Timeline for Why is there no symplectic version of spectral geometry?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2019 at 21:31 | comment | added | AlexArvanitakis | @ChristianBueno I think that formula is wrong. The correct formula is $\Delta f=\sqrt{g}^{-1} \partial_i(\sqrt{g} g^{ij}\partial_j f)$. | |
| Jun 26, 2018 at 6:31 | comment | added | Christian Bueno | @Hadrian Quan I've never seen that formula $\Delta_g(x^ix^j)=g_{ij}(x)$ for the metric in terms of the Laplacian before. I can see why this would be true for normal coordinates about $x$ (which is a little circular), but not more generally. Do you have a reference? | |
| Oct 26, 2017 at 12:49 | comment | added | Hadrian Quan | This is true, but this only demonstrates that the metric is not determined by the spectrum of the $\Delta_g$ alone. What I said above still holds. It would be very convenient if linear operators on Hilbert spaces could be understood entirely from the data of their spectrum, but this is too much too hope for in general. | |
| Oct 25, 2017 at 11:43 | comment | added | Vít Tuček | There are many examples of isospectral manifolds which are not isometric. en.wikipedia.org/wiki/Isospectral#Isospectral_manifolds | |
| Oct 21, 2017 at 15:43 | history | answered | Hadrian Quan | CC BY-SA 3.0 |