Consider $1\in \mathbb Z \subset \widehat {\mathbb Z}$(it's the topological generator of $\widehat {\mathbb Z}$). Let $l$ be a prime. Suppose there is a continuous representation $\rho:\widehat {\mathbb Z}\to \operatorname{Aut}_{\mathbb Q_l}(V)$ for some finite-dimensional vector space $V$ over $\mathbb Q_l$.
Denote $u=\rho(1)$, then I need to prove:
$\rho$ is semisimple $\Leftrightarrow$$u$ is semisimple.
I know the semisimplicity of $\rho$ is equivalent to: $V$ is a direct sum of irreducible representations, and the semisimplicity of $u$ is that $u$ is diagonalizable.
But how can we connect these two properties? Could you give some hints for me? Thanks.
Make the following two ideas rigorous:
If $u$ is semisimple, then every element in the image of the representation is semisimple. (Via topological generation.)
If two semisimple endomorphisms of a finite-dimensional vector space $V$ over a perfect field commute, then there is a decomposition of $V$ into subspaces which are irreducible simultaneously for both endomorphisms. Cf. e.g. Simultaneous semisimplicity of commuting endomorphisms, A family of commuting endomorphisms is semisimple if each element is semisimple, or Bourbaki Algebra VII §5 no. 8. --- And the group we are letting act here is abelian.
Note that if in 2 you replace "semisimple" with the stronger "diagonalisable" (see comment), then all those irreducible subspaces are one-dimensional (and the catchphrase is "commuting diagonalisable matrices are simultaneously diagonalisable").