Geodesics on the product of manifolds

Solution 1:

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.

Solution 2:

The connection of Levi Civita $\widetilde{\nabla}$ of 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 $N$ such that if $X=(X_1,X_2)$ and $Y=(Y_1,Y_2)$ are tangent fields on $M\times N$, then $\widetilde{\nabla}_XY=\nabla^1_{X_1}Y_1+\nabla^2_{X_2}Y_2$. Consequente, se $\gamma_1$ and $\gamma_2$ are geodesics of $M$ and $N$ respectively, it follows that $\nabla_{\gamma_1^{\prime}}^1\gamma_1^{\prime}=0$ and $\nabla_{\gamma_2^{\prime}}^2\gamma_1^{\prime}=0$. 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 $M\times N$.