I'm reading Stillwell's Naive Lie Theory. At end of Ch4 he said,
For readers acquainted with differential geometry, it should be mentioned that the exponential function can be generalized even beyond matrix groups, to Riemannian manifolds. In this setting, the exponential function maps the tangent space $T_P (M)$ at point $P$ on a Riemannian manifold $M$ into $M$ by mapping lines through $O$ in $T_P (M)$ isometrically onto geodesics of $M$ through $P$. The Riemannian manifolds $\mathbb S^1 = \{z \in \mathbb C : |z| = 1\}$ and $\mathbb S^3 = \{q \in \mathbb H : |q| = 1\}$, and their tangent spaces $\mathbb R$ and $\mathbb R^3$ , nicely illustrate the geodesic aspect of exponentiation. The exponential map sends straight lines through $O$ in the tangent space isometrically to geodesic circles in the manifolds (to $\mathbb S^1$ itself in $\mathbb C$, and to the unit circles $\cos \theta + u \sin \theta$ in $\mathbb H$, which are geodesic because they are the largest possible circles in $\mathbb S^3$ ).
I'm not familia with differential geometry at all, so probably should not ask here... But I just can't help wondering, how could the exponential function be isometrical?
For example, function $e^z$ maps the straight line $i\theta$ into $e^{i\theta}$, i.e. the unit circle $\mathbb S^1$. But the straight line has infinite length, while $\mathbb S^1$'s circumference is only $2\pi$, how could $e^z$ be an isometry? by isometrical it means the map shall keep the distance?
