If $\omega$ is a volume form, then $X\mapsto i_X\omega$ generates all $(n-1)$-forms

Question

I've read the following claim in a textbook (I don't think the context is relevant):

Let $M$ be a compact smooth manifold. If $\omega\in\Omega^n(M)$ is a volume form, then every $(n-1)$-form $\eta\in\Omega^{n-1}(M)$ is given by $\eta=i_X\omega$ for some vector field $X\in\Gamma(TM)$.

In order to convince myself, I tested the statement for $n=3$, only locally (i.e., in coordinate neighbourhoods), but couldn't do it generally.

The author, rather vaguely, says that $X\mapsto i_X\omega$ is an "isomorphism", but I don't know what this is supposed to mean here.

Any ideas?

Answer 1

This is mostly a pointwise statement. That $\omega$ is a volume form means that $\omega_p(X_1,\cdots,X_n) \neq 0$ if and only if $X_1,\cdots,X_n$ is a basis of $T_pM$. ($\omega_p$ is a non-zero "determinant".)

The map \begin{align*} \iota_p:T_pM &\to \Lambda^{n-1}_p(T_pM) \\ X &\mapsto \iota_X\omega \end{align*} is then injective, since $\omega(X_1,\cdot,\cdot,\cdots)=\omega(X_2,\cdot,\cdot,\cdots)$ implies $\omega(X_1-X_2,\cdot,\cdot,\cdots)=0$, and thus $X_1-X_2$ can't be completed to a basis. Therefore, it must be zero. By dimensional reasons, the map is then an isomorphism.

There is still the issue of smoothness when we go from the pointwise statement to the differential one. By taking local charts and trivializations, if $\omega=gdx_1\cdots dx_n$ and we are taking a differential form $\eta=\sum f_i dx_1 \cdots \widehat{dx_i}\cdots dx_n$, we are getting $X=\sum_i (-1)^{i+1}\frac{f_i}{g}\partial_i$ as our output, which is smooth.

Answer 2

Let $\phi:\Gamma(TM)\longrightarrow \Omega^{n-1}(M)$ be the map $X\mapsto i_X{\omega}$. We want to show that $\phi$ is an isomorphism. To show that it is surjective, locally suppose that the volume form is given by $$\omega = dx^1\wedge\cdots\wedge dx^n\,.$$ Let $X$ be a vector field locally given by $$X =\sum_{i=1}^n X^i\frac{\partial}{\partial x^i}\,,$$ where $\partial/\partial x^i$ is dual to $dx^i$ i.e. $dx^i(\partial/\partial x^j) = \delta^i_j$. Then using derivation property of the interior product: $$ i_X(\alpha\wedge\beta) = i_X\alpha\wedge\beta + (-1)^{\text{deg}(\alpha)}\alpha\wedge i_X\beta\,, $$ we have $$ i_X\omega = \sum^n_{i=1}(-1)^{i+1}X^idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n\,. $$ Any $\eta\in\Omega^{k+1}(M)$ can be written as $$ \eta = \sum^n_{i=1}f_idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n $$ Then setting $X^i = (-1)^{-i-1}f_i$ gives you the vector field you want.

I'll leave it to you to show that $\phi$ is linear and injective.