The explanations of this man are crystal clear. He makes complex subjects look easy and comprehensible. He makes hard aspects of the theory very clear. This man IS the book.
1:07:21 It's really important to highlight here that the map is from S^1 to R^2 and not from S^1 to the lemniscate (figure 8), as the lemniscate is not a smooth manifold (nor even a topological manifold). But R^2 IS a smooth manifold, and with an appropriate parametrization, we induce a well-defined tangent space at the intersection point of the lemniscate which is not present as a purely geometric object.
I’m thinking about the pullback story as a kind of communication: having someone call back from the manifold N to manifold M at the point p to tell them the parameters of the linear function in the cotangent space at p in manifold M. That response phone call happens after someone in M has pushed forward X to N. Still scratching my head though.
50:00 that drawing was very helpful, but only with the careful explanation of what the arrows mean: it’s a push forward of an operator on M at p, and the result of the pushing is symbolically represented on N as arrow at the image point in the direction of the image path on N. I think I got it.
I think is a better idea to call d_p "the differential operator at p on M" and define the "gradient" only when there is a symmetric non degenerate bilinear form defined on each T_p(M) whose expression in any chart are smooth enough, G. In that case there is a "natural" isomorphism between the corresponding vector-covector spaces i(G): T_pM--} T_p*M, and we can safely the gradient of f as the inverse image of d-(p)(f) by the isomorphism i(G). That makes the statements of classical vector analysis true, only with the (implicit) assumption of a G which permits the "identification" of vectors and covectors. The isomorphism is defined for each w on T_p(M) as i(G)(w)=u iff G(u,v)=(i(G)(w))(v) for each v on T_p(M). Everything is in place, and the name "gradient " in itself reminds you of the existence of the underlying bilinear form that makes the gradient of a function depend both on the function and on the bilinear form chosen.
I am confused about the tangent bundle. At roughly 1:41:00 there is a drawing of TM as a fiber bundle over M, and he gives an intuitive reasoning why TM should be 2*dim M-dimensional. My concern is with the notion that TM contains M at all. In fact, following the formalism as presented it is not even true that a point p lies in its own tangent space TpM. The closest thing we have is a "zero vector" or "velocity of the constant curve which stays at p for all parameter values", being a directional derivative operator which maps every smooth function f in C^\infty(M) to 0. A zero operator on C^\infty(M), at least in my mind, is a very different thing to a point p in M. So to me, p does not belong to TpM. So the disjoint union of all the TpMs does not contain any of the ps, and so TM does not contain M. Anyone can offer some insight? Is it just this intuitive argument for the dimension of TM which is a red herring? Or just the picture? Edit: Indeed it is the picture which is misleading, I guess that's why he said it's dangerous! After watching the rest of the video I don't believe it's claimed anywhere that M actually lies in TM, just the picture threw me off a bit. As far as I can tell, points in M are merely referred to in the construction of the chart maps to reference a basepoint in TM (what would *correspond* to a point in M, but is not *actually* a point in M). I will leave my comment here in case it is helpful for anyone else. If anyone more learned than myself has some input this is also welcome of course.
I'm a bit confused. He says that phi(K) is not a topological manifold, but it's an immersion of the Klein bottle into R3, which means that phi is a smooth map, does that not require that the target space be a manifold?
6:06, 10:36, 15:07, 39:27, 47:13 ((f phi) gamma = f (phi gamma)), 52:50, 58:19, 58:52, 1:03:53, 1:07:48, 1:14:36, 1:22:48, 1:24:45, 1:28:03, 1:31:15, 1:34:35, 1:39:46 (initial topology with open sets in TM defined by the preimage of all open sets in M under pi), 1:41:25, 1:46:51, 1:48:13 (utilize transition functions and transform components of X)
definition of gradient operator: ru-vid.comUgkxSmfAWGTJZtCvA9e9zgmybOcqQdmfToHO gradient operator as basis for co-vector space: ru-vid.comUgkxgNhOaNVCHjaEEiPrb9mFpEq9fSAlS2BF derivative, a.k.a. push-forward (subscript star, not shoulder start as in the video): ru-vid.comUgkx7u1vOs0MTule25U5LTf9PEFklCHHypaY pull-back(star on shoulder) ru-vid.comUgkxMSnGqZchv5X9PVC31FxnR-PFJhh_I5OY when smooth manifold can sit inside R^n : ru-vid.comUgkxDVZacfeI_OmMAYaL1A3DQEgtGKSDhqAP Wittnening theorem on immersion and embedding: ru-vid.comUgkxlXUuXzn0Ec7EgTKJeM4xbIutqrCN9RPj tangent bundle and vector fields: ru-vid.comUgkx6mBEvGVP_r_8AVArTUV1HQ4Yw_8XGcww
How can we define the tangent space to a set we don't know if it is a manifold? My question is motivated by the explanation of the difference between immersion and embedding because in the "8" shape the tangent vectors are defined at the intersection point at roughly 1:08:00. Do we use the tangent space of the manifold N to define the tangent vectors?
Thanks for your great lessons! One question: 18:01 minutes into the lecture you introduce d_p : C^{\infty} ---> T^*M but then write d_p(f)(X) = Xf where X in T_p M So shouldnt d_p then be defined as d_p : T_p M---> T^*M instead? Additionally Xf is clearly a vector so how can it be in T^*M ? Wouldn t it be possible to define d_p as d_p : T_p M---> T^*M where d_p(X) f = del_a( f * x^-1) dx^a(X) instead where dx^a(.) is a co-vector basis in the cotangent space? many thanks
Xf is a number not a vector defined as derivative of f in direction of x at point p. from d_p(f)(X) = Xf where X in T_p M you can see that d_p(f) is in T*_pM as it takes X(vector) and gives Xf(a number).Therefore d_p(f) is a covector for each f.Therefore, d_p is the given map. d_p : C^{\infty} -> T^*M
@@justanotherman1114 at 17:35 , he says d_pf(X)=0 for X being a tangent vector at the level set N={all p in M such that f(p)=constant}. I don't see why this should be true.
at 46:00 - isn't it a mistake to name the result of the pushforward also X? Since it has (phi º gamma) it means the curve is now R->N and not R->M. So it seems to me the result has to be a vector on a tangent space of N and not of M, so it is definitely not X anymore, is it?
If you wanted to name it, you're right in that it should be given a name like 'Y'. As I understand it, tanget vectors of some T_p(M) to a curve in M are not named, but called X_{gamma,p}, the name is in the curve gamma and the point p. So if the curve lies in N (as phi°gamma does) and the point is in N (as phi(p) does) then X_{phi°gamma,phi(p)} is an element of T_phi(p)(N). In other words, we do not change the letter X for tangent vectors of different manifolds, its clear that they are elements of a certain tangent space by their sub-indices
definitions are correct, the corrections are correct...but ..what is pusforward?....a guess..is a directional devitave, that applied to function f, gives what masses known as differential, so often , people forget that df, is a directional derivative...
They are indeed mapped to the same vector, but since the points on the circle are different - let us call them p and q - these two zero vectors belong to different tangent spaces, and are therefore mapped to this vector by two different applications : the derivative of Φ at p and the derivative of Φ at q. This does not contradict the injectivity of these two applications, although Φ itself is clearly not injective.
You misunderstood: he was saying it can't be bijective (to be an isomorphism) if it's surjective from a lower dim manifold to a higher dim manifold (i.e. non-trivially surjective and hence not bijective). (It's more intuitive if you start from injective from a higher dim manifold to a lower dim manifold.)
the claim of the pedagogy of this subyect is the lack of figures to illustrate of what is the matter in the botton line...to draw manifold could be difficult..but if the idea is to transmit effecitively the theme... a few simple drawing would do the job....see the videos of Robert Davie for the same subject...the difference is dramatic...
miguel aphan I agree to some extent. But I think the best is to see more than one exposure of this topic. He sometimes avoids drawings to avoid misunderstandings or oversimplification. On the other hand I find that his lectures are extremely precise and in a way “maximally” general. Others may make more drawings but sometimes a lot of stuff gets swept under the carpet with simple example drawings. I love drawings but here I’m forced to do often without, which resulted in paying much more attention to the formalism but also clearing quite a few misunderstandings I had about the subject. Best is to basically make your own drawings. Also, you can find online somewhere a complete script of these lectures - very helpful and self contained.
I learn best when there is something that I can see. I want a graph, sketch, picture, diagram, something more than an abstract collection of symbols. If you want to teach me an abstract something, then start with something concrete that I already know and can relate to, and then show me how your abstraction relates to the concrete subject matter that I know and am used to working with.
