Timeline for Can this problem be rephrased as optimization on a manifold?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2020 at 9:59 | comment | added | Liviu Nicolaescu | In this case minima/maxima exist. Such a point is a critical point, i.e., the differential of $f$ at such a point is zero. Find all the critical points (not easy in general) and then pick the critical point where $f$ has the smallest value (not easy). For a Random function will have many critical points and very few will be local minima. | |
| Jun 26, 2020 at 21:11 | comment | added | user8469759 | @LiviuNicolaescu I had a thought about this problem I don't think apart from compactness I cannot assume more. | |
| Jun 4, 2020 at 15:13 | comment | added | Liviu Nicolaescu | That will help. The more specific, the more helpful the answer. | |
| Jun 4, 2020 at 15:01 | comment | added | user8469759 | @LiviuNicolaescu I'll make my question as more specific as possible then. | |
| Jun 4, 2020 at 14:26 | comment | added | Liviu Nicolaescu | Regular submanifold is no assumption at all since by Whitney's embedding says that manifold can be properly embedded as a regular submanifold. Is this manifold compact? If it is not compact, then maybe there is no absolute maximum or there could be many local maxima and no absolute maxiumum, or there could be an absolute maximum. A special case of your question is: given a function $f:\mathbb{R}\to\mathbb{R}$ find its maximum. possible answers to this question are all possible in the more general case, and a bit more. | |
| Jun 4, 2020 at 13:23 | comment | added | user8469759 | And again, given my assumption (i.e. $f : \mathcal{M} \to \mathbb{R}$) in your list you can pick the scalar curvature. I think I've given already a stricter assumption already, i.e. regular submanifolds in $\mathbb{R}^n$, or isn't this enough? | |
| Jun 4, 2020 at 12:44 | comment | added | Liviu Nicolaescu | Here again, there are many flavors of curvature: the curvature tensor, the sectional curvatures, the Ricci curvature and the scalar curvature. Only the last is a scalar quantity and, usually, only scalar quantities can be optimized. Without any assumption on the manifold or the quantity you cannot expect a precise answer. | |
| Jun 4, 2020 at 12:32 | history | edited | user8469759 |
edited tags
|
|
| Jun 4, 2020 at 12:18 | comment | added | user8469759 | @LiviuNicolaescu As an example I'm trying to optimize curvature, I have this manifold and I want to find the point whose curvature is maximum (or a local optimizer). | |
| Jun 4, 2020 at 11:09 | comment | added | Liviu Nicolaescu | What are you trying to optimize? What is the role of the Riemann metric? The Fermat principle works on smooth manifold. If $x_0$ is a local minimum of $f$ then $x_0$ is a critical point of $f$, i.e., $df(x_0)=0$. Also at a critical point there is a well defined concept of Hessian, and if this Hessian is positive definite then that critical point is a local min. | |
| Jun 4, 2020 at 10:07 | review | First posts | |||
| Jun 4, 2020 at 10:27 | |||||
| Jun 4, 2020 at 10:04 | history | asked | user8469759 | CC BY-SA 4.0 |