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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?

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  • $\begingroup$ What do you mean by $g\mapsto \text{det}(g)$? For every linear representation that I can think of, the image of $\textbf{PGL}_n$ is contained in $\textbf{SL}_M$. $\endgroup$
    – Jason Starr
    Commented May 19, 2016 at 9:55
  • $\begingroup$ Are you taking $n=p-1$, so that the reduction of the determinant on $\textbf{GL}_{p-1}$ is a well-defined homomorphism $\textbf{PGL}_{p-1}(\mathbb{Z}_p) \to \mathbb{F}_p^\times$? If so, how do you extend this to $\textbf{PGL}_n(\mathbb{Q}_p)$? $\endgroup$
    – Jason Starr
    Commented May 19, 2016 at 9:58
  • $\begingroup$ For PGL_n, det takes values in H^1(k,mu_n)=k*/(k*)^n. But det is surjective and its target admits nontrivial homomorphisms to the circle when k is the p-adics so you've still got your topological character. $\endgroup$
    – Peter McNamara
    Commented May 19, 2016 at 10:15
  • $\begingroup$ @Jason and Peter: Yes sorry, I forgot to compose det with a character; I have changed the statement. $\endgroup$
    – Daniel Loughran
    Commented May 19, 2016 at 10:35
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    $\begingroup$ A version of @YCor's comment: Let $G$ be a simply connected absolutely simple $k$-group, where $k$ is a nonarchimedean local field (a $p$-adic field or the field of rational functions in one variable over a finite field). Assume that $G$ is isotropic (i.e., not isomorphic to $\mathrm{SL}(1,D)$ of a central division algebra $D$ over $k$). Then any nontrivial normal subgroup of $G(k)$ is central, hence finite. For a proof see the book by Platonov and Rapinchuk. Therefore, $G(k)$ admits no nontrivial homomorphisms into abelian groups. $\endgroup$
    – Mikhail Borovoi
    Commented May 19, 2016 at 11:16

1 Answer 1

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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 $S^1$.

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  • $\begingroup$ This example seems to contradict the Corollary to Theorem 21 in Riehm "The norm 1 group of a $p$-adic division algebra", which implies that the abelianization of $\mathrm{SL}(1,D)$ has order $q+1$. Am I missing something? $\endgroup$
    – Arkandias
    Commented Nov 15, 2018 at 16:02
  • $\begingroup$ @Arkandias: In Theorem 21 Riehm assumes that $D$ is not a dyadic quaternion algebra. $\endgroup$
    – Mikhail Borovoi
    Commented Nov 15, 2018 at 18:58
  • $\begingroup$ Yes, but the Corollary includes that case (see just above the statement, and most of the proof is about the dyadic quaternion algebra case). $\endgroup$
    – Arkandias
    Commented Nov 15, 2018 at 21:36
  • $\begingroup$ @Arkandias: Could you please explain how the corollary to Theorem 21 implies that the abelianization of ${\rm SL}(1,D)$ has order $q+1$? $\endgroup$
    – Mikhail Borovoi
    Commented Nov 16, 2018 at 13:44
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    $\begingroup$ @Arkandias: The maps $G(k)\to\mathbb{Z}_2/2\mathbb{Z}_2$ that I constructed is not a homomorphism! I will edit my answer tomorrow. $\endgroup$
    – Mikhail Borovoi
    Commented Nov 16, 2018 at 20:00

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