2
$\begingroup$

I'm now working on two exercises:

Exercise 1 Let $M^n$ be an oriented manifold without boundary, and $\alpha \in \Omega^{s}(M)$, $\beta \in \Omega^{n-s}(M)$ be differential forms on $M$. Let $X \in \mathcal{X}(M)$ be a smooth vector field on $M$ with compact support. Show that $$ \int_M \mathcal{L}_X(\alpha) \wedge \beta = - \int_M \alpha \wedge \mathcal{L}_X(\beta). $$

Exercise 2 Let $M^n$ be an oriented closed manifold (compact without boundary), and $g$ a Riemannian metric on $M$. If $X \in \mathcal{X}(M)$ be a smooth vector field on $M$ and $f \in C^{\infty}(M)$ be a smooth function on $M$. Show that $$ \int_M X(f) \cdot \Omega = - \int_M f \cdot \mathcal{L}_X(\Omega). $$ where $\Omega$ is the volume form of $(M,g)$.

I have been trying on the two exercises and meetting the same difficulty.

For 1: We know that $\mathcal{L}_X(\alpha) \wedge \beta + \alpha \wedge \mathcal{L}_X(\beta) = \mathcal{L}_X(\alpha \wedge \beta)$. So applying the integration on $M$ on both sides, it suffices to show that the integration vanishes: $$ \int_M \mathcal{L}_X(\alpha \wedge \beta) = 0. $$

For 2: We know that $\mathcal{L}_X(f \cdot \Omega) = X(f) \cdot \Omega + f \cdot \mathcal{L}_X \Omega$.So applying the integration on $M$ on both sides, it suffices to show that the integration vanishes: $$ \int_M \mathcal{L}_X(f \cdot \Omega) = 0. $$

Yet I don't know how to carry on to show that the two integrations are zero. It seems that there is some machinary that I'm not awareness of or not familiar with. So my question is:How to prove that the two integrations above is zero?

My attempts: I have tried to use Cartan's magic identity, namely $$ \mathcal{L}_X = i_X \circ \mathrm{d} + \mathrm{d} \circ i_X, $$ where $i_X$ is the interior multiplication. In both cases, the manifold $M$ has no boundary, so by Stokes' theorm, the integration on the summand $\mathrm{d} \circ i_X$ is zero. So it remains to show that the integration on the summand $i_X \circ \mathrm{d}$ is zero as well. I got stuck here. :(

Thank you all for commenting and answering!

$\endgroup$
2
  • $\begingroup$ The only thing you did not use is that $\alpha \wedge \beta $ and $\Omega$ are $n$-forms, thus have exterior derivative $0$ (i.e are closed). This will give you the answer. $\endgroup$
    – Didier
    Commented Dec 18, 2020 at 10:28
  • $\begingroup$ Thank you @DIdier_ ! $\endgroup$
    – Hetong Xu
    Commented Dec 18, 2020 at 11:26

1 Answer 1

4
$\begingroup$

For any top-form $\omega$, $$\int_M L_X(\omega) = \int_{M} d(i_X(\omega)) + i_X(d\omega) = 0, $$ as by Stokes' theorem, the integral of any exact top-form on a manifold without boundary is zero, and because $d\omega$ is zero, as it is an $(n+1)$-form on an $n$-dimensional manifold.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Hetong Xu
    Commented Dec 18, 2020 at 11:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .