MHB How Does Divergence Relate to the Volume Form in Riemannian Manifolds?

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Here's this week's problem!

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Problem. Let $X$ be a smooth vector field on an oriented Riemannian manifold $(M,g)$. Show that if $\nu$ is the volume form on $M$, then $d(i_X\nu) = (\text{div} X) \nu$.

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No one answered this week's problem. You can find my solution below.

Spoiler

Suppose $M$ is $n$-dimensional. In local coordinates,

$$\nu = \sqrt{G}\, dx^1 \wedge \cdots \wedge dx^n,$$

where $G = \operatorname{det}(g_{ij})$ and $g_{ij}$ is the Riemannian metric on $M$. So

$$i_X \nu = X^1\sqrt{G}\, dx_2 \wedge \cdots \wedge dx^n - X^2\sqrt{G}\, dx^1\wedge dx^3 \wedge \cdots \wedge dx^n + \cdots$$ $$\qquad + (-1)^{n-1}X^n \sqrt{G}\, dx^n\wedge dx^1 \wedge\cdots \wedge dx^{n-1}.$$

Hence

$$d(i_X \nu) = \frac{\partial}{\partial x^1}(X^1\sqrt{G})\, dx^1 \wedge \cdots \wedge dx^n - \frac{\partial}{\partial x^2}(X^2\sqrt{G})\, dx^2 \wedge dx^1 \wedge\cdots \wedge dx^n + \cdots$$ $$\qquad \quad + (-1)^{n-1} \frac{\partial}{\partial x^n}(X^n\sqrt{G})\, dx^n \wedge dx^1 \wedge \cdots \wedge dx^{n-1}$$
$$\qquad\quad = \left[\frac{\partial}{\partial x^1}(X^1\sqrt{G}) + \frac{\partial}{\partial x^2}(X^2\sqrt{G}) + \cdots + \frac{\partial}{\partial x^n}(X^n \sqrt{G})\right]dx^1 \wedge \cdots \wedge dx^n$$
$$\qquad \quad = \frac{1}{\sqrt{G}}\left[\frac{\partial}{\partial x^1}(X^1\sqrt{G}) + \frac{\partial}{\partial x^2}(X^2\sqrt{G}) + \cdots + \frac{\partial}{\partial x^n}(X^n\sqrt{G})\right]\nu$$
$$\qquad \quad = (\operatorname{div} X) \nu.$$