# Image of a path $\gamma$ is totally geodesic $\Rightarrow \gamma$ is a reparametrization of a geodesic?

Let $M$ be a Riemannian manifold. Assume $\gamma$ is a path in $M$ , such that it's image is a totally geodesic submanifold of $M$.

I am trying to prove (the seemingly trivial result) that $\gamma$ must be a reparametrization of a geodesic in $M$.

I came up with two possible explanations (described below). I would like to find a simpler argument.

First explanation:

Lemma(1): On a one dimensional Riemannian manifold, from any point there is only one path with constant speed $c$. (up to direction).

Proof:

Since any one dimensional manifold $M$ is locally diffeomorphic to $\mathbb{R}$ and the question is local in nature, we can assume $M = (0,1)$ , with some arbitary metric.

Assume $\beta,\alpha:I \to M=(0,1) \, , \, \alpha(0)=\beta(0)=p, \dot\alpha(0)=\dot\beta(0) \, , \, \forall t \, \|\dot\alpha(t)\|=\|\dot\beta(t)\|=1$.

Since $\|\dot\alpha(t)\|=1$, $\dot\alpha(t) \in \{1,-1 \}$. By continuity it does not change sign. Since we assumed $\dot\alpha(0)=\dot\beta(0) \,$, it follows that $\forall t \, \dot\alpha(t)=\dot\beta(t)$, so now integration gives us $\alpha(t)=\beta(t)$.

Corollary(1): On a one dimensional Riemannian manifold, any path with constant speed is a geodesic.

Proof:

Any geodesic has a constant speed, but from any point there is only one path with constant speed $c$. So this path must be a geodesic.

Proof of the claim:

Assume $S=\operatorname{Image} \gamma$ is a totally geodesic submanifold of $M$.

By corollary (1), if $\alpha$ is a constant speed reparametrization of $\gamma$, it's a geodesic in $S$. Hence it's a geodesic in $M$.

Second explanation:

Lemma(1): $S=\operatorname{Image} \gamma$ is totally geodesic iff (*) $\nabla_{\dot{\gamma}} \dot{\gamma} - \frac{g(\nabla_{\dot{\gamma}}\dot{\gamma}, \dot{\gamma} )}{g(\dot{\gamma}, \dot{\gamma})}\dot{\gamma} = 0.$

where $\nabla_{\dot{\gamma}} \dot{\gamma}$ is the usual covariant derivative (in M) along the path $\gamma$.

Proof:

Since the question is local, we assume $\gamma$ is injective, hence it's a invertible from it's image. So any path $\alpha$ in $S$, is of the form: $\alpha(s)=\gamma(\phi(s))$.

So, $\dot \alpha(s)=\dot \gamma(\phi(s)) \cdot \phi'(s)$

$\nabla_{\dot{\alpha}}^M \dot{\alpha}=\phi''(s)\cdot \dot \gamma(\phi(s)) + \big(\phi'(s)\big)^2 \cdot \nabla_{\dot{\gamma}}^M \dot{\gamma} (\phi(s))$

The covariant derivative along a path in a Riemannian submanifold $S \subseteq M$ is the projection on $TS$ of the covariant derivative in $M$. In our case this just amounts to projecting on $sp\{\dot\gamma(t)\} \subseteq T_{\gamma(t)}M$:

$\nabla_{\dot{\alpha}}^S \dot{\alpha}=\phi''(s)\cdot \dot \gamma(\phi(s)) + \big(\phi'(s)\big)^2 \cdot Pr\nabla_{\dot{\gamma}}^M \dot{\gamma} (\phi(s))$

So it's easy to see that:

$\alpha$ is a geodesic in $S$ (i.e $\nabla_{\dot{\alpha}}^S \dot{\alpha}=0$) $\Rightarrow$ $\alpha$ is a geodesic in $M$ (i.e $\nabla_{\dot{\alpha}}^M \dot{\alpha}=0$) iff $\nabla_{\dot{\gamma}} \dot{\gamma} - \frac{g(\nabla_{\dot{\gamma}}\dot{\gamma}, \dot{\gamma} )}{g(\dot{\gamma}, \dot{\gamma})}\dot{\gamma} = 0.$

Proof of the claim:

Let $\alpha(s)=\gamma(\phi(s))$ be a constant speed reparametrization of $\gamma$, so $|\dot \alpha(s)|=|\dot \gamma(\phi(s))| \cdot |\phi'(s)|=1$.

This implies (assuming $\phi'(s) >0$) $\phi'(s) =\frac{1}{\|\dot \gamma(\phi(s)) \|}=\frac{1}{\sqrt{g(\dot \gamma,\dot \gamma) \circ \phi}}$

Hence:

\begin{align} & \phi''(s)=-\frac{1}{\|\dot \gamma(\phi(s)) \|^2} \cdot \frac{1}{2\sqrt{g(\dot \gamma,\dot \gamma) \circ \phi}} \cdot \frac{d}{ds}\big( g(\dot \gamma,\dot \gamma) \circ \phi \big)= \\ & -\frac{1}{\|\dot \gamma(\phi(s)) \|^2} \cdot \frac{1}{2\sqrt{g(\dot \gamma,\dot \gamma) \circ \phi}} \cdot \Big( 2g(\nabla_{\dot{\gamma}} \dot{\gamma} ,\dot{\gamma}) \circ \phi \Big) \cdot \phi'(s) = \\ & - \big(\phi'(s)\big)^2 \cdot \frac{g(\nabla_{\dot{\gamma}}\dot{\gamma}, \dot{\gamma} )}{g(\dot{\gamma}, \dot{\gamma})} \circ \phi \end{align}

So finally,

$\nabla_{\dot{\alpha}} \dot{\alpha}=\phi''(s)\cdot \dot \gamma(\phi(s)) + \big(\phi'(s)\big)^2 \cdot \nabla_{\dot{\gamma}} \dot{\gamma} (\phi(s))= \big(\phi'(s)\big)^2 \cdot \Big( \big(\nabla_{\dot{\gamma}} \dot{\gamma} - \frac{g(\nabla_{\dot{\gamma}}\dot{\gamma}, \dot{\gamma} )}{g(\dot{\gamma}, \dot{\gamma})}\dot{\gamma} \big) \circ \phi \Big) =0$

As required.

In the spirit of your first argument, let $S \subseteq M$ be an embedded totally geodesic one-dimensional submanifold and choose $p \in S$. Let $\alpha \colon I \rightarrow S$ be a coordinate system around $p \in S$. In particular, $\alpha$ is an immersion and so we can compose $\alpha$ with a diffeomorphism and assume that $\alpha$ has constant speed. Then, computing in $S$ we have
$$0 = \frac{d}{dt} ||\dot{\alpha}(t)||^2 = 2 \left< \frac{D \dot{\alpha}}{dt}|_t, \dot{\alpha}(t) \right>$$
but since $S$ is one-dimensional and $\dot{\alpha}(t) \neq 0$, we must have $\frac{D\dot{\alpha}}{dt} \equiv 0$ showing that $\alpha$ is a geodesic in $S$ and thus in $M$.