Timeline for What is the Levi-Civita connection trying to describe?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2020 at 19:27 | comment | added | Gabe K | @MattF. I've made another question about ways to depict curvature. If you're interested I'd be happy to see an explanation of this diagram from a synthetic viewpoint. mathoverflow.net/questions/376796/… | |
| Nov 18, 2020 at 8:02 | comment | added | user44143 | Another attractive possibility is formalizing the pictures in Synthetic Differential Geometry, with $\epsilon^3=0$. It takes work to understand the algebra and logic of that approach, but it makes the geometry clear. | |
| Nov 16, 2020 at 18:50 | history | edited | Gabe K | CC BY-SA 4.0 |
Rewrote the section explaining the pictures
|
| Nov 16, 2020 at 15:59 | comment | added | Gabe K | @MattF. Thanks for making the change. To address that issue you can make $q=\exp_{\exp_p(\epsilon X)}(Tr(p,\epsilon X, \epsilon Y)).$ I'll edit that. | |
| Nov 16, 2020 at 15:56 | comment | added | user44143 | I edited the answer to use the abbreviation $q=\exp_{\exp_p(\epsilon X)}( \epsilon Y)$, but now I don't understand how that's defined, since $\epsilon Y$ is a vector in the tangent space at $p$ rather than at $\exp_p(\epsilon X)$. | |
| Nov 16, 2020 at 15:56 | comment | added | Gabe K | @RobertBryant That's a good idea. To make it fully rigorous you still need to parallel translate back to $p$, but that does seem simpler. To be honest, I was drawing a picture to give an intuitive idea without trying to be too formal about things, but it's helpful to have a good way to make everything precise. | |
| Nov 16, 2020 at 14:06 | comment | added | Robert Bryant | @GabeK: If that's all that's bothering you, why not just compare the results of following the two ways around the boundary of the exponentiated paralleogram from $p=exp_p(0)$ to $q = \exp_p(\epsilon(X+Y))$? I think that's still simpler than what you are trying to do. | |
| Nov 16, 2020 at 13:55 | history | edited | user44143 | CC BY-SA 4.0 |
introduced two abbreviations, changed all $P$'s to $p$'s, added one matching parenthesis just before the end of the long formula
|
| Nov 16, 2020 at 13:05 | comment | added | Gabe K | @RobertBryant That formula is much simpler. However, if you draw the schematic using a loop like that, it visually looks like you are taking four covariant derivatives of $Z$ rather than the commutation of two derivatives. Of course the formula I wrote down is totally unhelpful and should never be used, but the point is that it's possible to make the picture rigorous. | |
| Nov 16, 2020 at 10:13 | comment | added | Robert Bryant | @GabeK: Anyway, I think it would be easier to do this: Take the bounday of the actual parallelogram spanned by $\epsilon X$ and $\epsilon Y$ in $T_pM$, apply $\exp_p$ to it to get a closed $p$-based loop in $M$, parallel translate $Z\in T_pM$ around this loop to get a vector $\mathrm{Tr}(\epsilon X,\epsilon Y,Z)\in T_pM$, and compute that $$\lim_{\epsilon\to 0} \frac{\mathrm{Tr}(\epsilon X,\epsilon Y,Z)-Z}{\epsilon^2}=R(X,Y)Z.$$ The important thing about this particular formula is that the torsion does not appear. | |
| Nov 16, 2020 at 10:03 | comment | added | Robert Bryant | @GabeK: Unfortunately, it's hard to check your formula now because it goes off the page; MathJax doesn't render it fully on my machine. | |
| Nov 15, 2020 at 21:33 | history | edited | Gabe K | CC BY-SA 4.0 |
Fixed the formula for the curvature tensor.
|
| Nov 15, 2020 at 21:31 | comment | added | Gabe K | You're absolutely right. I thought I was being careful enough, but apparently not. I'll fix that and the other place where that mistake gets made. Hopefully there aren't too many more mistakes; that formula is pretty unwieldy. | |
| Nov 15, 2020 at 21:18 | comment | added | Robert Bryant | @GabeK: I hate to nitpick, but your final formula cannot be right. For example, given your definitions, the term $Tr(\exp_p( \epsilon X), \epsilon Y, Tr(p,\epsilon X,Z))$ does not make sense because $\epsilon Y$ is in $T_pM$, not $T_{\exp_p( \epsilon X)}M$. | |
| Nov 15, 2020 at 21:08 | comment | added | Gabe K | @RobertBryant I've edited the answer to address your point. The idea is that the diagram is infinitesimal in $X$ and $Y$ and rescaled to second order, so third order effects don't show up in the picture. It's also meant as a schematic, since the curvature and torsion live in $T_p M$ rather than the tangent space of a nearby point. | |
| Nov 15, 2020 at 21:01 | comment | added | Robert Bryant | @C.F.G: As I pointed out above, you have to take Nakahara's statement with a grain of salt. It is just not true that when the torsion vanishes identically, then the parallelogram described above closes exactly. In the torsion-free case, the error that measures the failure of closure is third order in $(X,Y)$, but it is not identically zero unless $R$ also vanishes identically. You can easily check this yourself for parallel translation on the $2$-sphere (or the hyperbolic disk) using the Levi-Civita connection, where you can do the calculation very explicitly. | |
| Nov 15, 2020 at 20:13 | history | edited | Gabe K | CC BY-SA 4.0 |
I added a discussion of how to interpret the pictures.
|
| Nov 15, 2020 at 18:54 | comment | added | C.F.G | @RobertBryant: I am not aware of details but this (and the second figure) is a paragraph of Nakahara's Book: Thus, the torsion tensor measures the failure of the closure of the parallelogram made up of the small displacement vectors and their parallel transports. | |
| Nov 15, 2020 at 17:44 | comment | added | Robert Bryant | @GabeK: Actually, one has to be rather careful about this: Even when the torsion is zero, the ' $XY$ parallelogram' described above will not generally close up when the curvature is nonzero. Thus, one cannot literally make sense of $R(X,Y)Z$ as a vector at the putative 'fourth vertex'. What is true is that for a general connection, what is drawn as the red difference in the second picture is an actual coordinate displacement that is essentially $T(X,Y)$, but, when the torsion is zero, this (nonzero) dispacement is essentially $R(X,Y)(X{+}Y)$, i.e., it is of total degree $3$ in $X$ and $Y$. | |
| Nov 15, 2020 at 17:43 | comment | added | Andrew NC | I have seen these diagrams before, and I would love for them to inform my intuition, but the vectors in these diagrams don't live in the same vector spaces, so I don't know how to understand them... | |
| Nov 15, 2020 at 17:24 | history | answered | Gabe K | CC BY-SA 4.0 |