# Orthogonal group over local field

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?

• 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. Mar 3, 2012 at 23:20
• Since people reading this page may have a similar question about the isotropic case, in which case $O_Q(V)$ is not compact, let me give a link to an MO page discussing that: mathoverflow.net/questions/370940/…. Oct 15, 2022 at 16:57
• If $(K,|\cdot|)$ is a nondiscrete normed field, $d>0$, $K^d$ is endowed with, say, the sup norm and $q$ is a quadratic form on $K^d$, then $\mathrm{(S)O}(q)$ is bounded (in the matrix algebra with operator norm) if and only if the inf of $|q(x)|$ for $x$ on the 1-sphere of $K^d$ is $>0$. ($\Leftarrow$ is obvious and $\Rightarrow$ follows from Witt's theorem. Non-discreteness is used to ensure that if the inf is zero and $q\neq 0$ then the inf of $q(x)$ for $x$ in the 1-sphere such that $q(x)\neq 0$ is also zero.)
– YCor
Oct 15, 2022 at 17:31
• @LSpice will you do the same for a link after my comment on the other page to come back to this page? I never remember how to create a link in comments that displays the title of the page in place of the URL. Oct 15, 2022 at 18:41
• @LSpice yes, thanks. Oct 15, 2022 at 20:28

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 nonzero $$c$$ in $$K$$ such that $$|v|_2 \geq |c|$$ 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 $$S_2 \subset (1/c) B_{\infty}$$, and $$S_2$$ is compact. (Hint: for $$v \in S_2$$, $$|v|_\infty = |a|$$ for some $$a \in K^\times$$, so $$(1/a)v \in S_\infty$$. Show $$|cv|_\infty \leq 1$$ for $$c$$ from the previous claim.)

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

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).

• @JimHumphreys's comment. Oct 15, 2022 at 18:39

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

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.