# The stabiliser group of an isotropic quadratic form over $\mathbb{Q}_p$ is non-compact?

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?

• 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 $\mathbb{Q}_p^*$.
– abx
Sep 5, 2020 at 14:52
• It works for an arbitrary nondiscrete normed field: if $q$ is isotropic then $SO(V)$ is unbounded.
– YCor
Sep 9, 2020 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.