7
$\begingroup$

Let $\Gamma$ be a lattice in $SL(2,\mathbb{R})$. Suppose that the trace field of $\Gamma$ is a totally real number field of degree $d$. This gives $d$ homomorphisms $\rho_i:\Gamma\to SL(2,\mathbb{R})$ (here $i=1,\ldots, d$), which act as Galois conjugation on traces. Here $\rho_1$ denotes the identity homomorphism.

Question 1: Can any of the $\rho(\Gamma_i)$ be discrete when $i\neq 1$?

Question 2: Can the group $\{(\rho_2(g), \rho_3(g), \ldots, \rho_d(g):g\in \Gamma\}$ ever be discrete in $SL(2,\mathbb{R})^{d-1}$?

A positive answer to the question 1 of course also gives a positive answer to question 2. I am most interested in the case when $\Gamma$ is not cocompact, and I would even be happy to know the answer when $\Gamma$ is a triangle group.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ In addition to Ian's answer: In the case of a triangle group the representation will never be discrete. This follows from rigidity of such groups. $\endgroup$
    – Misha
    Commented Jan 27, 2014 at 6:42
  • $\begingroup$ Ian: yes, I want to consider the invariant trace field. (Actually I had in mind a class of groups for which the two trace fields are the same.) $\endgroup$
    – Alex
    Commented Jan 27, 2014 at 15:35
  • $\begingroup$ Misha: Could you elaborate? What is the rigidity of triangle groups? $\endgroup$
    – Alex
    Commented Jan 27, 2014 at 15:35
  • $\begingroup$ I see. In the triangle group case each $\rho_i$ is a faithful representation of the triangle group to $SL(2,\mathbb{R})$ (which even sends parabolics to parabolics), and so if the image were discrete it would have to be the triangle group itself and the map would have to be as expected. This would be a contradiction if $i\neq1$. Correct? So this implies "No to question 1" for triangle groups. Does anyone have anything to say about question 2 for triangle groups? $\endgroup$
    – Alex
    Commented Jan 27, 2014 at 16:12
  • 1
    $\begingroup$ @IanAgol: Yes, I think that's right, although proving it would require some work. I think, the better question is to skip the relatively compact Galois conjugates in addition to $\rho_1$. $\endgroup$
    – Misha
    Commented Jan 28, 2014 at 8:36

1 Answer 1

Reset to default
7
$\begingroup$

You probably want to consider the invariant trace field $k$ instead of trace field $K$. The trace field $K$ is a multi-quadratic extension of $k$, and $Gal(K/k)$ acts trivially on the $PSL(2,\mathbb{R})$ representations of $\Gamma$ (it acts non-trivially on $\Gamma < SL(2,\mathbb{R})$, but with the same image in $PSL(2,\mathbb{R})$). For example, the lattice associated to the modular curve $X_0^+(N)$ will have trace field $\mathbb{Q}(\sqrt{N})$. This is an extension of the lattice $\Gamma_0(N)$ (which has integral trace) by the Fricke involution. The Galois automorphism sends the involution to its negative in $SL_2(\mathbb{R})$, but has the same image in $PSL(2,\mathbb{R})$, and therefore the same action on $\mathbb{H}^2$.

The answer to your question is yes for punctured torus groups. This follows from results of Brian Bowditch. A representation of the punctured torus group in $SL(2,\mathbb{R})$ corresponds to a non-zero real solution of the Markoff equation. In Proposition 4.11, it is shown that conversely a non-zero solution gives a discrete fuchsian representation. So if the trace field is totally real, all non-trivial Galois conjugates will be discrete fuchsian groups.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .