In this lectures notes Geometric wave equation by Christian Bär at page 17 he has
Definition 1.2.27. Let $\Omega$ be a starshaped with respect to $x$. We define the smooth positive function $\mu_{x}: \Omega \rightarrow \mathbb{R}$ by
$$
\mathrm{dvol}=\mu_{x} \cdot\left(\exp _{x}^{-1}\right)^{*}(\mathrm{dz}),
$$
where dvol is the Lorentzian volume form and dz is the standard Euclidean volume form on $T_{x} \Omega$. We call $\mu_{x}$ the local density function.
$\exp _{x}$ is the exponential map.
I am not understanding what he means by $\left(\exp _{x}^{-1}\right)^{*}(\mathrm{dz})$
My confusion is this for 1 dimensional for example $\mathrm{dvol}$ should act on vector $v$ say at a point $p$. But if we follow the usual definition of pullback we would have
$\mathrm{dvol}(v)= \mathrm{dz}\left(\exp _{x*}^{-1} v \right)$
Now if we treat the tangent space $T_xM$ at the point $x$ as a manifold than $exp _{x*}^{-1} v$ should be a vector in the tangent space of the tangent space $T_xM$.
How can $dz$ act on it?
Best Answer
$\exp_x\colon T_x\Omega \to \Omega$, so, restricting domain appropriately, $\exp_x^{-1}\colon\Omega\to T_x\Omega$. Since $dz$ is the volume form on $T_x\Omega$, it makes perfect sense to pull it back by $\exp_x^{-1}$ and get a top-degree form on $\Omega$.
To answer your question, never forget that the tangent space of a vector space (in this case, $T_x\Omega$) at any point is canonically isomorphic to the vector space. There is no problem.