Timeline for Can we "Curve" a manifold, as much as possible?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2017 at 13:54 | comment | added | Ali Taghavi | @SebastianGoette thank you for your very helpfull comment | |
| Sep 26, 2017 at 10:39 | comment | added | Sebastian Goette | @Omar The characteristic classes of the tangent bundle $TM$ and the (stable) normal bundle $TM^\perp$ do not depend on $n$. Because their sum is trivial, those of $TM$ can be computed from those of $TM^\perp$ and vice versa. But assume that $TM\oplus TM^\perp\cong\underline{\mathbb R^n}$ is an unstable realisation of the normal bundle. Then $\operatorname{Hom}(TM,\mathbb R^{n+1}/TM)\cong\operatorname{Hom}(TM,TM^\perp)\oplus T^*M$, which has different characteristic classes. The upshot is: the characteristic classes in question depend only on those of $TM$ and on $n$. | |
| Sep 24, 2017 at 21:20 | comment | added | Ali Taghavi | @Omar Thank you for your very helpfull comment and usefull information. | |
| Sep 24, 2017 at 15:37 | comment | added | Omar | The tangent bundle of $G(k,n)$ at $L$ is canonically isomorphic to $Hom(L,\mathbb{R}^n/L)$. So the pullback of the tangent bundle is isomorphic to $Hom(TM,\mathbb{R}^n/TM)$. The bundle does(I think so) depend on the embedding but if n is big enough it doesn't. The characteristic classes don't depend on n, they are algebraic combination of classes of $TM$, and $TM^\bot$. Concerning your first question, I think that the derivative of the map $\phi_{M,n}$ is Porteous's intrinsic derivative (see for example wisdom.weizmann.ac.il/~dnovikov/Manifolds5775/… page 149) | |
| Sep 23, 2017 at 20:49 | history | asked | Ali Taghavi | CC BY-SA 3.0 |