k-Chains and k-Cells
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To prove the Generalized Stokes’ Theorem, we’ll some way of defining a region and its boundary.
We want to prove everything in Vector Calculus by introducing objects known as differential forms. Up to this point, we’ve derived a formula for the volume element and formulas for the gradient, divergence, curl, Laplacian, and vector Laplacian in arbitrary coordinate systems. We were also able to prove any identity in Vector Calculus so long as it didn’t have integral in it. To handle identities with integrals, we need to prove the Generalized Stokes’ Theorem.
The Generalized Stokes’ Theorem says that you can convert integrals of the exterior derivative of some k-form ω over a region Ω to integrals of the k-form over the boundary of a region.
In other words, integrals “undo” derivatives just as they did back in your first Calculus class. We already know how to deal with k-forms and the exterior derivative, but we haven’t come up with a formal way of dealing with a region and its boundary, so we’re going to come up with three things:
- A way to define a simple region.
- A way to find the boundary of the simple region.
- A way to morph the simple region and its boundary to arbitrary regions.
These three things will allow us to prove the GST for everything except arbitrary manifolds.
Check Your Understanding
This article is mostly Topology, so you shouldn’t need too many prerequisites to read it. With that being said, there’s a lot of context for why we’re doing everything we’re doing in the previous articles.
Boundary of the Boundary
Use the formula for the boundary of a k-chain to show that the boundary of the boundary of a 2-cell and a 3-cell is zero.
k-Cells and the Pullback in Calculus III
Using ideas introduced in this article, explain why we can convert an integral over the filled in region shown in the image below
