# $a = e_1\wedge e_2 + e_3\wedge e_4$ is not decomposable

I'm reading about exterior algebra and wikipedia gives this example of a non decomposable 2 vector:

$a = e_1 \wedge e_2 + e_3\wedge e_4$

But it does not tell the reason. I know that decomposable means that it cannot be written as a product, but why?

The answer follows in the sentence after what you are looking at:

(This is a symplectic form, since α ∧ α ≠ 0.)

Meaning that if $a$ could be written as a single wedge product of $2$ vectors, then its wedge-square would be zero.

But $a\wedge a=(e_1 \wedge e_2 + e_3\wedge e_4)\wedge(e_1 \wedge e_2 + e_3\wedge e_4)=\\e_1\wedge e_2\wedge e_3\wedge e_4+e_3\wedge e_4\wedge e_1\wedge e_2=2(e_1\wedge e_2\wedge e_3\wedge e_4)\neq 0$

• You're right, I dismissed that phrase cause I didn't understand what it meant. But why it should be zero? – Guerlando OCs Oct 26 '17 at 20:25
• Because if $v\in V$ appears in a $k$ vector $a$, then $v\wedge v=0$ appears somewhere in $a\wedge a$. – rschwieb Oct 26 '17 at 20:27
• Why the last line of your equations is not 0? – Guerlando OCs Oct 26 '17 at 20:29
• @GuerlandoOCs I think you need to read the section about bases again. Any wedge product of linearly independent vectors is nonzero. That is one of the built-in features of the wedge product. – rschwieb Oct 26 '17 at 23:20
• @GuerlandoOCs if $a=v\wedge\cdots$ then $a\wedge a=(-1)^jv\wedge v\wedge\cdots=0\wedge\cdots=0$ after anticommuting things around. – rschwieb Oct 27 '17 at 0:04

I will give here a "direct proof", which does not rely on the "trick" of showing that $$a \wedge a \neq 0$$, where $$a = e_1 \wedge e_2 + e_3\wedge e_4$$.

Assume by contradiction that $$a = e_1 \wedge e_2 + e_3\wedge e_4 =v \wedge w$$, for some $$v,w \in V$$. Then, we have

$$v \wedge w \wedge e_3= e_1 \wedge e_2 \wedge e_3$$ $$v \wedge w \wedge e_4= e_1 \wedge e_2 \wedge e_4$$ $$v \wedge w \wedge e_1= e_3 \wedge e_4 \wedge e_1$$ $$v \wedge w \wedge e_2= e_3 \wedge e_4 \wedge e_2$$

which imply $$v \in \text{span}(e_1,e_2,e_3) \cap \text{span}(e_1,e_2,e_4)\cap\text{span}(e_3,e_4,e_1)\cap\text{span}(e_3,e_4,e_2)=\{0\}$$.

Since $$a\neq 0$$, we have reached a contradiction.

• I think it's worthy to elaborate on why $v \wedge w \wedge e_3= e_1 \wedge e_2 \wedge e_3$ implies $v \in \text{span}(e_1,e_2,e_3)$. – user26857 Jan 21 at 20:32