Timeline for A name for a pseudo-Riemannian manifold that admits no nonzero null vectors
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2023 at 11:12 | vote | accept | Martim Pereir | ||
| Oct 11, 2021 at 15:49 | comment | added | Pierre PC | @NicolasTholozan Although very true, I also like to think of them as degenerate metrics on the cotangent. Depending on what the OP has in mind, maybe their $T\mathbb R$ is actually a $T^*\mathbb R$. That said, even this case would not be considered a sub-Riemannian manifold by many authors. | |
| Oct 10, 2021 at 21:12 | comment | added | Nicolast | sub-Riemannian metrics are more like metrics that are "infinite in some directions" | |
| Oct 10, 2021 at 13:20 | comment | added | Pierre PC | This is not exactly what you are looking for, but maybe you'd be interested in sub-Riemannian geometry. | |
| Oct 9, 2021 at 9:22 | comment | added | Martim Pereir | @Pierre: Do the "metrics" considered in my example fit into some more general class? | |
| Oct 9, 2021 at 7:59 | comment | added | Martim Pereir | @Pierre: yes of course, thanks for pointing this out. | |
| Oct 8, 2021 at 18:08 | comment | added | Pierre PC | Your example is not considered a pseudo-Riemannian manifold, since at zero the metric is degenerate. | |
| Oct 8, 2021 at 12:21 | comment | added | Martim Pereir | Thank you for the answer! However are you sure that one can always construct a globally defined non-zero vector field. Take the manifold $\mathbb{R}$ and endow $T\mathbb{R}$ with a Riemannian metric that is semi-definite at $0$ and positive definite everywhere else (we can do so by multiplying the metric by a smooth function that vanishes at zero and only at zero). Then the bump fuction argument doesn't seem to work since the "support of the non-positive definitness" is a single point. | |
| Oct 8, 2021 at 9:30 | history | answered | Nicolast | CC BY-SA 4.0 |