# About the Fermi charts

In the book Topics in Differential geometry, Peter W. Michor defines the Fermi charts for a Riemannian manifold as follows.

Let $(M,g)$ be a Riemannian manifold. For simplicity, I assume that $M$ is geodesically complete. Let $c : 0 \in I \subset \mathbb{R} \to M$ a geodesic. Let $\lbrace \dot{c}(0) \rbrace^{\perp}$ denote linear space of tangent vectors in $T_{c(0)}M$ which are orthogonal to $\dot{c}(0)$ (for the metric $g_{c(0)}$). Define the mapping $\Theta : (t,v) \in I \times \lbrace \dot{c}(0) \rbrace^{\perp} \mapsto \mathrm{Exp}_{c(t)}\big( \mathrm{Pt}(c,t)v \big)$, where $\mathrm{Pt}(c,t)v$ denotes the parallel transport of $v$ from $T_{c(0)}M$ to $T_{c(t)}M$ along the curve $c$. The following result is given in the book :

The tangent map of $\Theta$ along $I \times \lbrace 0 \rbrace$ is given by :

$$T_{(t,0)}\Theta : (s,Y) \in \mathbb{R} \times \lbrace \dot{c}(0) \rbrace^{\perp} \mapsto s\dot{c}(t) + \mathrm{Pt}(c,t)Y$$

I tried to compute this tangent map myself but I couldn't. I considered a curve $\varphi : ]-\varepsilon,\varepsilon[ \to \mathbb{R} \times \lbrace \dot{c}(0) \rbrace^{\perp}$ such that : $\varphi(0) = (t,0)$ and $\dot{\varphi}(0) = (s,Y)$. But I do not see how to compute the derivative at $u=0$ of $(\Theta \circ \varphi)(u)$ (because $u$ appears in the base point of the Exponential and in the parallel transport). What would be the right way to compute this tangent map ?

You can use that $T_{(t,0)}\Theta$ is linear.
$T_{(t,0)}\Theta(s,0)=\frac{d}{du}|_0 \Theta(t+us, 0)=\frac{d}{du}|_0c(t+us)=s\dot c(t)$
$T_{(t,0)}\Theta(0,Y)=\frac{d}{du}|_0\Theta(t, uY)=\frac{d}{du}|_0\gamma_{(c(t), Pt(c,t)Y)}(u)= Pt(c,t)Y$.
• Thank you ! I have another question about $\Theta$. Is $T_{(t,0)}\Theta$ a linear isomorphism because $Pt(c,t)$ is a linear isometry ? – Pouteri Jul 23 '16 at 14:01