# Does it make sense to integrate a function and a differential form at the same time?

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

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

• 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 $M$. Sep 19, 2021 at 13:27
• @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. Sep 19, 2021 at 13:56
• What an odd question! Why can you not just take the sum of the two integrals? Sep 19, 2021 at 21:09

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 $$\int_U f\mathrm{d}vol_g + \omega = \int_{\varphi(U)} \left(f(x)\sqrt{\det g(x)} + A_{\omega,\varphi}(x)\right)\mathrm{d}\lambda.$$