Let $F\to M$ be a vector bundle and $E\subseteq F$ a subbundle. Suposse that $\nabla$ is a connection on $F$ s.t. preserves $E$, i.e. $\nabla_X(e)\in \Gamma E \quad \forall e\in \Gamma E, \ X\in\Gamma TM$, then define a connection on $E$ $\nabla^E_X=(\nabla_X)_{|E}$. Take $G=\frac{F}{E}$. Then we have the short exact sequences of vector bundles:

$$0\to E\to F\to G\to 0$$

and since $\nabla$ preserves $E $ then descendes to $G$ by the formula: $$\nabla^G_X(g)= \pi(\nabla_X (f)) \quad \text{where} \ \pi(f)=g$$

Then, we have a short exact sequences of vector bundles with connections:

$$0\to (E,\nabla^E)\to (F,\nabla)\to (G,\nabla^G)\to 0$$

And taking a splitting $\sigma:G\to E$ we can identify $F\cong E\oplus G$ And we can define $\widetilde{\nabla}= \nabla^E\oplus \nabla^G$. $\nabla$ and $\widetilde{\nabla}$ differ by the map $\gamma(g)=\nabla(\sigma(g))-\sigma(\nabla^G(g))$. It is, they are equal iff $\sigma$ intertwing the connections.

Questions:

- Given $F\to M,\ E\subseteq F$ and $\nabla$ preserving $E$. $\exists \sigma:G\to F$ splitting, s.t. $\gamma(g)=\nabla(\sigma(g))-\sigma(\nabla^G(g))=0$ ?
- If the answer of 1. is No. And if $\nabla$ is flat ?
- Or what happens in the particular case wehere M is a Riemannian manifold $F=TM$, $E=TN$ where $N$ is a totally geodesic submanifold ?