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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

$F^*(dV_\overline g)_p=det(DF_p)(dV_g)_p$

$\forall p\in M$ ?

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Please, don't get offended with my question, but did you tried to do the calculation? Maybe you could provide us with some of your efforts? – Tomás Jan 31 '13 at 11:15
Sorry... It's a nonsense! In abstract terms there is no reason why this is true. – Frankenstein Jan 31 '13 at 13:43

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 $$ (\Phi^* \omega)_x = \det(d \Phi(x)) \omega_{\Phi(x)} .$$

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