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\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}