Partial derivative of differential form

Question

In general, the component-wise partial derivative of a tensor field is not again a tensor field, since it doesn't has the correct transformation law. This is one of the reasons why one introduces covariant derivatives at the first place. Now, I stumbled over the following observation:

Let $M$ be a smooth manifold and $\omega_{\mu}dx^{\mu}\in\Omega^{1}(M)$ a $1$-form. Then, it is not too hard to see that

$$(\mathcal{L}_{\partial_{\nu}}\omega)_{\mu}=\partial_{\nu}\omega_{\mu}$$

for some fixed index $\nu$. This formula can be obtained by using for example Cartan's formula to calculate. In particular, since $\mathcal{L}_{X}$ maps a $1$-form to a $1$-form, we see that $\partial_{\nu}\omega:=(\partial_{\nu}\omega_{\mu})d x^{\mu}$ is a well-defined $1$-form. Similar calculations can be done for arbitrary $k$-forms. So, does this imply that partial derivative of forms are in fact again forms?

Answer 1

In a different frame the vector field $D\equiv \partial_\nu= \delta^\mu_\nu\partial_\mu $ will no longer have constant components $\delta^\mu_\nu$. You therefore need to include their transformation when you transform $\partial_\nu \omega_\mu$. This will bring you to the usual $({\mathcal L}_{D}\omega)_\mu = D^\nu\partial_\nu\omega_\mu + \omega_\nu \partial_\mu D^\nu $.

Answer 2

Yes, but also no.

Consider two charts $(U,x)$ and $(U,\bar x)$ on the same manifold $M$ (for simplicity, let us suppose both charts have the same domain).

Given $\omega\in\Omega^k(U)$, we define $\partial_\mu\omega:=\mathscr L_{\partial_\mu}\omega$.

First let's see how the Lie derivative behaves under multiplication by functions. For any smooth vector field $X\in\mathcal D(M)$ and smooth function $f\in C^\infty(M)$ we have $$ \mathscr L_{fX}\omega=\mathrm d(fX\rfloor\omega)+fX\rfloor\mathrm d\omega=\mathrm df\wedge X\rfloor\omega+f\mathrm dX\rfloor\omega+fX\rfloor\mathrm d\omega \\ =\mathrm df\wedge X\rfloor\omega+f\mathscr L_X\omega. $$

So then $$ \bar\partial_\mu\omega=(\frac{\partial x^\nu}{\partial \bar x^\mu}\partial_\nu\omega)=\mathrm d\frac{\partial x^\nu}{\partial\bar x^\mu}\wedge \partial_\nu\rfloor \omega+\frac{\partial x^\nu}{\partial\bar x^\mu}\partial_\nu\omega, $$ which shows that while the partial derivative gives a well-defined differential form (within the domain of the coordinate chart), this form is not related in a homogeneous way to the partial derivative with respect to another chart.