7
$\begingroup$

This question is inspired by the recent one about Gaussian measures over the reals:

What makes Gaussian distributions special?

I would be interested in a similar list of characterizations for the probability measure, on the field of $p$-adic numbers $\mathbb{Q}_p$, whose density with respect to the standard additive Haar measure is given by the indicator function of the ring of $p$-adic integers $\mathbb{Z}_p$.

$\endgroup$
8

1 Answer 1

4
$\begingroup$

Perhaps one answer is the appearance of the Gaussian in the oscillator (a.k.a. Weil) representation --- which makes sense even in non-zero characteristic. You're probably aware of the construction of this infinite dimensional representation for the double cover of $\mathrm{SL}_2(\Bbb{R})$. The story for $\mathrm{SL}_2 \big( \Bbb{F}_q \big)$ where $q$ a power of a prime strongly parallels the continuous story except, of course, it's finite:

Choose any non-square $\delta \in \Bbb{F}_q$ and form the quadratic extension $\Bbb{F}_q(\delta) \cong \Bbb{F}_q^2$. As usual, we identify elements of $\Bbb{F}_q(\delta)$ as linear combinations $z = a + \sqrt{\delta} b$ with $a, b \in \Bbb{F}_q$ subject the usual formulae for addition and multiplication; conjugation and norm are expressed as $\bar{z} = a - \sqrt{\delta} b$ and $\mathrm{N}(z) = a^2 - \delta b^2$ respectively. We'll need to choose (any) non-trivial additive character $\psi: \Bbb{F}_q \longrightarrow \Bbb{C}^*$ together with any multiplicative character $\chi: \Bbb{F}_q(\delta)^* \longrightarrow \Bbb{C}^*$ which generates the character group of $\Bbb{F}_q(\delta)^*$. Morally $\psi$ plays the role of the exponential function $\exp: \Bbb{R} \longrightarrow \Bbb{R}^*$ in this finite context. Now set

\begin{equation} W_\chi := \ \left\{ \begin{array}{l} \displaystyle \text{all functions} \ f: \Bbb{F}_q(\delta) \longrightarrow \Bbb{C} \ \ \text{such that} \\ \displaystyle f(wz) = \overline{\chi(w)} \, f(z) \ \text{whenever $\mathrm{N}(w)=1$} \end{array} \right\} \end{equation}

which is $q-1$ dimensional. The oscillator representation $\varrho_\chi: \mathrm{SL}_2(\Bbb{F}_q) \longrightarrow \mathrm{GL}(W_\chi)$ is determined by the action of the follow elements, which generate $\mathrm{SL}_2(\Bbb{F}_q)$:

\begin{equation} \begin{array}{ll} \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}} f(z) &\displaystyle = \ f(xz) \\ \displaystyle \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} 1 & \ \ y \\ 0 & \ \ 1 \end{pmatrix}} f(z) &\displaystyle = \ \underbrace{\psi \big( y \, \mathrm{N}(z) \big)}_{\text{the Gaussian $G_y(z)$}} \cdot f(z) \\ \displaystyle \displaystyle \varrho_\chi { \scriptstyle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}} f(z) &\displaystyle = \ -q \widehat{f}(-\bar{z}) \end{array} \end{equation}

where $\widehat{f}$ is the Fourier transform of $f$ with respect to the additive structure of $\Bbb{F}_q(\sqrt{\delta})$. Like the continuous Gaussian, $G_y(z)$ is an eigenfunction of this Fourier transform.

As far as I know, the construction that I've outlined (and which I learned from Amritanshu Prasad's online notes) can be carried out for $\mathrm{SL}_2\big( \Bbb{Z}/q\Bbb{Z} \big)$ where $q= p^N$ is still a power of a fixed prime $p$. Furthermore a coherent choice of additive and multiplicative characters $\psi_N$ and $\chi_N$ can be made for each $N \geq 1$ so that these oscillator representations agree with the inverse system

\begin{equation} 1 \longleftarrow \Bbb{Z}/p\Bbb{Z} \stackrel{\ \text{mod $p$}}{\longleftarrow} \Bbb{Z}/p^2\Bbb{Z} \stackrel{\ \text{mod $p^2$}}{\longleftarrow} \Bbb{Z}/p^3\Bbb{Z} \longleftarrow \, \cdots \end{equation}

thus allowing the oscillator representations $\varrho_{\chi_N}$ to be exported to the $p$-adic integers $\Bbb{Z}_p$, which will inherent some kind of Gaussian-like function $\varprojlim G_{\bf y}$ where ${\bf y}= (y_1,y_2,y_3, \dots)$ and $y_N = y_{N+1} \, \text{mod $p^N$}$ for each $N \geq 1$.

$\endgroup$
8
  • 1
    $\begingroup$ I would follow this up with the observation that Young's convolution inequality $\| f * g\|_r \leq c_{p,q} \| f \|_p \, \| g \|_q$ with ${1 \over p} + {1 \over q} = 1 + {1 \over r}$ is maximized when $f$ and $g$ are Gaussians. One might ask whether or not the same characterization holds for the $\Bbb{F}_q$-Gaussians $G_y$. $\endgroup$ Commented Mar 3, 2021 at 4:06
  • 1
    $\begingroup$ In a similar vein, the Babenko-Beckner inequality $\| \widehat{f} \|_q \leq \tilde{c}_{p,q} \| f \|_p$ with ${1 \over p} + {1 \over q} = 1$ and $1 < p \leq q$ is maximized by Gaussians. Does this characterize $\Bbb{F}_q$-Gaussians ? Is there some analogue at least ? $\endgroup$ Commented Mar 3, 2021 at 4:33
  • 1
    $\begingroup$ Thank you for a very nice answer! I remember reading a long time ago about Segal-Shale-Weil and the metaplectic representation, Mehler's formula etc. Do you know if one gets the Fourier transform as a particular value of the corresponding unitary representation, like at $t=\frac{\pi}{4}$ if I remember correctly for the Oscillator? If so are the eigenfunctions determined, like Hermite functions in the Archimedean case? $\endgroup$ Commented Mar 3, 2021 at 14:58
  • 1
    $\begingroup$ I also hope someone will write an answer along the line of your comments, namely some charaterization of the indicator function of $\mathbb{Z}_p$ as some kind of maximizer/minimizer. $\endgroup$ Commented Mar 3, 2021 at 14:59
  • 1
    $\begingroup$ If $\delta = -1$ is a non-square in $\Bbb{F}_q$ then the Fourier transform is indeed a special value of the action the subgroup $\mathrm{SO}_2(\Bbb{F}_q)$ consisting of all $2 \times 2$ matrices of the form $\begin{pmatrix} a & b \\ \delta b & a \end{pmatrix}$ with $a^2- \delta b^2 =1$. As far as Hermite functions are concerned, wouldn't you need a good candidate for a laplace operator $\Delta: \Bbb{C} \big[\Bbb{F}_q (\sqrt{ \delta }) \big] \longrightarrow \Bbb{C} \big[\Bbb{F}_q (\sqrt{ \delta }) \big]$ as well as an appropriate $\Bbb{C}$-valued potential function to play the role of $|z|^2$ ? $\endgroup$ Commented Mar 3, 2021 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.