[Qiaochu's question](https://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a) on a discrete analogue of harmonic function theory reminded me of some thoughts I had a long time ago about the relationship between cubical cohomology and de Rham cohomology.


The main reason de Rham cohomology is often used as a first introduction to algebraic topology is due to the few technical prerequisites. You can set up the differential graded ring structure with very little work. One of the important results is the Stokes theorem,


$$\int\_C d\omega = \int\_{\partial C} \omega,$$


where $\omega$ is a $p$-form and $C$ is a $(p+1)$-chain. Conceptually, it says the differential operator $d$ is adjoint to the boundary operator $\partial$ with respect to the non-degenerate bilinear pairing $\int$.


It is sometimes remarked that you could turn around and use this result to justify defining $d$ as the adjoint of $\partial$. Indeed, that is how simplicial cohomology is usually defined in terms of simplicial homology. But it seems clear that the better analogy is with cubical cohomology:


Establishing homotopy invariance and the cup product for singular simplicial cohomology is a bit technical. In contrast, it's easy with singular cubical cohomology: $I^m\ I^n$ = $I^{m+n}$, so cubical prisms and products are manifestly cubical, no subdivision needed. Demonstrating these properties for de Rham comology can be done with similar ease. The advantages and disadvantages of cubes over simplices have been discussed here [before](https://mathoverflow.net/questions/3656/cubical-vs-simplicial-singular-homology).


The analogy goes further. If you write out products and exterior derivatives of differential forms in coordinates and compare them with the corresponding formulas in cubical cohomology in a lattice-like neighborhood, they are exactly analogous in the same way that freshman calculus and the calculus of finite differences are analogous. If you have trouble wrapping your head around the geometrical interpretation of concepts in de Rham cohomology like products and differentials and especially more advanced concepts in Hodge theory, the analogy with cubical cohomology is very enlightening (or at least it was for me).


My question is whether anyone has worked out this relationship in detail. Maybe in a context where infinitesimals have a concrete existence as in algebraic geometry and synthetic differential geometry.



Cubical cohomology and de Rham cohomology
M. Carmen Minguez in [this article](http://archive.numdam.org/article/CTGDC_1988__29_1_59_0.pdf) constructs a homomorphism between de Rham and Cubical Singular Cohomology without showing that is an isomorphism. This is done in the context of Synthetic Differential Geometry.


In general Synthetic Differential geometers seem to be quite aware of the cubical setup, probably because differential forms naturally get a cubical structure if defined via infinitesimals. This is very visible in chapter IV, section 1 of Moerdijk/Reyes' [book](http://books.google.com/books?id=OSzvAAAAMAAJ&q=moerdijk+reyes+smooth+infinitesimal+analysis&dq=moerdijk+reyes+smooth+infinitesimal+analysis&hl=de&ei=HOtPTIzOIsGPOIq2iKcB&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCsQ6AEwAA) on synthetic differential geometry. In remark 1.8 (p. 145) they mention a bridge to cubical (co)homology - then, however, they choose to establish the isomorphism between de Rham and *simplicial* singular cohomology.


