Elements of $\wedge^2V$ expressible in the form $v_1\wedge v_2$

Question

If $V$ is a complex vector space, then an element $w\in \wedge^2V$ is of the form $v_1\wedge v_2$ for some $v_1,v_2\in V$ iff $w\wedge w=0$ in $\wedge^4V$. Could anybody give some intuition/show why this is so?

Answer 1

Assuming $V$ to be finite dimensional throughout, then clearly if $ v = v_1\wedge v_2 \in \Lambda^2(V) $ then $$ v\wedge v = v_1\wedge v_2\wedge v_1\wedge v_2 = 0 $$ for repeated factors. For the converse, we can do induction on dimension. If $\dim(V) = 2$ then $\dim(\Lambda^2(V)) = 1 $, hence every element is in the decomposable form as it is spanned by 1 element. If $\dim(V) = 3 $, for any $ a \in \Lambda^2(V)\setminus \{0\} $ we define $ A : V \rightarrow \Lambda^3(V) $ as $A(v) = a\wedge v $. As $\dim(\Lambda^3(V)) = 1 $ so $\dim(\ker(A)) \geq 2 $. If $w_1,w_2 $ be linearly independent vectors of the kernel and $ w_1,w_2,w_3 $ span $V$ then we have for $ c_1,c_2,c_3 \in \mathbb{C}$ $$ a = c_1(w_2\wedge w_3)+c_2(w_3\wedge w_1)+c_1(w_1\wedge w_2) $$ Thus as $ a\wedge w_1 = c_1(w_2\wedge w_3\wedge w_1) = 0 $ hence $ c_1 = 0 $ and similarly $ a\wedge w_2 = 0 $ implies $ c_2 = 0 $ hence $ a = c_3w_1\wedge w_2 $. Now we assume it to be true for $\dim(V) \leq n-1 $ then for $\dim(V) = n $ if $ v_1,v_2,....,v_n $ is a basis, $V'= span(v_1,...,v_{n-1}) $, then for $ a \in \Lambda^2(V) $ we can write it as $$ a = \sum_{i,j}c_{ij}(v_i\wedge v_j) = w\wedge v_n + a' $$ where $w = \sum^{n-1}_{i=1} c_{in}v_i \in V' $ and the rest of the expression $ a' \in \Lambda^2(V') $. Now using the condition for converse we have $ 0 = a\wedge a = 2(w\wedge a'\wedge v_n) + (a'\wedge a') $. Hence the following expressions being independent of $v_n $ we have $$ w\wedge a' = 0 = a'\wedge a' $$ By induction $ a'\wedge a' = 0 $ implies $ a' = w_1\wedge w_2 $ and hence $ w\wedge w_1\wedge w_2 = 0 $. So they are lienarly dependent i.e. there exists $ c, c_1,c_2 \in \mathbb{C} $ such that $$ cw+ c_1w_1 + c_2w_2 = 0 $$ Now if $ c=0 $ then $ w_1 $ and $ w_2 $ are linearly dependent hence $ a' = w_1 \wedge w_2 = 0 $ and $ a = w\wedge v_n $. Otherwise WLOG $ c = 1 $ and $ w = c_1w_1 + c_2 w_2 $ hence according to the 3-dimensional case above $a$ is in the decomposable product form. Thus by induction we have the converse for any $ a \in \Lambda^2(V) $.