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Say $p$ and $\ell$ are distinct prime numbers.

Let $G$ be a pro-$p$-group which acts continuously on a finite-dimensional $\mathbb{Q}_\ell$-vector space $V$.

Assume that the action of $G$ on $V$ is unipotent, i.e. $\exists n$ such that $(\sigma - 1)^n = 0$ for all $\sigma \in G$.

Does it follow that the action of $G$ on $V$ is trivial?

Alexander Gruber
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Evariste
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1 Answers1

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Consider $f:G\to H=\text{GL}_n(\mathbf{Q}_\ell)$ continuous. Then $f(G)$ is a pro-$p$-group (as a quotient of $G$). On the other hand, $H$ has an open subgroup $U$ which is pro-$\ell$, namely some open finite index subgroup of $\text{GL}_n(\mathbf{Z}_\ell)$ (if I remember correctly, the kernel of reduction modulo $\ell$ works if $\ell>2$, and of reduction modulo 4 if $\ell=1$). It follows that $f(G)\cap U$ is both pro-$p$ and pro-$\ell$, hence trivial, so $f(G)$ is discrete, hence finite.

Now assume in addition that the action is unipotent. The unipotent group in $\text{GL}_n(\mathbf{Q}_\ell)$ is torsion-free. So $f(G)$ is torsion-free and finite, hence trivial.

YCor
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  • Just a small remark : it is better to take $U$ as being the kernel of $\text{GL}n(\mathbf{Z}\ell) \to \text{GL}n(\mathbf{F}\ell)$, because it is a pro-$\ell$-group, while $\text{GL}n(\mathbf{Z}\ell)$ doesn't seem to be a pro-$\ell$-group to me. – Watson Sep 26 '18 at 10:15
  • @Watson thanks, you're right (I even think that one has to consider kernel mod 4 when $\ell=2$). – YCor Sep 26 '18 at 10:26