Today I Learned

February 22nd

Today I learned about the $p$-adic Gaussian from Terrence Tao . I think this is an instance of Pontryagin duality, but I don't know enough theory to say for sure. Anyways, the main character in our story is (fittingly) the character $\QQ_p\to\CC$ by\[e_p(x)=e^{2\pi i\{x\}_p},\]where $\{x\}_p$ is the $p$-adic fractional part of $x\in\QQ_p.$ Because $\QQ_p$ is locally compact (and abelian), we get an additive Haar measure, and we normalize it so that $\int_{\ZZ_p}dx_p=1.$ From this character we get a Fourier transform, which we denote\[\hat f(s)=\int_{\QQ_p}f(x_p)e_p(-sx_p)\,dx_p.\]Note that we're assuming stuff that I can't prove currently.

We're trying to build the machinery we had in Riemann's proof of the functional equation of $\zeta.$ We have an $\exp$ in $e_p,$ and now we would like to build a Gaussian $G_p$ like $G_\infty=e^{-\pi x^2}$ for $\RR.$ This function should be its own Fourier transform, so\[G_p(s)=\int_{\QQ_p}G_p(x_p)e_p(-sx_p)\,dx_p.\]We claim that the indicator function on $\ZZ_p$ works.

We're going to be obnoxiously careful because integration on $\QQ_p$ is somewhat new to me. We focus on the integral. Note $x_p\in\QQ_p\setminus\ZZ_p$ gives $G_p(x_p)=0,$ vanishing entirely, so we're left integrating $x_p\in\ZZ_p,$ which looks like\[G_p(s)\stackrel?=\int_{\ZZ_p}e_p(-sx_p)\,dx_p.\]Because we have the words Fourier and exponential and indicator floating around, we hope to turn this into a sum of roots of unity/discrete Fourier transform. In particular, we need to make the integral better-behaved. Fix $s=p^{-\nu}t$ with $t\in\ZZ_p\setminus p\ZZ_p,$ and we split up $\ZZ_p$ by$\pmod{p^\nu}.$ (These sets countably and disjointly partition $\ZZ_p.$) This looks like\[\int_{\ZZ_p}e_p(-sx_p)\,dx_p=\sum_{k\in\ZZ/p^\nu\ZZ}\int_{x_p\equiv k}e_p(-sx_p)\,dx_p.\]Now, for $x_p\equiv k\pmod{p^\nu},$ we get to write $x_p=k+\ell p^\nu$ for $\ell\in\ZZ_p,$ implying\[\{sx_p\}_p=\left\{p^{-\nu}tk+t\ell\right\}_p=\left\{p^{-\nu}tk\right\}_p.\]In particular, this is constant with respect to $k,$ so we are now proving\[G_p(s)\stackrel?=\sum_{k\in\ZZ/p^\nu\ZZ}e^{2\pi i\{-p^{-\nu}tk\}_p}\int_{x_p\equiv k}dx_p.\]At this point, we can remove the integral. The measures of $\{x_p\in\ZZ_p:x_p\equiv k\}$ must be equal for each $k$ because adding $1$ cyclically shifts them, and the Haar measures is fixed under such shifts. Because the total of all these $p^\nu$ cosets is $1,$ each individual coset must have measure $1/p^\nu.$ So it remains to prove\[G_p(s)\stackrel?=\frac1{p^\nu}\sum_{k\in\ZZ/p^\nu\ZZ}e^{2\pi i\{-p^{-\nu}tk\}_p}.\]Note $\left\{-p^{-\nu}tk\right\}_p=(-tk\bmod{p^\nu})/p^\nu,$ but $p\nmid t,$ so $-tk$ ranges over $\ZZ/p^\nu\ZZ$ as $k$ does. Thus, we want to show\[G_p(s)\stackrel?=\frac1{p^\nu}\sum_{k\in\ZZ/p^\nu\ZZ}e^{2\pi ik/p^\nu}.\]Only now do we remark that $s\in\ZZ_p$ implies that $\nu=0,$ in which case the sum degenerates to just a $1.$ Otherwise $s\in\QQ_p\setminus\ZZ_p,$ implying $\nu\ge1,$ so we are summing the $p^\nu$th roots of unity, which does come out to $0.$ So our Gaussian works. $\blacksquare$