$\newcommand{\M}{M}$
$\newcommand{\N}{N}$
$\newcommand{\TM}{TM}$
$\newcommand{\TN}{TN}$
$\newcommand{\TstarM}{T^\*M}$
$\newcommand{\Ga}{\Gamma}$


Let $\M,\N$ be smooth manifolds, $\phi:\M \to \N$ be a smooth map. Let $\nabla$ be a **symmetric** connection on $\TN$, and let $X,Y \in \Ga(\TM)$. 


Then the following holds:
 $$ \nabla^{\phi^\*\TN}\_X d\phi(Y) - \nabla^{\phi^\*\TN}\_Y d\phi(X) = d\phi([X,Y]) $$


where $\nabla^{\phi^\*\TN}$ is the pullback connection induced on $\phi^\*\TN$ by $\phi,\nabla$.


It is not hard to prove this by choosing local coordinates on $M,N$.




---


(This result comes up naturally in many different scenarios, for instance when calculating the first variation of the Dirichlet's integral)


**Question:**


Is there an "invariant" (coordinate-independent) proof of this result?


I do not think the "coordinates-proof" is simple enough. Perhaps there is an elegant way to show this result is indeed obvious at a glance?


The problem seems to be that the pullback-connection is defined by a characterising property which is local in nature (action on pullback sections), but I wonder if there is a way to bypass it somehow.



Does "symmetry" of a pullback connection should be obvious?
$\newcommand{\id}{\operatorname{Id}}$


Well, there is a natural way to view this "pullback-symmetry":


**Exterior derivative commutes with pullbacks:**


Let $f:M \to N$ be a smooth map, $E$ a vector bundle over $N$ with a connection $\nabla$. Then, there is a pullback operation: $ \Omega^k(N,E) \stackrel{f^\*}{\to} \Omega^k(M,f^\*E)$.


We have the commutative diagram
$$\begin{matrix}
 \Omega^k(N,E) \stackrel{d\_{\nabla}}{\longrightarrow} \Omega^{k+1}(N,E) \cr
 \downarrow f^\* \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \downarrow f^\* \cr
 \Omega^k(M,f^\*E) \stackrel{d\_{f^\*(\nabla)}}{\longrightarrow} \Omega^{k+1}(M,f^\*E) \cr
\end{matrix}$$


Now look at $\id\_{TN} \in \Omega^1(N,TN)$. The symmetry of the connection on $N$ is exactly $d\_{\nabla^{TN}} (\id\_{TN})=0$.


Considering $df=f^\*(\id\_{TN}) \in \Omega^1\big(M,f^\*(TN)\big)$, we get that


$$ d\_{\nabla^{f^\*(TN)}}(df)= d\_{\nabla^{f^\*(TN)}}(f^\*(\id\_{TN}))=f^\*(d\_{\nabla}\id\_{TN})=f^\*0=0$$. 


$d\_{\nabla^{f^\*(TN)}}(df)=0$ is exactly the "symmetry statement" mentioned in the question.


Of course, it remains to prove the above commutation property. For $k=0$ this is exactly the definition of the pullback connection. It seems that the general claim still requires coordinates, since we need to use the characterising property of the pull-back connection, which is local.


(I tried a direct approach using the invariant formula of the exterior covariant derivative, but this failed).


