# 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.

• Why should the determinant not be globally defined?
– user515010
Oct 5, 2020 at 5:41
• I think it is not invariant under coordinate changes? Oct 5, 2020 at 11:46
• It's hard to tell what's going on here because you've used some undefined notations. What's $g$? What's $g^{-}$? Oct 5, 2020 at 14:56
• It is easier if you include the reference. Where did you see this? Oct 8, 2020 at 18:34

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.
• The coordinate-independent object you're alluding to throughout this answer is called the Riemannian volume form. Using partitions of unity to patch together charts, it provides a well-defined measure on the manifold, from which you can derive $L^p$ spaces in the usual way. Oct 12, 2020 at 10:30