distributive law of differential form

Question

Suppose we have two $p$-form $\omega_{1},\omega_{2}$ and a $q$-form $\lambda$ where $$\omega_{1}=\sum_{I_{1}}b_{I_{1}}(x)dx_{I_{1}}\\\omega_{2}=\sum_{I_{2}}b_{I_{2}}(x)dx_{I_{2}}\\\lambda=\sum_{J}c_{J}(x)dx_{J}$$ where $I_{1},I_{2},J$ are increaseing indices in $\{1,\dots,p\}$ and $\{1,\dots,q\}$ respectively.

If $$\int_{\Phi}\omega=\int_{\Phi}\omega_{1}+\int_{\Phi}\omega_{2}$$ for every $p$-surface $\Phi$, we define $$\omega=\omega_{1}+\omega_{2}$$

Now if we define $$\omega\wedge \lambda=\sum_{I,J}b_{I}c_{J}dx_{I}\wedge dx_{J}$$

How do we use the definition above to prove $$(\omega_{1}+\omega_{2})\wedge \lambda=\omega_{1}\wedge \lambda+\omega_{1}\wedge \lambda$$

Answer 1

First, your notation is pretty horrible. Write \begin{align*} \omega_1 &= \sum b_I\,dx_I \quad\text{and}\\ \omega_2 &= \sum b'_I\,dx_I, \end{align*} with the same (increasing) multiindices and different coefficients.

Now show that it follows from your integration over $p$-surface definition that $\omega_1+\omega_2 = \sum (b_I+b'_I)\,dx_I$. (You will want to choose the $p$-surfaces to be small subsets of the coordinate $p$-planes. The usual way to proceed would be to assume you had some point $p$ and some increasing subset $I_0$ where the formula fails to hold.)