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Let $(X, \omega)$ be a translation surface, i.e., a Riemann surface with a homomorphism $1$-form. A central object is the group of affine automorphisms $\text{Aff}^+(X, \omega)$: homeomorphisms of $X$ that preserve the zeros of $\omega$ and are affine in the flat charts of $\omega$. The kernel of the derivative map $D: \text{Aff}^+(X, \omega) \to \text{SL}(2, \mathbb{R})$ is the group of translation automorphisms $\text{Aut}(X, \omega)$, or the holomorphic automorphisms of $X$ that preserve $\omega$.

Questions: Suppose you know that $\text{Aut}(X, \omega)$ or $\text{Aff}^+(X, \omega)$ is trivial. (This is generically the case.) I imagine that $X$ could have automorphisms that do not preserve $\omega$, but how could you detect them for a given translation surface $(X, \omega)$? Can you ever certify that a translation surface $(X, \omega)$ has no automorphisms, affine or otherwise?

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  • $\begingroup$ I suspect there isn't a solution to this problem in the terms you were hoping for. But there are often solutions for generic surfaces in families, using monodromy of the family. The point is that if one can consistently define an automorphism over a family of surfaces, the eigenspaces of this automorphism define flat subbundles. Often knowledge of monodromy can be used to rule this out. This can often by applied to, ex, certain GL(2,R) orbit closures of translation surfaces. $\endgroup$
    – Alex Wright
    Commented Sep 17 at 18:15

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I interpret your question as follows: Suppose that a Riemann surface $X$ is given as a flat surface (say, by gluing together euclidean polygons by local isometries, satisfying a few nice conditions). Then:

How does one prove that $X$ has exactly one conformal automorphism?

One approach would be to uniformise $X$ - that is, find (an approximation of) the unique hyperbolic metric in the conformal class of $X$. Perhaps this is done by numerical methods (as the general problem is highly transcendental). One then finds (say) all points in the hyperbolic metric of maximal injectivity radius. One then builds Voronoi domains about these. One then enumerates the combinatorial isomorphisms of the resulting tiling of $X$. This group contains all conformal automorphisms of $X$.

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    $\begingroup$ Thanks for your response. While I see this will work, I was hoping for a solution that stayed in the flat picture. (This is what I have the ability to compute with). A related equation: can you see from the equation of a Riemann surface, e.g. $y^2 = p(x)$, that it has no conformal automorphisms? $\endgroup$
    – Sam Freedman
    Commented Jun 3, 2023 at 17:55
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    $\begingroup$ Hmm. I assume you mean "other than the identity and the hyperelliptic involution $\tau(x, y) = (x, -y)$". I asked a colleague; she suggested I look in the book Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, and Harris. On page 45, exercise F-3, we find the following: Let $C$ be the curve determined by $y^2 = p(x)$. Suppose that $C$ has genus $g > 1$ and $p$ has $2g + 2$ distinct roots. Then $\tau$ is central in $\mathrm{Aut}(C)$. Also, $\mathrm{Aut}(C) / \langle \tau \rangle$ is isomorphic to the group of automorphisms of $\mathbb{CP}^1$ permuting the roots of $p$. $\endgroup$
    – Sam Nead
    Commented Jun 4, 2023 at 6:14
  • $\begingroup$ Sadly, I don't know how to prove the "hard" direction of the last sentence. This is because I don't know how to make automorphisms of algebraic varieties "concrete". Hyperbolic surfaces are much nicer in this regard! $\endgroup$
    – Sam Nead
    Commented Jun 4, 2023 at 6:52
  • $\begingroup$ Fair enough! The reason I brought in the algebraic language is because I know of some flat surfaces for which we know the explicit equation of the curve and the 1-form on that curve yielding the flat picture. So I was hoping to work backwards. $\endgroup$
    – Sam Freedman
    Commented Jun 4, 2023 at 18:54

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