Today I Learned

March 1st

Today I learned that local fields $F$ have character group $\widehat F$ isomorphic to themselves, from the MIT notes I've been reading. This is somewhat similar to the proof $\QQ_p\cong\widehat{\QQ_p}$ I did yesterday, but this is different in topological ways.

To set up, we fix any nontrivial character $\psi\in\widehat F.$ One exists by Pontryagin duality: $\widehat{F}\cong\{e\}$ means that $F\cong\widehat{\smash{\widehat F}\vphantom{1^]}\,}\cong\{e\},$ but $F\not\cong\{e\}.$ Then we define the map $\Psi:F\to\widehat F$ by\[\Psi_a:=(x\mapsto\psi(ax)).\]Before continuing, we check that this is well-defined. In particular, $\Psi_a$ is still a character because $\Psi_a(0)=\psi(a\cdot0)=\psi(0)=1.$ And $\Psi_a$ is continuous because it is the composite of the continuous maps $x\mapsto ax$ and $x\mapsto\psi(x).$ Anyways, we now claim the following.

This means that we have show $\Psi$ is both an isomorphism of groups and a homeomorphism of topological spaces.

We begin by getting the easier stuff out of the way first.

To show that $\Psi$ is a homomorphism, we have to show that $\Psi_{(a+b)}=\Psi_a\Psi_b$ for $a,b\in F.$ Well, plugging into the definitions, we see\[\psi((a+b)x)=\psi(ax+bx)=\psi(ax)\psi(bx),\]which is what we wanted. Now to show that $\Psi$ is injective, it suffices to show that it has trivial kernel. Well, $\psi$ is nontrivial (note we have used this condition here), so some $\psi(g)\ne1.$ Then for $a\ne0,$ we have $a\in F^\times,$ so\[\Psi_a\left(a^{-1}g\right)=\psi\left(aa^{-1}g\right)=\psi(g)\ne1.\]Thus, $\Psi_a$ is not the identity, so we see $\Psi$ has trivial kernel. $\blacksquare$

It would be nice, thematically speaking, to show surjectivity now to complete the proof that $\Psi$ is a group isomorphism. However, this will turn out to follow cleanly from a topological argument, so we postpone doing so.

Because $\Psi$ is injective, note that we can pull back the open sets in $\Psi(F)$ to make a topology on $F.$ To be explicit, we our open sets in the subspace $\Psi(F)$ have subbasis\[\big\{\psi_a\in\widehat F:\psi_a(K)\subseteq U\big\}\]for compact $K$ and open $U,$ and now we violently push this backwards to\[V(U,K):=\{a\in F:\psi_a(K)\subseteq U\}=\left\{a\in F:aK\subseteq\psi^{-1}(U)\right\}.\]This gives us a funny topology on $F,$ and we want to show that the normal topology on $F$ is the same as this one. That is, if an open set $U$ is open in the normal topology if and only if it's open in the funny topology, then we can push the funny topology back through to $\Psi(F)$ so that $U$ is open in the normal topology if and only if $\Psi(U)$ is open in $\widehat F$ and conversely.

The normal topology on $F$ is generated by open balls $B(a,r).$ To equate topologies, it suffices to focus on (sub)basis elements: we show that each $\bigcap_\bullet V(U_\bullet,K_\bullet)$ contains open ball neighborhoods of any individual element of a basis element $\bigcap_\bullet V(U_\bullet,K_\bullet),$ and each open ball contains some basis element $\bigcap_\bullet V(U_\bullet,K_\bullet)$ containing the center of the ball. Visualize this by populating funny open basis elements with balls and populating balls with funny open basis elements.

Quickly, we show this is enough. Indeed, then for any open ball $B(a,r),$ we write\[B(a,r)=\bigcup_{u\in B(a,r)}\left(\bigcap_{k=1}^{n_u}V(U_{u,k},K_{u,k})\right),\]where $u\in\bigcap_\bullet V(U_{u,\bullet},K_{u,\bullet})\subseteq B(u,r-|a-u|)\subseteq B(a,r).$ Thus open balls can be expressed as a union of funny basis elements, and anything open in the normal topology is open in the funny topology. And conversely, each basis element $\bigcap_{k=1}^nV(U,K)$ contains open ball neighborhoods of any individual element, so we can write\[\bigcap_{k=1}^nV(U,K)=\bigcup_{v\in V}B(a_v,r_v),\]where $v\in B(a_v,r_v)\subseteq\bigcap_{k=1}^nV(U,K).$ Thus, the open balls also form a subbasis of the funny topology, and anything open in the funny topology is open in the normal topology.

Continuing, the topological group structure actually lets us focus only around $0.$ Indeed, it will be enough to claim the following.

Indeed, suppose we knew this. To show that each funny basis element $V$ contains open ball neighborhoods around each element of $v,$ then we note $-v+V$ is an a funny open set (by homogeneity) containing $0.$ So fixing a finite intersection of some $V(U_\bullet,K_\bullet)$ to be a basis element around $0$ in $V,$ each $V(U_\bullet,K_\bullet)$ contains $0$ and thus some $B(0,r_\bullet)$ by lemma. Taking the minimum of the $r_\bullet$ lets us assert $B(0,r_\bullet)\subseteq-v+V,$ so\[B(v,r_\bullet)\subseteq V.\]To show that each open ball $B(a,r)$ contains a funny basis element $V$ around the center $a,$ we note that $B(0,r)$ contains some funny open set $V(U,K)$ around $0,$ and then $a+V(U,K)$ is still a funny open set now containing $a.$ Thus, we can extract a basis element of $a+V(U,K)$ containing $a$ to finish.

Now we show the (sub)lemma. To show that each $V(U,K)$ containing $0$ contains an open ball centered at $0,$ we note that this means exhibiting an $\varepsilon$ for which\[|a| \lt \varepsilon\implies aK\subseteq\psi^{-1}(U)\]after unraveling the definitions. Note that none of the previous arguments about basis elements have used anything about the topology on $F,$ but we note now that $K$ is compact requires that $K$ is bounded, so we're essentially trying to squeeze a small set into $\psi^{-1}(U).$ Well, $\psi^{-1}(U)$ is open by continuity, and it contains $0,$ so fix some\[B(0,r)\subseteq\psi^{-1}(U).\]Because $K$ is bounded, we can say $k\in K$ implies $|k|\le m$ for some $m,$ which now lets us finish with $\varepsilon=r/m.$ Indeed, if $|a| \lt r/m,$ then any $ak\in K$ satisfies\[|ak|=|a|\cdot|k| \lt \frac rm\cdot m=r,\]so $ak\in B(0,r)\subseteq\psi^{-1}(U).$ So we finish this part.

Conversely, we now show that each open ball $B(0,r)$ around $0$ contains some $V(U,K)$ containing $0.$ Unpacking definitions, we will exhibit compact $K$ and open $U$ containing $1$ such that\[aK\subseteq\psi^{-1}(U)\implies|a| \lt r.\]We are forcing $1\in U$ so that $a=0$ works. We note that if $\psi$ is trivial, then this is false, so we need to introduce $x\in G$ so that $\psi(x)\ne1.$ Now, the main trick here is to take a closed ball for $K,$ which are compact, and this will let us move this to a size condition. Namely, we take $U=S^1\setminus\{\psi(x)\},$ and\[K=\{k\in G:|k|\le|x|/r\}.\]The visual here is that $a\in F$ is permitted as long as $aK$ doesn't "blow up'' too much to hit $x.$ Anyways, we show this works; fix $a\in F,$ and we show the contrapositive that\[|a|\ge r\implies aK\not\subseteq\psi^{-1}(U).\]We see $|x/a|\le|x|/r,$ so $x/a\in K,$ implying that $x\in aK.$ However, $\psi(x)\notin U,$ so $x$ witnesses $aK\not\subseteq\psi^{-1}(U).$ This completes the proof that $\Psi$ is a homeomorphism onto its image. $\blacksquare$

We are now almost ready to show that $\Psi$ is surjective. To do this cleanly, we pick up some theory about the "annihilator.'' In a general locally compact abelian group $G,$ we fix a subgroup $H\subseteq G.$ Then the annihilator $H$ denoted $H^\perp\subseteq\widehat G$ is defined as\[H^\perp:=\big\{\chi\in\widehat G:\chi(H)=\{1\}\big\}.\]We note that $H^\perp$ is a subgroup of $\widehat G$ ($\chi_1(H)=\{1\}$ and $\chi_2(H)=\{1\}$ certainly implies $(\chi_1\chi_2)(H)=\{1\}$), and in fact writing\[H^\perp=\bigcap_{h\in H}\big\{\chi\in\widehat G:\chi(h)=1\big\},\]we see that $H^\perp$ is in fact closed. Indeed, it suffices to show that the intersected sets are closed, for which we note that the complement of these sets is\[\big\{\chi\in\widehat G:\chi(\{h\})\in S^1\setminus\{1\}\big\}.\]This complement is open because we're using the compact-open topology on $\widehat G.$

With the introductory remarks complete, we claim the following.

This lemma is quite remarkable, but we won't spend time admiring it. We note that $\perp$ always outputs closed subgroups, so at least this restriction to closed subgroups makes sense.

We start by showing that $\perp$ is inclusion-reversing because this is easier. For closed subgroups $A\subseteq B$ of $G,$ then we note $\chi\in B^\perp$ if and only if $\chi(B)=\{1\}.$ But now $A\subseteq B,$ so certainly\[\chi(A)=\{1\}.\]It follows $\chi\in A^\perp,$ so we see $B^\perp\subseteq A^\perp.$

To show that $\perp$ is bijective, we actually show that $\perp$ is kind of involutive, which is stronger. That is, we fix a closed subgroup $A$ of $G$ and show $\left(A^\perp\right)^\perp=A,$ in the sense that these are equal under the Pontryagin duality correspondence. (The technicality is that $\left(A^\perp\right)^\perp$ lives in the character group of $\widehat G,$ which is isomorphic but not equal to $G.$) Anyways, unpacking the definitions, we are interested in\[\left(A^\perp\right)^\perp=\Big\{\chi\in\widehat{\smash{\widehat G}\vphantom{1^]}\,}:\chi\left(A^\perp\right)=\{1\}\Big\}.\]Using Pontryagin duality, we can parameterize characters of $\widehat G$ with $a\in G$ by the map $\chi\mapsto\chi(a).$ So we are actually interested in showing that\[A\stackrel?=\left\{a\in G:\chi\in A^\perp\implies\chi(a)=1\right\}.\]This is equivalent to showing that\[A\stackrel?=\bigcap_{\chi(A)=\{1\}}\{a\in G:\chi(a)=1\}.\]So essentially we have to show that we can determine a closed subgroup $A\subseteq G$ by looking entirely at the character group of $G.$ Note that surely $A$ is a subset of the right-hand side, so the difficult part is showing that if $g\in G$ has $\chi(g)=1$ for each $\chi\in A^\perp,$ then $g\in A.$

We do this by modding out $G$ by $A.$ Because $G$ is abelian, $A$ is normal, and because $A$ is closed, the quotient space is still Hausdorff. Additionally, $G/A$ inherits being locally compact from $G.$ Indeed, focus on the projection map $\pi:G\to G/A$ which is open and continuous by construction. Then for any $gA\in G/A,$ we have some compact neighborhood of $g$ named $K_g\subseteq G.$ Then\[\pi^{-1}(K_g)\subseteq G/A\]is also compact because an open cover of one makes an open cover of the other by $\pi.$

Now, considering the space $G/A$ means that it will suffice to show that\[\bigcap_{\chi\in\widehat G'}\{a\in G:\chi(a)=1\}\stackrel?=\{e\}\]in an arbitrary locally compact abelian group $G'.$ Indeed, this is enough because then if $\chi(g)\in A$ for each $\chi\in A^\perp,$ then $\chi\circ\pi^{-1}$ is a character on $G/A$ for which\[\left(\chi\circ\pi^{-1}\right)(gA)=A.\]Additionally, all characters on $G/A$ are of this form because $\chi:G/A\to S^1$ can be pulled back to $\chi\circ\pi:G\to S^1,$ which satisfies $\chi=\chi\circ\pi\circ\pi^{-1}.$ Anyways, the point is that all characters take $gA$ to the identity element $A,$ so we must actually have $gA=A,$ implying $g\in A.$

Anyways, we now show that\[\bigcap_{\chi\in\widehat G'}\{a\in G:\chi(a)=1\}\stackrel?=\{e\}.\]This follows quickly from Pontryagin duality. Suppose $g$ is in the left-hand set so that we want to show $g=e.$ Well, under Pontryagin duality, $g$ turns into a character on $\widehat G,$ taking $\chi\mapsto\chi(g).$ But then we're taking each $\chi$ to $1$ directly, so $g$ corresponds to the identity mapping! So indeed $g=e.$

Because $\perp$ isn't actually involutive but only under a canonical isomorphism, we provide the details for why we now know $\perp$ is bijective. We do this clumsily because I can't be bothered to find the correct way. For injectivity, suppose $A$ and $B$ are closed subgroups of $G$ for which $A^\perp=B^\perp.$ However, then we see\[A=\bigcap_{\chi\in A^\perp}\{a\in G:\chi(a)=1\}=\bigcap_{\chi\in B^\perp}\{b\in G:\chi(b)=1\}=B.\]For surjectivity, fix a closed subgroup $C$ of $\widehat G,$ and we note that\[C=\bigcap_{a\in C^\perp}\big\{\chi\in\widehat G:a(\chi)=1\big\}.\]However, Pontryagin duality lets us equivalently associate $a\in C^\perp$ characters of $\widehat G$ with elements of $G.$ So now elements $a\in C^\perp$ look like elements $a\in G$ for which $\chi(a)=1$ for each $\chi\in C$; i.e., $\{a\}^\perp\supseteq C.$ Thus,\[C=\bigcap_{\{a\}^\perp\supseteq C}\big\{\chi\in\widehat G:\chi(a)=1\big\}.\]Moving the intersection into a single set, we are essentially asserting that\[C=\left\{a\in G:\{a\}^\perp\supseteq C\right\}^\perp.\]It remains to show that the inner set is a closed subgroup of $G.$ Well, it consists of $a\in G$ for which $\chi(a)=1$ for each $\chi\in C.$ This condition is closed under group multiplication and inversion, so we have a subgroup. Additionally, we can write the set as\[\bigcap_{\chi\in C}\{a\in G:\chi(a)=1\},\]which shows that it is closed, for $G\setminus\{a\in G:\chi(a)=1\}=\chi^{-1}(S^1\setminus\{1\})$ is open. $\blacksquare$

Having studied the annihilator sufficiently, we are now ready to complete the proof of the theorem.

Recall $F$ is complete with respect to its metric. So considering a convergent sequence in $\Psi(F),$ we can map this sequence back into $F,$ note that it must converge in $F,$ and then map the limit point back into $\Psi(F).$ (This only works because we know $\Psi$ is a homeomorphism onto its image.) Thus $\Psi(F)$ is closed.

To finish, we note that showing $\Psi(F)=\widehat F$ is equivalent to showing $\Psi(F)^\perp=\widehat F^\perp,$ from the theory developed above. We also know $\widehat F^\perp$ only consists of the trivial character, so we have to show that $\Psi(F)^\perp$ also only consists of the trivial character. Expanding out definitions, we want to show\[\Psi(F)^\perp=\Big\{x\in\widehat{\smash{\widehat F}\vphantom{1^]}\,}:\chi\in\Psi(F)\implies x(\chi)=1\Big\}\stackrel?=\{e\}.\]Using Pontryagin duality, this is the same as showing\[\Psi(F)^\perp=\{x\in F:\chi\in\Psi(F)\implies\chi(x)=1\}\stackrel?=\{0\}.\]However, $\Psi(F)$ is parameterized by $F,$ so the condition on $x$ is that $\Psi(a)(x)=1$ for all $a\in F.$ But $\Psi(a)(x)=\Psi(x)(a),$ so we require that $\Psi(x)$ be a trivial character, which requires $x=0$ because $\Psi$ is known to be injective. Thus, we are done here. $\blacksquare$