here) there exists a Haar measure $\mu$, a positive continuous linear functional $$ \mu : C_c(G) \to \R$$ on the space of continuous real function on $G$ with compact support, which is left invariant: $$ \mu (f)=\int_G f(x)d\mu(x) = \int_G f(g+x)d\mu(x), \fa g \in G.$$ This can also be seen in the language of measure theory as a regular $\sigma$-additive function on the $\sigma$-algebra of Borel $\mathcal{B}$ $G$, generated by the open sets $U$ of $G$. I will loosely call vol($U$)=$\mu$(U)=$\mu$($\chi_U$) the volume or measure of the set $U$. Left invariance means $\vol(U)=\vol(g+U)$ for all $g \in G$. We say a measure $\mu$ is regular if for all $U \in \mathcal{B}$, $$\mu(U)=\sup\{\mu(K):K \subset U, K~{\rm is~compact}\}=\inf\{\mu(K):U \subset K, K~{\rm is~compact}\}.$$ This Haar measure is unique up to a positive constant: Let $\mu$ and $\nu$ be two Haar measures on a locally compact group $G$, then there exists a positive constant $c$ such that $\mu=c\nu$. \section{Modulus} Let $K$ be a locally compact topological field, and let $\mu$ be a Haar measure on the additive group $K$. Now let $\alpha$ be an automorphism of $K$, and define the measure $\alpha(\mu)$ by $$\alpha(\mu)(U)=\mu(\alpha(U)), \quad \fa U \in \mathcal{B}.$$ Then $\alpha(\mu)$ is invariant and is thus also a Haar measure on $K$. Thus we have $\alpha(\mu)=m(\alpha)\cdot\mu$, where $m(\alpha)$ only depends on $\alpha$, and is independent of the choice of Haar measure, since Haar measures are proportional. This is called the {\it modulus} of the automorphism $\alpha$. Thus there is a modulus function $$m:{\rm Aut}(K) \to \R_{>0}.$$ Specifically look at automorphisms of of the form $x \mapsto ax$ for any $a \in K^*$. Then I will denote my $m(a)$ the modulus of this automorphism. Thus by definition, $$\vol(aU) = m(a) \cdot \vol(U), \quad \fa a \in K^*, U \in \mathcal{B}.$$ Now note that for $a,b \in K^*$, $$m(ab)\vol(U)=\vol(abU)=m(a)\vol(bU)=m(a)m(b)\vol(U),$$ so $m(ab)=m(a)m(b)$, and thus $m:K^* \to R_{>0}$ is a homomorphism. We generally extend the modulus so that $m:K \to R_{\geq 0}$ defining $m(0)=0$. \begin{claim} There is a positive constant $C$ such that $$m(x+y)\leq C\max(m(x),m(y)), \quad \fa x,y\in K,$$ where $$C=\sup_{m(z)\leq 1}{m(z+1)}$$ is the smallest such constant. \end{claim} \begin{proof} Let $x,y \in K$. If $x=y=0$ then the statement is trivial. Suppose $y\neq 0$ and $m(x)\leq m(y)$. Let $z=xy\ii$ thus $$m(z) = m(x)m(y\ii) = m(x)m(y)\ii \leq 1.$$ Now we have, $$m(x+y) = m((xy\ii+1)y) = m(z+1)m(y) = C'\max(m(x),m(y)).$$ Where $C' \leq C$ and letting $y=1$ shows that C is the smallest such constant. \end{proof} By the properties the modulus $m$ has exhibited so far, $m$ is a generalized absolute value. \begin{defin} A generalized absolute value on a field $K$ is a homomorphism $f:K^* \to R_{\geq 0}$ with $f(0)=0$ defined such that for some $C > 0$, $$f(x+y) \leq C\max(f(x),f(y)) \quad \fa x,y \in K.$$ \end{defin} If $C=1$ this is the non-Archimedean absolute value. An absolute value obeying the triangle inequality, $|x+y|\leq |x|+|y| \leq 2\max(|x|,|y|)$, and has $C=2$, but in fact the converse is also true. \begin{proof} Notice that $$f(a_1+a_2+a_3+a_4) \leq C\max(f(a_1+a_2),f(a_3+a_4)) \leq C^2\max_{1\leq i\leq 4}f(a_i),$$ or by induction, for $n=2^r$, $$f(a_1+\ldots+a_n)\leq C^r\max f(a_i) = 2^r\max f(a_i) = n\max f(a_i).$$ If $n$ is not a power of two, fill it up with $a_i=0$ for $n \leq i \leq 2^r$, then $$f(a_1 + \ldots a_n) \leq 2n\max f(a_i).$$ Now for any $x,y \in K$, \begin{eqnarray*} f((x+y)^n) &\leq& f\left(\sum_{i=0..n}{\left(n \atop i\right)x^iy^{n-i}}\right)\\ &\leq& 2(n+1)\sum{f\left(\left(n \atop i\right) \right)f(x)^if(y)^{i-1}}\\ &\leq& 2(n+1)\sum {2\left(n \atop i\right)f(x)^if(y)^{i-1}}\\ &=&4(n+1)(f(x)+f(y))^n. \end{eqnarray*} Thus $$f(x+y) \leq 4^{1/n}(n+1)^{1/n}(f(x)+f(y)) \to f(x)+f(y).$$ \end{proof} The square of the regular absolute value has $C=4$, $$|x+y|^2 = (|x|+|y|)^2 \leq (2\max(|x|,|y|))^2 = 4\max (|x|^2,|y|^2).$$ Thus any generalized absolute value with $C\leq 2$ makes a field into a metric space. So some power of a generalized absolute value will be a metric on a field. Thus our locally compact field $K$ is a metric space. And the topology generated by $m$ is the same as the topology on $K$. Now I will bring in a result from the theory of topological groups. \begin{claim} Let $H$ be a locally compact subgroup of a Housdorff topological group $G$. Then $H$ is closed. \end{claim} Applying this to our scenerio, our locally compact field $K$ is closed in its completion (both are metric spaces by the comments above) Thus our locally compact field $K$ is also complete. Now we can begin to classify by the value of the constant $C$. We also need to know: \begin{claim} Any discrete subfield of a non discrete locally compact field of charactoristic 0 is finite. \end{claim} \begin{proof} This involves first showing that the modulus $m$ is continuous, then showing that closed balls defined by $m$ are compact. Then you have $\{a^n\} \to 0$ in $K$ iff $m(a)<1$. Now let $F$ be a discrete subfield. Choose $a \in K$ with $m(a)>1$, then $m(a^{-n}) =m(a)^{-n} \to 0$, thus $a^{-n} \to 0$ blah blah... \end{proof} {\bf Case 1:} Suppose $C=1$. Then $K$ has a non-Archimedeann absolute value. Since $K$ has charactoristic 0 and is not discrete by hypothesis, $\Q \subset K$ is not discrete, and so the absolute induces a non-trivial non-Archimedean absolute value onto $\Q$, so by Ostrowski's Theorem this is equivalent to a $p$-adic absolute value. Since $K$ is complete, it contains the completion of $\Q$ under this absolute value, i.e. $\Q_p \subset K$. Now with some more work one can show that a locally compact normed space over $\Q_p$ is finite dimensional. {\bf Case 2:} Suppose $C>1$. Thus once more, $K$ induces a non-trivial absolute value onto $\Q$, which is not non-Archimedean. So by Ostrowski's Theorem this is equivalent to the regular absolute value on $Q$ and again, since $K$ is complete, $K$ contains $\R$. So $K$ is a real vector space over $\R$, one can show the only two such possibilities are $\R$ and $\C$. \end{document}