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

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

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).

• I think personnally that mistakes are instructive. If you want to edit, you should add a sentence like "there is a sign mistake, $\Omega$ should be blablabla, see the answer below" but keep the original post. Of course, this is a matter of taste. If you want just to correct the sign, just do so, and I will cancel the beginning of my answer Commented Oct 1, 2020 at 10:40