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?

  • $\begingroup$ 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 May 19 '16 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 May 19 '16 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 May 19 '16 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 May 19 '16 at 10:35
  • 2
    $\begingroup$ 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 May 19 '16 at 11:16

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 $S^1$.

  • $\begingroup$ 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 Nov 15 '18 at 16:02
  • $\begingroup$ @Arkandias: In Theorem 21 Riehm assumes that $D$ is not a dyadic quaternion algebra. $\endgroup$ – Mikhail Borovoi Nov 15 '18 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 Nov 15 '18 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 Nov 16 '18 at 13:44
  • 1
    $\begingroup$ @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 Nov 16 '18 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.