- #1

- 17

- 0

If it would be necessary to go via an explanation of the Riemannian curvature tensor, I would also like to understand that (bearing in mind I have studied very little Riemannian geometry).

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 17

- 0

If it would be necessary to go via an explanation of the Riemannian curvature tensor, I would also like to understand that (bearing in mind I have studied very little Riemannian geometry).

- #2

- 489

- 0

If most of this paragraph makes sense to you, and you have questions, let me know. I would imagine Cartan connections would be very difficult to pick up with no prior exposure to Riemannian geometry.

- #3

- 17

- 0

If most of this paragraph makes sense to you, and you have questions, let me know. I would imagine Cartan connections would be very difficult to pick up with no prior exposure to Riemannian geometry.

I 'know about' the Levi-Civita connection in that I know what it is and the fundamental lemma, but I have not studied them in any particular depth. I learned about Cartan connections via Klein geometry, i.e. homogeneous spaces and Lie groups/algebras, and a principal bundle with a Cartan connection being a generalisation of the former.

How are the structural equations defined with the Levi-Civita connection? Do you mean the equations of the fundamental lemma? Just to clarify, the equation (on a Lie group) I was referring to was

[tex]d\omega_G+\frac{1}{2}[\omega_G\wedge\omega_G]=0[/tex]

where [tex]\omega_G[/tex] is the Maurer-Cartan form of the Lie group. Since my first post, I have seen an alternative definition of curvature of an Ehresmann connection in Kobayashi/Nomizu as [tex]d\omega[/tex] composed with the horizontal projection, but I am not sure how this relates to Cartan connections.

Last edited:

- #4

- 489

- 0

It is my understanding that you can define all of these things analogously for any connection. However, in the absence of a metric, I think you have to

To answer your original question more succinctly, I believe they've defined it this way to be consistent with the Levi-Civita connection (in which case the curvature of the connection actually measures the curvature of the manifold).

- #5

- 17

- 0

How are the connection forms defined by the Levi-Civita connection, and how does the curvature form agree with the Riemannian curvature tensor (I assume you mean this when you refer to the curvature of the manifold)? Or is this an involved question for which I'm just going to have to read a book on Riemannian geometry (any recommendations)? Thanks for your help btw.

- #6

- 489

- 0

[tex]\nabla_X E_i = \omega_i^j(X) E_j [/tex].

You can define this for any connection. Now the curvature two forms are defined as follows: compute the connection 1-forms using the Levi-Civita connection, and define the two forms to be

[tex]\Omega_i^j = \frac{1}{2} R_{kli}^j \phi^k \wedge \phi^l [/tex].

Here we're using the curvature tensor.

In this context, Cartan's second structural equation says that

[tex]\Omega_i^j = d\omega_i^j - \omega_i^k \wedge \omega_k^j [/tex].

This might look slightly different from how you've seen it due to sign ambiguity in the curvature tensor, and ambiguity in the way different authors define the wedge product. Authors in Riemannian geometry tend to make it as difficult as possible to read each others' books.

If you want to learn more, I'd suggest do Carmo (Riemannian Geometry) or Lee (Riemannian Manifolds).

- #7

- 17

- 0

Share:

- Replies
- 34

- Views
- 661

- Replies
- 13

- Views
- 1K

- Replies
- 1

- Views
- 748

- Replies
- 0

- Views
- 321

- Replies
- 10

- Views
- 2K

- Replies
- 1

- Views
- 984

- Replies
- 2

- Views
- 172

- Replies
- 5

- Views
- 2K

- Replies
- 5

- Views
- 1K

- Replies
- 42

- Views
- 5K