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Let $f$ be an isometry (i.e a diffeomorphism which preserves the Riemannian metrics) between Riemannian manifolds $(M,g)$ and $(N,h).$

One can argue that $f$ also preserves the induced metrics $d_1, d_2$ on $M, N$ from $g, h$ resp. that is, $d_1(x,y)=d_2(f(x),f(y))$ for $x,y \in M.$ Then, it's easy to show that $f$ sends geodesics on $M$ to geodesics on $N,$ using the length minimizing property of geodesics and that $f$ is distance-preserving.

My question,

Is it possible to derive the result without using the distance-preserving property of isometries, by merely the definition?

What I have found so far;

Let $\gamma : I \to M$ be a geodesic on $M$, i.e. $\frac{D}{dt}(\frac{d\gamma}{dt})=0,$ where $\frac{D}{dt}$ is the covariant derivative and $ \frac{d\gamma}{dt}:=d\gamma(\frac{d}{dt}).$ Let $t_0 \in I,$ we have to show that $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})=0$ at $t=t_0,$ or $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0.$

We also know that

$ \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}= \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}.$

Since $\frac{d}{dt} \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=2 \langle \frac{D}{dt}(\frac{d \gamma}{dt}|_{t=t_0}),\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=0,$ therefore

$\frac{d}{dt} \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}=$

$2\langle \frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle _{f(\gamma(t_0))}=0.$

How can I conclude from $\langle\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}) \rangle _{f(\gamma(t_0))}=0$ that $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0?$

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    @Zhen Lin: so, basically you mean $\frac{D}{dt}(df)=df(\frac{D}{dt})?$2012-02-15

2 Answers 2

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Your calculation looks like an attempt to prove the naturality of the Levi-Civita connection, the fact that @Zhen Lin implicitly points to. In the settings of the question it can be stated as $ \nabla^g_X{Y}=f^* \left( \nabla^{(f^{-1})^* g}_{\operatorname{d}f(X)} \operatorname{d}f(Y) \right) $

Notice also that in fact you are using two different connections: one for vector fields along $\gamma \colon I \rightarrow M$ induced from $\nabla^g$ on $M$, and another one for vector fields along $f \circ \gamma \colon I \rightarrow N$ induced from $\nabla^h$ on $N$. Due to the naturality property they agree, and it may be helpful to distinguish $D^g_t:=\nabla^g_{\frac{d}{dt}{\gamma}}$ and $D^h_t:=\nabla^h_{\frac{d}{dt}(f \circ \gamma)}$ in the present calculation. Indeed, using $\frac{\operatorname{d}}{\operatorname{d}t}(f\circ \gamma)=\operatorname{d}f(\frac{\operatorname{d}}{\operatorname{d}t}\gamma) $ we get $ D^h_t{\frac{\operatorname{d}(f\circ\gamma)}{\operatorname{d}t}} = f_* \left( D^g_t{\frac{\operatorname{d}\gamma}{\operatorname{d}t}} \right) =0 $

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    @Nicolo' This is a good question, really. Sorry, I am traveling at the moment and don't have enough time. I admit that the first equation looks awkward, and I don't remember why I put it in that form that time. To clarify, $f$ is an isometry, so its differential $f_*$ has an inverse $f^*$, and $(f^{-1})^* g = h$. And yes, I am just applying this naturality property in the answer.2013-07-01
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It is actually an exercise in the Lee's book, I try to do it by following the hint. First, you have to understand the naturality of Riemannian connection, then everything will be clear. I like using $\nabla_{\frac{d}{dt}}$ instead of $D_t$ here.

First, Define an operator $\varphi^*\tilde{\nabla}_{\frac{d}{dt}}:\mathcal{J}(\gamma)\rightarrow\mathcal{J}(\gamma)$ by $(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V=\varphi^*(\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V))$. Suppose $V$ is a vector field along $\gamma.$ Show this operator satisfies three properties in the Lemma 4.9. in Lee's Riemannian Geometry.

(1) Linearity over $\mathcal{R}$:$(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})(aV+bW)=a(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V+b(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})W$ (2) Product rule: $(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})(fV)=\frac{df}{dt}V+f(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V.$ (3) If $V$ is extendible, then for any extension $\tilde{V}$ of $V$, $(\varphi^*\tilde{\nabla}_{\frac{d}{dt}})V=(\varphi^*\tilde{\nabla}_{\dot{\gamma}(t)})\tilde{V}.$ Then by uniqueness, the operator we defined above is just the unique operator $\nabla_{\frac{d}{dt}}$. that is $\nabla_{\frac{d}{dt}}V=\varphi^*(\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V)) \ \text{or} \ \ \varphi_*(\nabla_{\frac{d}{dt}}V)=\tilde{\nabla}_{\frac{d}{dt}}(\varphi_*V).$

Now If $\gamma$ is the geodesic in $M$ with initial $p$ and initial velocity $V$, i.e., $\nabla_{\frac{d}{dt}}\dot{\gamma}(t)=0$. Obviously $\varphi\circ\gamma$ is a curve in $\tilde{M}$ with initial point $\varphi(p)$ and initial velocity $\varphi_*V$, moreover, it is also the geodesic since $\tilde{\nabla}_{\frac{d}{dt}}\dot{\varphi}(\gamma(t))=\tilde{\nabla}_{\frac{d}{dt}}\varphi_*\dot{\gamma}(t)=\varphi_*(\nabla_{\frac{d}{dt}}\dot{\gamma}(t))=0.$