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 $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0?$

  • 6
    $\begingroup$ There's practically nothing to prove. The fact that $f$ is an isometry means that it will commute with all the operations of a Riemannian manifold - including covariant differentiation. $\endgroup$
    – Zhen Lin
    Feb 15, 2012 at 0:38
  • 1
    $\begingroup$ @Zhen Lin: so, basically you mean $\frac{D}{dt}(df)=df(\frac{D}{dt})?$ $\endgroup$ Feb 15, 2012 at 0:55

2 Answers 2


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 $$ D^h_t{\frac{\operatorname{d}(f\circ\gamma)}{\operatorname{d}t}} = f_* \left( D^g_t{\frac{\operatorname{d}\gamma}{\operatorname{d}t}} \right) =0 $$

  • $\begingroup$ Hi, I would like to ask a couple of things regarding your answer: i. are you assuming the natural property in your last equation? if so, how can it be proved from the definition of isometry as preserving induced metric? ii. what is the meaning of upper * and $(f^{-1})^*g$ in your first equation? $\endgroup$
    – jj_p
    Jun 20, 2013 at 10:40
  • $\begingroup$ @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. $\endgroup$ Jul 1, 2013 at 14:56

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 $$\tilde{\nabla}_{\frac{d}{dt}}\dot{\varphi}(\gamma(t))=\tilde{\nabla}_{\frac{d}{dt}}\varphi_*\dot{\gamma}(t)=\varphi_*(\nabla_{\frac{d}{dt}}\dot{\gamma}(t))=0.$$


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