TITLE: Characters of simply connected semsimple algebraic groups over local fields
QUESTION [9 upvotes]: 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?

REPLY [6 votes]: 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
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 $S^1$.