# Note on Lie derivatives and divergences

$\newcommand{\cd}{\nabla} \newcommand{\pd}{\partial} \newcommand{\Lie}{\mathcal{L}}$

One of Saul Teukolsky’s favorite pieces of advice is if you’re ever stuck, try integrating by parts. If we’re working with a covariant derivative $\cd_a$, and we have some tensor quantities $S, T$ under an integral, then every calculus student knows that we can move the derivative from one to the other,

\begin{align} \int_R S (\cd_a T) \sqrt{|g|} d^nx = \int_R (- \cd_a S) T \sqrt{|g|} d^nx + \textrm{boundary term.} \end{align}

(Of course, this only makes sense if your integral itself makes sense—so $\cd, S, T$ together make the integrand a scalar.) Here $R$ is the integration region, and the boundary terms above come from Stokes’ theorem, converting the integral of a total divergence in $R$ into the integral of a flux through the boundary $\pd R$.

Sometimes in differential geometry, instead of dealing with a metric-compatible covariant derivative $\cd_a$, we’re dealing with a Lie derivative $\Lie_v$ along a vector field $v$. Wouldn’t it be convenient, then, if we could integrate by parts with Lie derivatives? That is, do we have the property that

\begin{align} \label{eq:Lie-int-by-parts-q} \int_R S (\Lie_v T) \sqrt{|g|} d^nx \overset{?}{=} \int_R (- \Lie_v S) T \sqrt{|g|} d^nx + \textrm{boundary term?} \end{align}

After all, integrating by parts comes from rearranging the Leibniz rule $\cd_a(S\, T) = (\cd_a S) T + S (\cd_a T)$. But the Lie derivative, being a derivation, also satisfies the Leibniz rule,

\begin{align} \label{eq:Lie-Leibniz} \Lie_v(S\, T) = (\Lie_v S) T + S (\Lie_v T) . \end{align}

But in reality all this gets us is the formula

\begin{align} \int_R S (\Lie_v T) \sqrt{|g|} d^nx = \int_R (- \Lie_v S) T \sqrt{|g|} d^nx + \int_R \Lie_v(S \, T) \sqrt{|g|}d^nx \,. \end{align}

If we want this to become Eq. \eqref{eq:Lie-int-by-parts-q}, then we need the final term to be a total divergence. The sufficient condition for this to be a total divergence is that $v^a$ must be divergence-free. We can show this by detouring back through covariant derivatives. First, notice that for this integrand to make sense, we need all the indices in the product $ST$ to be contracted so that it is a scalar. Now the action of a Lie derivative on a scalar is the same as the action of that vector on the scalar, and can also be expressed in terms of any covariant derivative,

\begin{align} \Lie_v(S \, T) = v(S \, T) = v^a \pd_a (S \, T )= v^a \cd_a (S \, T) \,. \end{align}

Notice that when $\cd_a v^a = 0$, we can pull it inside the derivative,

\begin{align} \cd_a v^a = 0 \quad \Longrightarrow \quad v^a \cd_a (S \, T) = \cd_a ( v^a S \, T) \,. \end{align}

When this is the case, we can turn the last term into a boundary integral,

\begin{align} &\cd_a v^a = 0 \quad \Longrightarrow \nonumber\\ &\int_R S (\Lie_v T) \sqrt{|g|} d^nx = \int_R (- \Lie_v S) T \sqrt{|g|} d^nx + \int_{\pd R} S T v^a n_a \sqrt{|\gamma|} d^{n-1} x \,, \end{align}

where $n_a$ is the unit normal to the boundary $\pd R$ and $\sqrt{|\gamma|} d^{n-1}x$ is the proper “area” element.

## Special case: Killing vector field

A Killing vector field (KVF) is a special case of a divergence-free vector field. If $v$ is a KVF, it generates an isometry of the metric,