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Let $V$ be a vector space with basis $\{v_1,v_2,v_3,v_4,v_5\}$ and $W$ be the subspace of $\wedge^2(V)$ generated by $\{v_1 \wedge v_3, v_1 \wedge v_4, v_1 \wedge v_5, v_2 \wedge v_3, v_2 \wedge v_4, v_2 \wedge v_5 \}$. An element in $\wedge^2(V)$ is said to be decomposable if it is of the form $u \wedge v$ for $u,v \in V$. I need to find out the set of all decomposable elements in $W$. Of course the basis vectors of $W$ are decomposable but my guess is that the set is the image of $\mathbb P^1 \times \mathbb P^2$ under the Segre embedding but I have no clue how to show this.

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  • $\begingroup$ Dear icmes, I have corrected a typo in your question: you wrote indecomposable but meant decomposable in line 4. By the way, I have no idea what the Segre map has to do with your question. Care to explain ? $\endgroup$ – Georges Elencwajg Mar 24 '17 at 11:28
  • $\begingroup$ Note that $\mathbb P(W) \cong \mathbb P^5$ and the Segre embedding $\mathbb P^1 \times \mathbb P^2$ lands in $\mathbb P^5$. I guess the image of this embedding is the decomposable elements in $\mathbb P(W)$. $\endgroup$ – icmes Mar 24 '17 at 14:19
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Over a field of characteristic $\neq2$ the condition for a $2$-vector $w\in W$ to be decomposable is $w\wedge w=0$, as you can find for example on page 48 of this handout by Nigel Hitchin.
In your case write $$w=x_1v_1 \wedge v_3 +x_2 v_1 \wedge v_4+x_3 v_1 \wedge v_5+x_4 v_2 \wedge v_3+x_5 v_2 \wedge v_4+x_6 v_2 \wedge v_5$$ You will obtain $$w\wedge w=(x_2x_4-x_1x_5)v_1\wedge v_2\wedge v_3\wedge v_4\\+(x_3x_4-x_1x_6)v_1\wedge v_2\wedge v_3\wedge v_5\\+(x_3x_5-x_2x_6)v_1\wedge v_2\wedge v_4\wedge v_5 $$ So the condition $w\wedge w=0$ for a $2$-vector $w\in \wedge^2 W$ to be decomposable is that it lie on the intersection of the three quadric cones in $\wedge^2 W$ given by the equations $$x_2x_4-x_1x_5=0,\quad x_3x_4-x_1x_6=0,\quad x_3x_5-x_2x_6=0$$

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