Timeline for Symplectic blow-up
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12 at 10:38 | comment | added | ChoMedit | Is this construction independent to the choice of $I$? If so, how could we show that? | |
| Jun 14, 2013 at 22:26 | vote | accept | Mohammad Farajzadeh-Tehrani | ||
| Jun 14, 2013 at 17:09 | comment | added | Mohammad Farajzadeh-Tehrani | you are correct, sorry. | |
| Jun 14, 2013 at 16:49 | comment | added | Mike Usher | More generally, if you work in a unitary trivialization you can check that $-d\theta$ restricts to each fiber as the standard symplectic form, and that, on $TE|_X = TX\oplus E$, all elements of the $TM$ summand include trivially into $d\theta$ (it helps somewhat that the connection is unitary). So $\Omega$ indeed does coincide with $\omega$ along $X$. | |
| Jun 14, 2013 at 16:47 | comment | added | Mike Usher | If X is a point, so E is the standard $R^{2n}$, then $L=\sum (x_{i}^{2}+y_{i}^{2})/4$, $dL = \sum(x_i dx_i+y_i dy_i)/2$, and so $dL\circ J = \sum(y_i dx_i-x_idy_i)/2$, which is the negative of a primitive for the standard symplectic form. So by my calculation $\theta$ is $-\frac{1}{2}\sum r_{i}^{2}d\theta_i$. | |
| Jun 14, 2013 at 14:52 | comment | added | Mohammad Farajzadeh-Tehrani | I guess some sort of trace formula should be involved in the definition of $\theta$. | |
| Jun 14, 2013 at 14:50 | comment | added | Mohammad Farajzadeh-Tehrani | Lets X be a point. Then your formula says $\Omega = - d\theta$ (where $\theta= dL \circ J$ or $\theta= L dL \circ J$ if my point is correct) and you want this to be $\omega$ at that point. But a simple calculation shows your $\theta$ is $d\theta_1+\cdots+d\theta_n$ where $r_1,\theta_1,\cdots,r_n,\theta_n$ are polar coordinates on $C^n$. | |
| Jun 14, 2013 at 14:17 | comment | added | Mohammad Farajzadeh-Tehrani | Tanks, your definition of \theta looks completely natural, but: 1- I think your \theta is missing a factor of L behind (i.e. it should be LdL(Iv). 2- Forgetting that, how do you show that along X they coincide? Did you actually check that or you guess they should. | |
| Jun 14, 2013 at 1:56 | history | answered | Mike Usher | CC BY-SA 3.0 |