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Let $O(n)$ denote the $n$ - orthorgonal group consisting of all $n \times n$ real matrices $A$ with $A^{t}A = E_{n}$. Let $e_{n}$ is the column vector $(0,...,1)$ then $Ae_{n}$ is a unit vector in $\mathbb{R^{n}}$, hence lie in the sphere $S^{n-1}$. I want show that

$$p : O(n) \to S^{n-1}$$ $$A \mapsto Ae_{n}$$

is a locally trivial bundle with fiber $O(n-1)$. So I consider the open cover $U = S^{n-1} - e_{n}$ and $V = S^{n-1} - (-e_{n})$. I think this cover will work but I can't explicit the trivialization ( homeomorphism ) on each open set $U$ or $V$, i.e

$$\phi_{U} :U \times O(n-1) \to p^{-1}(U)$$

$$p\phi_{U}(u,A) = u \forall (u,A) \in U \times O(n-1)$$

And similiar with $V$. I tried to define a map $\alpha_{U} : U \to p^{-1}(U)$ and then define $\phi_{U}(u,A) = \alpha(u) \begin{pmatrix}A & 0 \\0 & 1 \end{pmatrix}$. In order to do that I need $\alpha(u)e_{n}=u$ and $\alpha(u) \in O(n)$. Can someone help me this point ?

In general, how can I know that there is a trivial fibre bundle of Stiefel manifold without using other concepts in theory of bundle ?

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  • $\begingroup$ One way you could define $\alpha(u)$ is to extend $u$ to an orthonormal basis $v_1,\dotsc v_{n-1}, u$ of ${\mathbb R}^n$, and then declaring $\alpha(u)$ to be the matrix $A =[v_1\,v_2\,\dotsm\,u]\in {\textrm O}(n)$. You need to do this in a smooth, consistent fashion on all of $U$ though. Maybe start with the basis $e_1,\dotsc, e_n$ and then apply a sequence of rotations (acting on planes in a fixed order) until the last vector matches $u$? $\endgroup$ – Glare Feb 18 '18 at 8:47
  • $\begingroup$ What is " sequence of rotations "? $\endgroup$ – Gankedbymom Feb 18 '18 at 9:21
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    $\begingroup$ You can do this by constructing an explicit trivialization by using the fact that for any $u\in U$ there exists a unique (shortest) spherical arc $a_p$ connecting $-e_n$ to $p$ and that $a_p$ depends smoothly on $p$. Alternatively, use Ehresmann's lemma: en.wikipedia.org/wiki/Ehresmann%27s_lemma $\endgroup$ – Moishe Kohan Feb 18 '18 at 14:15
  • $\begingroup$ @Gankedbymom My idea was that you can get from $e_n$ to $u$ by rotating only in $e_ie_j$-planes. So apply a rotation (a special orthogonal transformation) in the $xy$-plane that fixes everything else, then in the $yz$-plane, then in the $zw$-plane, etc. You can always get from one point on the sphere to another in this fashion, and if you require that the rotations are applied in a fixed order (I always start with $xy$, then $yz$, then etc.) you obtain a consistently defined special orthogonal matrix taking $e_n$ to $u$ that varies smoothly in $u$. $\endgroup$ – Glare Feb 18 '18 at 23:34
  • $\begingroup$ That being said, Moishe Cohen's method is much cleaner. There is no reason to take a piecewise concatenation of arcs connecting $e_n$ to $u$ (which is essentially what I suggest) when I can take the unique shortest one. $\endgroup$ – Glare Feb 18 '18 at 23:36

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