Say we have two one-forms $\alpha = \alpha_1 dx^1+…+\alpha_n dx^n$ and $\beta = \beta_1 dx^1+…+\beta_n dx^n$ and $\gamma = \alpha \wedge \beta$. How many independent components will $\gamma$ have, dependent on the dimension $n$?
Some of my jabs at the problem:
I know that $\gamma$ lives in a space of two forms of dimension $n(n-1)/2$ and this is thus an upper bound on its number of independent components.
On the other hand, it is created from $\alpha,\beta$, which have $2 n$ components altogether. Not all of them will show in the product, though, the exterior product is invariant to rescalings $\alpha \to c \alpha,\; \beta \to c^{-1} \beta,\; c\neq0$, and rotations $\alpha \to \cos(t)\alpha – \sin(t) \beta, \;\beta \to \cos(t)\beta + \sin(t) \alpha$. So the product can have at most $2n-2$ independent components. This gives me two upper bounds of the independent number of components $\gamma$, $2n -2$ and $n(n-1)/2$.
To make this more complicated, in dimension $n=4$ one can show that the property $\gamma\wedge\gamma = 0$ means that there are exactly five independent components of $\gamma$ (which saturates neither of the upper bounds given above). The conditions $\gamma\wedge…\wedge\gamma=0$ will generally provide $[n/2]$ new constraining equations (some of which have quite a lot of components).
I believe that the number of independent components of $\gamma$ for $n>2$ is $n$ in odd dimension and $n+1$ in even dimension, but I have not been able to prove this.
Best Answer
Ok, so apparently the solution to the problem is well known, I just did not know the correct keywords under which to search. I have gained the knowledge to formulate my answer from this and this answer.
The set of decomposable 2-forms $\gamma$ can be understood as the set of two-dimensional planes intersecting the origin in $n$-dimensional with different normalizations of their surface elements. This makes the dimension of their space equal to the dimension of the Grassmanian $Gr(2,n)$ plus one. This means that the independent number of components of $\gamma$ (the dimension of the manifold of decomposable forms in $\Lambda^2(V)$) is $2(n-2)+1 = 2n-3$.