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I have a short question concerning the answer to this question. There it is used that the maps $$U(n)/O(n)\longrightarrow S^1,\quad A\longmapsto \det(A)^2$$ as well as, later, $$SU(n)\longrightarrow SU(n)/SO(n)$$ are fibrations. After learning the definition of a fibration, and looking over several examples and related questions here, unfortunately I'm still wondering how to prove this? (For example, I see that $(\cdot)^2\colon S^1\rightarrow S^1$ is a fibration as it is a covering map, and that the composition of fibrations again is a fibration, but what about the $\det$-part and the quotients?)

Any help is highly appreciated, thanks a lot in advance!

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It is a general theorem that if a Lie group $G$ acts smoothly and properly on a manifold $M$, then the orbit space $M/G$ has the structure of a smooth manifold such that the quotient map $M\rightarrow M/G$ inherits the structure of a locally trivial principal $G$-bundle. See for example Lee's book "Introduction to Smooth Manifolds", chapter 21.

A closed Lie subgroup $H$ of a finite dimensional Lie group $G$ will act smoothly and properly on $G$. Therefore the quotient $G\rightarrow G/H$ is a principal $H$-bundle, and in particular a fibration.

The point is that the property of a map $p:E\rightarrow B$ being a fibration is a local property. If $p$ is a fibration and $(U_a)$ an open cover of $B$, then each restriction $E|_{U_a}\rightarrow U_a$ is a fibration. Conversely, if $B$ can be covered by open sets $U_a$ in such a way that each restriction $E|_{U_a}\rightarrow U_a$ is a fibration, then $p:E\rightarrow B$ can be shown to be a fibration. The details of this are worked out (by you) in Strom's 'Modern Classical Homotopy Theory', chapter 13. He also gives references to the original papers by Hurewicz and Dold that treat this approach.

One way to prove all this is by using a partition of unity. Any paracompact space admits a partition of unity, and any CW complex is paracompact, so in particular any smooth manifold is paracompact (since it is a CW complex by basic Morse theory).

As for the $det$ part, you can in fact identify this map with another quotient. The gropu $U(n)$ acts transitively on $S^1$ through the determinant, $A\cdot z=det(A)z$, and the stabiliser of the unit $I\in U(n)$ is $SU(n)$. Hence there is a homeomorphism (diffeomorphism in fact) $U(n)/SU(n)\cong S^1$ induced by $det$.

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  • $\begingroup$ Thanks a lot for your answer! $\endgroup$ – user103697 Mar 27 '18 at 12:26

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