Relation between lie derivative of a vector field and associated 1-form in a Lorentzian manifold

Question

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...

Answer 1

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.

Answer 2

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.