TITLE: Why are anisotropic tori compact?
QUESTION [13 upvotes]: I'll do my best to formulate everything in the modern language.  Let $F$ be a local field. A torus $S$ is an $F$-group scheme of finite type such that $S \times_F \overline{F}$ is isomorphic to a finite product of copies of $\mathbb{G}_{m,\overline{F}}$ as $\overline{F}$-group schemes, where $\mathbb{G}_{m,\overline{F}} = \textrm{Spec } \overline{F}[T,T^{-1}]$.  A character $\chi$ of $S \times_F \overline{F}$ is a morphism of $\overline{F}$-group schemes $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$.  
We say that $\chi$ is defined over $F$ if there exists a morphism of $F$-group schemes $\alpha: S \rightarrow \mathbb{G}_{m,F}$ such that $\chi = \alpha \times 1_{\overline{F}}$.  If it exists, $\alpha$ is unique.
The trivial character is the morphism $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$ coming from the $\overline{F}$-algebra homomorphism $\overline{F}[T,T^{-1}] \rightarrow \overline{F}[T_1^{\pm 1}, ... T_n^{\pm 1}]$, $T \mapsto 1$.  It is always defined over $F$.
We say that $S$ is anisotropic if the trivial character is the only character which is defined over $F$.
There is a natural way of topologizing the $F$-rational points $S(F)$ of $S$, such that it is a locally compact topological group.  See for example here (http://math.stanford.edu/~conrad/papers/adelictop.pdf).  
I have read that $S(F)$ is compact if and only if $S$ is anisotropic, but I have never seen an explanation of this.  What is the most natural way to realize this result?

REPLY [9 votes]: Sorry for adding a second proof, but I think I have now a much clearer one.
Let $k$ be a local field.
Let $\mathbf{T}$ be a $k$-anisotropic torus, that is $\mathbf{T}$ contains no $k$-subgroup $k$-isomorphic to $\mathbf{G}_m$.
It is standard that $T$ also doesn't have a non-trivial $k$-homorphism to $\mathbf{G}_m$.
We argue to show that $T=\mathbf{T}(k)$ is compact.
We assume as we may that $T$ is Zariski dense in $\mathbf{T}$.
We fix a non-trivial irreducible $k$-representation $\mathbf{T}\to\text{GL}_n$.
As mentioned above, $n>1$. We use the fact that the $T$-orbits on $\mathbb{P}^{n-1}(k)$ are locally closed to find a point $x\in \mathbb{P}^{n-1}(k)$ with a $T$-closed orbit. We let $S$ be the stabilizer of $x$ in $T$ and $\mathbf{S}$ be the Zariski closure of $S$ in $\mathbf{T}$. $\mathbf{S}$ is a proper $k$-subtorus. By an induction argument we may assume that $S=\mathbf{S}(k)$ is compact.
We deduce that $T$ is compact.

For completeness, let me add the remark that in charactersitic 0 the same proof yields the fact that $\mathbf{G}(k)$ is compact when $\mathbf{G}$ is a $k$-anisotropic reductive group.
The only thing to observe is that every $k$-subgroup of $\mathbf{G}$ is again reductive, what enables one to make an inductive argument as above. This follows from Jacobson-Morozov theorem: $\mathbf{G}$ has no unipotent $k$-subgroup.