Timeline for Does any warped product metric admit a function with hessian proportional to the metric?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2023 at 3:58 | comment | added | Matheus Andrade | @WillieWong it does. Thanks again! | |
| Dec 1, 2023 at 4:12 | comment | added | Willie Wong | @MatheusAndrade I posted a full proof of the warped product decomposition at the other question. Hopefully that helps you. | |
| Dec 1, 2023 at 3:06 | comment | added | Matheus Andrade | I see. Thanks for the clarifications! | |
| Dec 1, 2023 at 3:00 | comment | added | Willie Wong | The above is not a full proof; it is an argument providing a moral justification. The key is that the base manifold embeds as a totally geodesic submanifold of the warped product. So even if the level sets of $\varphi$ factor, the fact that these leaves have extrinsic curvature indicates that you cannot group part of it together into the base. | |
| Dec 1, 2023 at 2:54 | vote | accept | Matheus Andrade | ||
| Dec 1, 2023 at 2:54 | comment | added | Matheus Andrade | @WillieWong thanks! But it's been a long day for me and I'm still having some trouble putting your arguments together. It's probably obvious but I don't see why the level sets being totally umbilical in addition to the gradient vector field being geodesic implies the base is one-dimensional. I did try looking up the references in that link but it hasn't helped me. | |
| Dec 1, 2023 at 2:44 | comment | added | Willie Wong | And when $\psi = 0$ the example of $\mathbb{R}^n$ shows that it can also be a warped product over a higher dimensional base, but it is still definitely a warped product over a one dimensional base. | |
| Dec 1, 2023 at 2:42 | comment | added | Willie Wong | @MatheusAndrade in the accepted answer to the question you linked to, there is a reference. But here's a quick discussion. Suppose $\nabla^2\varphi = \psi g$, let $X = \nabla\varphi$ the gradient vector field, multiplying both sides by $X$ you find $\nabla_X X = \psi X$ and hence $X$ is geodesic. Let $Y,Z$ be such that $Y(\varphi) = Z(\varphi) = 0$, then $\langle \nabla_Y X,Z\rangle = \psi \langle Y,Z\rangle$ tells us that the level sets of $\varphi$ are totally umbilic. So unless $\psi = 0$ you cannot expect the base to have higher dimensions. | |
| Dec 1, 2023 at 1:52 | comment | added | Matheus Andrade | Thanks! But why must the base be one-dimensional? I can't see the reason. | |
| Dec 1, 2023 at 1:13 | history | answered | Jeffrey Case | CC BY-SA 4.0 |