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Let $S_g$ be a compact topological surface of genus $g$. I know there is the correspondence

$\{$Abelian differentials on compact Riemann surfaces of genus g$\}\leftrightarrow\{$ Translation surfaces on $S_g\}$

Given a collection of $n$ vectors $v_1,\dots,v_n$ in $\mathbb{R}^2$ there is a natural way to construct a polygon and this polygon will represent naturally a flat surface with trivial linear holonomy (of course if I want it to be a translation surface on $S_g$ I must check that the condition on the genus by Gauss-Bonnet theorem). Then one vector $v_j$ (which is the side of the polygon between the vertices $p_j$ and $p_{j+1}$) will be the integral $\int_{p_j}^{p_{j+1}}dz=\int_{\rho_j}\omega$ where $\rho_j\in H_1(S_g,\{p_1,\dots,p_k\},\mathbb{C})$, $\{p_1,\dots,p_k\}$ singular points of $\omega$.

My question is: how do I do the opposite construction, that is how do I obtain a polygon from a couple $(X,\omega)$?

It is my understanding that it all depends in the choice of a base $\{\rho_j\}_j$ of $H_1(S_g,\{p_1,\dots,p_k\},\mathbb{C})$, then the vectors which are the sides of the polygon will be the integrals $\int_{\rho_j}\omega$. But then my question becomes: How do I determine the base $\{\rho_j\}_j$? If it is always possible to obtain it, is it uniquely determined?

Thank you very much

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  • $\begingroup$ i think this might be possible for all but a finite number of possibilities. would this be acceptable? $\endgroup$ – JMP Jul 4 '15 at 6:15
  • $\begingroup$ as i understand it, any polygon must enclose all $\omega$-forms, and so is unique up to deformation. is this not therefore a 'smallest symmetric shape that contains the convex hull' problem? $\endgroup$ – JMP Jul 5 '15 at 6:43
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How do I determine the base $\{\rho_j\}_j$?

There are (infinitely) many possibilities for the base. You will need to review the notion of "natural coordinates" for an abelian differential to see why there is at least one base (see, for example, page 25 of "Flat surfaces" by Zorich). Once you have one base, you can use that to construct a polygon $P$ of the kind you want. Now, add arcs to the base until you have cut $P$ into a collection of triangles. (This larger set will not be linearly independent in $H_1$, but that does not matter.) Now that you have a triangulation, you can "flip" edges of the triangulation to get new triangulations. After doing enough flips, and taking a subset of the edges, you can arrive at a new base, totally different from the old one.

Exercise - do this for the square torus.

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