Talk:Tangent bundles of bundles (Ex)

Solution 0.1. There is a short exact sequence of vector bundles over $E$$<wikitex>; {{beginthm|Solution}} There is a short exact sequence of vector bundles over $E$ $<script type="math/tex">E$

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

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

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

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

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

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

\to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to 0$$ where $i$ is the inclusion and $j$ is defined by $j(e,v)=(e, d\pi(v))$. We choose a Riemannian metric on $TE$, then the orthogonal projection to $T_{\pi}E$ gives a splitting of the exact sequence. Which implies that $$ TE \cong \pi^*TB \oplus T_{\pi}E $$ {{endthm}} {{beginthm|Solution}} 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. {{endthm}} {{beginthm|Solution}} 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) $. {{endthm}} {{beginthm|Solution}} 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. {{endthm}}E

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

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

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