Let $(M,g)$ be a Riemannian Manifold and suppose that $J(x,y)$ is the volume density in normal coordinates based at $x$. In particular this means that we can write the Riemannian volume form as
$$ dv_g = J(x,y)dy \quad J(x,y) = \sqrt{\det g_{ij}(y)}$$
Denoting the exponential map based at $x$ by $exp_x$ I've been told that one can show that
$$J(x,y) = |\det D_{exp_x^{-1}(y)}exp_x|$$
but I don't think this is the case. I think that the roles of the exponential map and the exponential map inverse should be switched.
Question: Is it true that for $y$ in this geodesic normal coordinate chart based at $x$ that
$$J(x,y)dy = \left(exp^{-1}_x\right)^*(dx)$$
where $dx$ is canoncial volume measure on $T_xM \simeq \mathbb{R}^n$?
If this is true then $$J(x,y) = |\det\left( D_y exp_x^{-1}\right)|$$ which is different from the formula above.
Best Answer
Just to (hopefully) clarify notation and what not, I'll reiterate a few points, and try and rephrase your statement with a bit more explicit detail about coordinates.
Let $M$, and $N$ be $n$-dimensional manifolds. Top forms in $\Lambda^n M$ can be written as $f\ dx^1\wedge\dots\wedge dx^n$, where $f\in C^\infty M$, and likewise in $N$. Let $x^i$ and $y^i$ be local coordinates on $M$, and $N$, respectively, and let $\Phi:M\to N$ be a (local) diffeomorphism. We can write down the transformation law for the pullback of a top form in coordinates. $$ \Phi^*(f\ dy^1\wedge\dots\wedge dy^n)=\det(d\Phi)(f\circ\Phi)dx^1\wedge\dots\wedge dx^n $$ Where $\det(d\Phi)$ denotes the determinant of the Jacobian in the chosen local coordinates.
We ca write the Riemannian volume element as $\omega_g$. It is defined (pointwise) by $$ \omega_g|_p=\left.d\nu^{\bar{1}}\wedge\dots \wedge d\nu^{\bar{n}}\right|_{p} $$ Where $\nu^{\bar{i}}$ are oriented orthonormal coordinates at $p$ (bars just for clarity). That is, we can consider $\hat{\psi}:T_pM\to\mathbb{R^n}$ to be an orthonormal basis (i.e. global chart) for $T_pM$, and let $\psi=\hat{\psi}\circ\exp_p^{-1}$ be the chart induced by the exponential map and this basis. We can also choose another aribitraty oriented chart $\varphi$ containing $p$ with coordinates $x^i$. Using the pullback formula with the transition function $\Phi=\hat{\psi}\circ\exp_p^{-1}\circ\varphi^{-1}$, we have $$ \Phi^*\left(d\nu^{\bar{1}}\wedge\dots\wedge d\nu^{\bar{n}}\right)=\det(d\Phi)dx^{1}\wedge\dots\wedge dx^{n} $$ This allows us to compute the coordinate representation of $\omega_g$ (though only at $p$; this need not apply even on a neighborhood). We are in a sense computing the determinant of $d\exp_p^{-1}$ (and not $d\exp_p$), but composed with charts so that it is a sensible operation.
The $\sqrt{\det(g_{ij})}$ formula comes from the fact that pulling back the metric (i.e. applying the chain rule) gives us a relation between $\det(d\Phi)$ and $\det(g_{ij})$. $$ g_{ij}=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=g\left(\frac{\partial\nu^{\bar{k}}}{\partial x^i}\frac{\partial}{\partial \nu^{\bar{k}}},\frac{\partial\nu^{\bar{l}}}{\partial x^j}\frac{\partial}{\partial \nu^{\bar{l}}}\right)=\frac{\partial\nu^{\bar{k}}}{\partial x^i}\frac{\partial\nu^{\bar{l}}}{\partial x^j}g\left(\frac{\partial}{\partial \nu^{\bar{k}}},\frac{\partial}{\partial \nu^{\bar{l}}}\right)=d\Phi^{\bar{k}}_i\ d\Phi^{\bar{l}}_j\ g_{\bar{k}\bar{l}} $$ Taking the determinant of both sides at $p$ and using $g_{\bar{k}\bar{l}}=\delta_{\bar{k}\bar{l}}$ gives the desired relation.
Does this clarify things at all?