Timeline for Are square tiled surfaces dense in the moduli space of translation surfaces?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2023 at 21:00 | comment | added | Sam Nead | Looks good to me! I have been taught to say “differentials with rational coordinates are square-tiled and also are obviously dense”. Your proof is a nice way to get rid of that unpleasant word ‘obviously’. | |
| Jun 6, 2023 at 17:32 | comment | added | Mohith Nagaraju | @SamNead, Could you please check whether the below pictorial proof is valid? | |
| May 24, 2023 at 11:47 | comment | added | Sam Nead | 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!). | |
| May 24, 2023 at 9:35 | comment | added | Mohith Nagaraju | @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? | |
| May 23, 2023 at 19:50 | history | edited | Sam Nead | CC BY-SA 4.0 |
Added warning that some of my comments are wrong.
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| May 23, 2023 at 19:23 | comment | added | Sam Nead | And these annoying Riemann surfaces $X$, which are never square-tiled... well, their fibres $Q(X)$ are sort of like curves in the plane which don't go through any rational points. Like, say, the line $y = x + \pi$ in $\mathbb{R}^2$. | |
| May 23, 2023 at 19:22 | comment | added | Sam Nead | @MohithRaju - I deduce that square-tiled need not be dense in $Q(X)$. But perhaps that is ok. We don't need square-tiled surfaces in every fibre. We just need them close to every quadratic differential. That is still true, because it is true in period coordinates. | |
| May 23, 2023 at 19:17 | comment | added | Sam Nead | Well, crud. I have definitely made a mistake. There is no way your second consequence holds. (Reason - there are only countably many topological branched-over-one-point covers of the torus. Each of these gives a curve (complex dimension one) in moduli space. But in higher genus moduli space has dimension at least two - it is not a countable union of curves.) | |
| May 23, 2023 at 13:29 | comment | added | Mohith Nagaraju | 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. | |
| May 23, 2023 at 13:27 | comment | added | Mohith Nagaraju | @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? | |
| Apr 18, 2016 at 18:42 | vote | accept | Nuxil | ||
| Apr 18, 2016 at 0:32 | comment | added | Sam Nead | 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. | |
| Apr 16, 2016 at 16:08 | comment | added | Nuxil | Thank you! I think I got it now: it works because the scaling won't be isotopic to the identity? | |
| Apr 16, 2016 at 13:51 | comment | added | Sam Nead | 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 $Q(S)$. | |
| Apr 16, 2016 at 13:48 | comment | added | Sam Nead | 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... | |
| Apr 16, 2016 at 8:30 | comment | added | Nuxil | 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 | |
| Apr 15, 2016 at 22:44 | history | answered | Sam Nead | CC BY-SA 3.0 |