Theorem 4-10 Calculus on Manifolds by Spivak

Question

From Calculus on Manifolds By Michael Spivak,

If $\omega$ is a $k$-form and $\eta$ is a $l$-form then $d(\omega$ $\wedge \eta$)=$d\omega \wedge \eta+(-1)^k\omega \wedge d(\eta)$

In proof they have written: The formula is true if $\omega=dx^{i_1}\wedge...\wedge dx^{i_k}$, and $\eta=dx^{j_1}\wedge...\wedge dx^{j_l}$, since all the terms vanish. The formula is easily checked when $\omega$ is a $0$-form. The general formula may be derived from $(1)$ and these two observations.

My question is how should I checked the formula for $\omega$ is a $0$-form? I know a $0$-form $\omega$ will be a function $\omega:\mathbb{R}^n \to \mathbb{R}$.

Answer 1

Let $\omega, f : \mathbb{R^n} \rightarrow \mathbb{R}$ be $0$-forms and let $\eta = f dx^{j_1} \wedge...\wedge dx^{j_l} $ be an $l$-form. Then

\begin{equation} d(\omega \wedge \eta) = d(\omega\wedge (f dx^{j_1} \wedge...\wedge dx^{j_l})) = d((\omega f) dx^{j_1} \wedge...\wedge dx^{j_l}) = d(\omega f) \wedge (dx^{j_1} \wedge...\wedge dx^{j_l}) \end{equation}

Now, with summation implied over the index $i$, \begin{equation} d(\omega f) = \frac{\partial(\omega f)}{\partial x^i} dx^i = (\frac{\partial \omega}{\partial x^i} f + \frac{\partial f}{\partial x^i} \omega)dx^i= f(\frac{\partial \omega}{\partial x^i} dx^i) + \omega (\frac{\partial f}{\partial x^i} dx^i)= f (d\omega) + \omega (df) \end{equation}

Hence,

\begin{equation} \small{d(\omega f) \wedge (dx^{j_1} \wedge...\wedge dx^{j_l}) = (f (d\omega) + \omega (df)) \wedge (dx^{j_1} \wedge...\wedge dx^{j_l}) = d\omega \wedge \eta + \omega (df \wedge (dx^{j_1} \wedge...\wedge dx^{j_l}))} \end{equation}

But, \begin{equation} \small {d\omega \wedge \eta + \omega (df \wedge (dx^{j_1} \wedge...\wedge dx^{j_l})) = d\omega \wedge \eta + (\omega) d\eta = d\omega \wedge \eta + \omega \wedge d\eta = d\omega \wedge \eta + (-1)^0 \omega \wedge d\eta} \end{equation}

Hence, the formula holds for $k = 0$.