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.


  1. Given $F\to M,\ E\subseteq F$ and $\nabla$ preserving $E$. $\exists \sigma:G\to F$ splitting, s.t. $\gamma(g)=\nabla(\sigma(g))-\sigma(\nabla^G(g))=0$ ?
  2. If the answer of 1. is No. And if $\nabla$ is flat ?
  3. Or what happens in the particular case wehere M is a Riemannian manifold $F=TM$, $E=TN$ where $N$ is a totally geodesic submanifold ?
  • 1
    $\begingroup$ 2: flat connection $=$ representation of $\pi_1$. If the group is infinite, there's no reason why the category should be semisimple. (Just take a nontrivial Jordan block.) $\endgroup$ – Alex Degtyarev Oct 1 '16 at 0:06
  • $\begingroup$ 3: Also, it seems to me that, in the Riemennian case, the splitting is given by the orthogonal complement. $\endgroup$ – Alex Degtyarev Oct 1 '16 at 0:50
  • $\begingroup$ Maybe I got lost with the terminology, but it seems to me that if 1) holds then in each fiber of $F$ there is a complement of $E$ invariant by the holonomy group. So now it is enough to know examples of linear connections whose holonomy group have invariant subspaces without invariant complements e.g. Lorentzian geometry, take the holonomy group if a indecomposable but not irreducible Lorentzian manifold. If the connection is flat then the problem becomes a topological one i.e. if $M$ is connected and simply connected then 1) is true under the flatness of $\nabla$. $\endgroup$ – Holonomia Oct 1 '16 at 10:49

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