My question:

Is $e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 V$ not completely decomposable if $e_1$, $e_2$, $e_3$, $e_4$ is a basis for $V$?


closed as off-topic by user26857, graydad, user91500, Paramanand Singh, Claude Leibovici Jul 15 '15 at 7:21

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  • 1
    $\begingroup$ Write down two general vectors in $V$ in terms of your basis, and wedge them together. Setting that equal to your given element of $\bigwedge^2$ gives 10 linear equations in 8 variables. Do they have a solution? $\endgroup$ – Schemer Jul 14 '15 at 22:24

1) The trick is to find the space $W\subset V$ of vectors $v=ae_1+be_2+ce_3+de_4\in V$ killing $\omega=e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 V$ i.e. such that $v\wedge \omega=0$.
The criterion for $\omega$ to be decomposable is that $W$ have dimension $2$, the grading in which we take the wedge product $\bigwedge ^2V$ (the theory predicts that we will have $\dim W\leq 2)$ .
Since here $W=0$ the criterion tells that $\omega$ is indecomposable.
The criterion is proved in , for example, Theorem 8.10, page 31 of Keith Conrad's splendid handout here.

2) Another method is to remark that for a decomposable element $\eta$ we trivially have $\eta \wedge \eta=0$.
Since $\omega\wedge \omega=2 e_1 \wedge e_2 \wedge e_3 \wedge e_4 $, we again conclude that $\omega$ is not decomposable.

3) You are quite right to think that most $k$-vectors are indecomposable.
In the case at hand $\mathbb P(\bigwedge ^2 V)=\mathbb P^5$ and the non-zero decomposable vectors yield only a $4 $-dimensional hyperquadric $Q\subset \mathbb P^5$ in that $5$-dimensional space: the Klein quadric.