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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...


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Since $\chi(t)$ is a geodesic, $\nabla^2_{\dot{\chi(t)}}\dot{\chi(t)} = 0$, so yes, $(\gamma, \chi)$ is a geodesic. In fact $\nabla_{\dot{\chi}} \dot{\chi} = 0$ is the definition of being a geodesic. Your curve $\Gamma$ is also a geodesic with the metric $ag_1 + bg_2$. To see this, note that if $\gamma$ is a geodesic of $g_1$, it's also a geodesic of $ag_1$ for any $a$. Now just use the argument for $g_1 + g_2$. – Jason DeVito Dec 14 '12 at 1:01
I thought geodesics were when the second fundamental form $\nabla_{\dot\Gamma}\dot\Gamma - \nabla^{(\pi)}_{\dot\Gamma}\dot\Gamma$ vanished (with $\pi$ the projection on the tangent space)... what is the difference then ? – WhitAngl Dec 14 '12 at 1:09
When the second fundamental form vanishes, you have a totally geodesic submanifold. Certainly a 1-d totally geodesic submanifold is a (portion of a) geodesic, but there are many more examples. For example, $\mathbb{R}^2\subseteq \mathbb{R}^3$ (usual embedding) is a totally geodesic submanifold. – Jason DeVito Dec 14 '12 at 1:11
Give me some time to work out your actual proposed counterexample, and I'll get back to you. – Jason DeVito Dec 14 '12 at 1:13
Thanks! if you're right with the definition of the geodesic, then I understand why it should be zero. But I thought geodesics were 1d autoparallel submanifolds, and that by definition their second fundamental form vanished since the second fundamental form measures a curvature... – WhitAngl Dec 14 '12 at 1:15
up vote 7 down vote accepted

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.

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thanks! However I don't seize the difference between geodesic and totally geodesic in this case (wiki tells me "Totally geodesic submanifold is a submanifold such that all geodesics in the submanifold are also geodesics of the surrounding manifold" which doesn't really help me). Does totally geodesic mean that the curve is a geodesic with non constant speed (like in your $t^3$ example) ? – WhitAngl Dec 14 '12 at 14:37
(the reason I don't understand wiki's definition in this context is that I have a 1D submanifold, and if all geodesics in this 1D submanifolds are also geodesics in the surrounding manifold, then I am under the impression that my 1D submanifold is a geodesic in the surrounding manifold) – WhitAngl Dec 14 '12 at 20:53
If the image of a curve $c$ is totally geodesic, then $c$ itself is almost a geodesic - it may not have constant speed. Thus, after suitable reparameterization, it will be a geodesic. That is the only difference between a geodesic and a curve with totally geodesic image. – Jason DeVito Dec 14 '12 at 23:02
Thanks a lot !! – WhitAngl Dec 14 '12 at 23:19

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 $M\times N$.

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