1
$\begingroup$

Are there p-adic analogues to spherical harmonics? In the case of $K = \mathbb{R}$, the spherical harmonics form a basis to $L^2 [SO(3)]$ where

What happens in the $p$-adic case? Is there sphere still a compact manifold? By sphere I mean:

$$ S^2 = \{ (x,y,z) \in \mathbb{Q}_p^3: x^2 + y^2 + z^2 = 1 \}$$

In order to have harmonic analysis, what does it mean to integrate over the 2-sphere in this case? What is an example of an element of $L^2$ ?

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ The $p$-adic sphere is not compact. For $p\equiv 1\mod 4$, there is a square root of $-1$ in $\mathbb{Q}_p$, and we get a sequence of points $(p^{-n},\sqrt{-1}p^{-n},1)\in S^2$ going off to infinity. For $p\equiv 3\mod 4$ it takes a little playing around with Hensel's lemma. $\endgroup$
    – Julian Rosen
    Commented May 27, 2016 at 15:31
  • $\begingroup$ If I were to guess a generalization that might work, I would try thinking of $SU(2)$ as the maximal compact subgroup of $SL_2(\mathbb{C})$, and $S^2$ as the homogenous space $\mathbb{CP}^1$. The maximal compact subgroup of $SL_2(\mathbb{Q}_p)$ is $SL_2(\mathbb{Z}_p)$, so I might think about harmonic analysis for $SL_2(\mathbb{Z}_p)$ acting on $\mathbb{P}^1(\mathbb{Q}_p)$ (which is compact). I imagine this is well understood. $\endgroup$
    – David E Speyer
    Commented May 27, 2016 at 15:35
  • $\begingroup$ @DavidSpeyer I am trying to show solutions to $x^2 + y ^2 + z^2 = n$ over $\mathbb{Z}$ equidistribute over the 2-sphere as $n \to \infty$. This is too difficult. E.g. it is harder than Lagrange's 3-squares theorem. It can be solved using automorphic forms, although it relies on some challenging estimates. I am willing to take a bet there have been simplifications over the past 30 years. Here I am looking at an approach using $p$-adic analysis. $\endgroup$
    – john mangual
    Commented May 28, 2016 at 16:28

1 Answer 1

Reset to default
6
$\begingroup$

First, the group $G=SO(3,\mathbb Q_p)$ acts transitively on that quadric surface, by Witt's theorem. For $p\not=2$ this group is non-compact. The isotropy group of a point on the quadric is a subgroup $H$ isomorphic to $O(2)$, split or not depending on $p$ mod $4$. Then the harmonic analysis on $G/H$ can be addressed by fairly standard methods, although it's more complicated since $G$ is non-compact. (The irreducible admissible repns of $G$ containing a vector fixed under a ("special") maximal compact, a.k.a. "spherical repns", are classified by Borel-Casselman-Matsumoto as being unramified principal series.)

There's also the isogeny of $SL_2$ to (split) $SO(3)$, ...

Improve this answer
$\endgroup$
3
  • $\begingroup$ My, what a nice collection of notes you have. What does "split" mean here? Why are $O(2, \mathbb{Q}_p)$ and $SO(3,\mathbb{Q}_p)$ split. And what is "isogeny"? $\endgroup$
    – john mangual
    Commented May 27, 2016 at 22:30
  • 1
    $\begingroup$ $SO(\mathbb Q_p)$ is "split" means that it is isomorphic to $\mathbb Q_p^\times$, while if it is non-split, then it is isomorphic to the norm-one elements in a quadratic extension of $\mathbb Q_p$. If the quadratic form being preserved by the group is $x^2+y^2$, then when there's a $\sqrt{-1}$ the group is split, otherwise not, because of the way $x^2+y^2$ factors. For $SO(3)$ with quadratic form $x^2+y^2+z^2$, there is always a non-trivial $0$ unless $p=2$, so inside $SO(3,\mathbb Q)$ there's always a split $SO(2)$. Isogeny is a homomorphism with finite kernel and cokernel. $\endgroup$
    – paul garrett
    Commented May 28, 2016 at 15:06
  • $\begingroup$ $SL_2$ or $GL_2$ should keep me busy for a long time. I also found these notes of Bump, suggesting this topic is pretty well-understood. Keep in mind though my goal is not to be terribly general or modern. My goal is to solve a problem using the limited, fairly old-fashioned methods available too me. $\endgroup$
    – john mangual
    Commented May 28, 2016 at 16:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .