Timeline for What's the correct notion of determinant of a bilinear pairing?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2017 at 7:22 | history | edited | CommunityBot |
replaced http://mathematics.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Jul 31, 2013 at 8:27 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
corrected non-rendering mathjax; removed tag hep-th
|
| Dec 24, 2009 at 18:16 | history | undeleted | Theo Johnson-Freyd | ||
| Dec 24, 2009 at 8:00 | history | deleted | Theo Johnson-Freyd | ||
| Dec 1, 2009 at 21:25 | history | edited | Philipp Lampe |
edited tags
|
|
| Oct 8, 2009 at 19:54 | answer | added | Ilya Nikokoshev | timeline score: 0 | |
| Oct 8, 2009 at 13:39 | answer | added | Andrew Stacey | timeline score: 2 | |
| Oct 8, 2009 at 0:20 | comment | added | Theo Johnson-Freyd | Here's a way to test your definition. Let V be the space of smooth derichlet functions on [0,T]. I very firmly believe that the determinant of the pairing \int_0^T a f'(t) g'(t) dt (i.e. take derivatives and integrate, with a factor of a>0) should be T/a, where the "volume form" is "\int_0^T dx(t) dt". Via the usual pairing \int f(t) g(t) dt, my pairing is related to the map -a d^2/dt^2, which has zeta-function-regularized determinant 2T/\sqrt{a}. | |
| Oct 8, 2009 at 0:15 | history | asked | Theo Johnson-Freyd | CC BY-SA 2.5 |