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$\DeclareMathOperator\PSL{PSL}$ Let $ \mathbb{Z} $ be the ring of integers and $ \mathbb{R} $ the field of real numbers. Let $ \Sigma_g $ be a surface of genus $ g \geq 2 $. Let $ \pi_1(\Sigma_g) $ be the fundamental group of the surface. There are many way to embed $ \pi_1(\Sigma_g) $ into $\PSL_2(\mathbb{R}) $. There are, however, no ways to embed $ \pi_1(\Sigma_g) $ into $ \PSL_2(\mathbb{Z}) $. Given some $ g \geq 2 $, is there a good way (an algorithm) to find a real algebraic integer $ \alpha $ such that $ \pi_1(\Sigma_g) $ embeds in $ \PSL_2(R) $? Here $ R $ is the ring $$ R:=\mathbb{Z}[\alpha] $$ (Preferably $ \alpha $ is just a (real) quadratic extension. That is, $ \alpha $ is the root of some polynomial $ x^2+bx+c $ where $ b^2-4c \geq 0 $ and $ b,c \in \mathbb{Z} $.)

History of the question: The original question claimed that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ and asked for embeddings into $ \PSL_2(\mathbb{F}) $ where $ \mathbb{F} $ is a finite degree field extension of $ \mathbb{Q} $. The claim that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ is false. In fact there are many such embeddings. The first edit of the question fixed this and asked instead for an embedding into $ \PSL_2(R) $ for $ R $ a finite rank extension of $ \mathbb{Z} $ by algebraic integers. The current version of the question is the second edit.

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  • $\begingroup$ A solution for $g=2$ gives a solution for every other $g$ by taking covering spaces. $\endgroup$
    – Will Sawin
    Dec 11, 2021 at 19:22
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    $\begingroup$ How do you prove the claim about PSL(2,Q)? $\endgroup$ Dec 11, 2021 at 20:36
  • $\begingroup$ BTW, degree 2 fields are easy to find, but my guess is that one can also do this rationally. $\endgroup$ Dec 11, 2021 at 23:11
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    $\begingroup$ 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). $\endgroup$ Dec 13, 2021 at 7:46
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    $\begingroup$ 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. $\endgroup$ Dec 13, 2021 at 7:52

2 Answers 2

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For the new question an answer is given by arithmetic Fuchsian groups. For example it is well-known that the reflection group associated with the regular right-angled pentagon in $\mathbb H^2$ contains every surface group as it contains the fundamental group of the non-orientable surface of Euler characteristic -1 as an index-4 subgroup. On the other hand it is an index-10 subgroup in the (2,4,5)-triangle group $\Delta$ (by dividing the pentagon into triangles from the center with vertices on the pentagon's and on the middles of its edges). The latter is known to be arithmetic by a result of Takeuchi (see for example item 6 in the table in section 13.3 of Maclachlan--Reid https://zbmath.org/?q=an%3A1025.57001). Its trace field is $\mathbb Q(\sqrt 5)$, which means that $\Delta$ will be realisable as a subgroup of $\mathrm{PSL}_2$ over a quadratic extension $F$ of $\mathbb Q(\sqrt 5)$, and contained in the ring of integers $\mathbb Z_F$. There are many choices for such an extension, for example one can take $F = \mathbb Q(\sqrt 2, \sqrt 5)$ since the quaternion algebra ramifies only at primes dividing 2. So in principle you get an embedding of $\Delta$, and hence of any surface group, into $\mathrm{PSL}_2(\mathbb Z_F)$. You can ask Sage to compute an integral basis for the latter to get a complete answer to your question.

Another possibility to avoid the arithmetic machinery would be to use hyperbolic geometry to compute generators for the triangle group directly and see where they lie, this has probably been done but I don't know a reference (and I'm not sure how to choose the walls as to get something which would lie where it should, i.e. $\mathbb Z_F$).

This may be not optimal, maybe all surface groups embed into $\mathrm{PSL}_2$ over a real quadratic field. For example the (2, 4, 6)-triangle group is arithmetic with trace field $\mathbb Q$, it embeds into $\mathrm{PSL}_2(\mathbb Z[\sqrt 2])$ if I'm not mistaken (see the table in Maclachlan--Reid) so its surface subgroups do as well. However I don't know which genera will be realised this way.


EDIT For the new question a positive general answer is given by computations of John Voight which he recorded in https://arxiv.org/pdf/0802.0911.pdf. The paper gives a complete list of all "Shimura curves" (a particular family of hyperbolic 2-orbifolds) whose underlying surface has genus at most 2. In particular his table 4.1 indicates that the curve associated to the full unit group of the unique maximal order in the $\mathbb Q$-quaternion algebra of discriminant 26 has genus 2 and no singularities, so the image of this group in $\mathrm{PSL}_2(\mathbb R)$ is isomorphic to the surface group of genus 2 and it is contained in $\mathrm{PSL}_2(\mathbb Z[\sqrt 26])$.

This representation should be computable explicitely using software developed by Voight and others, that i'm not too familiar with (i don't think it's available in Sage currently).

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  • $\begingroup$ Yes, all surface groups embed over real quadratic fields, bit the proof is different from what you wrote. However, what's wrong with quoting Takeuchi and getting a faithful representation over Q? $\endgroup$ Dec 14, 2021 at 13:36
  • $\begingroup$ oh wow that comment about being able to embed some surface groups in $ PSL(\mathbb{Z}[\sqrt{2}]) $ has really piqued my curiosity. Now I'm changing the question to ask for embedding into a quadratic extension of $ \mathbb{Z} $. Obviously your answer is super awesome and I'll accept it if no one wants to talk about quadratic extensions of the integers. But I want to leave the (new edited) question open a little longer to see if I can find out more knowledge (and I upvoted you of course!) (and I think I might get that Maclachlan-Reid book now for the holidays!) $\endgroup$ Dec 14, 2021 at 13:41
  • $\begingroup$ @Moishe Kohan : the point is to get an embedding into the integral points $PSL_2(\mathbb Z_F)$, this is not possible for $F=\mathbb Q$ since all subgroups of $PSL_2(\mathbb Z)$ are virtually free. $\endgroup$ Dec 14, 2021 at 16:26
  • $\begingroup$ @MoisheKohan: Your comment "all surface groups embed over real quadratic fields, bit the proof is different from what you wrote" comes across as somewhat disrespectful. This answer gives a proof. You may prefer a different proof. Neither is "the" proof. $\endgroup$
    – HJRW
    Dec 14, 2021 at 16:29
  • $\begingroup$ @JeanRaimbault: Oh, I see, this is different from the original question, where OP asked only about an embedding in $PSL(2,F)$, $F$ a number field. $\endgroup$ Dec 14, 2021 at 16:31
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Here is, at Moishe Kohan's request, an optimal answer to the original question (which asked for a representation over any number field; as i noted in the comments there exists plenty of surface group representations with rational coefficients, as follows from a theorem of Takeuchi).


To prove the existence of single a $\mathrm{PSL}_2(\mathbb Q)$-representation of a surface group is quite simpler than the full proof of Takeuchi's theorem (which is not super hard itself in the torsion-free case). Of course it suffices to prove it in genus 2. To do so observe first that it is possible to find two matrices $A_1, B_1 \in \mathrm{SL}_2(\mathbb Q)$ such that they generate a discrete subgroup of $\mathrm{SL}_2(\mathbb R)$ and

$$ A_1B_1A_1^{-1}B_1^{-1} = \left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) $$

for some $a \in \mathbb Q \setminus \{0, \pm 1\}$ and the quotient $\langle A_1, B_1 \rangle \backslash \mathbb H^2$ is a one-holed torus (the set of matrices $A, B$ satisfying the last condition is an open set, so rational matrices are dense there, and we can always conjugate to diagonalise the commutator over $\mathbb Q$). Geometrically, in the half-plane model, this means that a fundamental domain for the action of $\langle A_1, B_1 \rangle$ on the convex hull of its limit set is a polygon with one side on the geodesic $c$ the geodesic from 0 to $\infty$ and the adjacent sides are orthogonal to $c$; the boundary component of the quotient torus (the convex core of $\langle A_1, B_1 \rangle \backslash \mathbb H^2$, let us call it $T$) is the image of $c$.

Now let $\sigma$ be the reflexion in $c$, and $A_2, B_2$ be the conjugates of $A_1, B_1$ by $\sigma$. Then $\langle A_2, B_2 \rangle \subset \mathrm{PSL}_2(\mathbb Q)$, and the quotient of $\mathbb H^2$ by $\Gamma = \langle A_1, B_1, A_2, B_2 \rangle$ is the double of $T$, that is a genus 2 surface. Hence $\Gamma$ is a genus 2 surface group inside $\mathrm{PSL}_2(\mathbb Q)$ which is discrete in $\mathrm{PSL}_2(\mathbb R)$.


Takeuchi's argument for the full theorem is similar to this, except using Calabi--Weil rigidity instead of the geometric argument, and being more precise about commutators. It is also possible to prove it using Fenchel--Nielsen coordinates.

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  • $\begingroup$ Could you explain why having a one-holed torus quotient is an open condition? $\endgroup$
    – HJRW
    Dec 15, 2021 at 19:19
  • $\begingroup$ Oh, I see it now — it’s equivalent to the endpoints of the axes on the boundary mutually separating. $\endgroup$
    – HJRW
    Dec 15, 2021 at 19:23

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