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

Question

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?

Answer 1

This can be proved the same way in the $p$-adic case as in the real case!

Assume that $Q \colon V \to \mathbb Q_p$ is isotropic. It is a basic fact that $V$ contains a hyperbolic plane $H$. (For a proof we can use the comment above by abx. Let $Q(v) = 0$ where $v \not= 0$. Writing $B$ for the bilinear form associated to $Q$, there's $w$ such that $B(v,w) \ne 0$ by nondegeneracy, and we can scale $w$ to make $B(v,w) = 1$. Then $u := w - \frac{1}{2}Q(w)v$ satisfies $Q(u) = 0$ and $B(u,v) = 1$, so $Q(xu + yv) = 2xyB(u,v) = 2xy$, so the $\mathbb Q_p$-span of $u$ and $v$ is a hyperbolic plane in $V$.)

Since $Q$ is nondegenerate on hyperbolic planes in $V$, $V = H \oplus W$ where $W = H^\perp$ is the orthogonal complement to $H$. For each $c \in \mathbb Q_p^\times$, the mapping $g_c \colon H \to H$ where $g_c(x,y) = (cx,(1/c)y)$ in the basis $\{u,v\}$ is in the special orthogonal group of $Q\rvert_H$. Therefore $g_c \oplus \operatorname{id}_W \in {\rm SO}(Q)$. Since the coordinates of $g_c$ as $c$ varies over $\mathbb Q_p^\times$ are unbounded, $\operatorname{SO}(Q)$ is not compact.

Answer 2

It works over an arbitrary nondiscrete normed field $K$ and without assuming non-degeneracy (excluding the case when the space is 1-dimensional and the form zero).

If $q$ is an isotropic quadratic form on a finite-dimensional vector space over $K$, then $\mathrm{SO}(q)$ is unbounded (and hence noncompact), except if the dimension is $1$ and the form $0$.

The easy argument was already given in characteristic $\neq 2$:

a) if the form is degenerate, choose $x$ in the kernel. Then since one excludes the 1-dimensional case, there is a linear map $f$ whose image is $Kx$ and vanishing on $Kx$. Then $\mathrm{Id}+\lambda f$ is in the special orthogonal group for every $\lambda\in K$, which is therefore not compact.

b) if the form is nondegenerate, choose $x\neq 0$ with $B(x,x)=0$. Choose $y_0$ with $B(x,y_0)\neq 0$. Then find $y$ of the form $y_0+\lambda x$ with $B(y,y)=0$. So $B(x,x)=B(y,y)=0$ and $B(x,y)\neq 0$. Since $(x,y)$ generates a nondegenerate plane its orthogonal is a complement. Then for every $t\in K^*$, the diagonal map $x\mapsto tx$, $y\mapsto t^{-1}y$, identity on the orthogonal, is in the special orthogonal group, which is therefore unbounded.

---

As mentioned by @LSpice one needs to reformulate the proof to allow characteristic 2.

Here we have a quadratic form $q$ (i.e. $q(\lambda x)=\lambda^2q(x)$ for all $x$ and $b_q(x,y)=q(x+y)-q(x)-q(y)$ is bilinear), whose kernel is by definition $\{x:\forall y:q(y+x)=q(y)\}$. Isotropic means that $q^{-1}(\{0\})$ is not reduced to zero.

(a) adapts in a straightforward way. For (b) one chooses $x\neq 0$ with $q(x)= 0$. Then one chooses $y_0$ such that $q(y_0+x)\neq q(y_0)$. Then $q(y_0+tx)$ is affine in $t$ and nonconstant, hence vanishes hence one can find $y$ such that $q(y)=0$ and such that $q(x+y)\neq 0$ (observe that on the plane $Kx+Ky$, $q$ vanishes exactly on $Kx\cup Ky$).

The orthogonal $H_x$ of $x$ (resp. $H_y$ of $y$) for $b_q$ is a hyperplane meeting $Kx+Ky$ in $Kx$ (resp. $Ky$), hence their intersection $H= H_x\cap H_y$ has codimension 2 and is a complement subspace to $Kx+Ky$. Then for $t\in K^*$, the endomorphism $x\mapsto tx$, $y\mapsto t^{-1}y$, identity on $H$, is in the special orthogonal group, which is therefore unbounded.

---

Note that the argument consists in finding a copy of the additive group in case (a) and of the multiplicative group in case (b).

It seems that in general:

• there's no copy of the additive group exactly when either the form is anisotropic, or the form is nondegenerate in dimension $\le 2$, or the form is zero in dimension 1.

• there's no copy of the multiplicative group exactly when the kernel has dimension $\le 1$ and the form is anisotropic modulo the kernel.