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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

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

2 Answers 2

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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.

EDIT: There is some nonsense (due to me) in the comments below. Caveat emptor!

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    $\begingroup$ 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
    Commented Apr 18, 2016 at 0:32
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    $\begingroup$ @SamNead the statement 'square-tiled surfaces are dense in the fiber over $[X]$' implies that every Riemann surface $X$ admits a square-tiled surface structure. I presume this discussion implies that the above is true. Is there an easier/direct way to see that every Riemann surface $X$ admits a square-tiled surface structure? $\endgroup$
    – Mohith Nagaraju
    Commented May 23, 2023 at 13:27
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    $\begingroup$ An interesting corollary of the statement 'every Riemann surface $X$ admits a square-tiled surface structure' is that every Riemann surface $X$ admits a branched cover over the square torus, branched over at most one point of the square torus. $\endgroup$
    – Mohith Nagaraju
    Commented May 23, 2023 at 13:29
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    $\begingroup$ @SamNead Thanks, this makes things clear. Could you provide a good reference or two which covers some of the background more rigorously and talks about these aspects? Nuxil mentions "An introduction to Veech surfaces" by Pascal Hubert and Thomas Schmidt. Is there any other? $\endgroup$
    – Mohith Nagaraju
    Commented May 24, 2023 at 9:35
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    $\begingroup$ I learned a lot from Alex Wright's article "Translation surfaces and their orbit closures: An introduction for a broad audience". On page two he references many more introductions; I've heard good things about the papers by Yoccoz and Zorich (and I am sure that the others are also very good!). $\endgroup$
    – Sam Nead
    Commented May 24, 2023 at 11:47
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Here is a more pictorial proof.

First, any translation surface can be obtained by choosing a suitable polygon and identifying pairs of parallel sides via translations.

Second, a tetris polygon is a polygon whose sides are all equal and are parallel to either the $x$-axis or $y$-axis.

Third, a tetris-like translation surface is a translation surface obtained from a tetris polygon after identifications of pairs of parallel sides via adequate translations.

Note that all tetris-like translation surfaces are square-tiled surfaces.

Figure 1 shows two examples of tetris-like translation surfaces.

Figure 1

Now to show square-tiled surfaces are dense among translation surfaces, it suffice to find tetris-like translation surfaces arbitrarily close to every translation surface.

Say, for e.g., we are given the translation surface $X$ obtained by identifying opposite sides of a diamond (See Figure 2).

Figure 2

Next, Figure 3 shows a tetris-like translation surface that is close $X$.

Figure 3

Now by making the vertical and horizontal sides smaller and bringing the jagged line closer to the diagonal edge of $X$, we get a tetris-like translation surface that is arbitrarily close to $X$.

More generally, any polygon can be approximated by a tetris polygon. Thus, tetris-like translation surfaces are dense in the space of all translation surfaces.

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