2-form whose self-wedge does not vanish?

Question

I know that any 2-form is decomposable if and only if its self-wedge vanishes.

Is there an element $β ∈ A_2(R^n)$ such that $β ∧ β \neq 0$.

Obviously, this $\beta $ must be indecomposable, but I cant think of an example.

Answer 1

Its self-wedge will be a 4-form, so we'd better take $n \geq 4$. Then

$$ \beta=dx_1 \wedge dx_2 + dx_3 \wedge dx_4 $$ should do the trick.

Answer 2

Symplectic geometry exists, so the answer is yes. Take the standard symplectic form on $R^{2n}$ for example (for $n=2$ this is Micah's answer).

https://en.wikipedia.org/wiki/Symplectic_manifold

Wikipedia makes the same observation in its discussion of decomposable vectors. https://en.wikipedia.org/wiki/Exterior_algebra