Let $(M, g)$ and $(N, h)$ be pseudo-Riemannian manifolds with Levi-Civita connections $\nabla^g$ and $\nabla^h$, respectively. Let $\varphi : (M, g) \to (N, h)$ be an isometry. Show that $\varphi_* \left(\nabla^g_X Y \right) = \nabla^h_{\varphi_*X}\varphi_*Y$ for all $X, Y ∈ \mathfrak{X}(M)$.
could someone show me how to prove this? For now, I know that maybe I should prove that the inner product of the left side is equal to the right side. And I am complete lost after that...
If $\varphi$ is a a diffeomorphism, then $\varphi_* : \mathfrak{X}(M) \to \mathfrak{X}(N)$ is a bijection. Use Koszul's formula, which uniquely determines the Levi-Civita connection: \begin{align} 2h\left(\nabla^h_{\varphi_*X}{\varphi_*Y},\varphi_*Z\right) =& (\varphi_*X)\cdot h\left(\varphi_*Y,\varphi_*Z \right) + (\varphi_*Y)\cdot h\left(\varphi_*X,\varphi_*Z \right)\\ &- (\varphi_*Z)\cdot h\left(\varphi_*X,\varphi_*Y\right) +h\left([\varphi_*X,\varphi_*Y],\varphi_*Z\right) \\ & - h\left([\varphi_*X,\varphi_*Z],\varphi_*Y \right) - h \left([\varphi_*Y,\varphi_*Z],\varphi_*X \right). \end{align} Use the naturality of the Lie bracket (i.e $[\varphi_*X,\varphi_*Y] = \varphi_*[X,Y]$), the fact that $\varphi$ is an isometry (i.e $h=\varphi_*g$) and the chain-rule to show that the RHS is equal to $$ \varphi_* \left(2g\left(\nabla^g_XY,Z \right) \right) $$ and conclude with uniqueness of the Levi-Civita connection.