If I want to compute with arbitrary 2-forms on $\mathbb{R}^4$, I wonder if each form must contain all 4 variables e.g $x, y, z, t$.
If so, I figure that two-forms in 4 dimension will correspond to the multi-index $$I = \{(1,2), (1,3), (1,4), (2,3), (2,4), (3,4) \}.$$ That is a set of $$\binom{4}{2}$$ strictly increasing 2-tuples if I have understood correctly. Given that the definition of $k$-forms is $$\omega = \sum a_I dx_I,$$ I pick arbitrary elements from $I$ to form 2-forms. Example $$\omega = dx_1 \wedge dx_2 + dx_1 \wedge dx_3$$ and $$\tau = dx_2 \wedge dx_3 + dx_3 \wedge dx_4.$$
The wedge product $\omega \wedge \tau$ is trivial because each term contain $dx_i \wedge dx_i = 0$.
EDIT: Can I choose how many basis elements to include in the linear combinations i.e. the $k$-forms?
Best Answer
Well, you can certainly compute products like $\omega \wedge \tau = 0$ without all four indices appearing.
If you want $\alpha \wedge \beta \ne 0$ for 2-forms $\alpha, \beta$ on $4$-space, then you do, indeed, need to have all four indices appear somewhere -- two of them in $\alpha$ (and possibly in $\beta$ as well), and the complementary pair in $\beta$ (and possible in $\alpha$ as well). Even so, that won't guarantee that $\alpha \wedge \beta \ne 0$.
For instance, if $$ \alpha = dx_1 \wedge dx_2 + dx_3 \wedge dx_4\\ \beta = dx_1 \wedge dx_2 - dx_3 \wedge dx_4 $$ then we have $\alpha \wedge \beta = 0$, despite all four indices appearing.