Timeline for Lie groups bundle
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2013 at 9:23 | comment | added | Michael Murray | An example to think about would be $U(2)/T \to U(2)/N_{U(2)}(T)$ which is $S^2 \to \mathbb{R}P_2$. The Weyl group is $\mathbb{Z}_2$ and $-1$ acts by the antipodal map on $S^2$. | |
| Jan 25, 2013 at 2:53 | comment | added | Michael Murray | Note too Oscar that in your case if $H=T$ then $K/H$ is a (finite) group, the Weyl group of $G$ and your fibre bundle is a principal bundle for the Weyl group. | |
| Jan 25, 2013 at 2:52 | comment | added | Michael Murray | Once you understand the associated bundle construction all that is missing is checking the map I have described is a fibre bundle map which is straightforward. | |
| Jan 25, 2013 at 1:18 | comment | added | S. Carnahan♦ | Check Steenrod's Topology of Fibre Bundles. | |
| Jan 24, 2013 at 23:21 | comment | converted from answer | Trakrendal | I don't understand how you build the associate fibre bundle and how you prove that there is an isomorphism of fiber bundle. | |
| Jan 24, 2013 at 23:19 | comment | added | David Roberts♦ | Oscar - yes, this is true, as $N_G(T)$ is a closed subgroup, which is what you need for Michael's first sentence to hold. This is a standard result: the free action is obvious, and one shows that the quotient map is a submersion by consider the map on Lie algebras - this gives local sections hence local triviality. You need the subgroup to be closed so that the quotient is a manifold. | |
| Jan 24, 2013 at 15:33 | comment | added | Oscar1778 | In my situation $G=U(n)$ and $K=N_{G}(T)$ is the normalizer of $T$ in $U(n)$ and $T$ is a maximal torus in $U(n)$ (i.e. the subgroup of diagonal matrix). So I ask you if $U(n) \rightarrow U(n)/N_{G}(T)$ is a principal $N_{G}(T)$-bundle (i.e. $N_{G}(T)$ acts freely on U(n)). Thaks. | |
| Jan 24, 2013 at 13:54 | history | answered | Michael Murray | CC BY-SA 3.0 |