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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?

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    $\begingroup$ 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
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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.

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