Covariant derivatives - Lec 25 - Frederic Schuller

This is from a series of lectures - "Lectures on the Geometric Anatomy of Theoretical Physics" delivered by Dr.Frederic P Schuller

[@tim-701cca]

Thank you again! I found that this series is excellent since I learnt a lot math and felt comfortable about principle bundle now, and especially I like the ways you present math. Your teaching greatly enhance my learning skills. You teach very well in this difficult subject and really present the main idea to me. I will work more to make everything precise to me and I hope I can fully understand these concept and present these in the future.

[@tim-701cca]

At 5:16, the construction of covariant derivative starts from connection on principle bundle-> parallel transport on P -> parallel transport on associated bundle -> covariant derivative on associated bundle. But at 1:12:27, it depends on connection one form and linear left action. It seems that we can construct covariant derivative without using the parallel transport. Therefore, parallel transport should also be defined using covariant derivative.

[@brendawilliams8062]

Thankyou

[@からすみ-m8g]

I don't understand the definition of the exterior derivative for module-valued differential forms. (Which is used in the definition of the exterior covariant derivative. ) It seems the definition for the ordinary differential forms isn't applicable... Is it defined in the previous lectures?

[@daviddiegocastro8215]

In previous lectures, it is defined the exterior derivative for F-valued differential forms, where F is a finite dimensional vector space. This includes the case of the connection one form (F = T_eG). The definition is exactly the same as for ordinary differential forms. For instance, for an F-valued one-form $\omega$, it would be $d\omega(X,Y) := X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$. The crucial point here is that for a finite dimensional vector space, F, an expression like X(\omega(Y)) boils down to the limit of the differential quotient (f(t_0+t) - f(t_0))/t, for f:= \omega(Y) and this makes sense because given a basis for F, there is an isomorphism to R^{dim F}. Even for an infinite dimensional vector space the above definition would still make sense if the vector space is a Banach space (equipped with a norm that makes it complete). However, for a generic module, taking derivatives is not possible because you can't define differential quotients in the first place: an expression like (f1 - f0)/t might be ill-defined because the inverse of t, 1/t, is not guaranteed to exist (the underlying scalar set is just a ring and not a field or a division ring).

[@jackozeehakkjuz]

At the end, the local representation S of the F-valued function \phi is said to be a local F-valued function too. But didn't we just wrote as a (desired) property (on the left-top blackboard) that the covariant derivative would be linear with functions? Is it that the covariant derivative acts differently on F-valued functions that on C^{infinity}(M) functions? Or is it that the local representation S is not an F-valued function but it is also a section to the associated bundle? Or am I completely lost?

[@richiedadhley1206]

The linearity property of the covariant derivative I think you are referring to is to do with the 'lower' entry (i.e. the T not the s). The T is an element of the tangent space to M, and so is part of a vector space so can be multiplied by a C^{infinity}(M) function. This is not true for s, which is a local section that maps points in M to points on F. This is the same behaviour as sigma and so you can see their relation. As he explains, the covariant derivatives is defined to act on F-valued sections, which s is. Also as he explains, if you took the frame bundle with the associated tangent bundle, then the section s will give you (locally) a vector at each point in M. From here you can then think of the covariant derivative as a method of 'differentiating the field from the section in the direction of the lower field, T'. Hopefully this helps!

[@bars3073]

Can we obtain Lie group by using exponential map on Lie algebra ?

[@alxjones]

The map exp: g -> G is defined so that it takes elements from the Lie algebra to elements of the Lie group. However, one is not always able to fully recover G from g via this map as it may not be surjective. In general, exp g will be a subset of G, and if exp is not surjective, a proper subset. Then some amount of G will be missing from the space recovered as exp g. You can see some sufficient conditions for the surjectivity of exp here: en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Surjectivity_of_the_exponential

[@jackozeehakkjuz]

In which books can I find this approach to the construction of the covariant derivative?

[@intergalakti176]

Try Nakahara - Geometry, topology and physics, the approach is mostly the same but it still might help to see it fromo a slightly different point of view

[@kapoioBCS]

also you can check Frankel the geometry of physics , but with a (very?) weird notation :P

[@mrnarason]

Introduction to smooth manifolds by lee Books is in this drive drive.google.com/folderview?id=0B9XbEQh3jB9pNkhSVzVoYmRtTHM

[@amiltonmoreira2341]

Naber "Topology, Geometry, and Gauge Fields Foundations" , section 6.8. A book that cover the lecture is Ishan "Modern differential geometry for physicist " .The proof omitted in this lecture you can find there. Only the covariant derivative which is treated differently

[@nelange]

@@mrnarason Broken link.

[@elsakkk2528]

I don’t understand 🤪 I’m korean

[@eranlavi6929]

Inattantive cameraman/woman