Orbits of the maximal compact subgroup on the light cone for $p$-adic groups

Question

It is known that if $Q$ is an indefinite non-degenerate quadratic form on $ \mathbb{R}^n$ with $n \ge 3$, then any maximal compact subgroup $K$ of the orthogonal group $SO(Q)$ acts transitively on the projectiviziation of the light cone. In other words, if $Q(x)=Q(y)=0$ for non-zero vectors $x,y \in \mathbb{R}^n$, then there exists $g \in K$, with $gx= \lambda y$ for some $ \lambda \neq 0$. My question concerned the $p$-adic situation: suppose $Q$ is non-degenerate quadratic form on $ \mathbb{Q}_p^n$ which is isotropic, i.e. the equation $Q(x)=0$ has a non-zero solution $x \in \mathbb{Q}_p^n$ and $n \ge 3$. It is well-know that there may be more than one conjugacy classes of the maximal compact subgroups. My question is: is there at least one such maximal compact subgroup that acts transitively on the projectiviziation of the light cone?

Answer 1

Let $(V,(x,y)\mapsto x.y)$ be a bilinear space. Let $L$ be a lattice on $V$, and let $G$ be its stabilizer in $O(V)$.

To an isotropic line $\ell$ in $V$ one can associate the ideal $I(\ell):=(\ell\cap L).L$, and of course, two lines $\ell_0$ and $\ell_1$ belong to a same orbit under $G$ only if $I(\ell_0)=I(\ell_1)$.

Now, when $p$ is odd, the maximal compact groups are the stabilizers of the lattices $L$ that have a decomposition : $$L\simeq H^k\perp <p>\otimes H^{r-k}\perp A$$ where

• $r$ is the Witt index of $V$,
• $k$ is an integer in $[0..r]$,
• $H$ is the hyperbolic plane over $\mathbf Z_p$ ($(x,y)\mapsto x_1y_2+x_2y_1$),
• $<p>\otimes H$ is the hyperbolic plane scaled by $p$ ($(x,y)\mapsto px_1y_2+px_2y_1$) and
• $A$ is a maximal integral anisotropic lattice.

Thus we see that there are only two favourable cases : $k=0$ and $k=r$. In both cases, a little more work will show that the action of $G$ is indeed transitive on the isotropic lines of $V$.

When $p=2$ one has to be a little more carefull in the description of the maximal compact groups, but the result is the same.

Reference : P. Garrett, Buildings and Classical Groups.

Answer 2

For ${\rm O}(Q)$, you can solve your poblem using Bruhat-Tits theory. I claim that if $K$ is a special maximal compact subgroup of ${\rm O}(Q)$, then $K$ acts transitively on the isotropic lines. Here is a proof.

The maximal parabolic subgroups of the reductive ${\mathbb Q}_p$-algebraic group ${\rm O}(Q)$ are the stabilizers of lines in ${\mathbb Q}_p^n$ generated by isotropic vectors (isotropic lines). Now note that ${\rm O}(Q)$ acts transitively on the isotropic lines. Indeed if $L_1$, $L_2$ are such lines they are isometric as quadratic spaces and by Witt's theorem, there exists $g\in {\rm O}(Q)$ such that $gL_1 =L_2$. We must prove that we can take $g$ in $K$. Let $P$ be the maximal parabolic subgroup fixing $L_1$ and let $B\subset P$ be a Borel subgroup. Since $K$ is special, we have the Iwasawa decomposition ${\rm O}(Q) = KB$, whence ${\rm O}(Q)=KP$. Then write $g=kp$, $k\in K$, $p\in P$ to obtain $L_2 =kL_1$. Q.E.D.

For ${\rm SO}(Q)$, one needs the transitivity of the action on the isotropic lines. I cannot find the argument.