Geodesics on the product of manifolds 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!
 A: The Levi–Civita connection $\widetilde{\nabla}$ of a product $M\times N$ is given by $\widetilde{\nabla}=\nabla^1+\nabla^2$ where $\nabla^1$ is the connection of Levi Civita of $M$ and $\nabla^2$ is the connection of Levi Civita of $N$. It means that if $X=(X_1,X_2)$ and $Y=(Y_1,Y_2)$ are tangent vector fields on $M\times N$, then $\widetilde{\nabla}_XY=\nabla^1_{X_1}Y_1+\nabla^2_{X_2}Y_2$.
It follows that, if $\gamma_1$ and $\gamma_2$ are geodesics of $M$ and $N$ respectively, then $\nabla_{\gamma_1^{\prime}}^1\gamma_1^{\prime}=0$ and $\nabla_{\gamma_2^{\prime}}^2\gamma_2^{\prime}=0$. Hence, if $\gamma=(\gamma_1,\gamma_2)$, we obtain
$$\widetilde{\nabla}_{\gamma^{\prime}}\gamma^{\prime}=\nabla_{\gamma_1^{\prime}}^1\gamma_1^{\prime}+\nabla_{\gamma_2^{\prime}}^2\gamma_2^{\prime}=0.$$
Therefore $\gamma$ is a geodesic of $M\times N$.
A: The key fact is the condition that $\gamma$ is a geodesic is strictly stronger than the condition that the image of $\gamma$ is totally geodesic.
This is clear from the equations - saying $\gamma$ is a geodesic means $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$ while the image of $\gamma$ being totally geodesic means $$\nabla_{\dot{\gamma}} \dot{\gamma} - \frac{g(\nabla_{\dot{\gamma}}\dot{\gamma}, \dot{\gamma} )}{g(\dot{\gamma}, \dot{\gamma})}\dot{\gamma} = 0.$$
If the image of $\gamma$ is totally geodesic it does follow that by reparamaterizing $\gamma$, it does become a geodesic.
Now, suppose $\Gamma = (\gamma_1(t), \gamma_2(t))$.  If $\gamma_1$ and $\gamma_2$ are both geodesics (meaning $\nabla_{\dot{\gamma}_i} \dot{\gamma}_i = 0$ for $i = 1,2$), then by your above equations, $\Gamma$ is a geodesic.
On the other hand, if $\gamma_1$ is a geodesic and $\gamma_2$ merely has totally geodesic image, then $\Gamma = (\gamma_1(t), \gamma_2(t))$ need not be totally geodesic.  Your equations show exactly why you shouldn't expect it to be, but here's a concrete counterexample.
Consider $M_1 = M_2 = \mathbb{R}$.  Let $\gamma_1(t) = t$ and let $\gamma_2 = t^3$.  Then $\gamma_1$ is a geodesic and the image of $\gamma_2$ is totally geodesic because there is no normal direction at all.  (If this is too trivial, take $M_1 = M_2 = \mathbb{R}^2$, $\gamma_1(t) = (t,0)$ and $\gamma_2(t) = (t^3, 0)$ - the rest of the argument will work in either case).  Then in $M_1\times M_2$, the image of $\Gamma(t) = (\gamma_1(t), \gamma_2(t))$ is the graph of a cubic polynomial, so is NOT a straight line, so is not a geodesic at all.  Nor is the subset totally geodesic.
