Principal fibre bundles - Lec 19 - Frederic Schuller

[@rajanalexander4949]

What an outstanding lecture by an outstanding professor. He knows when to "zoom in", and when to "zoom out". He balances friendly delivery with mathematical rigour. The fact that he can lecture so well for over two hours without notes is simply legendary.

[@ayan849]

you are the best teacher Prof. Schuller!

[@jmafoko]

a good teacher always starts by giving a motivation and general(rough) sketch of where we are going. schuller does exactly that. from the start you really want to know the subject matter in detail

[@kyubey3166]

Summary of the lecture. 0:00 Introduction and motivation 5.1 Lie group actions on a manifold 8:15 Definition of a left G-action 11:30 Example: action from a representation 14:11 Definition of a right G-action 16:38 Observation: right action from a left action 25:48 Definition of equivariant maps (structure preserving maps for Lie group actions) 33:45 Definition of orbits, orbit space, stabilizers and free actions 48:34 Observation: for a free G-actions, orbits are diffeomorphic to G 5.2 Principal fibre bundles 53:16 Definition of a principal G-bundle 1:02:31 Important example: frame bundle 1:22:00 Definition of a principal bundle map 1:27:24 Generalization of a principal bundle map 1:38:22 Lemma: principal bundle maps with respect to the same group and base space are diffeomorphisms 1:40:28 Christmas tree 1:54:39 Definition of a trivial bundle 2:06:10 Theorem: a principal bundle is trivial iff there exists a smooth section

[@____uncompetative]

Thank you.

[@aieousavren]

THANK YOU! ❤ This is invaluable. I watched the whole lecture and then I wanted to go back to the definition of a principal bundle, so this is very helpful so I can jump right back without seeking around. Really really appreciate it!

[@manmeetworld]

I've been reading about Dr. Shuller and finally came to youtube to my amazement there he is.

[@thephysicistcuber175]

The observation near the 50:00 mark isn't complete: I'm pretty sure the orbit space is to be seen as an _immersed_ submanifold rather than an _embedded_ submanifold. Example: if R acts on the torus T^2 by sending t to the translation with vector t*(1, sqrt (2)) then all orbits are dense curves, and thus not embedded.

[@fractalfan]

It is an embedded submanifold in the case where G is compact.

[@thephysicistcuber175]

@@fractalfan Was compactness of G assumed in these lectures? Because I'm pretty sure GL(n) has been mentioned multiple times.

[@fractalfan]

@@thephysicistcuber175 No, it wasn't, but I just wanted to bring up a case where it is embedded.

[@millerfour2071]

10:50, 20:48, 31:54, 39:16, 43:10, 44:46 (see last lecture), 46:48, 52:33 (each orbit of a free action, an equivalence class, is isomorphic to G), 55:30, 57:23, 1:04:20, 1:09:43, 1:14:46, 1:19:30, 1:31:00, 1:42:49, 1:47:30, 1:49:59, 1:51:12, 1:54:03, 1:58:12, 2:00:31, 2:13:00 (diagram commutes so pi=(pi in trivial bundle)(u)), 2:15:55, 2:22:09, 2:23:42 (apply right action with the inverse of one and impose free action condition), 2:29:10 (principle fiber bundle (free action/equivalence class based on orbit), bundle map (no need for group), principle bundle map (between 2 principle fiber bundle with one extra condition on bundle map)), 2:32:23 (Watching this on the day of Xmas :) cheers)

[@stephengibert4722]

These lectures are wonderfully clear! Thanks very much Prof. Schuller

[@daviddiego2570]

These lectures are great! The insight and deep understanding of the topic by Prof. Schuller are astonishing, one can really learn a lot by following his lectures. I have a doubt about some little theorem that was left as an exercise maybe two lectures before this one. The problem was to prove that if the action of a Lie group G on a manifold M is free, then G is diffeomorphic to any orbit O_p, on the manifold. I am a bit confused by the statement of the theorem. If G is to be diffeomorphic to O_p, it is necessary that O_p has some differentiable structure and, the natural structure to assume is for it to be a sub manifold of M. However, this is not the general case. For instance, the action of the Lie group (R,+) (the real line with the usual addition) on the torus, given by f(t;alpha,beta) = (e^{i2pi (alpha +t)} , e^{i2pi (beta +r t)}) for some irrational 0

[@daviddiegocastro8215]

@Matthew Raymond Thanks for your reply! Yes, after a quick wiki search I found about proper action of a Lie group, G, on a manifold, M. Restricting to free and proper actions, I was then able to show the "embeddingness" of G in the manifold M.

[@vivgm5776]

It is not clear to me why at 24:00, the left action and right action on component and basis vector are not in the other way around. It seems that G acts on the left on components. Could someone explain ?

[@luocheng030]

Thank you very much for the lectures, Prof Schiller. I follow this series and the GR series until this point. They are all great.

[@jackozeehakkjuz]

With "the GR series" do you mean that WE heraeus winter school of gravity and light playlist called "central lecture course"?

[@luocheng030]

Jackozee Hakkiuz Yes

[@jackozeehakkjuz]

Thank you.

[@priyanshubansal6776]

Well, great lecture! However, there is still ambiguity in the definition of the principal fiber bundle regarding the choice of action: if the fundamental action was the left one, then why did he choose the right one?

[@mauricesangermann1320]

The wikipedia definition of principal fibre bundles requires the group acton to be free AND transitive, while Prof. Schuller only requires it to be free. But he seems to use the transitivity, when he proves the bijectivity of principal bundle maps. Has he simply forgotten to add this requirement or am I missing something?

[@simonrea6655]

The wikipedia definition only requires the action to be *fibre-wise* free and transitive. This condition is hidden in the bundle isomoprhism in Prof Schuller's definition, since G acts transitively on the orbits. By the way, he does mention the fibre-wise transitivity at 1:20:25 You can also check out my notes for the course at mathswithphysics.blogspot.com

[@danielplacido8746]

@@simonrea6655 once we go to quotient projection bundle, fibre preservation and transitivity is clear (there). But why does these properties translate back to the original bundle?

[@andrea.dibiagio]

If the G-action does not act transitively on the fibers of (E,pi,M) then there would be points in the same fiber that are not in the same orbit. Since Bundle isomorphisms map fibers to fibers, there could be no isomorphism to (E,rho,E/G).

[@pedidep]

In case anyone else got confused, the dimension of the frame bundle is dim(LM)=2 dim(M). The whole argument, which was given in aside, shows us this.

[@zoltankurti]

No. The base space obviously has dimension dim(M). The preim(p) of any p in M is diffeo to G, and in this case G is GL(dim(M),R), which has dimensions dim(M)^2. The dimension of the bundle is dim(M)+dim(M)^2. You can also think about this as providing a dim(M) basis vectors at each point in M, at least the sections do that, and at each point you need dim(M)^2 numbers to do that if you are using lets say coordinate induced basis vectors.

[@pedidep]

@@zoltankurti Yes!!!! Thank you very much for clarification.

[@zoltankurti]

@@pedidep no problem!

[@dpr265]

Thankfully I just learned some basics of categories so that I can understand the definitions easily.

[@kharanshu2854]

Confusion at 58:30 : If rho is a surjection, how does inverse of rho exist? Let's assume Schuller means preimage of E/G wrt rho (which is E) must be diffeomorphic to G. But then for the SO(2) example, how is R^2 \ {(0,0)} diffeomorphic to S^1 ?

[@tim-701cca]

rho is a quotient map, it may not be injective. Note that only the orbit is diffeomorphic to G. Only SO(2, R) is diffeomorphic to S^1

[@kharanshu2854]

@@tim-701cca aha! makes sense. danke!

[@GunsExplosivesnStuff]

1:15:51 I believe this defines a left action, so act by the inverse of g.

[@TenzinLundrup]

Question from a lay person: g |> p means and element of G acting on an element of M and sending it to another element (say q) of M, i.e., q = g |> p?

[@hdbrot]

Yes.

[@HyperFocusMarshmallow]

At 1:33:45 prof. Shuller writes down a ρ-map from SL(2,C) to SO(1,3), but shouldn’t it be in the other direction following previous notation? Is there something wrong there? If so does anyone know the fix?

[@abhisheksarkar566]

Thank you very much Sir, great leactures

[@isaacdiaz1798]

I was not clear for me why dim LM=dim M+ (dim M)^2 Can someone explain? My best regards

[@albertwood8836]

Isaac Díaz As a topological space, a fibre bundle looks locally like the Cartesian product; in this case, M×F where F is a fibre. Now dim (M×F) =dim(M) + dim(F), and F is isomorphic to GL(dim(M),R) as mentioned in the lecture, which has dimension dim(M)^2.

[@praveenxavier8143]

I think that the definition of the principle bundle as: a bundle (E,\pi,M) that is bundle isomorphic to the bundle (E,\pi',E/G), is missing something, namely the fact that u: E \to E, has to be such that: for all p,g, there must exist a g' such that u(p ightaction g)=u(p) ightaction g' . This condition can be shown to imply that the fibre of \pi is 'preserved' under right action by G (preserved meaning that: for all p,g \pi(p ightaction g)=\pi(p)), and that the same said action is transitive on the fibre of \pi (two conditions which turn up in the Wikipedia article, but are not stated in the lecture). Also note that the missing condition is satisfied if one requires that u be, Identity-map-on-G-equivariant in the language of the ho equivariance that was discussed earlier in the lecture. That condition reads: \forall p,g, u(p ightaction g)=u(p) ightaction g

[@shanborlangbynnud2320]

No, he is not using that the group action is transitive, infact he use that two elements belong to the same fiber means they are related under the quotient P\to P/G .

[@kyubey3166]

I think you are right. I've tried to figure out how to prove injectivity at 1:45:00 but there is this missing part that you pointed out.

[@kyubey3166]

@@shanborlangbynnud2320 I don't see why that should be true. According to his definition of a bundle isomorphism, two elements p_1 and p_2 belong to the same fiber \pi(p_1) = \pi_(p_2) if and only if u(p_1) and u(p_2) are related under the quotient, but how does this imply that p_1 and p_2 are also related under the quotient?

[@John-er7we]

You are right, Praveen. In order to obtain equivalent definitions we need that the f: M \to E/G is given by: f(x) = [p], where p is some point in \pi^{-1}(x). This f needs to be well defined, and a diffeomorphism. And the u: E -> E needs to be the identity map. With these requirements, the definition is equivalent with the one from wikipedia.

[@peterpalumbo3644]

You could probable consider time and anti--time as non vanishing vector fields.

[@thephysicistcuber175]

Shouldn't you require the bundle morphism in the definition of a principal G-bundle to respect the G-action somehow? Like require u to be G-equivariant.

[@kapoioBCS]

I dont see why the preim(G_p)=G_p Any ideas?

[@danielplacido8746]

Why is the \Chi map smooth?

[@xrhsthsuserxrhsths]

I have this question too.. if you find any answers please notify me!!

[@neelmodi5791]

in the case of two principal bundles where the base manifold is the same M, what if there is no u that constitutes a principal bundle map with id:M->M, but there does exist u together with some diffeo f:M->M with (u,f) a bundle map. can we still prove that u is a diffeo? edit: reason for asking is that the above statement appears to be true as mentioned near the beginning of lecture 22, but i cant seem to prove it

[@marcvaillant4477]

I'm not sure you even need 𝑓 to be more than smooth in order to prove that 𝑢 is a diffeomorphism (though certainly the bundles will not end up being isomorphic without 𝑓 being a diffeomorphism ). How about this. Suppose that (𝑃, 𝜋, 𝑀) and (𝑄, 𝜋', 𝑀) are principal G bundles with principal bundle map 𝑢:𝑃 ➝ 𝑄 smooth, 𝑓:𝑀➝𝑀 a diffeomorphism. Then certainly 𝑓∘𝜋:𝑃 ➝𝑀 is a smooth projection so that (𝑃, 𝑓∘𝜋, 𝑀) is a principal G bundle and (𝑢, id) is a principal bundle map between (𝑃, 𝑓∘𝜋, 𝑀) and (𝑄, 𝜋', 𝑀). Now, invoking the lemma, we must have that 𝑢 is a diffeomorphism, and therefore (𝑢,𝑓) is a principal bundle isomorphism. Let me know what you think.

[@HotPepperLala]

I think he means [Epsilon] = G, not with pho^{-1} since [Epsilon] is the orbit space.

[@marcvaillant4477]

The notation is abused somewhat. He is using [epsilon] to denote the set of points in the orbit, and to denote a **point** in the quotient space E/G. Recall that E/G is quotient space obtained by "gluing" together the elements of each orbit so that each orbit is now considered a point in E/G. You get back to the elements which comprise the orbit via the inverse of the canonical projection, i.e. rho^{-1}([epsilon]) is the collection of elements of E in the orbit of epsilon.

[@amir-sb1sg]

Or any specific lecture note does he follow?

[@jacobrose4805]

Someone made a PDF of the lecture notes mathswithphysics.blogspot.com/2016/07/frederic-schullers-lectures-on-quantum.html

[@manimusicka2]

The camera movements in this particular session are very distracting unfortunately. Seems like someone different is filming this session

[@gregorspv]

2:20:16 Geblobbed! Love it.

[@NgocAnhNguyen-mf8jr]

at 1:16:00 when he talk about frame bundle, does he mean that frame bundle is principal GL(dim(M))-bundle if dim(M)>2? If so, could anyone please explain what happens if dim(M) is 1 or 2? Many thanks!

[@philippg6023]

No, it works in all dimensions. He was just reasoning why the action is free.

[@naghshargh6353]

u are a great prof. But I have to say that a frame is not only a basis it is an ordered basis. thanks

[@zoltankurti]

He always used an ordered set of basis, so there is no problem there.

[@____uncompetative]

There seems to be a lot of cross-over between this and Eric Weinstein's work in progress _Geometric Unity_ which aspires to be a _Unified Field Theory._ I will subscribe to this channel in hopes I can understand a bit more about Gauge Theory.

[@4Drow]

In which lecture is general bundle (the triple) defined, aswell as description of a bundle pullback and so on is given? Thanks.

[@4Drow]

Nevermind, found that already. The order confused me, as in the class I'm attending, the order is very different.

[@vinbo2232]

Q: why is the section on S^2 with limited vanishing points not called section? Thanks.

[@marcvaillant4477]

Vector fields that vanish at a point are still sections of the tangent bundle, because the fibres (tangent spaces) are vector spaces, so the 0 vector is an element of each tangent space. However, the fibres of the **frame bundle**, ordered bases of the tangent spaces, are not vector spaces . I.e. you need two non-zero linearly independent vectors to build a basis at a point. So, to construct a section of the frame bundle, you need two linearly independent non-vanishing vector fields. Since a smooth vector field on 𝑆² must vanish at least one point, you will not be able to construct such a section.

[@giuliocasa1304]

At minute 20:23 he says that the inverse is important because it flips the elements of the product: but in the end I see a double flip acting, so I guess that a definition without the inverse would have worked the same. Why is the inverse really needed? (I guess it's more a convention and because you want that left and right actions are opposite one another in this case). BTW Let me add that all these videos are a unique resource for me to understand the concept of principal bundle and more generally of Differential Geometry, that would have been otherwise impossible just from reading wikipedia or arxiv or the likes. So thank you, first of all!

[@37metalgearsolid]

Nope. Give it a try. I.e. define your right action for a given left action, as: p < g := g > p. Then, compute, (p < g1) < g2 = g2 > (g1 > p) = (g2 g1) > p = p < (g2 g1) As you can see, we do NOT have (p < g1) < g2 = p < (g1 g2)!

[@giuliocasa1304]

@@37metalgearsolid Thank you for taking the time to answer and explain, now I have understood.

[@Maria-dk2fv]

what is excessibly mathematical?

[@md2perpe]

A comment about disjoint unions: The disjoint union of sets A(x) where x runs over some set M consists of all pairs (x, a) such that a belongs to A(x). Therefore the map at 1:10 actually takes (x, (e_1, ..., e_d)) to x. en.wikipedia.org/wiki/Disjoint_union

[@revooshnoj4078]

1:40:28 XD

[@leander905]

yes

[@kyubey3166]

yes

[@amir-sb1sg]

What's the text?

[@BobSaget-et6ln]

The book he's used the past is "Modern Differential Geometry for Physicists" by Chris J Isham (he identifies it in some of his older lecture notes.)

[@danielplacido8746]

Why is the action of PFB fiber preserving? He's invoking it all the time but I can't figure why it's true.

[@bendonahoo8563]

That follows from drawing a commutative diagram for the isomorphism between a principal G-bundle P over M, and the corresponding (due to definition of PFB) isomorphic principal G-bundle P over P/G

[@shubhamkadian9345]

Even I have the same doubt.

[@billf7585]

Because its isomorphic by definition to a bundle whose base space is the orbit space of the action by G. So if two points are in the same fiber, then the projection map followed by the smooth map f: M -> P/G implies they are in the same orbit. I'm not sure if I answered exactly what you were asking, but as Ben said, it follows from the commutativity of the diagram.

[@shlomobachar4123]

I am watching this for meditative purposes. I don’t understand a word (even if I have an engineering degree and had some courses in math and physics)…😂

[@peterpalumbo3644]

Entropy forward and backward could be considered an absolute since without it there may be nothing else except frozen space.

[@leewilliam3417]

Mmmmm😊

[@geraldpysniak6228]

table of contents guy