a smooth field of 1-vectors in 3-D
Posts tagged with differential forms
The reason is that the matrix of the exterior derivative is equivalent to the transpose of the matrix of the boundary operator. That fact has been known for some time, but its practical consequences have only been understood recently.
[S]uppose you know the boundary of each
k-cell in a cell complex in terms of
(k−1)-cells, i.e., the boundary operator. Then you also know the exterior derivative of all discrete differential forms (i.e., cochains). So, you know calculus. Smooth or discrete.
A beautiful depiction of a 1-form by Robert Ghrist. You never thought understanding a 1→1-dimensional ODE (or a 1-D vector field) would be so easy!
What his drawing makes obvious, is that images of Phase Space wear a totally different meaning than “up”, “down”, “left”, “right”. In this case up = more; down = less; left = before and right = after. So it’s unhelpful to think about derivative = slope.
BTW, the reason that ƒ must have an odd number of fixed points, follows from the “dissipative” assumption (“infinity repels”). If ƒ (−∞)→+∞, then the red line enters from the top-left. And if ƒ (+∞)→−∞, then the red line exits toward the bottom-right. So no matter how many wiggles, it must cross an odd number of times. (Rolle’s Thm / intermediate value theorem from undergrad calculus / analysis)
Found this via John D Cook.