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?



Orthogonal group over local field
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.


