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Could anybody provide a motivated sketch of why the isomorphism classes of the differentiable rank $k$ real vector bundles over the sphere $S^q$ are given by$$\text{Vect}_k(S^q) \simeq \pi_{q - 1}(\text{O}(k))/\mathbb{Z}_2,$$and the isomorphism classes of the complex vector bundles are given by$$\text{Vect}_k(S^q, \mathbb{C}) \simeq \pi_{q - 1}(\text{U}(k))?$$I have seen the proof in Bott-Tu, but I was wondering if anybody could supply their own motivations in a sketch.

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closed as off-topic by BS., Fernando Muro, Alex Degtyarev, Mark Grant, Chris Gerig Mar 10 '16 at 16:14

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  • $\begingroup$ Any bundle is trivial over a (conractible) hemisphere. Gluing two such trivial bundles is an element of $\pi_{q-1}$ of the structure group. $\endgroup$ – Alex Degtyarev Mar 10 '16 at 10:58
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This is general for a suspension of a space. A suspension is the union of two cones $SX=CX_+\cup CX_-$. A cone is contractible, hence a vector bundle trivializes over each cone. The only information about the vector bundle is contained in how the two trivializations are glued along $C_-X \cap C_+X\cong X$. This can be seen as a map $X\rightarrow GL(k)$. In fact only the homotopy class of this map matters. For the sphere $S^q=SS^{q-1}$, this corresponds to a homotopy class of maps $S^{q-1}\rightarrow GL(k)$. But $GL(k)$ deformation retracts to $O(k)$. Lets assume $q>1$. Now $O(k)$ has two connected components. As the sphere $S^{q-1}$ is connected if $q>1$, it lands in one of the two components. Hence the bundle is orientable. If the bundle admits an orientation reversing automorphism, we get two different homotopy classes representing the same vector bundle. I believe this is where the $\mathbb{Z}/2\mathbb{Z}$ comes from.

The story for complex vector bundles is similar but a bit simpler as $U(k)$ is connected. This is called the clutching construction. Details are in Hatcher's notes on Vector bundles and $K$-theory, or Atiyah's $K$-theory for example.

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    $\begingroup$ Where does the "mod $\mathbb{Z}_2$" in the case of real vector bundles come from? $\endgroup$ – Greg Graviton Mar 10 '16 at 10:06

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