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
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.
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.
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?
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.
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.
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.
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
2:24 vector field
19:29 ring
25:10 R-module
32:13 D-module has a basis
47:15 Zorn's lemma
1:35:45 free module
1:49:33 (p,q) tensor field
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?
linear independence - basis requirement - I could then specify a vector in two or more ways
But what if at each point only two of the vector fields in the basis were nonzero?
any set containing the zero vector fails to be linearly independent
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.
@Casey Wojcik In two dimensions... yes. But the space of vector fields in S^2 is not finite dimensional!
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.
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?
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.
if it is in B, then it is in span of B. trivial case.
Or he could have written (B \ {v}) u {v}, right?
What book do you recommend to follow these lectures about tensor space theory and topological manifolds and bundles?
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.
"Differential Geometry of Manifolds" by 'Stephen Lovett'
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".
Yes, that's right. An alternative I like more is to quantify x only over P\{m}.
It can be formulated as "if 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.
For the definition of total order, doesn't reflexivity follow immediately from totality? For all a in the set, either a
26:33 A moth is flying by checking the math, trying to figure out if R-modules are edible.
In 1:50:26, I'm confused since it is not said whether the space of sections of TM is indeed finitely-generated.
The theorem at 1:40:26 implies it is.
"If you're not fully prepared, you're not prepared"
Corollary: If a stop is not full, then it is not a stop.
visitor @26:32
in 1:3918, the professor talked about 'finitely generated', what does this mean since we are not guaranteed there must be basis for module?
Since the module is free, we are guaranteed a basis, by definition. Finitely generated simply means that the generating set is finite.
Thankyou
Mind blown🙀 1:50:17
a haa!
1:53:15
This reminds me of a great math joke...
What's yellow and equivalent to the axiom of choice in ZF set theory?
Zorn's lemon 🤣🤣🤣
1:09:21 Professor Schuller just wants to be free, like everybody else, and this guy won't let him. RIP American dream. SMH
me till 1:51:00 much of the lecture is pretty much pointless, i've seen this in algebra/linear algebra b4
me at 1:51:01 holy shit fuck