4
$\begingroup$

Let $F$ be a finite extension of $\mathbb{Q}_p$ and let $\rho\colon G_F \to GL_n(\mathbb{C})$ be a continuous representation, where $G_F$ is the Galois group of $\bar{F}$ over $F$. Then $\rho$ is in fact continuous for the discrete topology on $GL_n(\mathbb{C})$.

I know how to prove this in a somewhat clumsy way which uses that $G_F$ is profinite. Is there a succinct proof that only uses that $G_F$ is totally disconnected? Does this statement hold for any totally disconnected group?

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ It doesn't hold for arbitrary totally disconnected groups. For example, take the exponential map ℚ→GL_1(ℂ). $\endgroup$
    – Anton Geraschenko
    Commented Nov 11, 2009 at 17:56
  • 5
    $\begingroup$ I think the standard proof goes like this: GL_n(C) has no small subgroups (in the sense that there's an open neighbourhood of the identity with the property that the only subgroup in this neighbourhood is the trivial group) and G_F has a basis of neighbourhoods of the identity consisting of open subgroups (because that's what Galois groups look like). So rho^{-1}(U) is open, for U a no-small-subgroup open set, and hence contains a subgroup H, and so rho(H) is trivial and so on and so on. Note that Q does not have a basis of neighbourhoods of 1 consisting of subgroups. $\endgroup$
    – Kevin Buzzard
    Commented Nov 11, 2009 at 18:53
  • 1
    $\begingroup$ Thanks Kevin. You should leave this as an answer, not a comment! $\endgroup$
    – David Zureick-Brown
    Commented Nov 11, 2009 at 19:17
  • $\begingroup$ It's not an answer to the question, it's an answer to something else ;) It's Anton who answered the question! $\endgroup$
    – Kevin Buzzard
    Commented Nov 11, 2009 at 20:30

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .