Let $T$ be a torus over a $p$-adic field $F$. Let $q = f(F/\mathbb{Q}\_p)$, and normalize the absolute value $| \cdot |$ on $F$ so that a uniformizer has value $\frac{1}{q}$.


Let $X(T)\_F$ be the group of rational characters of $T$ which are defined over $F$. Define a homomorphism


$$H\_T: T(F) \rightarrow \textrm{Hom}\_{\mathbb{Z}}(X(T)\_F,\mathbb{Z})$$


by $$H\_T(t)(\chi) = \log\_q |\chi(t)|$$


Giving the free abelian group $\textrm{Hom}\_{\mathbb{Z}}(X(T)\_F,\mathbb{Z})$ the discrete topology, it is easy to see that $H\_T$ is continuous: the kernel is open as a finite intersection of open subgroups $\chi^{-1}\mathcal O\_F^{\ast}$, as $\chi$ runs through a basis of $X(T)\_F$.


It follows that $\textrm{Ker}(H\_T)$ contains every compact subgroup of $T(F)$, because if $E$ is any compact subgroup of $T(F)$, then $H\_T(E)$ is isomorphic to a compact, hence finite, subgroup of $\mathbb{Z}^n$, and every nontrivial subgroup of $\mathbb{Z}^n$ is infinite.


On the other hand, I have read that $\textrm{Ker}(H\_T)$ itself is compact. How can one see this?



The unique maximal compact subgroup of a torus
There might be an easier way to see this but here is one way. 


Any $F$-torus $T$ can be written as $T = T\_s T\_a$ with finite intersection, where $T\_s$ (resp. $T\_a$) is the unique maximal split (resp. anisotropic) subtorus of $T$ (Springer, LAG, Proposition 13.2.4). Further, the co-characters all come only from the split part; indeed (Springer, LAG, Proposition 13.3.3(iii)), $T$ is anisotropic iff any $F$-morphism $\mathbb G\_m \to T$ is constant. Since $F$ is locally compact, $T\_a(F)$ is compact (Springer, somewhere). Thus it suffices to restrict $H\_T$ to $T\_s$ where the kernel is the $\mathcal O\_F$-points of $T\_s$, a compact group. Up to finite index, the kernel is a product of $T\_a(F)$ with $T\_s(\mathcal O\_F)$ and is hence compact. 


