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Dec 18, 2021 at 23:32 vote accept Ian Gershon Teixeira
Dec 15, 2021 at 19:04 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 15, 2021 at 8:22 answer added Jean Raimbault timeline score: 4
Dec 14, 2021 at 13:33 history edited Ian Gershon Teixeira CC BY-SA 4.0
added 26 characters in body
Dec 14, 2021 at 13:14 history edited Ian Gershon Teixeira CC BY-SA 4.0
edited title
Dec 14, 2021 at 7:33 answer added Jean Raimbault timeline score: 4
Dec 14, 2021 at 4:36 comment added Ian Gershon Teixeira @JeanRaimbault My mistake the question contains an incorrect assertion as asked. I'll change it from $ \mathbb{Q} $ to $ \mathbb{Z} $
Dec 14, 2021 at 4:26 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 13, 2021 at 7:52 comment added Jean Raimbault if I remember it correctly Takeuchi's proof is essentially algorithmic ; you take any representation, then you find rational matrices which are close to the images of the generators and satisfy the surface group relation, and you conclude by applying the Calabi--Weil rigidity theorem which tells you that this gives a faithful discrete representation.
Dec 13, 2021 at 7:46 comment added Jean Raimbault Indeed representations with $\mathbb Q$-coefficients are dense in the set of $\mathrm{PSL}_2(\mathbb R)$-representations for a surface group, this is due to Takeuchi (zbmath.org/?q=an%3A0204.39801) ; a generalisation to Fuchsian groups with torsion is given by Maclachlan and Waterman (zbmath.org/?q=an%3A0552.20028).
Dec 11, 2021 at 23:11 comment added Moishe Kohan BTW, degree 2 fields are easy to find, but my guess is that one can also do this rationally.
Dec 11, 2021 at 20:36 comment added Moishe Kohan How do you prove the claim about PSL(2,Q)?
Dec 11, 2021 at 19:22 comment added Will Sawin A solution for $g=2$ gives a solution for every other $g$ by taking covering spaces.
Dec 11, 2021 at 19:01 history edited LSpice CC BY-SA 4.0
a -> an
Dec 11, 2021 at 15:35 history edited YCor CC BY-SA 4.0
formatting
Dec 11, 2021 at 15:11 history asked Ian Gershon Teixeira CC BY-SA 4.0