Timeline for Is there a relationship between tensor (or form) bundles and iterated tangent/cotangent bundles on a manifold?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2013 at 16:31 | vote | accept | Kensmosis | ||
| Jan 9, 2013 at 15:29 | answer | added | Robert Bryant | timeline score: 7 | |
| Jan 9, 2013 at 6:10 | comment | added | Pavel Safronov | The relative tangent bundle to $p: TM\rightarrow M$ is $p^* TM$, so you have an exact sequence $0\rightarrow p^* TM\rightarrow TTM\rightarrow p^*TM\rightarrow 0$. For smooth manifolds it is split, hence $TTM\cong p^*TM\oplus p^*TM$. This is different from, say, (the pullback of) $T^{(2,0)}M$, which is $TM\otimes TM$. | |
| Jan 8, 2013 at 23:28 | comment | added | Martin Brandenburg | Well by definition $\Lambda^p M = T^{(n,0)}(M) / \Sigma_n$ (wrt to the sign action), but probably this is not what you want? | |
| Jan 8, 2013 at 21:01 | history | asked | Kensmosis | CC BY-SA 3.0 |