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Calculus is the dual of topology

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Calculus is the dual topology, in what sense?

The exterior derivative is dual of the boundary operator.

In the discrete case we can be more specific.

Suppose we know the (local) topology of a cell complex (or a simplicial complex etc). This simply means that we can express the boundary of each $k$-cell in the complex in terms of $(k−1)$-cells. This data comprises the matrix of the boundary operator. But the matrix of the exterior derivative is equal to the transpose of the matrix of the boundary operator! This means that we now know the exterior derivatives of all discrete differential forms on the complex. It is fair to say then that we know calculus on the complex.

This fact has been known for a very long time, but the practical consequences have only become to be understood more recently, and they are quite significant, specifically for modelling physics with PDEs.

I used to describe this as "Calculus is topology", but

  • "is" means "is isomorphic to" here, and
  • not naturally isomorphic as vector spaces because
  • the arrows of the operators are reversed, plus
  • the dual has the ring structure, and, furthermore,
  • over the integers the whole thing becomes much more complicated.

So, to put our conclusion in other words,

calculus is built on top of topology with algebra.

Strictly speaking, what you get isn't the calculus we study as freshmen...


Related to this idea is: Calculus / algebra = topology.

See also Topology, Algebra, and Geometry, Math is an art.