Associated fibre bundles - Lec 20 - Frederic Schuller 

Summary of the lecture 5.3 Associated bundles 1:01 Definition of associated bundles 13:24 Example a): tangent bundle as an associated bundle 32:13 Example b): tensor bundles as associated bundles 42:38 Example c): tensor density bundles 58:08 Associated bundle maps 1:11:21 Trivial associated bundles 1:17:34 Important theorem: sections and F-valued functions on P 1:24:05 Restriction and extension of bundles 1:34:35 Theorem: extensions are always possible, restrictions require some conditions

17:50 Note that the left-action of GL(d) on R^d should be defined as g > f = g^a_b f_b . As defined now (with the inverse of g rather than g itself), it is a right-action. The inverses come later when building the equivalence relation. Prof corrects this at 56:10.

at 1:20:35 I think it is important to note that phi cannot be just any map, it must be G-equivariant in the sense that phi(pphi(p) otherwise the proof cannot be established.

u are great Prof

At 1:20:22 I guess the function on the principal fibre bundle f should be defined as f: P→F: f(pf(p) to have a well defined section (constructed later in the proof) on the associated bundle P xG F.

For the definition of the associated (p, q) - tensor bundle @ 33:00, don't you need the tensor product of R^d and R^d* and not the cartesian product? Otherwise the dimensions doesn't add up: (p+q)*d vs. d^(p+q)

at around 1:23:12, for well definition to hold, it must be true that $\phi(p\triangleleft g)=g^{-1}\triangleright \phi(g)$.

10:07, 19:00, 24:14 (elements in an equivalence class are different representations of the same tangent vector with different basis from frame bundle), 30:05, 36:35, 47:09, 55:13, 1:06:46, 1:14:21, 1:16:33, 1:19:08, 1:28:30, 1:37:58

I think the diagram drawn for denoting the triviality of the principal fibre bundle at around 1:13:28 is not appropriate

I have to say he is good, I would like him to explain the foldy wouthuysen transformation in detail from the mathematical point of view

Does anybody know of any textbooks or literature I could use to augment these lectures? I was feeling really good for the first sixteen lectures or so, but I'm getting really lost in the discussion of bundles. I need to get my hands dirty with problems.

Since the determinant of a transformation operation is the change of the size of the object.

1:29:03 there is an inconsistency with the notes. According to the notes the bundle map has to go from P' to P, rather than the other way around. Indeed I was stuck trying to prove that every H-principal bundle can be extended to a principal G-bundle, whereas it looks a lot simpler if the definition of an extension has the map going the other way around.

The problem I think, is that in the definition of the associated bundle, he introduces a left action wigh I think is ill-defined. To understand the expression "inv(g)>p" you need to know previously which kind of action ">" represents. If , as we presume ">" is a left action , the one given , that we can call ">>" will be defined by g>>p= inv(g)>g. But it, s easy to verify that the action ">>" is in fact a right action. It´s the "specular image " of the left action ">". So, I think in the definition of the associated bundle, I think, he should have written "g" and not"inv(g)".

In the important theorem at 1:20, I am having trouble showing well-definition of the given s_phi=[p, phi(p)]. For this to be well defined it needs to be the same for any p and p' in pi^-1(x). If this is the case then there exists a g in G such that p'=pphi(p) = phi(p

At 15:29 it is a *right* action, it should read < : LM x GL(d,R) --> LM

Around 1:32:59 he has the right idea but makes incorrect claims. - We can reduce from the general linear group GL(d,R) to the orthogonal group O(d,R) if and only if the manifold M admits a Riemannian metric. And it turns out every manifold does admit a Riemannian metric. - We can reduce from GL(d,R) to SO(d,R) if and only if M admits a Riemannian metric *and M is orientable*. Notice the extra orientability condition; this is NOT always satisfied. Similarly for the remarks at 1:34:08, the correct statements are - We can reduce from GL(d,R) to the Lorentz group O(1,d-1) if and only if M admits a Lorentzian metric. - We can reduce from GL(d,R) to SO(1,d-1) if and only if M admits a Lorentzian metric *and is orientable*. Again, note the role of orientability. Finally, for completeness, here’s a theorem characterizing existence of Lorentzian metrics: the following statements are equivalent: (1) M admits a Lorentzian metric (2) M admits a time-orientable Lorentzian metric (3) M admits a nowhere vanishing vector field (4) M is either non-compact, or else M is compact and has vanishing Euler characteristic. See see O’Neill, page 149 for a proof. The equivalence of the first three is not too bad, also (3) (4) is a ‘standard’ but somewhat difficult topological result (I’m sure one can find a proof in Steenrod’s book on Fiber bundles).

Gonzalez Matthew Lopez Deborah Moore Patricia

I think there is a consistency problem. At 5:52, the definition of the equivalence relation is made acting from the left on the element from the fibre F, using the inverse of the element that acts from the right on the element of the total space of the principal bundle P. But at 17:45 when he's defining the left action of GL(d,R) on the fibre, then again he uses the inverse group element of the g that acts upon the fibre element. So the definition of u at 23:36 does not work because at the end the transformation provided by g is flipped by the left action but again flipped by the definition of the equivalence class. One of them should not be made using the inverse, whether the equivalence class or the definition of the left-action. Please correct me if I'm wrong.