Timeline for If there exists a function on a Riemannian manifold such that its Hessian matrix is the identity matrix?
Current License: CC BY-SA 4.0
14 events
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| Feb 10, 2023 at 10:52 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor typo fixes
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| Feb 10, 2023 at 9:48 | answer | added | Vladimir S Matveev | timeline score: 4 | |
| Feb 8, 2023 at 18:10 | comment | added | Robert Bryant | @RomainGicquaud: Actually, there are many (incomplete) Riemannian manifolds $(M^n,g)$ that admit a function $f$ that satisfies $\mathrm{Hess}(f) = g$. For example, let $(N^{n-1},h)$ be any Riemannian manifold, let $M= \mathbb{R}^+\times N$, and let $g = \mathrm{d}r^2 + r^2\,h$ be the usual cone metric for $(N,h)$. Then $f = \tfrac12 r^2$ has $\mathrm{Hess}(f) = g$. However, $(M,g)$ won't be complete, and including a point for $r=0$ gives a complete smooth metric only if $(N,g)$ is a unit sphere in $\mathbb{R}^n$. | |
| Feb 8, 2023 at 14:31 | comment | added | Romain Gicquaud | It is known that the existence of a (non trivial) function satisfying $\mathrm{Hess V} = V g$ characterize the hyperbolic space (as far as I remember this is proven by X. Wang in "On the uniqueness of the AdS spacetime"). I would not be surprised if the existence of a function satisfying $\mathrm{Hess f} = g$ characterize the Euclidean space. | |
| S Feb 8, 2023 at 11:50 | history | edited | gmvh | CC BY-SA 4.0 |
Corrected grammar and spelling, added top-level tag
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| S Feb 8, 2023 at 11:50 | history | suggested | Ievgeni | CC BY-SA 4.0 |
exsit => exist
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| Feb 8, 2023 at 11:44 | review | Suggested edits | |||
| S Feb 8, 2023 at 11:50 | |||||
| Feb 8, 2023 at 11:40 | history | edited | Davidi Cone | CC BY-SA 4.0 |
deleted 13 characters in body
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| Feb 8, 2023 at 11:39 | history | edited | Davidi Cone | CC BY-SA 4.0 |
added 98 characters in body
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| Feb 8, 2023 at 11:28 | comment | added | Davidi Cone | Sorry! Let $ \nabla $be the Levi-Civita connection of $(M, g) $. For the differentiable function $v$ on $(M, g) $, define $ \nabla v $ as the gradient of $v$. Use $\nabla^ 2 v $ to represent the Hessian matrix of $v $. Locally, it can be expressed as $ \nabla_ {i j} v=\nabla_ i\left(\nabla_j v\right)-\Gamma_ {i j}^k \nabla_ k v$. | |
| Feb 8, 2023 at 10:40 | comment | added | abx | Please define first what you call the Hessian matrix. | |
| Feb 8, 2023 at 10:07 | review | Close votes | |||
| Feb 15, 2023 at 3:05 | |||||
| S Feb 8, 2023 at 9:39 | review | First questions | |||
| Feb 8, 2023 at 18:59 | |||||
| S Feb 8, 2023 at 9:39 | history | asked | Davidi Cone | CC BY-SA 4.0 |