Exercise 15.30 (p.389):
Suppose (M, g) and ($\tilde{M}, \tilde{g}$) are positive-dimensional Riemannian manifolds with or without boundary, and $ F: M \to \tilde{M}$ is a local isometry. Show that $ F^*\omega_{\tilde{g}} = \omega_{g}$.
$\omega_{g}$ and $\omega_{\tilde{g}}$ are the Riemannian volume forms on M and $\tilde{M}$ - which are assumed to be oriented. (Evaluated on positively-oriented orthonormal frames, they give a value of +1 on their respective manifolds.) Shouldn't the above read $ F^*\omega_{\tilde{g}} = \pm\omega_{g}$? In particular, if F is an orientation-reversing local isometry, the volume form on $\tilde{M}$ gets pulled back to the negative of its counterpart on M - right? Or am I missing something in the definitions?
Thanks a lot.
You are right. Consider $(\mathbb{R},dx\otimes dx)$. Then $x\mapsto -x$ is an isometry but reverses the orientation.
Some isometries do preserve the orientation. They are called "orientation preserving isometries".