Timeline for Is there a relationship between the Jacobian at a point and the curvature at that point?
Current License: CC BY-SA 3.0
17 events
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| Sep 5, 2017 at 11:16 | comment | added | user9563 | Ah of course, sorry I couldn't explain myself properly there. I'm actually interested in the Euler characteristic, and I was wondering if there's a way to use the Gauss-bonnet theorem to get at that, for which I would need an estimate of local curvature. That's the full story. Thinking about this just made me wonder how much information about the local curvature can be extracted from the estimated Jacobian. | |
| Sep 5, 2017 at 10:58 | comment | added | Tobias Fritz | Concerning the Takens embedding, en.wikipedia.org/wiki/Takens%27_theorem states explicitly that it "does not preserve the geometric shape of structures in phase space". Hence whatever notion of curvature you use for the embedding manifold, it will not reflect the geometrical properties of real-world dynamical systems that you're after. | |
| Sep 5, 2017 at 10:57 | comment | added | Tobias Fritz | The circle has curvature only because you're thinking of it as embedded in Euclidean space. This concept of curvature is not invariant under symplectic (=canonical) transformations, and therefore is not something that you can talk about at the level of symplectic manifolds. | |
| Sep 5, 2017 at 10:22 | comment | added | user9563 | Thanks again. I don't know anything about symplectic manifolds so I am a little confused. Talking simple mindedly, the Hamiltonian $H=\frac{1}{2}(x^2+y^2)$ describes a circle in the phase space. But the circle has curvature right? $\frac {1}{R}$ or $\frac{1}{\sqrt{2} H}$ in this case. As for the context of my question. I'm interested in analyzing geometrical properties of real world dynamical systems, from some experimental data I can reconstruct a phase space (Takens embedding), and can estimate local Jacobians. So I was wondering under what conditions can I also extract the curvature? | |
| Sep 5, 2017 at 8:54 | comment | added | Tobias Fritz | @user9563: Exactly, $F$ being a Hamiltonian vector field on a symplectic manifold would be one possible relation to a geometrical structure. However, since a symplectic manifold doesn't have any notion of curvature, again your question doesn't really make sense in this case. Perhaps you should elaborate on how your question came up and what application you have in mind. | |
| Sep 5, 2017 at 7:28 | comment | added | user9563 | @TobiasFritz I see, thank you for the answer (and your patience). I want to understand more what you meant by "since you have not assumed anything about how $F$ relates to the geometry". Can you give an example of how $F$ can relate to the geometry? Like when $F$ is derived from a Hamiltonian? | |
| Sep 5, 2017 at 6:44 | comment | added | Tobias Fritz | @user9563: So your manifold is Riemannian or otherwise comes equipped with a connection? Then the answer is no, the Jacobian does not tell you anything about the curvature, since you have not assumed anything about how $F$ relates to the geometry. If you don't assume anything, then you also can't conclude anything. | |
| Sep 5, 2017 at 5:15 | comment | added | user9563 | @PiyushGrover Curvature of the manifold that the trajectories lie on. So for example if the trajectories of $\dot{\textrm{x}}=F(\textrm{x})$ lie on a torus then does the Jacobian evaluated at a point $\textrm{x}_i$ (or some transformation) contain any information about the curvature obtained by doing parallel transport around $\textrm{x}_i$. | |
| Sep 5, 2017 at 5:05 | comment | added | user9563 | @TobiasFritz Thanks for making me think about it a little more. I mean the curvature one would get by doing parallel transport around a point. | |
| Sep 4, 2017 at 15:20 | comment | added | Piyush Grover | curvature of what ? | |
| Sep 4, 2017 at 14:33 | comment | added | Tobias Fritz | @user9563: But what do you actually mean by curvature? The usual notion of curvature is the curvature of a connection, but it's not clear which connection you'd be referring to in your context. | |
| Sep 4, 2017 at 12:21 | comment | added | Neal | One reference for the theorem that every nonzero vector field is locally a coordinate field is an early chapter of F. Warner's book Introduction to Differentiable Manifolds and Lie Groups | |
| Sep 4, 2017 at 11:06 | comment | added | user9563 | I'm wondering whether the Jacobian (or some transformation) contains any information about the local curvature. No, I wasn't aware of that theorem. Looking it up now! | |
| Sep 4, 2017 at 11:03 | history | edited | user9563 | CC BY-SA 3.0 |
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| Sep 4, 2017 at 10:39 | comment | added | Tobias Fritz | What kind of curvature are you referring to, in the absence of a metric or a connection? And do you know the theorem stating that every nonzero vector field is locally a coordinate vector field? | |
| Sep 4, 2017 at 10:35 | review | First posts | |||
| Sep 4, 2017 at 10:52 | |||||
| Sep 4, 2017 at 10:34 | history | asked | user9563 | CC BY-SA 3.0 |