Grant Sanderson

Really beautiful video by @00aleph00 about homology and cohomology, and how it offers another way to think about the derivative. "The derivative is the opposite of the boundary"

*er, read that as why a 2d _complex_ vector space has a richer structure than a 4d real vector space.

So somewhere along the way, the plane of (v, Rot(v)) is the same as the plane of (v, i*v). At that point, v is an eigenvector. I think this works. Do you find it satisfying? Needlessly confusing? Is there a way to clean it into a more succinct view?

But that means i*v must be in that same plane as well. (Surely there's a quicker way to see that...)

But in fact this means they must all be in a 2d subspace. Why? Well, v is orthogonal to Rot(v), and also to i*v. If Rot(i*v) is confined to be in the same 3d subspace as {v, Rot(v), i*v}, then because it's orthogonal to i*v it must be in the same plane as {v, Rot(v)}.

This determinant is positive at θ=0, and negative at θ=pi, so must be 0 somewhere along the way, so at that point all four must be linearly dependent (in the same 3d subspace).

Take the 4-tuples of vectors (v_θ, Rot(v_θ), i*v_θ, Rot(i*v_θ)), and consider their orientation in the following sense: Treat them as 4d real vectors and consider the sign of the determinant of a matrix with these vectors as columns.

Is there a way to make this less fuzzy?

It seems hardly coincidental that at the halfway point, v_{pi/2} = [1, i], we hit a pint where {v_θ, Rot(v_θ)} sits on a spanning plane, which is to say v_{pi/2} is an eigenvector of Rot.

So in some (admittedly vague) way, the pair (v_θ, Rot(v_θ)) swaps orientation as θ ranges from 0 to pi.

However, they form a basis of the xy-plane (in a real-vector sense) with a different orientation.

At the beginning, Rot(v_0) = Rot([1, 1]) = [1, -1], so (v_0, Rot(v_0)) "spans" the xy plane in the real number sense of the word span. Similarly, v_pi = [1, -1] gets taken to Rot(v_pi) = [1, 1], so (v_pi, Rot(v_pi)) also span-in-the-real-sense the xy plane.

In this view, the family v_θ = [1, e^{i*θ}] forms a circle around the point (1, 0, 0) which is perpendicular to the x-axis.

If you want to picture this, we might imagine the 3d slice of the full vector space that looks like [x, y + zi] for values x, y and z, each one pictured as the point (x, y, z) in 3d space.

The idea is to consider the family of vectors v_θ = [1, e^{i*θ}] for various angles θ, along with how Rot acts on these vectors.

Again, my apologies for being too lazy to draw pictures for all this right now, it's hard to overstate how much time it would take to do it effectively. Instead, give your mind's eye a little massage, because we're about to ask a lot from it (poor thing.)