# Is the pullback of a connection compatible with a bundle metric also compatible with the pullback metric?

Let $M$ be a smooth manifold, and $E\to M$ a smooth vector bundle. Then if we have a smooth morphism $f:N\to M$ of smooth manifolds, then we obtain the following commutative diagram, $$\require{AMScd} \begin{CD}f^*E @>>> E \\ @VVV @VVV \\ N @>f>> M \end{CD}$$ where $f^*E$ is the pullback of $E$. Therefore, if we have some bundle metric $g\in E^*\otimes E^*$, we can define the pullback bundle metric $\tilde g:=f^*g\in f^*E^*\otimes f^*E^*$ in the natural way, and for any connection $$\nabla:\Gamma(\mathrm TM)\otimes\Gamma(E)\to\Gamma(E)$$ we can define the pullback connection $$f^*\nabla:\Gamma(\mathrm TN)\otimes\Gamma(f^*E)\to\Gamma(f^*E)$$ as follows:

Let $(x^i)$ be coordinates on $N$ mapping into coordinates $(y^j)$ on $M$, and let $E$ be locally trivial with basis $(e_k)$ on $M$, then if $\nabla_{\partial/\partial y^i}e_j=\Gamma_{ij}^k e_k$, then if we set $$\tilde\Gamma_{ij}^k=\frac{\partial f^l}{\partial x^i}\Gamma_{lj}^k\circ f$$ then we can define $\tilde\nabla$ by letting $\tilde\nabla_{\partial/\partial x^i} e_j=\tilde\Gamma_{ij}^k$.

Now, suppose $\nabla$ is compatible with $g$, that is to say, for $a,b\in\Gamma(E)$ and $X\in\Gamma(\mathrm T M)$, $$X\langle a,b\rangle = \langle\nabla_Xa,b\rangle+\langle a,\nabla_Xb\rangle$$ then I wish to prove that for $a,b\in\Gamma(f^*E)$ and $X\in\Gamma(\mathrm T N)$, $$X\langle a,b\rangle = \langle\tilde\nabla_X a,b\rangle+\langle a,\tilde\nabla_Xb\rangle$$ but is this true? I've been having trouble proving this even for $X=\frac{\partial}{\partial x^i}$ and $a=e_j$, $b=e_k$. In this case, I obtain that

$$\frac{\partial}{\partial x^i}\langle e_j,e_k\rangle = \frac{\partial g_{jk}}{\partial x^i} = \frac{\partial g_{jk}}{\partial y^l}\frac{\partial f^l}{\partial x^i}$$ and

$$\langle\tilde\nabla_{\partial/\partial x^i}e_j,e_k\rangle + \langle e_j,\tilde\nabla_{\partial/\partial x^i}e_k\rangle =\tilde\Gamma_{ij}^l g_{lk} + \tilde\Gamma_{ik}^l g_{lj} = \Gamma_{mj}^l\frac{\partial f^m}{\partial x^i}g_{lk} + \Gamma_{mk}^l\frac{\partial f^m}{\partial x^i}g_{lj}$$ but I don't see why these two expressions should be equal.

• Note that it is wrong to think of $\nabla$ as a map $\Gamma(TM)\otimes\Gamma(E)\to\Gamma(E)$, as it is only tensorial in the $\Gamma(TM)$ part. – Amitai Yuval Jun 18 '17 at 2:52

Every connection is determined by its parallel transport, and a connection is compatible with a given metric if and only if the associated parallel transport preserves the metric. Hence, we wish to show that if parallel transport with respect to $\nabla$ preserves $g$, then parallel transport with respect to $f^*\nabla$ preserves $f^*g$. This follows immediately from the following characterization of parallel transport with respect to $f^*\nabla$:
Let $\gamma:[0,1]\to N$ be a smooth path. Then we have $f\circ\gamma:[0,1]\to M$, and the fibers of $f^*E$ over the endpoints of $\gamma$ are identified with the fibers of $E$ over the endpoints of $f\circ\gamma$. Parallel transport with respect to $f^*\nabla$ along $\gamma$ is identical to parallel transport with respect to $\nabla$ along $f\circ\gamma$. (This is actually one of the ways to define the pullback connection).
You have already obtained the answer. From your last expression using the identity $$g_{kl}\Gamma^{l}_{mj}+g_{jl}\Gamma^{l}_{mk}=\partial_{y^m}g_{jk}$$ one gets $$(g_{kl}\Gamma^{l}_{mj}+g_{jl}\Gamma^{l}_{mk}){\partial f^m\over\partial x^i}={\partial f^m\over\partial x^i}{\partial g_{jk}\over\partial y^m}$$ Which is the expected answer. You can check the identity by using the standard form for Christoffels $$\Gamma^{i}_{jk}={g^{il}\over 2}({\partial g_{lk}\over\partial y^j}+{\partial g_{lj}\over\partial y^k}-{\partial g_{jk}\over\partial y^l})$$