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I am trying to prove that the Euler equation for a Riemannian 3-manifold $M$ $$ \frac{\partial u}{\partial t} + \nabla_{u} u = - \operatorname{grad} p \qquad \text (1) $$ is equivalent to $$ \frac{\partial u^\flat}{\partial t} + \mathcal L_u u^\flat = \frac{1}{2} d(u^\flat(u)) - dp \qquad \text (2) $$ where $u$ is a (velocity) vector field, $p$ is a function on $M$ (pressure term), $\mathcal L_u$ is the Lie derivative along $u$, and $\flat$ is the musical isomorphism from the tangent bundle to the cotangent bundle.

By applying $\flat$ to (1) and recalling that $\operatorname{grad} p = (dp)^\sharp$ one easily arrives to $$ \frac{\partial u^\flat}{\partial t} + (\nabla_{u} u)^\flat = - dp \qquad \text (3) $$ which is close already! Now the only step remaining is to show that \begin{equation} (\nabla_{u} u)^\flat = \mathcal L_u u^\flat - \frac{1}{2} d(u^\flat(u)) \qquad (4) \end{equation}

I have arrived to the following local expression for the LHS in terms of the metric $(g_{ij})$, a local frame $(\partial_i)$ and its coframe $(dx^i)$, thus considering $u=u^i \partial_i$: $$ (\nabla_{u} u)^\flat = g_{lm} \left( u^j\partial_j(u^m) + u^i u^j \Gamma^m_{ij} \right) dx^l \qquad \text (5) $$

My doubt is how can I arrive to the same expression for the RHS of (4). Thanks a lot for your help!

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    $\begingroup$ Do you have a coordinate formula for the Lie derivative of a 1-form? Should be fairly straightforward from there. $\endgroup$
    – Anthony Carapetis
    May 18, 2020 at 5:33
  • $\begingroup$ @AnthonyCarapetis I don't. And I don't consider anything else on u: $u=u^i\partial_i$ $\endgroup$
    – Txordi
    May 18, 2020 at 7:58
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    $\begingroup$ Well, what's your definition of $\mathcal L_u \theta$ when $\theta$ is a one-form? I could answer this question but it's only of use to you if it starts somewhere you're familiar with. $\endgroup$
    – Anthony Carapetis
    May 18, 2020 at 8:31
  • $\begingroup$ Sorry @AnthonyCarapetis I was a bit slept before and I didn't understand. I am working with the Cartan's formula: $\mathcal L_{X}\omega = \iota_X d \omega + d \iota_X \omega$, where $\iota_X(\omega) = \omega(X,\cdot)$ is the interior multiplication. With this, the RHS of (4) reads as $\iota_u d u^\flat + \frac{1}{2} d(u^\flat(u))$. I tried to express it as in (5) but I did not quite reached it, even when considering that $\Gamma_{cab} = \frac{1}{2}\, \left(\partial_{b}g_{ca} + \partial_{a}g_{cb} - \partial_{c}g_{ab}\right)$. $\endgroup$
    – Txordi
    May 18, 2020 at 9:21

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