bgoated01
a year ago
I’m working on a PhD adjacent to computational differential geometry, and while I’ve made a lot of progress on the computation, I still don’t have much intuition for k-vectors and k-forms. I love coming across articles like this that help me build intuition. The article for which this is the second part was really helpful, but I’m going to have to come back to this second part a few times to fully grasp it. I also loved this quote from one of the articles listed as a source (but with a broken link, I found it at http://yaroslavvb.com/papers/notes/piponi-on.pdf): “Think of a vector as a pin, and a one-form as an onion. You evaluate a one-form on a vector by counting how many onion layers it goes through.”
Edit: This one also looks good: https://math.uchicago.edu/~may/REU2018/REUPapers/Bixler.pdf
enriquto
a year ago
> I still don’t have much intuition for k-vectors and k-forms
The references that you cite are great. Also Tristan Needham's book "Visual Differential Geometry and Forms", or the classic "Gravitation" where this visual language was fist exposed.
Besides this continuous interpretation, you may also enjoy further insight via the discrete case, called "discrete exterior calculus", or "discrete differential geometry" or even "graph signal processing".
The idea is the following. The base space is a graph, consisting of vertices, edges joining pairs of vertices, triangles joining triplets of edges, and so on.
Now, k-forms on the graph are, by definition, functions defined over the k-cliques of the graph. Thus:
0-forms are functions defined on the vertices
1-forms are functions defined on the edges
2-forms are functions defined on the triangles
...
k-forms are functions defined on the (k+1)-cliques (i.e. complete subgraphs of k+1 vertices)
Now the exterior derivative of a k-form is the (k+1)-form obtained by computing differences over each k-subclique.
For example, if f is a "scalar-field" or 0-form, its differential df is the 1-form defined by (df)(a,b)=f(b)-f(a) for each edge (a,b). If "F" is a 1-form, then dF is the 2-form given by (dF)(a,b,c)=F(a,b)+F(b,c)-F(c,a). And so on. Thus, it is clear intuitively that ddf=0, because all the edge differences cancel out when you traverse the edges of a triangle.
If your graph has a notion of duality (a mapping between k-cliques and (n-k)-cliques, for example when the graph is planar), then you can define the laplacian Δ=dd and all its associated goodies.
Furthermore, you can easily define the integral of a k-form over a k-chain (a subset of k-cliques), and the boundary of a k-chain as the (k-1)-chain obtained by differentiating its indicator function. The the discrete version of Stokes theorem becomes an algebraic triviality.
I recall feeling immense pleasure when I first learned about all this. Hoping you feel something similar!