# How many independent components are there in the exterior product of two one forms?

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.

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 $$2(n-2)+1 = 2n-3$$.