Question

Would anyone be able to tell me how to prove that the orthogonal group over a local field for an anisotropic quadratic form is compact?

Answer 1

I am lazy, so I'll write this out as a sequence of claims without proofs.

Let $K$ be the local field and let $| \ |$ denote the absolute value on $K$. Let $V$ be the vector space with anisotropic form $\langle \ , \ \rangle$. Choose an arbitrary basis $e_1$, ..., $e_n$ of $V$. Define functions $| \ |_{\infty}$ and $| \ |_2$ from $V \to \mathbb{R}$ as follows: $$\left| \sum a_i e_i \right|_{\infty} = \max(|a_i|).$$ $$| v |_2 = | \langle v,v \rangle|^{1/2}.$$

Claim: The unit $\infty$-ball, $B_{\infty}:=\{ v \in V: |v|_{\infty} \leq 1 \}$, is compact.

Claim: The unit $\infty$-sphere, $S_{\infty} := \{ v \in V: |v|_{\infty} = 1 \}$, is compact.

Claim: The function $| \ |_2$ is continuous.

Claim: There is a positive constant $r>0$ such that $|v|_2 \geq r$ on $S_{\infty}$. (This is the step that uses anisotropy.)

Claim: Define the unit $2$-sphere by $S_2 := \{ v \in V: |v|_{2} = 1 \}$. Then $S_2 \subset (1/r) B_{\infty}$, and $S_2$ is compact.

Claim: The orthogonal group embeds as a closed subspace of $S_2^n$, and is hence compact.

Answer 2

The proof for the archimedean case is given in our own Pete Clark's notes., see Theorem 1.

Answer 3

To address Jim Humphreys' comment concerning a unified argument covering all (non-archimedean) local fields, see the proof by Gopal Prasad (in "An elementary proof of a theorem of Bruhat-Tits-Rousseau and of a theorem of Tits", Bull. SMF 110) that a connected reductive group $G$ over a henselian non-trivially valued field $k$ has $G(k)$ bounded (equiv. compact, when $k$ is locally compact) if and only if $G$ is $k$-anisotropic.

Relevance: the special orthogonal group of an anisotropic non-degenerate quadratic form over a field $K$ is $K$-anisotropic as a connected semisimple algebraic group over $K$ (so the question posed falls into the context of Prasad's argument). Indeed, arguing by contradiction, suppose there is a nontrivial split $K$-torus in the special orthogonal group. This leads to a nontrivial $K$-rational zero of the quadratic form by considering the weight space decomposition for the action of such a split torus on the $K$-vector space in question (any nontrivial element in a single weight space is such a $K$-rational zero).

Answer 4

At least in the case of an algebraic extension $K$ of a $\mathbf Q_p$ with ring of integers $A$, you can also easily check (or read in O'Meara) that if a form $q$ on a vector space $V$ over $K$ is anisotropic, then the set of vectors $x$ with $q(x)\in A$ is an $A$-lattive $L$ on $V$. Thus $O(V,q)=O(L,q\vert_L)$ is compact.