10:13, 11:44, 21:00, 26:00, 34:10, 40:19, 44:53 (for S^2 you cannot choose a vector field basis that is smooth globally with S multiplication over C^\inf(M) point wise. We can have 3 smh vector fields each has two vanishing points where the other 2 are not, as “basis”, but the 0 field can be expressed with different sets of non 0 coefficients, specifically changing for all the p on M other than the 6 vanishing pts, because tangent space has dim 2. However, it is not the case that one can be written as a linear comb of the other two by “dividing” the coefficient like with R, because C^\inf(M) does not have multiplicative inverse for all non everywhere 0 function, but at all 6 vanishing points all coefficients must evaluate to 0 causing all of them not invertable. This is consistent with non existent of basis. 1:29:48), 52:53, 56:06 (upper bound only makes sense for totally ordered subset, definition for an upper bound changed later 1:02:20, you can have elements in P that cannot be compared at all, or only comparable to a subset of all elements that are comparable to some element 1:08:38. Also implies there could be multiple upper bounds, since u can come from P and the fourth condition could not be applied to 2 non comparable upper bounds (probably should stick with the modified def), and maximal elements (upper bounds that are not comparable and all non comparable elements)), 57:52, 1:12:08, 1:18:22, 1:26:27, 1:34:08, 1:39:20, 1:43:34, 1:49:11, 1:50:10, 1:54:27
He takes such special care around 38:25 to not call the theorem "The Hairy Ball Theorem." I can only assume it's because they are just as immature on that side of the pond as we are in America.
For the issue at 1:26:00, one just has to take v in S but not in the subspace generated by B. The contradiction will show that there is no such a thing. That is, B is a generator.
The definition of "maximal element" is missing something. It should be "m is maximal if there is no x such that (m leq x AND m not equal x)", in other words "there is nothing STRICTLY greater than m".
I'm just wondering if you use a 4-dim tangent bundle such that you make of each 4-d tangent vector a quaternion, if you then still have just a module or is it an actual field? It would be some 4-dimensional manifold where each point has a quaternion fiber. One nice property should be that this is also a Lie group: (quaternions are diffeomorphic to SU2). Later added ( I may revisit this): I'll leave this comment here, but by the end of the lecture I kind of realized I was barking up the wrong tree. I don't really know what kind of bizarre thing it would be if we had a Tanget module with basis e_1, ..., e_d and the coefficient functions would all be quaternions, q^1, ...., q^d. I thought about this as i recalled that the quaternion group is also a division ring. But what the heck I created here doesn't make a lot of sense to me, now that I think about it.
I don't know if you already know this but there are some lecture notes by Schuller: mathswithphysics.blogspot.com/2016/07/lectures-on-geometric-anatomy-of.html Also, in the end of these notes, he has a relatively long list of books for the different topics.
Regarding 43:00 I don't quite understand why there is no vector field basis for Gamma(TS^2): Even if there is no way to do it with two vector fields in the basis, surely there must be a way to do it with 3 or more vector fields in the basis?
You are totally correct! Suppose there was a basis with more than two fields in it. The set where two of them (say X and Y) is non-zero is an open set. The set where at least one of the others (different from X and Y) is non-zero is also an open set. Since we are in a connected set, those must intersect. Therefore, you would have at least three of them non-zero in an open set. Now, using a bump function you can create a field with non unique expansion.
may be a silly question, but why can we not always make the choice Q is trivial at 1:45:00? specifically, why is there no trivial C^infinty(S^2)-module? is this a corollary of the hairy ball theorem?
