Timeline for What is the best way to draw curvature?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2020 at 14:29 | comment | added | Gabe K | @RobertBryant That's a good way to explain it. It's also a nice way to start thinking about the Theorema Egregium (i.e. the geodesic curvature of the loop vs the curvature inside), so that's an added benefit. | |
| Nov 20, 2020 at 14:28 | comment | added | Robert Bryant | Corrections to the above: "equitorial" should have been "equatorial" and, for the last sentence: "Does this capture the desired relationship between 'local holonomy' (i.e., parallel transport) and curvature?" [I was distracted by running up against the character limit.] | |
| Nov 20, 2020 at 13:53 | comment | added | Robert Bryant | About the second picture (which, by the way, is bad; the transport along the equitorial segment is visibly not parallel): The OP doesn't like it because it uses 'global' information about the sphere rather than 'local' information. How about this instead: "On any closed oriented surface, parallel translation around any loop (large or small) rotates vectors counterclockwise by an amount equal to the Gauss curvature averaged over the region enclosed by the loop (counted with signed multiplicity)." Does this capture the desired relationship between 'local holonomy' and parallel transport? | |
| Nov 20, 2020 at 12:38 | comment | added | J W | For pedagogical questions, you might be interested to know of the existence of matheducators.stackexchange.com. | |
| Nov 20, 2020 at 7:21 | answer | added | Nitin Nitsure | timeline score: 3 | |
| Nov 19, 2020 at 17:20 | comment | added | Gabe K | @DeaneYang There's probably no one picture that captures everything about curvature. However, having multiple pictures that show different aspects is helpful, at least for my intuition. | |
| Nov 19, 2020 at 16:41 | comment | added | Deane Yang | Ultimately, the symmetries of local geometric tensors are all consequences of the property that partial derivatives commute. | |
| Nov 19, 2020 at 16:40 | comment | added | Deane Yang | Is it really reasonable to expect a picture that shows the symmetry properties of the full curvature tensor? I usually prefer simpler pictures like the ones shown by @MohammadGhomi for sectional curvature, which is a function on the set of 2-planes and view the Riemann tensor as the natural linear algebraic extension of the sectional curvature to the full space of $2$-tensors. This is analogous to how I think of the Hessian of a function geometrically. Offhand, I also don't know how to show geometrically that the Hessian is symmetrc. | |
| Nov 19, 2020 at 15:23 | comment | added | Mark Wildon | It defies summarising, and doesn't go into the tensor formulation, but still I recommend the discussion of curvature in the chapter on curved space in the Feynmann lectures. feynmanlectures.caltech.edu/II_42.html | |
| Nov 19, 2020 at 14:21 | comment | added | Robert Bryant | @RobertMastragostino: Actually, I'm puzzled by your comment. After all $T^\nabla(X,Y)$ does not involve any derivatives, neither of $\nabla$, $X$, nor $Y$. The formula for $T^\nabla(X,Y)$ is deliberately constructed so that all of the derivatives in the individual terms cancel out in the sum. Even when one takes $X$ and $Y$ to be coordinate vector fields, $T^\nabla(X,Y)$ can be nonzero. Thus, $[X,Y](p)$ should not show up in a picture of $T^\nabla(X,Y)(p)=T^\nabla(X(p),Y(p))$, and the black arrows need not be 'flow lines' of $X$ and $Y$; all that extraneous stuff will wash out in the end. | |
| Nov 19, 2020 at 13:40 | comment | added | Robert Bryant | @GabeK: Hmmm. So your goal is to draw a picture that would make it intuitive that two vector-valued $3$-forms are equal? Such a picture would have to be something completely trivial on a surface, since, in that case the first Bianchi identity is just $0=0$. Then, in $3$-dimensions, it would have to be something involving the parallelepiped generated by the vectors $X$, $Y$, and $Z$. (Or maybe something more complicated, since visualizing $T(X,Y)$ alone involves a pentagon.) | |
| Nov 19, 2020 at 13:40 | comment | added | Steven Gubkin | @RobertMastragostino Can you flesh out your comment into an answer, with the picture you have in mind? I think yours is the only comment/answer which actually addresses the question asked. | |
| Nov 19, 2020 at 11:42 | answer | added | Sebastian | timeline score: 6 | |
| Nov 19, 2020 at 7:08 | answer | added | Mozibur Ullah | timeline score: -4 | |
| Nov 19, 2020 at 3:31 | history | became hot network question | |||
| Nov 18, 2020 at 22:16 | comment | added | Gabe K | Sorry, I meant to say "without considering torsion or geodesics." Parallel transport is central to the entire idea of curvature. | |
| Nov 18, 2020 at 22:06 | answer | added | Gabe K | timeline score: 13 | |
| Nov 18, 2020 at 21:58 | comment | added | Gabe K | @RobertBryant What you are saying is correct that curvature is a more general invariant, and that you can discuss it without considering torsion or parallel transport. However, the reason to have a picture where you can see both curvature and torsion is to understand things like the first Bianchi identity. $$\mathfrak{S}(R(X, Y) Z)=\mathfrak{S}\left(T(T(X, Y), Z)+\left(\nabla_{X} T\right)(Y, Z)\right)$$ | |
| Nov 18, 2020 at 21:58 | answer | added | Joseph O'Rourke | timeline score: 17 | |
| Nov 18, 2020 at 21:48 | comment | added | Robert Bryant | You are confounding separate things: Curvature is defined for any linear connection on any vector bundle, so interpreting $R(X,Y)s$ for $s$ a section and $X$ and $Y$ vector fields should not involve geodesics. Rather, one should, after verifying that $R(X,Y)s)$ is tensorial and multilinear, take $X$ and $Y$ to be coordinate vector fields, where the paralleogram is obvious. Second, if you want to interpret torsion for the tangent bundle (which is the only place it makes sense), have a look at mathoverflow.net/questions/133342/… | |
| Nov 18, 2020 at 21:15 | answer | added | Mohammad Ghomi | timeline score: 40 | |
| Nov 18, 2020 at 20:17 | answer | added | Ian Agol | timeline score: 16 | |
| Nov 18, 2020 at 19:46 | comment | added | Robert Mastragostino | If X and Y don't commute then their flows don't form a parallelogram like that. You need a pentagon that adds an [X,Y] side at the far end. Then both pictures are about how vectors rotate around loops, and the first picture comes about when you try to specify the loop by the flows of two vector fields. | |
| Nov 18, 2020 at 19:27 | history | asked | Gabe K | CC BY-SA 4.0 |