Let $(M,g)$ be an positively oriented $n$-dimensional Riemannian manifold with or without boundary and $\text{dvol}_g$ its riemannian volume form. Let $f$ be a continuous real-valued function on $M$ and $\omega$ be an continuous $n$-form on $M$. If $(U\subseteq M, \phi)$ is a chart on $M$ such that $\omega$ can be written as $\omega = A \text{d}x^1\wedge \cdots \wedge \text{d}x^n$ where $A: U \longrightarrow \mathbb{R}$ is continuous function. I know that the integration of these objects are really well-defined:

$$ \text{integral of $f$ on $U$} :=\int_U f \text{dvol}_g = \int_{\phi(U)}f(x) \sqrt{\text{det}(g_{ij})} \text{d}x^1 \cdots \text{d}x^n \tag1$$ and

$$\text{integral of $\omega$ on $U$} := \int_U \omega = \int_{\phi(U)} A(x) \text{d}x^1 \cdots \text{d}x^n \tag2$$

My question: does it make sense integrate both $f$ and $\omega$ over $U$? In other words, does it make sense

$$\int_U (f \text{dvol}_g + \omega) \stackrel{?}{=} \int_{\phi(U)} \left( f(x)\sqrt{\text{det}(g_{ij})} + A(x)\right)\text{d}x^1 \cdots \text{d}x^n $$

  • $\begingroup$ The fact is that if $M$ is oriented (with a volume form $\omega$), there is a natural bijection between $\mathcal{C}^{\infty}(M)$ and $\Omega^{\text{top}}(M)$, given by $f \mapsto f \omega$, and therefore, these two notions of integration coincide. In your specific case the volume form orienting $M$ is $\mathrm{dvol}_g$, and $f\mathrm{dvol}_g$ is indeed a top-form on $M$. $\endgroup$
    – Didier
    Sep 19, 2021 at 13:27
  • 1
    $\begingroup$ @Didier: Thanks for clarifying it to me. Initially I was wondering if once you define the volume form on a riemannian manifold, every form you integrate must have the $\sqrt{\det{g}}$ term. $\endgroup$ Sep 19, 2021 at 13:56
  • 3
    $\begingroup$ What an odd question! Why can you not just take the sum of the two integrals? $\endgroup$ Sep 19, 2021 at 21:09

1 Answer 1


Integrating a $n$-form over a $n$-dimensional oriented manifold $M$ is possible thanks to the formula $$ \int_U \omega := \int_{\varphi(U)} A_{\omega,\varphi}(x^1,\ldots,x^n)\mathrm{d}\lambda $$ where $\varphi\colon U\to\varphi(U)$ is a chart, $A_{\omega,\varphi}$ is the unique function on $\varphi(U)$ such that $\varphi_* \omega = A_{\omega,\varphi} \mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$, and $\mathrm{d}\lambda$ is the Lebesgue measure on $\varphi(U)$. A partition of unity argument allows us to integrate $\omega$ on all of $M$ (if $M$ is compact or if $\omega$ has compact support), independently of the considered charts. It is an intrinsic notion.

But there is no natural way to integrate functions over $M$. However, given a reference volume form $\omega_0$, there is a bijection between $\mathcal{C}^{\infty}(M)$ and $\Omega^n(M)$ given by $f\mapsto f\omega_0$. This allows us to integrate functions over $M$ if $M$ is compact or if $f$ has compact support, with the definition $$ I_{\omega_0}(f) := \int f\omega_0. $$ This notion of integration of functions on $M$ closely depends on the choice of $\omega_0$! It is not intrinsic.

In the case of a Riemannian manifold, there is a natural choice of volume form: the Riemannian volume form. In local coordinates, it can be expressed as $$ \mathrm{d}vol_g = \sqrt{\det g(x)}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n, $$ and therefore, in this precise context, integrating a function $f$ means integrating the top-form $f \mathrm{d}vol_g$, which reads in local coordinates $f(x) \sqrt{\det g(x)} \mathrm{d}x^1\wedge\cdots\wedge \mathrm{d}x^n$.

So it does not make sense to integrate "a function + a top-form": what makes sense is to integrate the sum of two top forms. It follows that if $f$ is a function, if $\omega$ is a top-form, and if $\omega_0$ is a reference volume form, the only thing that makes sense is $\int f\omega_0 + \omega$. If the reference volume form is given by a Riemannian metric, it reads, in local coordinates $$ \int_U f\mathrm{d}vol_g + \omega = \int_{\varphi(U)} \left(f(x)\sqrt{\det g(x)} + A_{\omega,\varphi}(x)\right)\mathrm{d}\lambda. $$


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