Let $k$ be a $p$-adic field, $T$ a torus over $k$, and $S$ an $k$-subtorus of $T$. If $\chi: S(k) \rightarrow \mathbb{C}^{\ast}$ is a smooth (resp. continuous) homomorphism, then does $\chi$ necessarily extend to a smooth (resp. continuous) homomorphism on $T(k)$?


Even in the special case where $T$ is split over $k$, I am not sure of the answer. One can find a complentary $k$-subtorus $S'$ of $T$ such that $T$ is the direct product of $S$ and $S'$. Then $S(k) \times S'(k)$ is isomorphic to a subgroup of finite index in $T(k)$, so one is reduced to the case of considering finite index subgroups of a finite product of copies of $k^{\ast}$.


We do have a homomorphism $H\_T: T(k) \rightarrow \textrm{Hom}\_{\mathbb{Z}}(X(T)\_k,\mathbb{Z})$ defined by 


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


whose kernel is the unique maximal open compact subgroup of $T(k)$, see for example [my previous question](https://mathoverflow.net/questions/269444/the-unique-maximal-compact-subgroup-of-a-torus). The same for $S(k)$. It might be possible to restrict a given character to $\textrm{Ker } H\_S$, which is then necessarily unitary, and look at the fact that $S(k)/\textrm{Ker } H\_S$ is a discrete finite rank free abelian group, and do something there.


If there is a good notion of an Ext functor in the category of locally compact abelian Hausdorff groups, I was also thinking it might be possible to look at a sequence like 


$$0 \rightarrow \operatorname{Hom}\_{\textrm{top-grp}}(T(k)/S(k),\mathbb{C}^{\ast}) \rightarrow \operatorname{Hom}\_{\textrm{top-grp}}(T(k),\mathbb{C}^{\ast})$$ $$\rightarrow \operatorname{Hom}\_{\textrm{top-grp}}(S(k),\mathbb{C}^{\ast}) \rightarrow \operatorname{Ext}^1\_{\textrm{top-grp}}(T(k)/S(k),\mathbb{C}^{\ast})$$


and look at $\operatorname{Ext}^1\_{\textrm{top-grp}}(T(k)/S(k),\mathbb{C}^{\ast})$ when $\mathbb{C}^{\ast}$ is alternatively viewed in its usual topology and the discrete topology.



For tori $S \subseteq T$, every character of $S(k)$ extends to a character of $T(k)$?
First, continuous characters on $T(k)$ are the same thing as smooth characters, by a no small subgroup argument. [This answer](https://mathoverflow.net/questions/293665/every-quasicharacter-of-an-open-subgroup-extends-to-a-quasicharacter-on-the-whol) shows that if I have an inclusion of abstract groups $H \subset G$, then every homomorphism of $H$ into $\mathbb{C}^{\ast}$ extends to a homomorphism of $G$ into $\mathbb{C}^{\ast}$. This is because $\mathbb{C}^{\ast}$ is a divisible abelian group. 


Write $S = S(k)$ and $T = T(k)$. Let $K\_S, K\_T$ be the maximal compact open subgroups of $S$ and $T$. Let $\chi$ be a character of $S$. The restriction of $\chi$ to $K\_S$ is a unitary character. By Pontryagin duality, $\chi$ extends to a continuous unitary character $\overline{\chi}$ of $K\_T$. Now we extend $\overline{\chi}$ to a continuous character of $K\_TS$ by setting $\overline{\chi}(ks) = \overline{\chi}(k)\chi(s)$. This is well defined, because $K\_T \cap S = K\_S$.


Finally, the character $\overline{\chi}$ on $K\_TS$ extends abstractly to a homomorphism on all of $T$. It is automatically continuous, because $K\_TS$ is open in $T$.


