Does "symmetry" of a pullback connection should be obvious?

Question

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

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 $M,N$.

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(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.

Answer 1

Here is a proof which is not very elegant, but avoids coordinates. First of all, you define $$\hat T(X,Y)=\nabla_X^{\phi^*TN}d\phi(Y)-\nabla_Y^{\phi^*TN}d\phi(X)-d\phi([X,Y])$$ and observe that this is a tensor. Hence, you can stick to the 2-dimensional case (generated by the (commuting) flows of two appropriate vector fields extending the tangent vectors $X$ and $Y$).

Here, you have to distinguish cases: Let $p\in M$ (M 2-dimensional)

1.case: $d_p\phi$ is surjective: this is the classical case of a submanifold and easily follows by the fact that the commutator of $\phi$-related vectorfields is $\phi$-related to the commutator of the vectorfields.

2.case: $p$ is in the boundary of case 1: This follows from the fact that $\hat T$ is obviously continous.

3.case: There is a neighbourhood of $p$ such that $d_q\phi$ has at most rank 1, i.e. it has atleast 1-dimensional kernel. Then, the satment follows from skew-symmetry in $X,Y$ of $\hat T$ and the fact that you can find a non-vanishing vectorfield $Y$ such that $d\phi(Y)=0$ by the definition of the connection $\nabla^{\phi^*TN}.$

Answer 2

$\newcommand{\id}{\operatorname{Id}}$

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).