Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
To avoid confusion, it's best to define what you mean here by "local field". Different sources use that term more narrowly or more broadly. Aside from that, Witt's work on quadratic forms over various kinds of fields showed that for non-archimedean local fields there are no anisotropic forms in 5 or more variables. Compactness of reducive algebraic groups which are anisotropic over a field is a theme which comes up in work of Borel-Tits and many others, but I don't know a unified argument covering all local fields. – Jim Humphreys Mar 3 '12 at 23:20

4 Answers 4

up vote 8 down vote accepted

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.

share|cite|improve this answer

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).

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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