# For tori $S \subseteq T$, every character of $S(k)$ extends to a character of $T(k)$?

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. 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.

• I am sorry, I had changed notation in the middle of typing. I intended $F = k$. – D_S Feb 22 '18 at 23:24

First, continuous characters on $T(k)$ are the same thing as smooth characters, by a no small subgroup argument. This answer 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 $T$.
• Why do you have your roundabout argument with $K_S$ and $K_T$, rather than simply extending directly from $H = S(k)$ to $G = T(k)$? – LSpice Feb 23 '18 at 16:51
• Oh, I see; smoothness. It might make more sense to take for $K_S$ the kernel of $\chi$. – LSpice Feb 23 '18 at 16:52