# Riemannian manifold: Integration by parts of Lie derivative

How can we apply integration by parts to the Lie derivative?

Background: In the Hamiltonian formulation of general relativity, we have the momentum constraint (using abstract index notation, $D_a$ is the covariant derivative on the 3-dimensional spatial slice with positive definite metric, $N^a$ is a vector field) $$C_{\vec{N}}=2\int\mathrm{d}^3x\,(D_aN_b)p^{ab}=\int\mathrm{d}^3x\,(\mathcal{L}_{\vec{N}}q_{ab})p^{ab}=-\int\mathrm{d}^3x\,(\mathcal{L}_{\vec{N}}p^{ab})q_{ab}$$ In this situation, $p^{ab}$ is a symmetric (2,0) tensor density of weight 1 and $q_{ab}$ the spatial metric which is just a symmetric (0,2) tensor field. Now I could prove the given equalities exploiting the fact that $p^{ab}$ is density 1 and $q_{ab}$ is the metric. This means the statement $$\int\mathrm{d}^3x\,(\mathcal{L}_{\vec{N}}q_{ab})P^{ab}=-\int\mathrm{d}^3x\,(\mathcal{L}_{\vec{N}}P^{ab})q_{ab}$$ holds in general for $q_{ab}$ being the metric and $P^{ab}$ being symmetric and of density 1. To prove this, I wrote everything in covariant derivatives and showed equality.

However, I'm wondering if there is a more geometrical way to see that this is true (maybe directly from Stokes' theorem). Furthermore, I'm wondering if the statement is always true or really just in the restricted situation that I could prove.

• Is $N^a$ just any vector field, or is it supposed to be, say, the shift vector? – Muphrid Feb 7 '14 at 5:25
• Of course, in the special situation of $C_{\vec{N}}$ being the momentum constraint, $\vec{N}$ gets the interpretation of the shift vector field - but this is only relevant if we reconstruct the 4-manifold. My question is completely general, so you can assume that $\vec{N}$ is supposed to be a completely arbitrary vector field on the 3-manifold. – LFH Feb 7 '14 at 17:47

First use that the Lie derivative satisfies the Leibniz rule, $$£_N(q_{ab} p^{ab})=(£_Nq_{ab})p^{ab}+q_{ab}£_Np^{ab}$$ to rewrite the integral as $$\int d^3x (£_N q_{ab})p^{ab}= \int d^3x\,£_N(q_{ab}p^{ab}) - \int d^3x\,(£_N p^{ab})q_{ab}$$ Now note that the first integrand on the RHS is the Lie derivative of a scalar density. Scalar densities of weight 1 are more naturally thought of as a differential form of rank $d$, where $d$ is the dimension of the surface you are integrating over, so $3$ in this case. The 3-form in this case is $$p^{ab}q_{ab}\tilde{\epsilon}_{ijk}$$ where $\tilde{\epsilon}_{ijk}$ is the alternating symbol. The above expression is a tensor because $\tilde{\epsilon}$ is a density of weight -1, and $p^{ab}$ is weight 1. Now we apply Cartan's magic formula for differential forms, which says $$£_N = i_N d + d i_N,$$ where $d$ is the exterior derivative, and $i_N$ denotes contraction with the vector $N$. So when the Lie derivative acts on a $d$ form, the $i_N d$ term vanishes because the exterior derivative of a $d$ form is a $d+1$ form, which vanishes for a dimension $d$ manifold. So the integral we are focusing on becomes $$\int_S di_N(p^{ab}q_{ab} \tilde{\epsilon}) = \int_{\partial S}i_N(p^{ab}q_{ab}\tilde{\epsilon}).$$ Here we have used Stoke's theorem to relate the integral of an exact form to an integral of a form on the boundary of the region of integration. If the surface $S$ has no boundary, or if you assume the fields die off at the boundary, your identity results.
• Thank you very much. I entirely agree and now see why it is obviously true in general: I can use integration by parts whenever I have a Lie derivative acting on a tensor density contracted with another tensor density such that the whole object is of weight 1 (which means it is equivalent to a N-form (in N dimensions) by contracting with the epsilon tensor). For a manifold with boundary we would get the additional term $\int_{\partial S}i_N(p^{ab}q_{ab}\tilde{\epsilon}_{c_{1\cdots}c_N})$. Great! – LFH Feb 22 '14 at 18:16