[enriquto]

I am intrigued by your ideas and wish to subscribe to your newsletter.

Lol, this is not the first time that I am mocked here :)

How does Stokes' theorem in a discrete setting just amount to associativity?

Manifolds are modeled by graphs, and calculus on manifolds becomes linear algebra using the matrices naturally associated to these graphs.

Consider a graph with n vertices and m edges.

The most important matrix is the oriented incidence matrix B, of size mxn, that has a single +1 and a single -1 on each row, indicating the vertices connected by the corresponding edge.

Scalar fields := functions defined over the vertices = vectors of R^n

Vector fields := functions defined over the edges = vectors of R^m

When you interpret matrices as linear operators:

    B     : R^n -> R^m is the gradient operator
    -B'   : R^m -> R^n is the divergence operator
    -B'.B : R^n -> R^n is the laplacian

A subset of the vertices is given by a binary vector M \in R^n. The integral of a scalar field f over M is M'.f

The outwards boundary of a subset M is -B.M. The flux of a vector field F through this boundary is (-B.M)'.F

Stokes theorem is thus the trivial identity (-B.M)'.F = M'.(-B'.F)

(I use a dot for matrix products because the star breaks the formatting. You can also define the divergence without the minus sign, but I like my laplacians to be negative-definite, it feels weird otherwise.)

[gjm11]

I didn't actually intend my comment as mockery. I'm very much for finding correspondences between discrete and continuous things. I just didn't see how to set things up so that Stokes turned into associativity.

I think this is in fact the same thing as the "true by definition" found in the second of my links above -- but I hadn't thought hard enough about how that cashes out concretely to see that you can express it as the associativity of a three-way product. Nice.

[enriquto]

stokes theorem is often written (in a continuous setting) as

    < dM, F > = < M, dF >

(the adjoint of the boundary operator is the exterior derivative). In the discrete case, these integrals are actually products of matrices. There is nothing too deep here.

Edit: regarding the "true by definition" issue, you can do one of two things. (1) Build the definition of the boundary and exterior derivative independently and then verify that Stokes theorem holds. Or (2) build the definition of only one of these two operators, and then define the other one as its adjoint. In that second case, Stokes theorem is true by definition. But it doesn't matter too much, these are just two different ways to write the same thing.