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Let $(M,g)$ be a Lorentzian manifold and $X$ and $u$ represent two vector fields in $M$ such that $\mathcal{L}_X u=0$, that is, $u$ is Lie transported along the integral curve of $X$. My question is: given the associated 1-form to $u$, $u^\flat$, is $\mathcal{L}_X u^\flat=0$?

I would say yes, but my common sense may be tricking me and I don't really know how to prove such a thing.

Edit: Following Jackozee Hakkiuz suggestion I tried using the Leibniz rule such that $$ X<u^\flat,u> ~=~ <\mathcal{L}_X u^\flat,u> + <u^\flat,\mathcal{L}_X u> ~ =~ <\mathcal{L}_X u^\flat,u>~. $$ But I'm stuck. I tried playing around with the expressions of the above quantities in a local coordinate system but I can't seem to find a relation that answers my question...

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    $\begingroup$ It depends on the metric. Hint: use the definition of $u^{\flat}$ and the Leibniz rule. Try it, it's not difficult. I'll come back later to see if you got it. If you didn't I'll help you :3 $\endgroup$ – Jackozee Hakkiuz Jul 6 '18 at 17:41
  • $\begingroup$ @JackozeeHakkiuz Thank you sir/madam. Not not look like I haven't tried, give me some hours. I will be in transit for a couple hours and only later can work out your hints $\endgroup$ – PML Jul 6 '18 at 17:47
  • $\begingroup$ Ok, right. :) I'll be back soon, then. $\endgroup$ – Jackozee Hakkiuz Jul 6 '18 at 18:00
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The definition of $u^{\flat}$ is $g(u,\cdot)$, so that it maps $u^{\flat}:v\mapsto g(u,v)$ for every vector field $v$.

This, plus the fact that the $L_X$ follows the Leibniz rule with respect to contraction, allows us to calculate

$$L_Xu^{\flat} = L_X(g(u,\cdot)) = (L_Xg)(u,\cdot) + g(L_Xu,\cdot)$$ By assumption, the second term vanishes, so it all boils down to whether $L_Xg=0$ or not.

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Let $X$ and $Y$ be two vector fields on $(M,g)$ that satisfy $$L_XY=[X,Y]=0.$$ By properties of the Lie derivative, and non-degeneracy, it follows that $0=\iota_{[X,Y]}g=L_X(\iota_y(g))+ \iota_Y(L_Xg)$ and so $$L_X(\iota_y(g)=- \iota_Y(L_Xg)$$

Now $Y^\flat= \iota_Y(g)=g(Y,\cdot)$. Thus

\begin{align*} L_XY^\flat&=L_X(\iota_Yg)\\ &=-\iota_Y(L_Xg) \end{align*}

Thus, you can claim that $L_XY^\flat=0$ if $X$ is a killing vector field.

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