Levi Civita connection on product of Riemannian Manifold

Question

I'm trying to prove that for a product Riemannian manifold, say $M=M_1\times M_2$ with product metric one has $\nabla_{X_{1}+X_{2}}^{M}\left(Y_{1}+Y_{2}\right)=\nabla_{X_{1}}^{M_{1}}\left(Y_{1}\right)+\nabla_{X_{2}}^{M_{2}}\left(Y_{2}\right)$ for $X_i,Y_i\in TM_i$ via Koszul formula. So i choose $Z=Z_1+Z_2$ and i get \begin{equation} 2g(\nabla^M _{X_1+X_2} Y_1+Y_2,Z)=2g(\nabla ^{M_1}_{X_1}Y_1,Z_1)+2g(\nabla ^{M_2}_{X_2}Y_2,Z_2)+ X_2g(Y_1,Z_1)+X_1g(Y_2,Z_2)+Y_1g(X_2,Z_2)+Y_2g(X_1,Z_1)-Z_1g(X_2,Y_2)-Z_2g(X_1,Y_1) \end{equation}

I'd like that those therms cancels each out so the first two therms give the thesis, right? but i cant see how this happens

Answer 1

Take, for example, $X_2 g(Y_1,Z_1)$. $\require{cancel}g(Y_1,Z_1)=g_1(Y_1,Z_1)+\cancel{g_2(Y_1,Z_1)}$ is constant in the second coordinate $p_2\in M_2$, and $X_2$ is a derivation on $M_2$. So $X_2 g(Y_1,Z_1)=0$. Similarly the other terms.