Let $G$ be a semisimple algebraic group over $\mathbb{Q}\_p$. Then by definition $G$ admits no non-trivial *algebraic* characters, i.e. homomorphisms $G \to \mathbb{G}\_m$.


However, it is quite possible that $G(\mathbb{Q}\_p)$ admits *topological* characters. E.g. take $G=\mathrm{PGL}\_n$ and consider the composition
$$\mathrm{PGL}\_n(\mathbb{Q}\_p) \to \mathbb{Q}\_p^\*/\mathbb{Q}\_p^{\*n} \to S^1, \quad g \mapsto \chi(\det(g)),$$
where $\chi: \mathbb{Q}\_p^\*/\mathbb{Q}\_p^{\*n} \to S^1$ is some character.


In this special case $G$ is adjoint, however. I can also do similar constructions for other adjoint groups. So I'm wondering whether this can also happen for simply connected $G$.



> 
> Let $G$ be a simply connected semisimple algebraic group over $\mathbb{Q}\_p$. Is any continuous homomorphism
>  $$G(\mathbb{Q}\_p) \to S^1$$
>  trivial?
> 
> 
> 



Characters of simply connected semsimple algebraic groups over local fields
As I have written in a comment, the answer is YES (any abstract homomorphism into an abelian group is trivial) when $G$ is an *isotropic*, simply connected, simple algebraic group over a nonarchmedean local field $k$. For a proof see the book by Platonov and Rapinchuk, Section 7.2, Theorems 7.1 and 7.6.
Note that any simply connected *anisotropic* simple group is isomorphic to $\mathrm{SL}(1,D)$, where $D$ is a central simple algebra over a finite separable extension $K$ of $k$.


However, the answer is NO when $k=\mathbb{Q}\_2$,
$G=\mathrm{SL}(1,D)$, and $D$ is the quaternion division algebra over $k$. 
EDIT of 18.11.2018: 
As Arkandias explains in his comments below, for the group $G=\mathrm{SL}(1,D)$ as above, it follows from the Corollary to Theorem 21 of Carl Riehm's paper [The norm 1 group of a p-adic division algebra](https://www.jstor.org/stable/2373336?origin=crossref&seq=1#metadata_info_tab_contents)
that the abelianization $G^{\rm ab}:=G/[G,G]$ is a group of order 3. Since the commutator subgroup $[G,G]$ is open and hence, closed in $G$, we see that $G$ admits a non-trivial continuous homomorphism to $S^1$.


