Maybe the question will be stupid, but I'm a beginner in riemannian geometry...
We have two riemannian manifolds $(M,g)$, $(\overline M,\overline g)$ and a diffeomorphism $F:M\rightarrow\overline M$ between them. If $dV_g$ and $dV_\overline g$ are the riemannian volume forms of $M$ and $\overline M$ respectively, is it true that
$F^*(dV_\overline g)_p=det(DF_p)(dV_g)_p$
$\forall p\in M$ ?
Recall that $$DF_p : T_p M \rightarrow T_{\bar{p}} \bar{M}$$ how do you define its determinant??
However, its true that if you consider an $n$-form $\omega$ on $\mathbb{R}^n$ and a smooth map $\Phi : \mathbb{R}^n \rightarrow \mathbb{R}^n$ there is a formula $$ (\Phi^* \omega)_x = \det(d \Phi(x)) \omega_{\Phi(x)} .$$