I'm recently looking about the classification theorem of local fields, starting from the definitions. One of the common definition is locally compact Hausdorff non-discrete topological field. The other is locally compact field completed with a non-trivial absolute value.
Given a field in the first definition, it's common to define a Haar measure on it which turns out inducing an absolute value in the second definition. Reversely any aboslute value induces a metric topology on the field.
My question is, are they inverse to each other? If so, any such topological field has an unique topology.
I dive into Weil's book ''Basic Number Theory'' and get the answer, which is yes. The following is an outline of the proof.
We will show one direction here ''the metric topology induced by the Haar measure agrees with the original topology''. The other direction can be shown by first classifying all the local fields and checking them explicitly.
Given a locally compact Hausdorff non-discrete topological field $(K,\tau)$, there exists a Haar measure $\mu$ which has value $1$ on some compact neighborhood $X$ of $0$ containing a open neighborhood $U$ of $0$, and induces an absolute value $\alpha:K\to \mathbb{R}_{\geq 0}$ s.t. $\alpha (s):=\frac{\mu(sX)}{\mu(X)}$ (true for any non-null measurable set $X$) and $\alpha(0):=0$.
It's easy to see $\alpha(ab)=\alpha(a)\alpha(b)$ from the left translation invariant property of Haar measure. The topology $\tau'$ on $K$ induced by $\alpha$ is generated by the open balls $B(x,r)$ with center $x$ and radius $r$. So it suffices to show that the open balls generates $\tau$ as well. By homogeneous property of topological field, it suffices to show $B(0,r)$s form a neighborhood basis of $0$ in $\tau$, i.e. $\forall r>0,B(0,r)\in \tau$ and $\forall U\in \tau$ s.t. $0\in U$,$\exists r>0$ s.t. $B(0,r)\subset U$.
Part I: $\alpha$ is continuous.
So we want to show $\forall \epsilon>0,\forall a\in K,\exists U\in \tau\text{ s.t. }a\in U,\forall x\in U,-\epsilon\leq \alpha(x)-\alpha(a)\leq \epsilon$. If so then $B(0,r)=\alpha^{-1}([0,r))$ is automatically open.
Note that $\alpha$ is continous if and only if $\alpha$ is upper semicontinuous and lower semicontinuous at all $a\in K$. We will show that $\alpha$ is continuous at $0$ and upper semicontinuous on the open subset $K^\times$ (every singleton is closed by Hausdorff).
Claim: If $\alpha$ is upper semicontinous on $K^\times$, then it is continuous on $K^\times$.
Proof: As $K^\times$ is open, $\alpha$ is uppser semicontinuous on $K^\times$ iff $\alpha|_{K^\times}$ is upper semicontinuous iff $\alpha^{-1}:K^\times \to \mathbb{R}_{>0}$ is lower semicontinous. Fix $\epsilon>0$ and $a\in K^\times$, $\exists U,V\in \tau$ s.t. $a\in U,a^{-1}\in V$, $\forall x\in U$ we have $$\alpha(x)\leq \alpha(a)+\epsilon$$ and $\forall x\in V^{-1}$ we have $$\alpha(a^{-1})^{-1}-\epsilon\leq \alpha(x^{-1})^{-1}$$ so $\alpha(a)-\epsilon\leq \alpha(x)\leq \alpha(a)+\epsilon$ for all $x\in U\cap V^{-1}\ni a$.$\square$
Now we deal with the equation $\alpha(x)\leq \alpha(a)+\epsilon$. We don't know much about $\alpha$ so we multiply $\mu(X)$ for some non-null measurable set $X$ which can be chosen later. So we just need $\mu(xX)\leq \mu(aX)+\epsilon$.
Note that Haar measure is by definition outer regular, i.e. it can be approximated from above by open sets. This is how we get $\epsilon$. So $\exists V\in \tau$ s.t. $aX\subset V$ and $\mu(V)\leq \mu(aX)+\epsilon$.
Hence we need to find the open subset $U\ni a$ s.t. $\forall x\in U,xX\subset V$ so that $\mu(xX)\leq \mu (V)$. Equivalently, enlarge from $aX\subset V$ to $UX\subset V$.
As the mutliplication map $\pi:K\times K\to K,(a,b)\to ab$ is continuous, $\forall x\in X,\exists U_x,X_x\in \tau$ s.t. $a\in U_x,x\in X_x$ and $U_x X_x\subset V$. Clearly $(\bigcap_{x\in X} U_x)X\subset V$ so $U:=\bigcap_{x\in X} U_x$ is our candidate.
Using locally compactness, pick compact $X$ containing a open neighborhood of $0$ which has measure $1$. There exists a finite subset $I\subset X$ s.t. $X\subset \bigcup_{x\in I}X_x$ hence $U:=\bigcap_{x\in I}U_x\ni a$ satisfies $UX\subset V$. Thus $\alpha$ is upper semicontinuous on $K$ (the proof works with $a=0$). As $\alpha(0)=0$ and $\alpha$ is non-negative, $\alpha$ is continuous at $0$. So $\alpha$ is continuous on $K$.
Part II: $B(0,r)$s form a neighborhood basis of $0$.
Fix a open $U\ni 0$, we want to find a radius $r$ s.t. $B(0,r)\subset U$. $B(0,r)$ is something checked by $\alpha$. Equivalently we need to show $\alpha^{-1}([0,r))\subset U$, equivalently $\alpha(U^c)\subset [r,\infty)$, i.e. bounded below by a positive number.
Recall that continuous image of compact set is compact, so bounded and closed in $\mathbb{R}_{\geq 0}$, moreover $0\notin U^c$ so $0\notin \alpha(U^c)$, if $U^c$ is compact then $\alpha(U^c)$ is bounded below by a positive element so we are done.
Now we use locally compactness, pick $X\ni 0$ compact which contains a open $V\ni 0$. And consider $U\cap V \subset X$. Rename $U\cap V$ as $U$. So $\alpha(U)\subset \alpha(X)$ is bounded above by $m$ for some $m>0$. Hence $U\subset \overline{B}(0,m):=\alpha^{-1}([0,m])$.
So $U^c=(\overline{B}(0,m)-U)\coprod \overline{B}(0,m)^c$ where $\overline{B}(0,m)-U$ is a closed subset of compact (to be shown in Part III) so compact. Both $\alpha(\overline{B}(0,m)-U)$ and $\alpha(\overline{B}(0,m)^c)$ is bounded below, the result follows.
Part III: $B_m:=\overline{B}(0,m)$ is compact.
Again we utilize locally compactness. $B_m$ is already closed so it suffices to show it lies in a compact set. There exists a compact neighborhood $X$ of $0$ which contains a open neighborhood $U$ of $0$, and we can deform it using translation and multiplication so that it maintains compact and covers $B_m$. As $\alpha$ is multiplicative, it is natural to consider multiplication by a big enough element.
To find an arbitrarily large element, it suffices to find an arbitrarily small non-zero element and then take inverse. As $K$ is not discrete, $\{0\}$ can not be open, $B(0,\epsilon)$ is always non-trivial. Hence $\alpha$ is unbounded.
We would like to show $B_m\subset tX$ for some large $t$ or equivalently $t B_m\subset X$ for some small $t$. Fix any non-zero $t\in B(0,1)$, $\{t^n\}_{n\geq 1}$ can be arbitrarily small. For every element $a$ in $B_m$ or actually any element, we expect $t^n a\to 0$ so the open neighborhood $U$ of $0$ must contain some $t^n a$.
Clearly any limit point of $\{t^n a\}$ is still a limit point under $\alpha$, which is $0$. As $\alpha$ is non-zero except at $0$, $0$ is the only possible limit point. If $a=0$ we are done, if $a\neq 0$ it's equivalent to consider $\{t^n\}$ using the homeomorphism $s\mapsto sa^{-1}$. And we could pick non-zero $t\in B(0,1)\cap U\cap W$ where $W$ is an open neighborhood of $0$ s.t. $WX\subset U$ (see above), which must be non-trivial since $\{0\}$ is not open. Therefore $\{t^n\}$ lies in $X$ which is compact so limit point compact (every infinite subset has a limit point), and it has a limit point which must be $0$.
For each $a\in B_m$ denote $n(a)$ the smallest integer s.t. $t^{n(a)}a\in X$, if $n(a)$ is bounded above by some $N$ then $B_m$ will be covered by $\bigcup_{i=0}^N t^{-i}X$ so compact. It suffices to consider $n(a)$ s.t. $n(a)\geq 1$, so $t^{n(a)-1}a\notin X$ and hence $t^{n(a)}a\notin tX$. So $t^{n(a)}a\in X\cap (tX)^c$. As $\alpha(a)\leq m$, to show $n(a)$ is bounded above, it suffices to show $\alpha(X\cap (tX)^c)$ is bounded below by a positive number.
Let $C=\overline{X\cap (tX)^c}$, then $C$ is a closed subset of $X$ hence compact. Also $X\cap (tX)^c\subset (tX)^c \subset (tU)^c$, $C$ is a closed subset of $(tU)^c$, so $0\notin C$. Therefore $\alpha(C)$ is bounded below by a positive integer. The result follows.