What is the determinant of a metric on a riemannian manifold? On a domain $\Omega \subset \mathbb R^{n}$ in the Euclidean space, suppose we have a smooth function $f$, and $g^{ij} \partial_{ij}f \geq h$, where $g^{ij}$ are smooth functions on $\Omega$ satisfying some positivity condition. In some maximum principle one needs to consider the quantity $\| \frac{h^{-}}{D^{\ast}}\|_{L^{n}(\Omega)}$, where $h^{-} = \max\{0, -h\}$(The precise statement of the maximum principle is not important here), where $D^{\ast} = (det g_{ij})^{\frac{1}{n}}$. My question is: On a smooth Riemannian manifold $M$, one can just take $g^{i j}$ to be the inverse of the component $g_{i j}$ of a Riemannian metric, but how do we define $\| \frac{h^{-}}{D^{\ast}}\|_{L^{n}(M)}$? The determinant of the metric is not globally defined there, so $\frac{h^{-}}{D^{\ast}}$ is not a well-defined function.
 A: To me it seems you take this factor in order to get a well-defined norm: on an $n$-dimensional mfd, you can only integrate $n$-forms (where $\dim(M)=n$). Anything else requires you to add some kind of determinant factor of $\sqrt{|g|}=\sqrt{|\det g_{ij}|}$ (formally, you could see this as an artifact of applying the Hodge star operator to a zero form).
In particular, when calculating the $p$-norm for a smooth function $f\in C^\infty(M)$, the function $|f|^p$ is still a function and not an $n$-form, so we need to include that determinant factor to get a coord-independent integral, i.e. we integrate the function $\sqrt{|g|}|f|^p=|\sqrt[2p]{|g|}f|^p$, and I would suspect that the factor of $\sqrt[2p]{|g|}=\sqrt[2p]{|\det(g_{..})|} = |det(g^{..})|^{-\frac{1}{2p}}$ is what you mean by $1/D^*$...
This is my guess, but I have also been wondering how to properly define $L^p$ spaces on manifolds...
TL;DR: while the determinant of the metric is not a globally defined (and very coordinate-dependent) object, it combines with $dx^1\wedge\dots\wedge dx^n$ to form a (more or less) coordinate-independent object (up to minuses...) that we can use to integrate functions. Since for $L^p$ norms we are not integrating functions but their $p$-th powers, we need to include the $p$-th root of the determinant factor inside the function if we wish to express the $L^p$ norm on the manifold by the $L^p$ norm on a chart.
