# Splitting of the tangent bundle of a vector bundle and connections

Let $\pi:E\to M$ be a smooth vector bundle. Then we have the following exact sequence of vector bundles over $E$:

$$0\to VE\xrightarrow{} TE\xrightarrow{\mathrm{d}\pi}\pi^*TM\to 0$$

Here $VE$ is the vertical bundle, that is kernel of the bunble map $\mathrm{d}\pi$. This sequence splits, so there is a bundle morphism $\sigma:\pi^*TM\to TE$ such that $\mathrm{d}\pi\circ\sigma=\mathrm{Id}$. It can be shown that the data of such a $\sigma$ is equivalent to a connection (covariant derivative) $\nabla$ on $E\to M$ (see for instance the volume 2 of Greub-Halperin-Vanstone). Also we know that $VE\cong\pi^*E$ so that we can write the following decomposition:

$$TE\cong\pi^*E\oplus\pi^*TM$$

My question is the following: considering $E$ as a manifold, I would like to consider differential forms on $E$, that is $\Gamma(\Lambda T^*E)$; is the De Rham differential linked to $\nabla$ in some way?

Thank you for your help. Any reference would be very useful.

The covariant derivative $\nabla$ is a map $\nabla: \Gamma(TM)\times \Gamma(E)\longrightarrow \Gamma(E)$ which is $\mathbb R$-bilinear and satisfies $$\nabla_{f\cdot X} s=f\cdot \nabla_X s\quad \textrm{and}\quad \nabla_\alpha (f\cdot s)=f\cdot \nabla_X s+\mathcal{L}_{X}(f)\cdot s,$$ for every $f\in C^\infty(M)$, $X\in\Gamma(TM)$ and $s\in \Gamma(E)$. We can then define $$d_\nabla: \Gamma(\Lambda^p T^*M\otimes E)\longrightarrow \Gamma(\Lambda^{p+1} T^*M\otimes E),$$ setting $$\begin{eqnarray*} d_\nabla \varepsilon(X_1, \ldots, X_{p+1})&&:=\sum_{j=1}^{p+1}(-1)^{j+1} \nabla_{X_j} \varepsilon(X_1, \ldots, \widehat{X_j}, \ldots, X_j)\\ &&+\sum_{i<j} (-1)^{i+j} \varepsilon([X_i, X_j], X_1, \ldots, \widehat{X_i}, \ldots, \widehat{X_j}, \ldots, X_{p+1}).\end{eqnarray*}$$ To define this I'm using the isomorphism:

$$\Gamma(\Lambda^p T^* M\otimes E)\simeq \mathsf{Hom}_{C^\infty(M)} (\Lambda^p \Gamma(TM), \Gamma(E))\simeq \mathsf{Alt}^p_{C^\infty(M)}(\Gamma(TM), \Gamma(E)).$$

If I'm not wrong, $d_\nabla^2=0$ if and only if $\nabla$ is a flat connection.

Up to this point everything is ok. Now, my guess is that you have a product:

$$\cdot:\Gamma(\Lambda^p T^*M)\times \Gamma(\Lambda^q T^* M\otimes E)\longrightarrow \Gamma(\Lambda^{p+q} T^*M \otimes E),$$ defined in the obvious way and it holds:

$$d_\nabla (\alpha\cdot \omega)=d_{\mathsf{dR}}\alpha\cdot \omega+(-1)^{|\alpha|} \alpha \cdot d_\nabla \omega.$$

Hope it helps.

P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.