Timeline for Integrals from a non-analytic point of view
Current License: CC BY-SA 2.5
13 events
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| Sep 13, 2010 at 4:37 | comment | added | Theo Johnson-Freyd | BTW: I think I was too cavalier, by which I means sloppy. There certainly can be more than one dimension worth of linear functionals out of the algebra of differential forms (with compact support) that vanish on exact forms and on all Lie derivatives. In particular, if the manifold is disconnected, I could rescale all forms on one of the pieces by whatever constant I want. | |
| Sep 13, 2010 at 4:35 | history | edited | Theo Johnson-Freyd | CC BY-SA 2.5 |
I said something false.
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| Sep 12, 2010 at 20:24 | comment | added | Vivek Shende | @Terry: Is there a version of the characterization-as-Haar measure which gives the global "integral over a manifold" as the "annihilator of de Rham d & lie derivatives wrt. all vector fields"? | |
| Sep 12, 2010 at 19:08 | comment | added | Theo Johnson-Freyd | @Vivek: I haven't read Berezin's original papers. For the theory of super integration, the book by Voronov, if memory serves, is decent, but I don't have it handy to check. Alternately, I think that Deligne and Mumford's chapter in QFT for Mathematicians describes super integration. | |
| Sep 12, 2010 at 19:06 | comment | added | Theo Johnson-Freyd | @Terry Tao: yes, precisely. And I should emphasize that I'm not a historian; I think that the fact that Lebesgue measure is the unique translation-invariant measure (up to scalar) is older than Berezin, but that it's Berezin who proposed to use this observation to generalize to a more algebraic treatment of integration. @Dan Petersen: I was being sloppy. There are, to my mind, different good notions of "distribution". There are some that let you integrate any smooth function, some for any continuous function with compact support, etc. | |
| Sep 12, 2010 at 17:48 | comment | added | Terry Tao | One can also view Berezin's observation as an infinitesimal version of the observation that Lebesgue measure is the (unique up to constants) Haar measure on R^n. (If a measure is translation invariant, then it will annihilate derivatives with respect to infinitesimal translations, i.e. partial derivatives.) To me, this is one of the most natural definitions of Lebesgue measure, though it probably does not qualify as "non-analytic" in the sense you are looking for. | |
| Sep 12, 2010 at 16:06 | comment | added | Johannes Hahn | @Dan: Of course you have to use some topology and of course $C_c^\infty(M)$ is the space on is interested in. If you use test functions with compact support, you'll get all distributions. If you use all smooth test functions, then you'll only get the distributions with compact support. | |
| Sep 12, 2010 at 15:26 | comment | added | Dan Petersen | Without knowing anything about these matters I am a little weirded out that you define a distribution to be any linear map $\mathcal{C}^\infty(M) \to \mathbb{R}$. Do you not need any continuity assumptions with respect to some appropriate topology? And do you assume M compact? If not, the constant function 1 will not even define a distribution in this sense unless you replace $\mathcal{C}^\infty(M)$ with, say, the algebra of compactly supported smooth functions. | |
| Sep 12, 2010 at 15:08 | comment | added | Dmitri Pavlov | @Vivek: The observation of Berezin is also known as Poincaré duality. See my answer for details. | |
| Sep 12, 2010 at 15:06 | comment | added | Dmitri Pavlov | @Harry: As I explain in my answer, you can easily complete the space of functions with compact support to the space of all integrable functions. Of course, not every smooth function with non-compact support is integrable. | |
| Sep 12, 2010 at 11:26 | comment | added | Vivek Shende | I really like the observation of Berezin! Do you have a reference? | |
| Sep 12, 2010 at 7:58 | comment | added | Harry Gindi | This only allows us to integrate smooth functions with compact support, right? | |
| Sep 12, 2010 at 7:41 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |