Hello,
The following question appears as a step in my proof. It seems easy but somehow I have not been able to prove this. I could solve few special cases though. Any help in this context is welcome.
Thank you.
Question:
Let $u,v\in \Lambda^2$ be such that $$ u\wedge u=-v\wedge v\neq 0. $$ Does there exist $\alpha,\beta\in\mathbb{R}$ such that $$ (\alpha u+\beta v)\wedge (\alpha u+\beta v)=0\text{ and }\alpha u+\beta v\neq 0? $$
I assume that you are working with a real vector space. Otherwise, you could take $v=iu$ and get a counterexample.
Let $u$ and $v$ in $\Lambda^2(V^\ast)$ satisfy $u^2 = - v^2 \not = 0$. Since $u^2\not=0$, it follows that, for any $a\in V^\ast$, the equation $a\wedge u = 0$ implies $a=0$, with a similar statement for $v$. For $x\in V$, I will let $i_x$ denote interior product with $x$.
Now, if $x\in V$ satisfies $i_x(u^2)=0$, then $i_xu\wedge u = 0$, so $i_xu=0$. Applying the same argument to $v$ and using $u^2 = - v^2$, one sees that $i_xu=0$ if and only if $i_xv=0$. Thus, $u$ and $v$ have the same kernel, and hence the same rank, and so we are reduced to the case in which $u$ and $v$ are symplectic forms (i.e., nondegenerate) on $V$.
In the case that the dimension of $V$ is $4$ (i.e., when $u^3 = v^3 = 0$), the result is trivial, so assume that the dimension of $V$ is bigger than $4$. Then for any $a\in V^\ast$, it follows that the $3$-forms $a\wedge u$ and $a\wedge v$ have only $a$ as a linear factor. However, if $x\in V$ is nonzero, it follows from our assumption that $i_xu\wedge u = - i_xv\wedge v$, so $i_xu$ and $i_xv$ must be linearly dependent (and nonzero) for all nonzero $x\in V$. From this, it easily follows that there is a nonzero constant $c$ such that $i_xv = c\ i_xu$ for all $x\in V$, which implies that $v = c\ u$, which, in turn, implies that $v^2 = c^2\ u^2 \not= - u^2$, so this case cannot happen.
