Let $\mathbb{Q}_p$ denote the $p$-adic integers. Let $V$ be a $\mathbb{Q}_p$-vector space and $Q : V \rightarrow \mathbb{Q}_p$ be a non-degenerate integral quadratic form. We say that the pair $(Q,V)$ is $\textbf{isotropic}$ if there exists $v \in V \setminus \{0\}$ such that $Q(v) = 0$. Let $$SO_Q(V) := \{ \sigma \in GL(V) : Q(\sigma x) = x \ \text{and} \ \det(\sigma) =1\}$$ be the special stabiliser group of the quadratic form $Q$. I would like to prove the following Lemma.

$\textbf{Lemma}$: Let $(Q,V)$ be isotropic. Then, $SO_Q(V)$ is not compact.

I know how to prove this for an isotropic quadratic form on a $\textbf{real}$ vector space. Anyone has a hint on how to prove it for $p$-adic vector spaces?

  • 4
    $\begingroup$ Choose $w\in V$ such that $Q(v,w)=1$, and put $u=w-\frac{1}{2}Q(w) v$. Then $u$ and $v$ span a hyperbolic plane; its automorphism group contains $\mathbb{Q}_p^*$. $\endgroup$ – abx Sep 5 '20 at 14:52
  • $\begingroup$ It works for an arbitrary nondiscrete normed field: if $q$ is isotropic then $SO(V)$ is unbounded. $\endgroup$ – YCor Sep 9 '20 at 10:20

This is the same proof as in the real case!

Assume that $Q$ is isotropic. Let $h$ be the polar form of $Q$. Then it is a basic fact that $V$ contains a hyperbolic plan $H$. This means that $H$ has a basis $(e_1 ,e_2 )$ satisfying $Q(e_1 )=Q(e_2 )=0$ and $h(e_1 ,e_2 )=1$, say. Since $H$ is non-degenerate it has an orthogonal complement $W$. For $x\in {\mathbb Q}_p^\times$ let $g_x=g_H \oplus {\rm id}_W$ be the endomorphism of $V$ such that $g_H\in {\rm End}(H)$ has matrix ${\rm diag}(x,x^{-1})$ in the basis $(e_1 ,e_2 )$. Then $g_x\in {\rm SO}(Q)$ and the coordinates of $g_x$ are not bounded as $x$ varies. Hence ${\rm SO}(Q)$ is not bounded.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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