Timeline for Reference for non-parallel harmonic $k$-forms
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2022 at 0:32 | comment | added | Robert Bryant | @C.F.G: Well, I chose 3 dimensions just because it was easier, but there are certainly examples in dimension 4 as well. Let $M^4$ be a compact orientable 4-manifold that is simply-connected, not a product, does not support a Kähler metric, and yet does have non-trivial Betti number $b_2$. (It's easy to construct such examples, for example, the oriented connected sum of $\mathbb{CP}^2$ with itself.) Then every metric $g$ on it will have non-zero harmonic $2$-forms (by deRham's Theorem), but none of them can be $g$-paralell. | |
| Jul 6, 2022 at 13:22 | comment | added | C.F.G | Dear prof. Bryant, Your argument mainly based on that the dimension of $M$ is odd, Do you think that one can construct an example in even dimensions? | |
| Feb 23, 2021 at 12:21 | comment | added | Robert Bryant | @C.F.G: I'm not sure what you mean. The deRham cohomology group is a topological invariant, but the choice of a form representing a given cohomology class is not a topological invariant, there is not even a choice of a form that is a diffeomorphism invariant, since the diffeomorphisms homotopic to the identity don't fix any form of positive degree other than the form that vanishes identically. | |
| Feb 23, 2021 at 11:10 | comment | added | C.F.G | That's seems reasonable but I compare it with homotopy(homology) groups like loops homotopy class that are invariant under homeo. but why these are not invariant under homeo intuitively? | |
| Feb 23, 2021 at 10:34 | comment | added | Robert Bryant | @C.F.G: There's no reason to believe that a form of positive degree that is harmonic for one metric would be harmonic for a different metric. If $\omega$ is closed (which doesn't depend on the metric), you are asking whether $\mathrm{d}(\ast_h\omega) = 0$ for a metric $h$, which is a non-trivial first-order system of pde for $h$ as long as $\omega$ is nonzero, and it is generally weaker than the equation $\nabla_h\omega = 0$ (which is also a first order equation for $h$). | |
| Feb 23, 2021 at 7:34 | vote | accept | C.F.G | ||
| Feb 23, 2021 at 6:25 | comment | added | C.F.G | Thanks professor. Do you know that these non-parallel forms are invariant under metric deformation? i.e. for a deformation of $g$ like $h$, $\nabla_h \omega\neq 0$ and $\Delta_h\omega=0$? (something like this post) | |
| Feb 23, 2021 at 1:12 | history | answered | Robert Bryant | CC BY-SA 4.0 |