Partial derivative of a differential form

Question

Given a differential form on a manifold $\alpha$, does it makes sense to have its partial derivative $\partial_{\mu}\alpha$? According to my understanding the only sensible derivative on the space of differential form is the exterior derivative $\text{d}$ so $\partial_{\mu}\alpha$ as such is not a well defined object. Can someone please confirm or refute my understanding?

Answer 1

You have to keep in mind the domains and codomains of your maps. Let ${M}$ be a smooth manifold. The partial derivative of a differential form $\omega\in\Omega^{n}({M})$ with ${n}>{0}$ is simply not defined, it would only be defined on the component functions of that differential form in a coordinate chart. The partial derivative operator is a map $\frac{\partial}{\partial{x}^{\mu}}:{C}^{\infty}({M})\rightarrow{C}^{\infty}({M})$, that satisfies the following conditions: \begin{align*} &(\mathrm{i}){\quad}\frac{\partial({a}\Phi({x})+{b}\Psi({x}))}{\partial{x}^{\mu}}={a}\frac{\partial\Phi({x})}{\partial{x}^{\mu}}+{b}\frac{\partial\Psi({x})}{\partial{x}^{\mu}}\\[0.5em] &(\mathrm{ii}){\quad}\frac{\partial(\Phi({x})\Psi({x}))}{\partial{x}^{\mu}}=\Phi({x})\frac{\partial\Psi({x})}{\partial{x}^{\mu}}+\Psi({x})\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\\[0.5em] &(\mathrm{iii}){\quad}\frac{\partial\Phi(\overline{x}({x}))}{\partial{x}^{\mu}}=\frac{\partial\overline{x}^{\nu}}{\partial{x}^{\mu}}\frac{\partial\Phi(\overline{x})}{\partial\overline{x}^{\nu}}\bigg\vert_{\overline{x}=\overline{x}({x})} \end{align*} The exterior derivative operator is a map $\mathrm{d}:\Omega({M})\rightarrow\Omega({M})$, where $\Omega({M})=\bigoplus_{{n}={0}}^{\mathrm{dim}({M})}\Omega^{n}({M})$, that satisfies the following conditions: \begin{align*} &(\mathrm{i}){\quad}\mathrm{d}({a}\omega+{b}\eta)={a}{\,}\mathrm{d}\omega+{b}{\,}\mathrm{d}\eta\\[0.5em] &(\mathrm{ii}){\quad}\mathrm{d}(\omega\wedge\eta)=\mathrm{d}\omega\wedge\eta+(-{1})^{\mathrm{deg}(\omega)}\omega\wedge\mathrm{d}\eta\\[0.5em] &(\mathrm{iii}){\quad}\mathrm{d}\mathrm{d}\omega={0} \end{align*} For a smooth function, the exterior derivative becomes the total differential $\mathrm{d}:{C}^{\infty}({M})\rightarrow\Gamma^{\infty}({M},{T}^{\star}{M}),{\,}{C}^{\infty}({M})\ni\Phi\mapsto\mathrm{d}\Phi=\frac{\partial\Phi({x})}{\partial{x}^{\mu}}\mathrm{d}{x}^{\mu}\in\Gamma^{\infty}({M},{T}^{\star}{M})$. For a differential form $\omega\in\Omega^{2}({M})$, the exterior derivative becomes \begin{align*} \mathrm{d}\omega&=\mathrm{d}(\omega_{\mu\nu}({x}){\,}\mathrm{d}{x}^{\mu}\wedge\mathrm{d}{x}^{\nu})\\[0.5em] &=\mathrm{d}\omega_{\mu\nu}({x})\wedge\mathrm{d}{x}^{\mu}\wedge\mathrm{d}{x}^{\nu}\\[0.5em] &=\frac{\partial\omega_{\mu\nu}({x})}{\partial{x}^{\varrho}}\mathrm{d}{x}^{\varrho}\wedge\mathrm{d}{x}^{\mu}\wedge\mathrm{d}{x}^{\nu}\\[0.5em] &=\frac{1}{{3}!}\bigg(\frac{\partial\omega_{\mu\nu}({x})}{\partial{x}^{\varrho}}-\frac{\partial\omega_{\nu\varrho}({x})}{\partial{x}^{\mu}}-\frac{\partial\omega_{\varrho\mu}({x})}{\partial{x}^{\nu}}\bigg)\mathrm{d}{x}^{\varrho}\wedge\mathrm{d}{x}^{\mu}\wedge\mathrm{d}{x}^{\nu}{\,}{.} \end{align*} Formally, we say that the partial derivative $\frac{\partial}{\partial{x}^{\mu}}$ is a derivation on the algebra of smooth function ${C}^{\infty}({M})$ and the exterior derivative $\mathrm{d}$ is an anti-derivation on the algebra of differential forms $\Omega({M})$.