I'm completely stuck proving the naturality of the pullback connection.
The strategy suggested is a follows:
We let $\phi: (M,g) \to (\tilde{M}, \tilde{g})$ be an isometry, with connections $\nabla$ and $\tilde{\nabla}$ respectively.
Define the "pullback connection":
$(\phi^* \tilde{\nabla})_X Y=\phi_*^{-1} ( \tilde{\nabla}_{\phi_* X}(\phi_* Y)). $ Now show that it is symmetric and compatible with the metric, and hence agrees with the unique Riemannian connection on $M.$
I am completely stuck with showing metric compatibility.
According to the definition of a symmetric connection, I need to show $(\phi^* \tilde{\nabla})_X \langle Y, Z \rangle= \phi_*^{-1} ( \tilde{\nabla}_{\phi_* X}(\phi_* \langle Y,Z \rangle)) = \langle\phi_*^{-1} (\tilde{\nabla}_{\phi_* X}(\phi_* Y)),Z \rangle + \langle Y,\phi_*^{-1} (\tilde{\nabla}_{\phi_* X}(\phi_* Z)) \rangle. $
I am willing to believe that this is just a matter of unwinding definitions. However, I don't even understand how $\phi_* \langle Y, Z \rangle$ should even be defined.
Could anyone help get me started by perhaps doing the first few terms of the computation to show me how it should go? (or else guide me to somewhere in the literature where this is done?)
Best Answer
You need to show that $Xg(Y,Z)=g((\phi^\ast\tilde\nabla)_XY,Z)+g(Y,(\phi^\ast\tilde\nabla)_XZ)$.
Try showing that for $p\in M$, $$g_p(((\phi^\ast\tilde\nabla)_XY)_p,Z_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\tilde\nabla_{\phi_\ast X}\phi_\ast Y)_{\phi(p)},(\phi_\ast Z))_{\phi(p)}\right)$$ and similarily $$g_p(Y_p,((\phi^\ast\tilde\nabla)_XZ)_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\phi_\ast Y)_{\phi(p)},(\tilde\nabla_{\phi_\ast X}\phi_\ast Z)_{\phi(p)}\right)$$
Add these terms and use the fact that $(\phi^{-1})^\ast g=\tilde g$ and the metric compatibility of $\tilde\nabla$.
You should get that there sum at $p$ is $$ \left((\phi_\ast X)(\tilde g(\phi_\ast Y,\phi_\ast Z)\right)(\phi(p)).$$ Then try showing that this is equal to $\left(Xg(Y,Z)\right)(p)$