Timeline for Why is this subset associated to a $2$-tensor dense?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2023 at 3:59 | vote | accept | Matheus Andrade | ||
| Aug 29, 2023 at 23:20 | comment | added | Matheus Andrade | @RobertBryant Thanks a lot! You're absolutely right. | |
| Aug 29, 2023 at 22:57 | answer | added | Romain Gicquaud | timeline score: 4 | |
| Aug 29, 2023 at 22:54 | comment | added | Robert Bryant | There certainly is something wrong with the second claim: If $S\equiv0$ and $\mathrm{dim}(M)>1$, then $M_S = M$ is open and connected (if $M$ is connected), but the eigenvalues of $S$ are not distinct. I think you are being careless about the statement of the second property. I think it should be something like "The $E_S$ distinct eigenvalues of $S$ on a connected component of $M_S$ are smooth". | |
| Aug 29, 2023 at 21:57 | comment | added | Romain Gicquaud | @RBega2: You are not providing a counterexample. In what you say, the complementary of the set $M_S$ is the frontier of the support of $\theta$. Outside the support of $\theta$, $S$ only has one eigenvalue (namely zero) but this set is open so it provides a neighborhood for all its points. | |
| Aug 29, 2023 at 21:42 | comment | added | RBega2 | There is no way this is true using only smoothness. For instance take a compactly supported one form $\theta$ and consider $\theta^2$. However, if $S$ satisfies a nice enough PDE, then this should be some sort of unique continuation result. | |
| Aug 29, 2023 at 21:07 | history | asked | Matheus Andrade | CC BY-SA 4.0 |