Timeline for Do eigenfunctions determine the geometry of a manifold? If so, do finitely many suffice?
Current License: CC BY-SA 4.0
7 events
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| Aug 30, 2022 at 13:06 | vote | accept | Elchanan Solomon | ||
| Jul 2, 2021 at 18:29 | comment | added | Robert Bryant | @WillSawin: I agree. That pointwise dimension bound exists; it grows at a rate proportional to $n^6$, though, which seems far too high. The family of ideals in $J^3_x$ that define a $Q(g)_x$ depends on $O(n^2)$ parameters, so one shouldn't need $O(n^6)$ parameters to pick one out. It seems harder to figure out whether one can prove 'global genericity' with some bound that depends on the geometry, as was done in the Bates article that was cited, but I agree that, if this kind of genericity holds, then compactness will prove that some finite subset of $\mathcal{E}(g)$ is sufficient. | |
| Jul 2, 2021 at 18:20 | comment | added | Will Sawin | This linear independence is given by the nonvanishing of some determinant and thus is an open condition, so if we can check that there are enough linearly independent such eigenvectors at any given point, then finitely many eigenfunction suffice in a neighborhood of a given point, and by compactness finitely many suffice overall. | |
| Jul 2, 2021 at 18:17 | comment | added | Will Sawin | I think we can bound the number of elements needed to determine $Q(g)_x$, if any do at all. Each jet in $J_x^3$ determines a vector in $\operatorname{Sym}^2 J_x^3$, and the homogeneous quadratic equations satisfied by the jet correspond to the perpendicular subspace to the vector. So if we have $\dim \operatorname{Sym}^2 J_x^3 -n $ linearly independent such vectors, then $Q(g)_x$ is determined as the perpendicular subspace to the space generated by these vectors. | |
| Jul 2, 2021 at 18:02 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 2 characters in body
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| Jul 2, 2021 at 17:47 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed some typos.
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| Jul 2, 2021 at 15:59 | history | answered | Robert Bryant | CC BY-SA 4.0 |