Let $M$ be an oriented $m$-manifold without boundary, $X\in \mathfrak X(M)$ be a vector field, and $\omega \in \Omega^m_c(M)$ be a compactly supported top-degree differential form.

Question: Do we always have $$\int_M \mathcal L_X\omega=0,$$ where $\mathcal L_X$ is the Lie derivative?

Motivation: The integration operation has a diffeomorphism invariance, i.e., if $F:M\to M$ is an orientation-preserving diffeomorphism, then $$\int_M \omega = \int_M F^*\omega.$$ Now, suppose that the diffeomorphism $F$ is infinitesimal, i.e., $F$ maps each $p\in M$ to a point that is infinitesimally separated from $p$. Such infinitesimal diffeomorphism can be heuristically represented by a vector field $X\in \mathfrak X(M)$, which is an 'arrow' from $p$ to $F(p)$.

Furthermore, the infinitesimal version of the pullback $F^*\omega$ is the Lie derivative $\mathcal L_X\omega$. Hence, I suspect that the infinitesimal statement of the diffeomorphism invariance of integral should be $$\int_M \mathcal L_X\omega=0.$$


1 Answer 1


$\def\intprod{\mathbin{\lrcorner}}$ Consider the Cartan magic formula $$\mathcal{L}_X\omega = X\intprod(d\omega) + d(X\intprod \omega),$$ where $d$ is the exterior derivative and $X\intprod \omega$ is the interior product of $X$ and $\omega$.

For a top-degree form $d\omega=0$, so we conclude by Stokes theorem that $$\int_M \mathcal{L}_X\omega = \int_M d(X\intprod \omega) = \int_{\partial M}X\intprod \omega = 0$$ since the manifold $M$ has no boundary.

  • $\begingroup$ Wow! Thanks for your simple and elegant answer! $\endgroup$
    – Laplacian
    Commented Oct 17, 2022 at 8:22

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