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 $0$.

May such an object be interpret as a discrete version of some topological construction, say, flat connection?

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?


Let $\Sigma$ be your surface.
I'll construct a new surface $\tilde \Sigma$ as follows. Take $\Sigma$ apart into individual triangles, and reglue them with a twist at each edge (the adjacency graphs for the faces of $\Sigma$, and for the faces of $\tilde\Sigma$ are the same). Note that the surfaces $\Sigma$ and $\tilde\Sigma$ could have different genus, and that $\tilde\Sigma$ could be non-orientable.

Your data is equivalent to having a piecewise linear map (not necessarily an embedding) of the universal cover of the new surface $\tilde\Sigma$ into $\mathbb R^n$, so that the vertices of the triangulation map to $\mathbb Z^n$.
That map is well defined up to an overall translation.

  • $\begingroup$ 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. $\endgroup$ – Daniil Rudenko Jul 8 '12 at 12:16
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    $\begingroup$ 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. $\endgroup$ – André Henriques Jul 8 '12 at 12:20
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    $\begingroup$ 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. $\endgroup$ – Daniil Rudenko Jul 8 '12 at 13:26
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    $\begingroup$ @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 $\mathbb R^n$. $\endgroup$ – André Henriques Jul 8 '12 at 13:36
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    $\begingroup$ 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. $\endgroup$ – Daniil Rudenko Jul 8 '12 at 14:48

This should be a comment.

Since the image is Abelian $\phi$ factors through the universal Abelian cover (much smaller than the universal cover) and your objects are equivalent to a map $\psi$ from the integer first homology of your surface to $Z^n$ along with the data of an equivariant map from the one skeleton of this cover to $R^n$ linear on edges and with vertices going to integer points.
Since this is equivariant this descends to a map $\tau$ from the surface to the torus $R^n$ mod $Z^n$.

In terms of flat connections $\psi$ is the monodromy action of the (Abelianization of) the fundamental group on the fiber ($R^n$).

In the case n=2 one characteristic class interpretation of your associated 2-cocycle class is as the pullback of the fundamental class of the torus $R^2$ mod $Z^2$ (classifying space of $Z$ x $Z$) via $\tau$.


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