Recovering the Lie derivative of a covector from Cartan's formula on 1-forms The coordinate expression of the Lie derivative of a covector field, $w$ with respect to a vector field $X$ is given by:
$$
\mathcal{L}_{X}w = \left(X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}} + w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}}\right)\mathrm{d}x^{\mu}
$$
But since a covector is also a 1-form, this should be recoverable using Cartan's formula:
$$
\mathcal{L}_{X}w = \left(\mathrm{d}\iota_{X} + \iota_{X}\mathrm{d}\right)w
$$
where $\mathrm{d}$ is the external derivative and $\iota_{X}$ is the internal produce with respect to $X$.
However, when I expand Cartan's formula, I get:
$$\begin{align}
\left(\mathrm{d}\iota_{X} + \iota_{X}\mathrm{d}\right)w_{\mu}\mathrm{d}x^{\mu} &= \left(\frac{\partial X^{\alpha}w_{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}}\right)\mathrm{d}x^{\mu} \\
&= \left(w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\alpha}}{\partial x^{\mu}} + X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}}\right)\mathrm{d}x^{\mu} \\
&= \left(X^{\alpha}\frac{\partial w_{\mu}}{\partial x^{\alpha}} + w_{\alpha}\frac{\partial X^{\alpha}}{\partial x^{\mu}}\right)\mathrm{d}x^{\mu} + X^{\alpha}\frac{\partial w_{\alpha}}{\partial x^{\mu}}\mathrm{d}x^{\mu}\\
&= \mathcal{L}_{X}w + X^{\alpha}\mathrm{d}w_{\alpha}\\
\end{align}$$
Most likely I've done something wrong (the extra term comes from the product rule on $\frac{\partial X^{\alpha}w_{\alpha}}{\partial x^{\mu}}$, but I'm not sure this the correct way to treat the gradient of $X^{\alpha}w_{\alpha}$), or there's some reason that $X^{\alpha}\mathrm{d}w_{\alpha}$ vanishes. $X^{\alpha}$ and $w_{\alpha}$ are in general not zero, so that implies the components of $w$ are closed, so:
$$
\mathrm{d}w_{\alpha} = 0
$$
but I can't think of any way to justify this, since in general they're just functions on a manifold and functions aren't necessarily closed (are they?).
 A: You incorrectly evaluated $\iota_X(dw)$. The correct formula for that is
$$\sum_{\alpha,\mu} X^\alpha\frac{\partial w_\mu}{\partial x^\alpha} dx^\mu - X^\mu\frac{\partial w_\mu}{\partial x^\alpha} dx^\alpha.$$
Note that the latter terms add up to $-\sum\limits_\mu X^\mu dw_\mu$.
A: You didn't quite evaluate $\iota_X \textrm{d}w$ correctly, thus $\textrm{d}w$ need not be zero as you suspect. I will write the solution out in more detail, and below I will denote partial derivatives $\partial/\partial x^\mu$ as $\partial_\mu$.
$\begin{split}
\mathcal{L}_Xw&=(\textrm{d}\,\iota_X+\iota_X\,\textrm{d})\,w_\mu \textrm{d}x^\mu \\
&= \textrm{d}(X^\mu w_\mu )+\iota_X(\partial_vw_\mu\textrm{d}x^v\wedge\textrm{d}x^\mu)\\
&=\partial_v(X^\mu w_\mu)\textrm{d}x^v+\iota_X(\partial_vw_\mu\textrm{d}x^v\wedge\textrm{d}x^\mu)\\
&=w_\mu\partial_vX^\mu \textrm{d}x^v +X^\mu \partial_v w_\mu\textrm{d}x^v+\iota_X(\partial_vw_\mu\textrm{d}x^v\wedge\textrm{d}x^\mu)\\
\end{split}$
Now recall how the interior product acts on wedge products. For a p-form $\alpha$, q-form $\beta$, and vector field $X$, we have
$\iota_X(\alpha\wedge\beta)=\iota_X(\alpha)\wedge\beta+(-1)^p\alpha\wedge\iota_X\beta$
Thus, the second term above expands into
$\begin{split}
\iota_X(\partial_vw_\mu\textrm{d}x^v\wedge\textrm{d}x^\mu)&=X^v\partial_vw_\mu\textrm{d}x^\mu+(-1)^1X^\mu\partial_vw_\mu\textrm{d}x^v\\
&=X^\mu\partial_\mu w_v\textrm{d}x^v-X^\mu\partial_v w_\mu \textrm{d}x^v
\end{split}$
Plugging this result in, we'll get two of the four terms to cancel out.
$\begin{split}
\mathcal{L}_Xw &=w_\mu\partial_v X^\mu \textrm{d}x^v +X^\mu \partial_vw_\mu\textrm{d}x^v\\
&\,\,\,\,+X^\mu\partial_\mu w_v\textrm{d}x^v-X^\mu\partial_v w_\mu \textrm{d}x^v\\
&=(w_\mu\partial_v X^\mu +X^\mu\partial_\mu w_v)\textrm{d}x^v
\end{split}$
Thus we recover the standard formula for the components of the Lie derivative of a covector field.
$(\mathcal{L}_Xw)_\mu = w_v\partial_\mu X^v +X^v\partial_v w_\mu$
