I'm reading the survey "An introduction to Veech surfaces" by Pascal Hubert and Thomas Schmidt.

At page 19 they state "In any fixed stratum, the set of square-tiled surfaces of that stratum is dense.". The reason should be that in the coordinates for the moduli space of translation surfaces given by the period map (integration of the 1-form on relative periods) the coordinates of the translation surfaces are exactly $\mathbb{Q}+i\mathbb{Q}$. So the assertion follows because of the density of $\mathbb{Q}$ in $\mathbb{R}$.

Well, it's hard for me to believe in Hubert and Schmidt's assertion.

Isn't it true that the square tiled surfaces are the "integer points" of the moduli space of translation surfaces? If I have a translation surface tiled by squares with the side of length one, isn't it true that the relative periods are contained in $\mathbb{Z}+i\mathbb{Z}$?

So my questions are:

Is Hubert and Schmidt's assertion wrong?

If so, which are the translation surfaces corresponding to rational points?

Thank you

  • 2
    $\begingroup$ Please give some more background. Is it clear that squares are supposed to have side length $1$? Which geometric structure is considered - conformal? $\endgroup$ Apr 15, 2016 at 18:49
  • $\begingroup$ they don't specify it $\endgroup$
    – Nuxil
    Apr 16, 2016 at 6:53

1 Answer 1


Is Hubert and Schmidt's assertion wrong?

No, they are correct. They allow squares where the sidelength is not equal to one.

If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other. Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.

  • $\begingroup$ I see... But isn't this true only in the space of translation surfaces $\mathcal{T}:=\{(X,\omega)\}$? $(X,\omega)$ and $(X,r\omega)$ are isomorphic as translation surfaces, so in the moduli space of translation surfaces $\mathcal{T}/Diff^+$ the rational multiples should be all the same point $\endgroup$
    – Nuxil
    Apr 16, 2016 at 8:30
  • $\begingroup$ Ok - I am not using language the way you want me to - for that I apologize. So I will back up a bit. Suppose that $S$ is a topological surface (closed, connected, oriented). Let $\hat{Q}(S)$ be the space of all pairs $(X, \omega)$ where (i) $X$ is a Riemann surface marked by $S$ and (ii) $\omega$ is a one-form. We form a quotient $Q(S)$ by taking $(X, \omega)$ to be equivalent to $(X', \omega')$ if there is a biholomorphic map $h {:}\, X \to X'$ that (a) commutes (up to isotopy) with the markings and (b) pulls $\omega'$ back to $\omega$. Now... $\endgroup$
    – Sam Nead
    Apr 16, 2016 at 13:48
  • $\begingroup$ there is a forgetful map from $Q(S)$ to Teichmuller space $T(S)$ obtained by forgetting $\omega$. Set $Q(X)$ to be the fiber of this map over the point $[X] \in T(S)$. I claim that $Q(X)$ is naturally homeomorphic to the vector space of one-forms on $X$. Square-tiled surfaces (where I allow scaling by a real number) are dense in that vector space, and thus in $Q(X)$, and thus in $Q(S)$. $\endgroup$
    – Sam Nead
    Apr 16, 2016 at 13:51
  • $\begingroup$ Thank you! I think I got it now: it works because the scaling won't be isotopic to the identity? $\endgroup$
    – Nuxil
    Apr 16, 2016 at 16:08
  • 1
    $\begingroup$ No. It works because scaling is not relevant. You can't "really" tell the difference between $\omega$ and $r\omega$. In a similar fashion - the union of the lines in $\mathbb{R}^2$ (through the origin and of rational slope) are dense. $\endgroup$
    – Sam Nead
    Apr 18, 2016 at 0:32

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