Commutator of connection and bundle endomorphism

Question

Let $E\rightarrow M$ be a vector bundle over a manifold $M$, equipped with a connection $\nabla^E$. Let $\Phi$ be a smooth endomorphism of $E$, that is a smooth section of the endomorphism bundle of $E$ over $M$.

Question: Let $X$ be a vector field on $M$. Is the commutator

$$[\nabla^E_X,\Phi]=\nabla^E_X\circ\Phi-\Phi\circ\nabla^E_X$$
always an endomorphism of $E$?

Thoughts: This is certainly true if $\Phi$ is multiplication by a function, as one can show using the Leibniz rule. The statement in question seems to be a natural analogue for the case of endomorphisms.

Answer 1

Any linear connection $\nabla^E$ has a unique extension to a tensor derivation on $E$, i.e. a connection (which I'll just write $\nabla$) on each tensor product $E\otimes\cdots\otimes E \otimes E^* \cdots \otimes E^*$ satisfying

• $\nabla \xi =\nabla^E \xi$ for $\xi \in \Gamma(E)$,
• the Leibniz rule $\nabla(\xi \otimes \zeta) = \nabla \xi \otimes \zeta + \xi \otimes \nabla \zeta$ and
• $\nabla(C(\xi)) = C(\nabla(\xi))$ for any contraction $C$.

I think this is a fact that is well worth your time to prove and internalize, and it makes your desired claim easy: Since an endomorphism is just a section of $E \otimes E^*,$ we thus have $$\nabla_X(\Phi(\xi)) = \nabla_X(C(\Phi \otimes \xi)) =(\nabla_X \Phi)(\xi) + \Phi(\nabla_X \xi),$$ so the the commutator is just $[\nabla_X^E, \Phi] = \nabla_X\Phi\in\Gamma(\operatorname{End}(E)).$

If you want to prove your claim more directly, here are a couple of approaches to try:

• Show directly that $[\nabla^E_X, \Phi] : \Gamma(E) \to \Gamma(E)$ is $C^\infty(M)-$linear, and thus an endomorphism by the bundle homomorphism characterization lemma. This should be a very neat calculation using the Leibniz rule.
• Fix a local frame for $E$ and work in components/coordinates. You should be able to manipulate $[\nabla^E_X, \Phi]\xi$ and end up with some linear function of the components $\xi^b$.