My question is about the answer to this post : Geodesics on the product of manifolds
If $(M_{1},g_{1})$, $(M_{2},g_{2})$ are Riemannian manifolds and if $\nabla^{1}$ (resp. $\nabla^{2}$) denotes the Levi-Civita connection on $M_{1}$ (resp. $M_{2}$), the following result is clear : if $\gamma_{1}$ (resp. $\gamma_{2}$) is a geodesic of $M_{1}$ (resp. $M_{2}$), then $(\gamma_{1},\gamma_{2})$ is a geodesic of $M_{1} \times M_{2}$. It follows from the previous post.
However, I am not convinced why the converse is true. If $\gamma=(\gamma_{1},\gamma_{2})$ is a geodesic of $M_{1} \times M_{2}$, is it true that $\gamma_{1}$ (resp. $\gamma_{2}$) is a geodesic of $M_{1}$ (resp. $M_{2}$) ? Let $\tilde{\nabla}$ be the Levi-Civita connection on $M_{1} \times M_{2}$. By definition, $\tilde{\nabla}_{\dot{\gamma}}\dot{\gamma} = \nabla^{1}_{\dot{\gamma_{1}}} \dot{\gamma_{1}} + \nabla^{2}_{\dot{\gamma_{2}}} \dot{\gamma_{2}} = 0$. Does that imply that $\nabla^{1}_{\dot{\gamma_{1}}} \dot{\gamma_{1}} = 0$ and $\nabla^{2}_{\dot{\gamma_{2}}} \dot{\gamma_{2}} = 0$ ?
Yes, because the components are linear independent. $\nabla^1$, e.g., maps the tangent space of the first factor into itself (in the sense that $\nabla^1_X Y\in TM_1$ if $X, Y$ are. This is assuming you are working with the product metric and corresponding Levi Civita connections, of course).