Let $E\to M$ be a vector bundle over a smooth (compact) manifold $M$. Let $\nabla^E$ be a connection on $E$. Given a smooth map $f\colon N\to M$, where $N$ is another smooth (compact) manifold, we have the pullback connection $\nabla^{f^*E}$ on the pullback vector bundle $f^*E$.
$\nabla^{f^*E}$ is uniquely determined by
$\nabla^{f^*E}_X(f^*s)=(\nabla^E_{df(X)}s)\circ f$
where $X\in TN$ and $s$ is a section of $E$.
Now, regarding the curvature, do we have
$R^{f^*E}(X,Y)s=R^E(df(X),df(Y))s$
for $X,Y\in TN$, $s$ a section of $f^*E$?
If yes, what is the best way to prove this identity? Can it be shown by just using the definition of curvature ($\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$) and a computation in local coordinates? To me that seems messy, but maybe there is some way to simplify the calcuations that I don't see.
Best Answer
You can use the fact that the pullback connection preserves parallel transport, in the sense that the pullback of a parallel transport is the same as the parallel transport of the pullback. You can then use the definition of curvature as the composition of parallel transports, which yields the same result. You are missing some pullbacks in your formula, however.
Here is an explicit calculation, where I'm using $\varphi$ instead of f
$ F^{\varphi^*\nabla}(X,Y)\varphi^*s= \varphi^*\nabla_X \varphi^*\nabla_Y \varphi^*s - \varphi^*\nabla_Y \varphi^*\nabla_X \varphi^*s -\varphi^*\nabla_{[X,Y]}\varphi^*s= \\ \varphi^*\nabla_X \varphi^*(\nabla_{d \varphi (Y)}s)-\varphi^*\nabla_Y \varphi^*(\nabla_{d \varphi (X)}s)-\varphi^*(\nabla_{d \varphi([X,Y])}s)=\\ \varphi^*\Big(\nabla_{d\varphi(X)}\nabla_{d\varphi(Y)}s-\nabla_{d\varphi(Y)}\nabla_{d\varphi(X)}s-\nabla_{[d \varphi(X),d\varphi(Y)]}s\Big)= \varphi^*(F^\nabla(d \varphi(X),d \varphi(Y))s) $
And the last step uses the naturality of the Lie bracket.