Is there any elementary (i.e. without using analytical methods like the theory of Riemann surfaces or more elaborate results from differential geometry) way to show that the universal covering of the compact oriented surface of genus $g>0$ is homeomorphic to $\mathbb R^2$?
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asked May 25, 2013 at 18:36
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5$\begingroup$ Yes! For the torus (genus $1$) you just have the quotient of $\Bbb R^2$ by the action of $\Bbb Z^2$ (acting by translations). The quotient is just that of a square, and it is the torus. A slight different action produces the Klein bottle (twisting along a coordinate). For bigger genus there is a similar construction in the hyperbolic plane $\Bbb H^2$. You have to choose a regular geodesic polygon with the necessary number of edges ($4g$ edges for genus $g$). There is a discrete group of isometries of $\Bbb H^2$ which acts by identifying the edges in the proper way to obtain the surface. $\endgroup$– Daniele ZuddasCommented May 25, 2013 at 19:08
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2$\begingroup$ This looks more like teaching than like research. $\endgroup$– Fernando MuroCommented May 25, 2013 at 22:15
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1$\begingroup$ Voted to close because the answer is immediate from the classification of surfaces (as Daniele explains). $\endgroup$– Igor BelegradekCommented May 26, 2013 at 2:48
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$\begingroup$ Daniele's construction is not so much an application of the classification of surfaces, and more an application of the Poincare polygon theorem, in order to obtain the universal covering map with domain $\mathbb{H}^2$ using the gluing pattern for the specific surface asked for by the OP. $\endgroup$– Lee MosherCommented May 26, 2013 at 13:00
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1$\begingroup$ Lee Mosher: this depends on how you define a compact surface of genus $g$. If the definition is "connected sum of $g$ tori" no classification is needed. If the definition is "compact oriented 2-manifold of Euler characteristic $2-2g$, then it is needed. I think, the polygon argument is easy, while the classification of surfaces is less so. $\endgroup$– Igor BelegradekCommented May 27, 2013 at 0:21
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1 Answer
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You can build a certain covering space of the surface $S$ rather explicitly as a nested union of closed discs $D_1 \subset D_2 \subset D_3 \subset \cdots$, each contained in the interior of the next, from which it follows that the union is $\mathbb{R}^2$ and, being simply connected, is therefore the universal covering space. To construct these discs, start by representing your surface in the usual fashion as the quotient of a $4g$-gon, with oriented 1-cells assigned labels, and with side pairings respecting orientations and labellings. The disc $D_1$ is a single copy of the $4g$-gon $Q$ with appropriate side labels, and the map $D_1 \to S$ is the one given by the quotient map. Then $D_2$ is constructed by attaching additional copies of $Q$ to the periphery of $D_2$, exactly as they ought to be: one copy of $Q$ attached to each edge of $D_1$, and additional copies of $Q$ attached to each vertex so as to fill out $4g$ copies of $Q$ attached around each vertex of $D_1$. Et cetera.
answered May 25, 2013 at 18:57