Let $M$ be a Riemannian manifold with metric $g$ and volume form $dV$. Let $L_X$ denote the Lie derivative taken over a vector field $X$. Suppose $X$ is a vector field over $M$ with compact support and $div(X) = 0$, where the divergence on a Riemannian manifold is defined as
$$d(X \lrcorner dV) = (div X) dV$$
where $\lrcorner$ denotes interior multiplication.
I would like to show that
$$\int L_Xf dV = 0$$
for all smooth functions $f$ over $M$.
My approach is to use Cartan's formula so that:
$$L_X f = X \lrcorner d(fdV) + d(X\lrcorner fdV) = d(X\lrcorner fdV) = (div X) fdV = 0.$$
Thus
$$\int_M L_Xf dV = \int_M 0 dV = 0.$$
The hint to this problem is to use the divergence theorem. Based on my work above I don't think is this necessary, is this correct? However if I were to use the divergence theorem, we have that
$$\int_m (div X)dV = \int_{\partial M} g(X,N) dV_{\tilde{g}}$$
where $N$ is the outward-pointing unit normal vector field along $\partial M$ and $\tilde{g}$ is the induced Riemannian metric on $\partial M$. I am new to Riemannian geometry and I am not sure how to compute the integral over the boundary where the integrand involves the metric on $M$. How would one compute this? How does this relate to the original integral containing the Lie derivative?
Best Answer
The Lie derivative of a function must be a function, not a top form. Here are the right computations. $\DeclareMathOperator{\Div}{div} \DeclareMathOperator{\lie}{\mathscr{L}} \DeclareMathOperator{\d}{d} \newcommand{\dx}{\d \!}$ \begin{align} (\lie_Xf) \dx V &= \lie_X(f\dx V) - f \lie_X\dx V & \text{by Leibniz rule}\\ &= X \lrcorner \d(f\dx V) + \d(X\lrcorner f\dx V) - f (\Div X) \dx V & \text{by Cartan's magic formula} \\ &= \d(X\lrcorner f\dx V) \end{align} the last equality being true since $f\dx V$ is a top form (hence $\d(f\dx V) = 0$) and $\Div X = 0$. The result now follows from Stokes' formula.