Talk:Tangent bundles of bundles (Ex)

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Solution 0.1. There is a short exact sequence of vector bundles over E

\displaystyle 0 \to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to 0

where i is the inclusion and j is defined by j(e,v)=(e, d\pi(v)). We choose a Riemannian metric on TE, then the orthogonal projection to T_{\pi}E gives a splitting of the exact sequence. Which implies that

\displaystyle  TE \cong \pi^*TB \oplus T_{\pi}E

Solution 0.2. A necessary condition for T_{\pi}E being the pullback of some bundle over B is that TF is trivial. This is seen by restricting the bundle to a point in B. On the other hand, obviously when E is a vector bundle or the bundle \pi \colon E \to B is trivial then T_{\pi}E is the pullback of a vector bundle over B. We don't know if this is true in general.

Solution 0.3. If \pi \colon E \to B is itself a smooth vector bundle, then \pi^*T_{\pi}E\cong \pi^*E, therefore TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E).

Solution 0.4. Denote the associated vector bundle by \xi with projection \pi_{\xi} \colon E(\xi) \to B, then TE \oplus \underline{\mathbb R} = i^*\pi_{\xi}^*(TB \oplus  \xi), where i \colon E \to E(\xi) is the inclusion of the sphere bundle into the vector bundle.

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