# Unipotent action of pro-$p$-group

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?

## 1 Answer

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.

• 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