Timeline for Delta distribution on manifolds
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20 at 11:45 | vote | accept | psl2Z | ||
| Oct 18 at 18:34 | history | became hot network question | |||
| Oct 18 at 16:10 | comment | added | Ben McKay | I would suggest using the language of currents, following de Rham's book Differentiable Manifolds: Forms, Currents, Harmonic Forms. | |
| Oct 18 at 15:36 | answer | added | Dmitri Pavlov | timeline score: 9 | |
| Oct 18 at 13:58 | comment | added | Igor Khavkine | This may be just a matter of terminology what you call "distribution density" is also sometimes (perhaps more often even) referred to as just a "distribution". In that sense your $\delta_p$ is already a "natural delta distribution", end of story. You can tensor distributions with the sections of any other vector bundle (including the volume form bundle). If that vector bundle has no canonical (not everywhere zero) sections, then you lose the "natural deltas" in the tensor product. | |
| Oct 18 at 10:59 | comment | added | Gro-Tsen | It seems to be that you've convincingly argued that the answer to your own question is no. If there were a “natural” way to define a distribution $δ_p$ for every $p$, then comparing (chartwise) it to the distribution density $δ_p$ would give a smooth density (alternatively: find the smooth density on $M$ so that the integral of the putative distribution $δ_p$ against this density gives $1$ at every $p$); and no such “natural” density exists, so neither can a “natural” distribution $δ_p$. I'm just rephrasing pretty much what you yourself explained: why does it not convince you? | |
| Oct 18 at 10:31 | history | asked | psl2Z | CC BY-SA 4.0 |