Timeline for Interpretation of Riemann tensor antisymmetry
Current License: CC BY-SA 3.0
9 events
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| Apr 3, 2013 at 0:29 | comment | added | Ryan Budney | @DAVID: here are my notes. The interpretation of the curvature tensor appears at the start of Chapter 4. I use bundle notation throughout so chapters 1,2 and 3 might be useful warm-up. rybu.org/?q=mathmultimedia | |
| Feb 11, 2013 at 8:48 | comment | added | Ryan Budney | This argument appears around page 703 in Marcel Berger's "Panoramic View of Riemannian Geometry". My version of it is an elaboration on a fairly standard perspective on the curvature tensor, but most DG texts kind of skimp on it. Berger states it but does not go into as much detail as I describe above. I'll have some DG lecture notes up on my website in summertime where these kinds of details will be spelled-out in elaborate detail -- I'm teaching a DG course this semester so I'm thinking about these kinds of expositional issues at the moment. | |
| Feb 11, 2013 at 7:40 | vote | accept | DAVID | ||
| Feb 11, 2013 at 7:38 | comment | added | DAVID | thanks Ryan . that is great .but I am really interested in reading more about what you said.where I can find these stuff? | |
| Feb 11, 2013 at 7:36 | vote | accept | DAVID | ||
| Feb 11, 2013 at 7:40 | |||||
| Feb 10, 2013 at 17:55 | comment | added | Ryan Budney | @Robert: That wasn't the "this" I was referring to. :) I suppose I wrote my response too late in the evening. What I was getting at was that this interpretation forces us to think of the values of $R(v,w)$ as tangent vectors in the group of linear automorphisms of $T_p N$. | |
| Feb 10, 2013 at 15:22 | comment | added | user21349 | In case the OP doesn't know some of the sophisticated terminology and notation in the answer (I don't), here's what I think is the same explanation done with crayons. Say the vectors $v$ and $w$ looked like this placed tail to tail: |____. Changing $w$ to $-w$ gives this: ____|. Parallel transport takes you around the new rectangle in the opposite handedness. Since tensors are linear, flipping the sign on $w$ flips the sign of the result. Therefore reversing the direction of parallel transport flips the sign. Swapping $v$ and $w$ also reverses the direction, so it must also flip the sign. | |
| Feb 10, 2013 at 14:10 | comment | added | Robert Bryant | @Ryan: Actually, the reason you give in your final paragraph is not related to the reason you give in the rest of your argument. One would have $R(v,w)=-R(w,v)$ for any connection whether it was Riemannian or not, and for the reason you give: Reversing the path will invert the holonomy transformation. However, only when you know that the holonomy preserves the metric will you know that it is an orthogonal transformation, so that the skew-symmetry of the tangent space to the orthogonal group comes into play, which is a different reason from the first, and what the OP is really asking about. | |
| Feb 10, 2013 at 11:12 | history | answered | Ryan Budney | CC BY-SA 3.0 |