Timeline for A Lagrangian connection and its algebraic interpretation
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2017 at 12:57 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added an explicit description of the condition in terms of the Levi-Civita connection.
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| Jun 14, 2017 at 8:37 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added a comment expanding on the answer.
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| Jun 14, 2017 at 8:29 | comment | added | Robert Bryant | @მამუკაჯიბლაძე: Yes, they are nearly always different. Just take the case of the connection $\nabla$ on $T^*\mathbb{R}$ that satisifes $\nabla(\mathrm{d}x) = 0$. Then $\nabla^*(\partial/\partial x) = 0$, but, if $g = f(x)\,\mathrm{d}x^2$ where $f$ is positive and nonconstant, then you'll find that $$\nabla'(\partial/\partial x) = f'(x)/f(x)\,\mathrm{d}x\otimes \partial/\partial x,$$ so $\nabla'\not=\nabla^*$ even in this simple case. | |
| Jun 13, 2017 at 22:17 | comment | added | მამუკა ჯიბლაძე | It was a surprise for me that $\nabla'$ and $\nabla^*$ may be different. Are there examples where it is easy to see that difference? | |
| Jun 13, 2017 at 19:49 | comment | added | Ali Taghavi | @RobertBryant Thank you very much for your answer. | |
| Jun 13, 2017 at 19:46 | vote | accept | Ali Taghavi | ||
| Jun 13, 2017 at 11:43 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added some comments and extra information
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| Jun 13, 2017 at 11:10 | comment | added | მამუკა ჯიბლაძე | Thank you, I believe in presence of your last comment no further clarification is necessary. | |
| Jun 13, 2017 at 11:02 | comment | added | Robert Bryant | @მამუკაჯიბლაძე: I guess the issue is whether one regards this identification at "natural". My point is that the question makes sense for connections on $T^*M$ in general, whether there is a Riemannian metric specified or not, and this is the "natural" version of the question. Of course, you are right that in the category of Riemannian manifolds, you can identify $TM$ with $T^*M$; it's just that the metric merely complicates the question. Nevertheless, I can answer the question in the form it was asked, if you like. | |
| Jun 13, 2017 at 9:36 | comment | added | მამუკა ჯიბლაძე | But the question is about Riemannian manifolds, so the identification of $TM$ with $T^*M$ is available, no? | |
| Jun 13, 2017 at 8:24 | history | answered | Robert Bryant | CC BY-SA 3.0 |