Today I Learned

February 28th

Today I learned the classification of characters over $\QQ_p$ as $\QQ_p\cong\widehat{\QQ_p},$ from Keith Conrad as usual. Fix $e_p(x):=e^{2\pi i\{x\}_p}.$ The following is what we're building towards.

In fact it is true that $\Psi_\bullet$ is an isomorphism of the groups as well as a homeomorphism of the two spaces.

We begin by showing that $\Psi_\bullet$ is well-defined. For any $a\in\QQ_p,$ we need to know $\Psi_a$ is a character. Well, $\Psi_a$ is homomorphic because\[\Psi_a(x+y)=e^{2\pi i\{a(x+y)\}_p}=e^{2\pi i\{ax\}_p}\cdot e^{2\pi i\{ay\}_p}=\Psi_a(x)\Psi_a(y).\]In particular, we know $\{x+y\}_p=\{\{x\}_p+\{y\}_p\}$ by expanding $x=x_0+x_\bullet/p^\bullet$ and $y=y_0+y_\bullet/p^\bullet$ for $x_0,y_0\in\ZZ_p$ and $x_\bullet,y_\bullet\in\ZZ/p^\bullet\ZZ.$ Further, $\Psi_a$ is continuous because if we have a Cauchy sequence $x_\bullet\to x$ in $\QQ_p,$ then\[\{x_\bullet\}_p\to\{x\}_p\]in $\RR.$ For example, $\{x_\bullet\}_p=\{x\}_p$ holds as soon as $|x_\bullet-x|\le1,$ which must happen eventually. It follows that $\Psi_a(x_\bullet)\to\Psi_a(x).$ Note how discrete this argument feels.

It also not terribly difficult to show that $\Psi_\bullet$ is an injective homomorphism. Showing show $\Psi_\bullet$ is homomorphic is similar to showing $\Psi_\bullet$ is well-defined. For $a,b,x\in\QQ_p,$ we see\[\Psi_{a+b}(x)=e^{2\pi i\{(a+b)x\}_p}=e^{2\pi i\{ax\}_p}\cdot e^{2\pi i\{bx\}_p}=\Psi_a(x)\Psi_b(x),\]which is what we wanted. Again, $\{x+y\}_p=\{\{x\}_p+\{y\}_p\}$ by expanding out fractional parts.

For injective, we need to know that distinct $a,b\in\QQ_p$ give distinct characters $\Psi_a\ne\Psi_b.$ Well, $a\ne b$ means $b-a\ne0,$ so $|b-a|_p \gt 0.$ If we let $|b-a|_p=p^{-\nu},$ then surely\[a\not\equiv b\pmod{p^\nu}.\]In particular, $\left\{a/p^\nu\right\}_p\ne\left\{b/p^\nu\right\}_p,$ so $\Psi_a\left(1/p^\nu\right)\ne\Psi_b\left(1/p^\nu\right).$ So $\Psi_a\ne\Psi_b,$ as desired.

To show $\Psi_\bullet$ is an isomorphism of groups, it remains to show that $\Psi_\bullet$ is surjective. The algebraic core of the argument is the following claim.

The idea here is to extract $c\in\ZZ_p$ as a Cauchy sequence. The interesting behavior of $\QQ_p$ occurs at negative powers of $p,$ so we note that for a power $p^\bullet,$ we have\[\chi\left(1/p^\bullet\right)^{p^\bullet}=\chi\left(p^\bullet\cdot1/p^\bullet\right)=1.\]It follows $\chi\left(1/p^\bullet\right)$ is a $p^\bullet$th root of unity, so we may define the sequence $\{c_\bullet\}$ to satisfy $\chi\left(1/p^\bullet\right)=e^{2\pi ic_\bullet/p^\bullet}.$ Noting that\[e^{2\pi ic_\bullet/p^\bullet}=\chi\left(1/p^\bullet\right)=\chi\left(1/p^{\bullet+1}\right)^p=e^{2\pi ic_{\bullet+1}/p^\bullet}\]implies that $c_{\bullet+1}\equiv c_\bullet\pmod{p^\bullet}.$ In particular $\{c_\bullet\}$ is a Cauchy/coherent sequence in $\ZZ_p,$ so it will converge to some $c\in\ZZ_p$ with $\left\{c/p^\bullet\right\}=c_\bullet/p^\bullet.$

Having extracted $c,$ we claim each $x\in\QQ_p$ has\[\chi(x)\stackrel?=e^{2\pi i\{cx\}_p}.\]This comes down to some technical computations involving $\{\bullet\}_p.$ Fix $x=y/p^\bullet$ for some $y\in\ZZ_p,$ and we can write $y=x_\bullet+x_0p^\bullet$ with $x_\bullet\in\ZZ/p^\bullet\ZZ$ and $x_0\in\ZZ_p$ so that $x=x_0+x_\bullet/p^\bullet.$ Combining this with $\chi(x_0)=1,$ we see\[\chi(x)=\chi(x_0)\chi\left(x_\bullet/p^\bullet\right)=1\cdot\chi\left(1/p^\bullet\right)^{x_\bullet}=e^{2\pi ic_\bullet x_\bullet/p^\bullet}.\]It remains to compare $c_\bullet x_\bullet/p^\bullet$ with $\{cx\}_p.$ Well, we can write $c=c_\bullet+c_0p^\bullet$ with $c_\bullet\in\ZZ/p^\bullet\ZZ$ and $c_0\in\ZZ_p$ so that\[cx=\underbrace{cx_0}_{\in\ZZ_p}+\underbrace{c_0x_\bullet}_{\in\ZZ_p}+c_\bullet x_\bullet/p^\bullet.\]It follows $\{cx\}_p=\left\{c_\bullet x_\bullet/p^\bullet\right\}_p,$ and because $c_\bullet,x_\bullet\in\ZZ,$ we have $\{cx\}_p=\left\{c_\bullet x_\bullet/p^\bullet\right\}.$ This completes the proof of the lemma. It is somewhat remarkable that we have not used continuity anywhere in this argument; I think $\chi(\ZZ_p)=\{1\}$ guarantees that $\chi$ is locally constant, which obviates the need for continuity. $\blacksquare$

At this point, we're a clever step away from getting any character $\chi.$

The main idea here is to show that $\chi$ should be $1$ on small inputs. This roughly comes from a difference in the way the topologies on $\QQ_p$ and $S^1$ behave, for $\QQ_p$ has subgroups of arbitrarily small norm while $S^1$ does not. To make this explicit, look at the open set\[U_1:=\left\{z\in S^1:|z-1| \lt 1\right\}\subseteq S^1.\]Because $\chi$ is continuous, $\chi^{-1}(U_1)$ is also open. Note $\chi(0)=1,$ so $0\in\chi^{-1}(U_1),$ so there's a basis element around $0$ named $p^\nu\ZZ_p\subseteq\chi^{-1}(U_1).$ This is our only use of the continuity of $\chi.$

So far everything done is topological, but now noting that $p^\nu\ZZ_p$ is a subgroup $\QQ_p$ requires $\chi\left(p^\nu\ZZ_p\right)\subseteq U$ to also be a subgroup. But looking at $U,$ the only available subgroup is $\{1\}$! Thus, we define $\chi_0(x):=\chi\left(p^\nu x\right)$ so that\[\chi_0\left(\ZZ_p\right)=\chi\left(p^\nu\ZZ_p\right)=\{1\}.\]In particular, the previous lemma now kicks in so that $\chi_0(x)=e^{2\pi i\{cx\}_p}$ for some $c\in\ZZ_p.$ It follows\[\chi(x)=\chi_0\left(x/p^\nu\right)=e^{2\pi i\left\{c/p^\nu\cdot x\right\}_p}.\]Thus we have shown $\chi$ has the desired form with constant $c/p^\nu\in\QQ_p.$ $\blacksquare$

While we're here, we remark that the above argument can show that $\Psi_\bullet$ is an open map.

Here $\widehat{\QQ_p}$ is endowed with the compact-open topology, which is created by the subbasis of sets\[\left\{\chi\in\widehat{\QQ_p}:\chi(K)\subseteq U\right\}\]for any compact $K$ and open set $U.$ It is a fact that $\widehat{\QQ_p}$ is a topological group under this topology.

Anyways, any open set in $\QQ_p$ can be decomposed into basis elements, and because function application preserves unions, it therefore suffices to show that basis elements of the topology of $\QQ_p$ map to open sets. Further, for any basis element $x+p^\nu\ZZ_p,$ we see\[\Psi\left(x+p^\nu\ZZ_p\right)=\Psi(x)\cdot\Psi\left(p^\nu\ZZ_p\right),\]so it suffices to show that $\Psi\left(p^\nu\ZZ_p\right)$ is an open set. Well, $a\in p^\nu\ZZ_p$ if and only if\[\Psi_a\left(p^{-\nu}\ZZ_p\right)=\Psi_1\left(ap^{-\nu}\ZZ_p\right)=\Psi(\ZZ_p)=\{1\}.\]But then we can substitute $\{1\}$ out with $U_1$ because $p^\nu\ZZ_p$ is a subgroup (which requires $\Psi\left(p^\nu\ZZ_p\right)$ to be a subgroup as above), so indeed\[\Psi\left(p^\nu\ZZ_p\right)=\left\{\chi\in\widehat{\QQ_p}:\chi\left(p^{-\nu}\ZZ_p\right)\subseteq U_1\right\}.\]So we see $\Psi$ is an open map. $\blacksquare$

To show that $\Psi$ is an isomorphism of topological groups, we need to complete the proof that $\Psi$ is a homeomorphism by showing $\Psi$ is continuous. I don't currently know how to do this.