Given two Riemannian manifolds $(M, g_1)$ and $(N, g_2)$, and geodesic curves $\gamma(t)$ in $M$ and $\chi(t)$ in $N$. Is the curve $\Gamma(t) = (\gamma(t),\chi(t))$ a geodesic in the product manifold $(M \times N, g_1 + g_2)$ ? Is it a geodesic if we now consider the product manifold $(M \times N, \alpha g_1 + \beta g_2)$ where $\alpha$ and $\beta$ are two positive (or zero) scalar constants ?
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EDIT: Although I feel it should be the case (geodesic), could you tell me if my counter-example is right:
Let's say $g_1$ is euclidean, and $g_2$ is not. I am interested in whether $\nabla_{\dot\Gamma}\dot\Gamma$ has only components along $\dot\Gamma$ (ie., $\Gamma$ is autoparallel). Let's call the Levi-Civita connections on $M$ and $N$, respectively $\nabla^1$ and $\nabla^2$. Since $g_1$ is euclidean, $\nabla^1_{\dot\gamma}\dot\gamma=0$. Projecting $\nabla_{\dot\Gamma}\dot\Gamma$ on $\dot\Gamma^\perp$ (to check whether its perp component is $0$), I get $\nabla^{(\pi)}_{\dot \Gamma}\dot\Gamma = (0, \nabla^2_{\dot\chi}\dot\chi - \frac{g_2(\nabla^2_{\dot\chi}\dot\chi,\dot\chi)}{g_1(\dot\gamma,\dot\gamma)+g_2(\dot\chi,\dot\chi)}\dot\chi)$. This second term isn't expected to be zero, right? Since $\chi$ is geodesic, only $\nabla^2_{\dot\chi}\dot\chi - \frac{g_2(\nabla^2_{\dot\chi}\dot\chi,\dot\chi)}{g_2(\dot\chi,\dot\chi)}\dot\chi)$ is zero...
Thanks!