For simplicity, consider simplicial homology in $\mathbb{R}^2$. It seems to me that oriented area is a cocycle, since it vanishes on simplicial cycles. The situation being Euclidean, it must consequently be a coboundary $\delta \phi$. What is this $\phi$?
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First you presumably mean simplicial cohomology? Second what do you mean when you say that oriented area is a cocycle? This is something that makes more sense in the DeRham complex but to me seems strange to say in this situation. Could you describe the map of 2-simplices that you are calling oriented area? – Sempliner Aug 21 '15 at 23:38
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This is strange to me too, so I'm trying to make sense of this. I do mean simplicial cohomology. The map I have in mind takes simplex $[v_0,v_1,v_2]$ to the determinant $(1/2)|v_1-v_0,v_2-v_0|$. This is just ordinary area for, say, clockwise oriented simplices, and $(-1)$ times the area for counterclockwise oriented simplices. – Bob Johns Aug 21 '15 at 23:44
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As Sempliner says, it's a good idea to think of this question in De Rham language.
Oriented area is just the differential $2$-form $dx\wedge dy$. Namely, it is given by$$S\mapsto\int_Sdx\wedge dy,$$where $S$ is a (regular enough) $2$-simplex. Yes, this $2$-form is closed and hence exact. As always, there are many primitives. For example, we have$$dx\wedge dy=d(ydx)=d(-xdy).$$So, the $\phi$ you are looking for can be for example$$\phi(S)=\int_Sydx,$$where $S$ is a $1$ simplex.
Amitai Yuval
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Great! Can you say a little more about "regular enough" or provide a reference? – Bob Johns Aug 22 '15 at 07:50
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@BobJohns "Regular enough" is relevant especially for singular homology. If you consider only linear simplices, that is, triangles with straight edges, all of those are regular enough. – Amitai Yuval Aug 22 '15 at 13:04