Find the reduced form of the 2-form in $\mathbb{R}^4 \otimes \mathbb{C}$

Question

Let us have a 4-dimensional vector space $V$. Let $U$ be all the forms $\Omega$ from $\Lambda^2 (V \otimes \mathbb{C})$ such that $\Omega \wedge \Omega = 0$

I heard that the condition $\Omega \wedge \Omega$ implies that the form can be represented as a wedge of two vectors $\Omega = v_1 \wedge v_2$

I have found such $\Omega = e_1\wedge e_2 + e_3 \wedge e_4 + i(e_1 \wedge e_3 + e_2 \wedge e_4)$

You see that wedge square is $0$. However I can't find the pair of vectors $ v_1 , v_2$. What are they?

I have $\Omega = e_1 \wedge(e_2 + i e_3) + (e_3 + i e_2) \wedge e_4$. But now I don't see how can I add something or multiply by a complex number to get those $v_1$ and $v_2$

A related question was issued in 2013 and I tried to apply the answer about tensor reduction,

Following the piece of advice got $\Omega = e_1 \wedge v + i v \wedge e_4 + 2 e_3 \wedge e_4$ where $v = e_2 + i e_3$. I don't get how to proceed from here.

UPD: As noted in the answer, the original tensor isn't reducible. The right one can be $\Omega = e_1\wedge e_2 - e_3 \wedge e_4 + i(e_1 \wedge e_3 + e_2 \wedge e_4)$.

Answer 1

You have a sign mistake. This should be $-e_3\wedge e_4$.

Indeed $\Omega\wedge \Omega=(e_1\wedge e_2 + e_3 \wedge e_4 + i(e_1 \wedge e_3 + e_2 \wedge e_4))\wedge(e_1\wedge e_2 + e_3 \wedge e_4 + i(e_1 \wedge e_3 + e_2 \wedge e_4))= e_1\wedge e_2\wedge e_3\wedge e_4+e_3\wedge e_4\wedge e_1\wedge e_2-(e_1 \wedge e_3 + e_2 \wedge e_4)\wedge(e_1 \wedge e_3 + e_2 \wedge e_4)=e_1\wedge e_2\wedge e_3\wedge e_4+e_3\wedge e_4\wedge e_1\wedge e_2-e_1\wedge e_3\wedge e_2\wedge e_4-e_2\wedge e_4\wedge e_1\wedge e_3.$

Note that $e_3\wedge e_4\wedge e_1\wedge e_2=(-1)^3e_4\wedge e_1\wedge e_2\wedge e_3=(-1)^6e_1\wedge e_2\wedge e_3\wedge e_4=e_1\wedge e_2\wedge e_3\wedge e_4$, that $-e_1\wedge e_3\wedge e_2\wedge e_4=e_1\wedge e_2\wedge e_3\wedge e_4$, and that $-e_2\wedge e_4\wedge e_1\wedge e_3=-e_4\wedge e_1\wedge e_2\wedge e_3=e_1\wedge e_2\wedge e_3\wedge e_4.$

Hence $\Omega\wedge \Omega=4 e_1\wedge e_2\wedge e_3\wedge e_4\neq 0$.

To answer your question, I will assume the correct sign, that is $\Omega=e_1\wedge e_2-e_3\wedge e_4+i(e_1\wedge e_3+e_2\wedge e_4) $

Passing to dual spaces, the question boils down to this one: let $\Omega:E\times E\to\mathbb{C}$ be an alternating bilinear map. Find linear maps $f_1,f_2:E\to\mathbb{C}$ such that $\Omega(e_i,e_j)=f_1(e_i)f_2(e_j)$ for all $i,j$

Here $\Omega(e_1,e_2)=1=f_1(e_1)f_2(e_2)$. In particular, $f_1(e_1)\neq 0$, and we may assume wlog that $f_1(e_1)=1$. Thus $\Omega(e_1,e_j)=f_2(e_j)$ for all $j$. We then get $f_2(e_1)=0$ (since $\Omega(e_1,e_1)=0$), $f_2(e_2)=1,f_2(e_3)=i,f_2(e_4)=0.$ Hence $f_2$ corresponds by duality to $v_2=e_2+ie_3$.

Similarly $\Omega(e_i,e_2)=f_1(x)f_2(e_2)=f_1(e_i)$ for all $i$. We get $f_1(e_1)=1,f_1(e_2)=0$, $f_1(e_3)=0,f_1(e_4)=-i$ ($\Omega$ is alternating, hence skewsymmetric). This corresponds to $v_1=e_1-ie_4$.

Now $v_1\wedge v_2=(e_1-ie_4)\wedge(e_2+ie_3)=e_1\wedge e_2+i e_1\wedge e_3-ie_4\wedge e_2+e_4\wedge e_3=e_1\wedge e_2-e_3\wedge e_4+i(e_1\wedge e_3+e_2\wedge e_4)=\Omega. $ (this confirms the sign mistake, by the way).