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Given compact Lie groups $H \subset K \subset G$, there is a fiber bundle $ \frac{K}{H} \rightarrow \frac{G}{H} \rightarrow \frac{G}{K}$. Do you have a simple proof of this?

flag	asked Jan 24 at 12:20 Oscar1778 71●1

Michael Murray · Answer 1 · 2013-01-24 13:54:29Z UTC

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If you are prepared to accept that $G \to G/K$ is a principal $K$-bundle there is an easy proof. You have that $K$ acts on the homogeneous space $K/H$ so you have an associated fibre bundle $$ \frac{G \times K/H}{K} \to G/K $$ with fibre $K/H$. The total space is actually $G/H$. You can construct a fibre bundle isomorphism from this to $G/H$ by $$ gH \mapsto [g, H]_K $$ where $[g, H]_K $ is the orbit under the $K$ action $(g, H)k = (gk, k^{-1}H)$. Why is this well defined ? You can check that $$ ghH \mapsto [gh, H]_K = [g, hH]_K = [g, H]_K $$ as $H \subset K$. It has an inverse which is $$ [g, kH]_K \mapsto gkH $$ This is also well-defined as $$ [gk_1, k_1^{-1}kH]_K \mapsto gk_1 k_1^{-1} k H = gk H $$

answered Jan 24 at 13:54

Michael Murray
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	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. – Oscar1778 Jan 24 at 15:33
	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. – David Roberts Jan 24 at 23:19
	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. – Michael Murray Jan 25 at 2:53
	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 $S^2$. – Michael Murray Jan 25 at 9:23

Trakrendal · Answer 2 · 2013-01-24 23:21:24Z UTC

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I don't understand how you build the associate fibre bundle and how you prove that there is an isomorphism of fiber bundle.

answered Jan 24 at 23:21

Trakrendal
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	Check Steenrod's Topology of Fibre Bundles. – S. Carnahan♦ Jan 25 at 1:18
	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. – Michael Murray Jan 25 at 2:52

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