Timeline for Is there a relationship between Fourier transformations and cotangent spaces?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2015 at 7:58 | vote | accept | Rogier Swierstra | ||
| May 7, 2015 at 14:50 | answer | added | Tobias Diez | timeline score: 3 | |
| May 3, 2015 at 7:00 | comment | added | YHBKJ | When $Q$ is a projective variety, I think this is related to the Mukai-flop. | |
| May 1, 2015 at 13:59 | comment | added | Igor Khavkine | No. Unless I'm misunderstanding it, your analogy is flawed for the reason I commented on earlier and also because in the formula that you give for $S[\phi]$ is only valid when $\phi$ solves the equations of motion, so the $p_i$ are not independent variables like in the Fourier transform. What is true is that the path integral is an oscillatory integral. But not all oscillatory integrals are Fourier transforms and neither is the path integral without the external source term. | |
| May 1, 2015 at 13:22 | comment | added | Rogier Swierstra | @IgorKhavkine But. Your answer to my path integral question doesn't address the point I was trying to make. In coordinates, writing $\theta = p_idq^î$ for the tautological one-form, then $S[\phi]=\int_\phi \theta = \int_\phi p_i dq^i$. I meant that the action-exponent itself reminds me of a Fourier term. Do you know if G&S mention this? | |
| May 1, 2015 at 13:15 | comment | added | Rogier Swierstra | @IgorKhavkine Thanks for the reference (available [online](www.freeinfosociety.com/media/pdf/2287.pdf)) which is very promising. | |
| Apr 30, 2015 at 13:04 | answer | added | Igor Khavkine | timeline score: 3 | |
| Apr 30, 2015 at 12:39 | comment | added | Igor Khavkine | Your path integral, as written maps $F$ to a number rather than another function, so a naive analogy with the Fourier transform fails. However, if you write it with an external source $J$, $\int \exp(\frac{i}{h}(S[\phi]+J\cdot\phi)) F(\phi) \mathcal{D}\phi$, where $J\cdot\phi = \int J(x) \phi(x) dx$, then the path integral, as a function of $J$, is essentially the Fourier transform of $\exp(\frac{i}{h}S[\phi]+\log F(\phi))$. I think that should make the formal analogy clear. | |
| Apr 30, 2015 at 12:28 | comment | added | Willie Wong | The first part of your question strongly reminds me of microlocal analysis. But unfortunately I am not able to make the connection precise. Hopefully someone who can will come along and explain. | |
| Apr 30, 2015 at 11:55 | review | First posts | |||
| Apr 30, 2015 at 11:57 | |||||
| Apr 30, 2015 at 11:53 | history | asked | Rogier Swierstra | CC BY-SA 3.0 |