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.

  • $\begingroup$ "semisimplicity of $u$ is that $u$ is diagonalizable". Are you sure that's the definition? (It makes a proof easier, but:) Since $\mathbb Q_l$ is not algebraically closed, a general vector space $V$ over it will have endomorphisms $f$ which are semisimple (which for me means, $V$ is semisimple as $\mathbb Q_l[f]$-module) without $f$ being diagonalisable (over $\mathbb Q_l$). In fact, every proper field extension of $\mathbb Q_l$ will give examples: take, as $f$, multiplication with any element not in the ground field. $\endgroup$ Aug 13, 2022 at 15:38
  • $\begingroup$ @TorstenSchoeneberg Thanks for pointing out! Yes, the definition should be changed as what you've said. Actually, semisimple is equivalent to "diagonalizable over the algebraic closure of $\mathbb Q_l$", rather than over $\mathbb Q_l$. $\endgroup$
    – Richard
    Aug 14, 2022 at 14:22

1 Answer 1


Make the following two ideas rigorous:

  1. If $u$ is semisimple, then every element in the image of the representation is semisimple. (Via topological generation.)

  2. 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").

  • 1
    $\begingroup$ Elaborating on 1: We know $\rho(\widehat{\mathbb Z})$ is the closure of $\rho(\mathbb Z)$ [this needs a proof]. But being semisimple is a closed condition, so if $\rho(\mathbb Z)$ consists of semisimple elements, so does $\rho(\widehat{\mathbb Z})$. $\endgroup$
    – Kenta S
    Aug 14, 2022 at 8:03

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