Baby Rudin theorem 10.20

There are the definitions which we need for the following theorem

There is the theorem:

Theorem 10.20

(a)If $\omega, \lambda $ are $k-$ and $m-$ forms, respectively of class $C^{1}$ in $E$, then $$d(\omega \land \lambda)=(d\omega)\land \lambda + (-1)^{k}\omega \land d\lambda$$

(b) If $\omega$ is of class $C^{1}$, then $d^{2} \omega=0$

The proof says that because of ($57$) and ($60$) $(a)$ follows if $(63)$ is proved for the special case $\omega$ = $f dx_I$, $\lambda$ = $g dx_j$ where $f,g$ $\in$ $\mathscr C'(E)$, $dx_I$ is a basic $k-form$ and $dx_j$ is a basic $m-form$.

I don't understand how does it follow from this special case ($\omega$ = $f dx_I$, $\lambda$ = $g dx_j$) while $\omega=\sum b_{I}(x)dx_{I}$ and $\lambda=\sum c_{J}(x)dx_{J}$

Any help would be appreciated.

Solution 1:

If you already accept the proof for some basic k-formm and m-form, then the general case if from that:

$$ \begin{aligned} & d\left(\sum_{I} b_{I}(x) d x_{I} \wedge \sum_{J} c_{J}(x) d x_{J}\right) \\ =& d\left(\sum_{I} \sum_{J} b_{I}(x) c_{J}(x) d x_{I} \wedge d x_{J}\right) \\ =& \sum_{I} \sum_{J} d\left(b_{I}(x) c_{J}(x) d x_{I} \wedge d x_{J}\right) \\ =& \sum_{I} \sum_{J}\left[d\left(b_{I}(x) d x_{I}\right) \wedge c_{J}(x) d x_{J}+(-1)^{k} b_{I}(x) d x_{I} \wedge d\left(c_{J}(x) d x_{J}\right)\right] \end{aligned} $$

and:

$$ \begin{aligned} & \sum_{I} \sum_{J} d\left(b_{I}(x) d x_{I}\right) \wedge c_{J}(x) d x_{J} \\ =& \sum_{I}\left[d\left(b_{I}(x) d x_{I}\right) \wedge\left(\sum_{J} c_{J}(x) d x_{J}\right)\right] \\ =& \sum_{I}\left[d\left(b_{I}(x) d x_{I}\right) \wedge \lambda\right] \\ =&\left[d\left(\sum_{I} b_{I}(x) d x_{I}\right)\right] \wedge \lambda \\ =&(d \omega) \wedge \lambda \end{aligned} $$

The other part is $(-1)^k \omega \wedge (d\lambda)$ can be proved in a similar way.