Timeline for Stokes's theorem for pseudo-differential forms
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 29, 2021 at 10:14 | comment | added | Liviu Nicolaescu | @Theone I'm glad it helped. I like that book myself. | |
| Jan 29, 2021 at 3:04 | comment | added | Theone | @LiviuNicolaescu, Thanks for the reference, too! That made everything very clear. It's also a very nicely written book. If you want to put this into an answer, I would approve it. | |
| Jan 29, 2021 at 2:57 | comment | added | Theone | @BertramArnold Thanks for the reference! I hadn't heard of supermanifolds before, but I'll take a look at them! | |
| Jan 28, 2021 at 15:01 | comment | added | Liviu Nicolaescu | The Stokes's theorem for pseudoforms in proved in T. Frankel's book The Geometry of Physics, 3rd Edition, 2012, page 117. | |
| Jan 28, 2021 at 10:46 | comment | added | David Roberts♦ | @BertramArnold that looks like good material for an answer! | |
| Jan 28, 2021 at 10:31 | comment | added | Bertram Arnold | While this fuss about orientations might seem pedantic, for more general objects such as supermanifolds there are no longer any "top-forms", and the pseudo-differential forms you define correspond to "integral forms", which are fundamentally different from differential forms (e.g. they don't form an algebra). Witten's paper "Notes on Supermanifolds and Integration" (arxiv.org/abs/1209.2199) explains this quite well, Section 3.5 about Stokes's theorem is especially relevant. | |
| Jan 28, 2021 at 10:28 | comment | added | Bertram Arnold | Any pseudo-differential $k$-form defines a $(n-k)$-de Rham current, i.e. a linear functional on compactly supported $(n-k)$-forms, and thus corresponds more to a dimension $(n-k)$ homology rather than a degree $k$ cohomology class. In particular, pulling it back to a general submanifold requires the normal bundle to be oriented. This is true for the boundary (take the inward or outward pointing normal vector field; the choice will cancel out) but not in general. Since Stokes's theorem can be proved in local orientable charts, the same proof works for pseudo-differential forms. | |
| Jan 28, 2021 at 9:27 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Jan 28, 2021 at 6:36 | comment | added | Theone | That's a fair question. The integral is just a real number. For the second part of your comment: maybe it all comes down to whether you can pull back a section of a differential pseudo-form. I don't know if it's possible either... | |
| Jan 28, 2021 at 6:30 | comment | added | David Roberts♦ | No, I meant: given a pseudodifferential form, is its integral a number or something more exotic? A section of $\Psi$ wasn't the right thing to suggest here. Also, to even do the integration over $\partial M$, you need to be pulling back of $\omega$ along the inclusion of the boundary, so this has to make sense... | |
| Jan 28, 2021 at 6:08 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Typo fixed.
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| Jan 28, 2021 at 6:08 | comment | added | Theone | The integral of a section of $\Psi$ (as far as I know) is only defined over $M$, or over subset of $M$. In this case, the intergral is a real number. That is an interesting point about restricting $\Psi$ to the boundary. But, does the restriction of the pseudoscalar bundle to the boundary canonically give the pseudoscalar bundle of the boundary? If not, I don't see how you can integrate a $(n-1)$-form on $\partial M$ and get a real number answer. | |
| Jan 28, 2021 at 5:41 | comment | added | David Roberts♦ | Presumably the integral is a section of $\Psi$ rather than a real-valued function? If so, one would need a map of bundles between the pseudoscalar bundle of the boundary, and the restriction of the pseudoscalar bundle of $M$ to the boundary, in order to compare the sections. | |
| Jan 28, 2021 at 5:36 | review | First posts | |||
| Jan 28, 2021 at 7:33 | |||||
| Jan 28, 2021 at 5:32 | history | asked | Theone | CC BY-SA 4.0 |