Timeline for Hodge theory in higher eigen-spaces?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2022 at 6:50 | comment | added | Sebastian Goette | While the dimensions of individual eigenspaces depend on the choice of metric, one can cook up invariants like Ray-Singer torsion by considering all these dimensions at once. To get something truely topological, one needs to twist with an acyclic flat vector bundle. | |
| Apr 11, 2022 at 1:19 | comment | added | Jesse Silliman | You may be interested in Bott, "Morse theory indomitable", which explains an idea of Witten's for how the eigenspaces of a deformed Laplacian, $\Delta_s$, can be related to Morse theory. | |
| Apr 11, 2022 at 0:03 | comment | added | paul garrett | As @WillSawin comments, the dimensions of eigenspaces of Laplace-Beltrami operators are not invariant under perturbations of metric. For example, the "square" torus has Laplacian eigenvalues essentially equal to sums of integer squares. The multiplicities are understandable via Gaussian integers. Slightly changing one of the lengths makes all the multiplicities $1$, instead (for incommensurate side lengths this follows from the obvious lazy proof.) :) | |
| Apr 10, 2022 at 22:03 | comment | added | Will Sawin | I meant the dimension wouldn't be invariant. Consider the case of a torus, with the flat metric. Eigenspaces can be computed by Fourier decomposition. | |
| Apr 10, 2022 at 21:24 | comment | added | Student | The space of harmonic forms $\mathcal{H}_g$ aren't invariant of the metric, but their dimensions are. Are the dimensions of $ker(\Delta - \lambda)$ at least invariant of metric? | |
| Apr 10, 2022 at 20:30 | comment | added | Will Sawin | This wouldn't be invariant of the metric which would make it, at least naively, hard to recover topological information from it. | |
| Apr 10, 2022 at 19:43 | history | asked | Student | CC BY-SA 4.0 |