Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with sg.symplectic-geometry
Search options not deleted
user 37103
Hamiltonian systems, symplectic flows, classical integrable systems
28
votes
Accepted
Is $\Bbb S^2 \times \Bbb S^4$ symplectic?
No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ …
- 5,169
0
votes
Momentum a cotangent vector
$p$ is the differential of L with respect to the second variable ($\dot q$), so it represents a linear functional on the tangent space at $q, \dot q$), given by
$$ (u,v)\to \frac{d}{dt} L(q,\dot q+tv …
- 5,169