Let $\left(M,g \right)$ be a smooth, oriented, compact, connected three dimensional Riemannian manifold (without boundary). Let $\xi, \eta$ be two divergence-free vectorfields on $M$. I found the following identity without further comments:
$\left< \xi, \nabla \times \left(\eta \times \xi \right) \right>=\left< \nabla \times \xi, \left(\eta \times \xi \right) \right>=\left< \eta, \left( \xi \times (\nabla \times \xi) \right) \right>$,
where $\nabla \times \xi$ is the (generalised) curl of a vectorfield, $\xi \times \eta$ the (generalised) cross product of vectorfields and $\left<\xi,\eta \right>=\int_M g(\xi,\eta)\omega$ and $\omega$ is the corresponding volume form on $(M,g)$.
Thanks to this thread I was able to show the first identity, invoking Stokes theorem. But how exactly do I get the second?
Any help is greatly appreciated!
The second equality here is a pointwise one: we have $$\def\curl{\nabla \times}g(\curl \xi, \eta \times \xi) = g(\eta, \xi \times \curl \xi).$$This is just the special case $\zeta = \curl \xi$ of the identity $$g(\zeta,\eta \times \xi) = g(\eta, \xi \times \zeta),$$ which expresses the cyclic symmetry of the (Riemannian-generalized) scalar triple product. Since this is a pointwise identity, we can reduce it to the Euclidean version $$\zeta\cdot(\eta \times \xi) = \eta \cdot (\xi \times \zeta)$$ simply by choosing an orthonormal basis for the tangent space.