most recent 30 from http://mathoverflow.net 2013-05-23T12:31:13Z http://mathoverflow.net/feeds/user/21620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101634/discrete-version-of-some-topological-object Daniil Rudenko 2012-07-08T10:09:24Z 2012-07-09T00:06:05Z

<p>Consider a triangulated orientable surface with the following data: on each edge a vector with integer coordinates is written so that for each triangle the sum of the vectors corresponding to three edges of its boundary is $0$.</p> <p>May such an object be interpret as a discrete version of some topological construction, say, flat connection? </p> <p>Since for each triangle sum of the vectors corresponding to its boundary is $0$, it is possible to assign to each triangle its "area"- the determinant of the matrix formed by any two vectors of the boundary of the triangle. I also wonder, if there exists some interpretation of determinants of these triangles. Maybe, the 2-cochain formed by them is representing some characteristic class?</p>

http://mathoverflow.net/questions/89180/triangulations-of-lattice-polygons Daniil Rudenko 2012-02-22T12:32:17Z 2012-04-19T16:28:29Z

<p>Let L be a 2-dimensional lattice and P- a lattice polygon. Suppose, it is triangulated into lattice tiangles. What are restrictions on their areas? For instance, can a lattice triangle of even area always be divided into lattice triangles of area 1? Is there any general approach to such questions? </p>

http://mathoverflow.net/questions/101634/discrete-version-of-some-topological-object/101637#101637 Daniil Rudenko 2012-07-08T14:48:29Z 2012-07-08T14:48:29Z

Your answer lead me to the answer on the first half of my question: for each representation of fundamental group there exists a corresponding flat connection. In our case the structure group would be $\mathbb{Z} \oplus \mathbb{Z}$. But what still puzzles me is interpretation of &quot;areas&quot;. It should be some 2-cocycle.

http://mathoverflow.net/questions/101634/discrete-version-of-some-topological-object/101637#101637 Daniil Rudenko 2012-07-08T13:26:13Z 2012-07-08T13:26:13Z

As I understood Your construction, You take any vertex of the surface and send it to zero. Then for each other vertex You consider a path, connecting it with initial one and sum all the vectors along the path. This gives coordinates of the image of the vertex. But it is not true that the sum of the vectors over any loop is 0. It is true for contractable loops only.

http://mathoverflow.net/questions/101634/discrete-version-of-some-topological-object/101637#101637 Daniil Rudenko 2012-07-08T12:16:47Z 2012-07-08T12:16:47Z

Thank you for the answer! Actually, I am specially interested in the case when vectors are of dimension two. In this case it seems to me that if the genus of the surface is not 0, than the map You suggest will not be well defined.

http://mathoverflow.net/questions/89180/triangulations-of-lattice-polygons/89183#89183 Daniil Rudenko 2012-02-22T13:21:32Z 2012-02-22T13:21:32Z

Yes, this observation gave rise to this question.