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Let $M$ be a complete connected Riemannian manifold and $r\in M$ and $p,q\in B(r;R)$ where $R$ is the convexity radius at point $r$. Let $u\in T_pM$ and $v\in T_qM$ and $\epsilon>0$ be such that $\exp_psu,\exp_qsv\in B(r;R)$ and $\Gamma:(-\epsilon,\epsilon)\times[0,1]\to M$ be defined by

$$\Gamma(s,t)=\exp_{\exp_psu}(t\exp^{-1}_{\exp_psu}\exp_qsv)$$

I want to calculate the variation field $V(t)=\partial_s\Gamma(0,t)$ at $t=0$ and $t=1$, that is

$$\partial_s\Gamma(0,t)=\frac d{ds}\Gamma(s,t)|_{s=0}$$

If $\gamma(s)=t\exp^{-1}_{\exp_psu}\exp_qsv$ ,then $$\frac d{ds}\Gamma(s,t)|_{s=0}=(\exp_{\exp_psu})_{*t\exp^{-1}_pq}(\dot\gamma(0))$$

if $\mu=\exp_qsv$, then $$\dot\gamma(0)=(t\exp^{-1}_{\exp_psu})_{*q}(\dot\mu(0))$$ So, $$V(0)=(\exp_{\exp_psu})_{*0}((0)_{*q}(\dot\mu(0)))$$ and $$V(1)=(\exp_{\exp_psu})_{*\exp^{-1}_pq}((\exp^{-1}_{\exp_psu})_{*q}(\dot\mu(0)))$$ but I don't know how continue from here. In this paper is claimed that $V(0)=u$ and $V(1)=v$ without any explanation. Can anyone give me some hint for calculating of $V(0)$ and $V(1)$, please?

1 Answers 1

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To calculate $V(t)$ at $t = 0$, we need to calculate the partial derivative $$\frac{\partial}{\partial s} \Gamma(t,s)|_{t=0,s=0} = \frac{d}{ds} \Gamma(0,s)|_{s = 0}.$$

Note that we have

$$ \Gamma(0,s) = \exp_{\exp_p(su)}(0) = \exp_p(su)$$

so $\Gamma(0,s)$ is a geodesic eminating from $p$ in the direction $u$. Hence,

$$ \frac{d}{ds} \Gamma(0,s)|_{s = 0} = \frac{d}{ds} \exp_p(su)|_{s = 0} = u. $$

Similarly,

$$ \Gamma(1,s) = \exp_{\exp_p(su)} \left( \exp_{\exp_p(su)}^{-1}(\exp_q(sv))\right) = \exp_q(sv)$$

so again, $\Gamma(1,s)$ is a geodesic eminating from $q$ in the direction $v$ and so $\frac{d}{ds} \Gamma(1,s)|_{s = 0} = v$.

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    Thanks, but I have a question, $$\frac d{ds}\exp_psu|_{s=0}=\lim_{s\to0}\frac{\exp_psu-p}s=\lim_{s\to0}\frac{\gamma_u(s)-p}s$$ How can I verify that this limit is equal to $u$? – 2017-02-05
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    $\gamma(s) = \exp_p(su)$ is a curve in a manifold, you can't just differentiate it like you wrote (there is no meaning to subtracting points). Pretty much by definition of $\exp_p$, the curve $\gamma(s) = \exp_p(su)$ is a geodesic which satisfies $\gamma(0) = p$ and $\dot{\gamma}(0) = u$. – 2017-02-05
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    More explicitly, $\exp_p(v)$ is defined by letting $\gamma \colon I \rightarrow M$ be a geodesic in $M$ with $\gamma(0) = p$ and $\dot{\gamma}(0) = v$ and then $\exp_p(v) = \gamma(1)$. That is, $\exp_p(v)$ is where the geodesic that starts at $t = 0$ at $p$ with velocity $v$ reaches when $t = 1$. Using the fact that a scalar reparametrization of a geodesic is a geodesic, this immediately implies that $\exp_p(sv)$ is the geodesic which starts at $p$ with velocity $v$. – 2017-02-05