Timeline for Discrete version of some topological object.
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2012 at 14:48 | comment | added | Daniil Rudenko | 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 "areas". It should be some 2-cocycle. | |
| Jul 8, 2012 at 13:37 | history | edited | André Henriques | CC BY-SA 3.0 |
fixed mistake
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| Jul 8, 2012 at 13:36 | comment | added | André Henriques | @Daniil. You're perfectly right: you caught a mistake. You get a homomorphism of $\phi:\pi_1(\tilde \Sigma)\to \mathbb Z^n$ and a piecewise linear map from the universal cover of $\tilde \Sigma$ to $\mathbb R^n$ such that if two points of the universal cover differ by the action of some element $g$ of $\pi_1(\tilde \Sigma)$, then their images differ by $\phi(g)$ in $\mathbb R^n$. | |
| Jul 8, 2012 at 13:26 | comment | added | Daniil Rudenko | 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. | |
| Jul 8, 2012 at 12:20 | comment | added | André Henriques | The map is well-defined, independently of the dimension, and it has nothing to do with the genus of the surface. The map might simply fail to be injective. | |
| Jul 8, 2012 at 12:16 | comment | added | Daniil Rudenko | 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. | |
| Jul 8, 2012 at 12:12 | vote | accept | Daniil Rudenko | ||
| Jul 8, 2012 at 10:31 | history | edited | André Henriques | CC BY-SA 3.0 |
added 400 characters in body
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| Jul 8, 2012 at 10:24 | history | answered | André Henriques | CC BY-SA 3.0 |