Determine wether a skew-symmetric tensor is decomposable or not.

Question

For $x_1,x_2,x_3,x_4$ elements of a vector space $V$, I would like to find out wheter the skew-symmetric tensor $11x_1 \wedge x_2 + 10x_1 \wedge x_3 + 17x_1 \wedge x_4 − 13x_2 \wedge x_3 − 10x_2 \wedge x_4 + 11 x_3 \wedge x_4$ of $\bigwedge^2V$ is decomposable or not, in other words wheter it can be expressed as $u_1 \wedge u_2$ for $u_1,u_2 \in V$.

I've attempted to reduce it denoting $u_1 = ax_1 + bx_2 + cx_3 + dx_4$ and similarly for $u_2$, however I didn't end up with a clear solution..

Does anyone have any idea how to know wether this skew-symmetric tensor is decomposable or not ?

Answer 1

I'll assume that $x_1,x_2,x_3,x_4$ are linearly independent. Set

$$ \alpha = 11x_1 \wedge x_2 + 10x_1 \wedge x_3 + 17x_1 \wedge x_4 − 13x_2 \wedge x_3 − 10x_2 \wedge x_4 + 11 x_3 \wedge x_4 $$

and assume that $\alpha = u_1 \wedge u_2$ for some $u_1,u_2 \in V$. This means that both $\alpha \wedge u_1 = u_1 \wedge u_2 \wedge u_1$ and $\alpha \wedge u_2 = u_1 \wedge u_2 \wedge u_2$ are zero. So let's try to solve the equation $\alpha \wedge u = 0$. Write $$ u = ax_1 + bx_2 + cx_3 + dx_4. $$ Then

$$ \alpha \wedge u = \left( 11c - 10b - 13a \right) x_1 \wedge x_2 \wedge x_3 + \left( 11d - 17b - 10a \right) x_1 \wedge x_2 \wedge x_4 + \left( 10d - 17c + 11a \right) x_1 \wedge x_3 \wedge x_4 + \left( 10c -13d + 11b \right) x_2 \wedge x_3 \wedge x_4 = 0 $$

iff $$ \begin{pmatrix} -13 & -10 & 11 & 0 \\ -10 & -17 & 0 & 11 \\ 11 & 0 & -17 & 10 \\ 0 & 11 & 10 & -13 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}. $$ Solving this linear system of equations we see that the space of solutions is two dimensional and $u$ solves $\alpha \wedge u = 0$ iff

$$ u \in \operatorname{span} \left \{ -10x_1 + 13x_2 + 11x_4, 17x_1 -10x_2 + 11x_3 \right \}. $$

Now we can check that

$$ \left( -10x_1 + 13x_2 + 11x_4 \right) \wedge \left( 17x_1 -10x_2 + 11x_3 \right) = \left( 10^2 - 13 \cdot 17 \right) x_1 \wedge x_2 + \\ \left( (-10) \cdot 11 \right) x_1 \wedge x_3 + \left( (-17) \cdot 11 \right) x_1 \wedge x_4 + \left( 13 \cdot 11\right) x_2 \wedge x_3 + \left( 10 \cdot 11 \right) x_2 \wedge x_4 + \\ \left( -11^2 \right) x_3 \wedge x_4 = \\ (-11) \cdot \left( 11x_1 \wedge x_2 + 10x_1 \wedge x_3 + 17x_1 \wedge x_4 − 13x_2 \wedge x_3 − 10x_2 \wedge x_4 + 11 x_3 \wedge x_4 \right) = \\ (-11) \alpha $$

and so we see that

$$ \alpha = \left( \frac{10}{11} x_1 - \frac{13}{11} x_2 - x_4 \right) \wedge \left( 17x_1 -10x_2 + 11x_3 \right) $$

is decomposable.