# Decomposition of the tangent bundle of a tensor product of vector bundles

If $$E \to B$$ is a smooth vector bundle, then the tangent bundle $$T E$$ is a vector bundle over both $$E$$ and $$T B$$. (If you like, the latter structure is the derivative of the vector bundle structure on $$E$$.) If $$F \to B$$ is another smooth vector bundle and $$\pi : T B \to B$$ is the obvious projection, I have seen it written (e.g. here) that there is a canonical isomorphism $$T(E \otimes F) \cong \pi^* E \otimes T F$$ as vector bundles over $$T B$$.

Supposing that we can do this, we can in particular take $$F = B \times \mathbb{R}$$ (the trivial rank 1 bundle over $$B$$) and obtain a vector bundle isomorphism $$\pi^* E \otimes_{TB} T(B \times \mathbb{R}) \to T E.$$ But since $$T \mathbb{R} \cong \mathbb{R}^2$$ canonically, there is a chain of canonical isomorphisms of vector bundles over $$T B$$ $$\pi^* E \otimes_{TB} T(B \times \mathbb{R}) \cong \pi^* (E \otimes T \mathbb{R}) \cong \pi^* (E \oplus E).$$ (To avoid doubt, $$E \otimes T \mathbb{R}$$ just means the fiberwise tensor product of the vector bundle $$E$$ with the ordinary vector space $$T \mathbb{R}$$.)

In other words, we have $$T E \cong \pi^*(E \oplus E)$$ over $$T B$$. Of course this is locally (noncanonically) true, but globally such an isomorphism would e.g. define a canonical connection on any vector bundle $$E$$ by giving horizontal lifts of arbitrary vector fields on $$TB$$ to $$TE$$.

I imagine, then, that the precise statement is that connections on $$E$$ induce isomorphisms $$T(E \otimes F) \cong \pi^* E \otimes T F$$; intrinsically, how do we define this map? The answer I linked above also made reference to a canonical isomorphism $$\rho : \pi^* E \otimes T F \to T E \otimes \pi^* F$$, which appears to me to be in the same situation.

I have explained the situation when there are connections on both $$E$$ and $$F$$ on MathOverflow over here. As I explain there, there is a surjective map $$\tau : TE \otimes_{TB} TF \to T(E \otimes F)$$ which has a kernel (since the fiber dimension of the source is twice that of the target). As I also explain there, a connection on $$E$$ induces a direct sum decomposition of $$TB$$-bundles $$TE \cong ZE \oplus_{TB} HE$$ (with $$ZE$$ and $$HE$$ of the same dimension.
In our case, this means that $$\tau$$ splits as a sum of maps $$\tau_Z : ZE \otimes_{TB} TF \to T(E \otimes F)\\ \tau_H : HE \otimes_{TB} TF \to T(E \otimes F).$$ The images $$\tau_Z(ZE \otimes_{TB} TF)$$ and $$\tau_H(HE \otimes_{TB} ZF)$$ are both equal to $$Z(E \otimes F$$), so the image of $$\tau_Z$$ is contained in the image of $$\tau_H$$. Since $$\tau$$ is surjective, by counting dimensions the only way this is possible is if $$\tau_H$$ is an isomorphism.
Of course, a connection on $$E$$ gives a corresponding horizontal lift isomorphism $$\operatorname{hlift} : \pi^*E \to HE$$, so (since $$\tau_H \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF}) = \tau \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF})$$) the composite $$\mu := \tau \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF}) : \pi^*E \otimes TF \to T(E \otimes F)$$ is the canonical isomorphism we desire.
By the way, if $$K_E : TE \to E$$ and $$K_F : TF \to F$$ are both connector maps over here I show that the canonical connector $$K_{E \otimes F}$$ for $$T(E \otimes F)$$ is $$K_{E \otimes F} \circ \tau = K_E \otimes \pi_F + \pi_E \otimes K_F.$$ Under the identification $$\pi^*E \otimes TF \cong T(E \otimes F)$$ above this means that the connector for $$\pi^*E \otimes TF$$ becomes \begin{align*} (X, e) \otimes \xi \in \pi^*E \otimes TF &\mapsto ( K_E \otimes \pi_F + \pi_E \otimes K_F) (\operatorname{hlift}(X,e) \otimes \xi)\\ &= K_E (\operatorname{hlift}(X,e)) \otimes \pi_F \xi + \pi_E (\operatorname{hlift}(X,e)) \otimes K_F \xi\\ &= 0 \otimes \pi_F \xi + e \otimes K_F \xi\\ &= (\pi_E \otimes K_F)((X, e) \otimes \xi). \end{align*}
In other words, the connector on $$\pi^*E \otimes TF$$ is just $$\pi_E \otimes K_F$$, with the connection on $$E$$ hidden in the identification $$\mu : \pi^*E \otimes TF \to T(E \otimes F)$$ itself.