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 $$\mathrm{PGL}_n(\mathbb{Q}_p) \to \mu_n \subset S^1, \quad g \mapsto \det(g).$$

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?

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?