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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?

Wikipedia: https://en.m.wikipedia.org/wiki/Exterior_algebra

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

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  • $\begingroup$ You're right, I dismissed that phrase cause I didn't understand what it meant. But why it should be zero? $\endgroup$ – Guerlando OCs Oct 26 '17 at 20:25
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    $\begingroup$ Because if $v\in V$ appears in a $k$ vector $a$, then $v\wedge v=0$ appears somewhere in $a\wedge a$. $\endgroup$ – rschwieb Oct 26 '17 at 20:27
  • $\begingroup$ Why the last line of your equations is not 0? $\endgroup$ – Guerlando OCs Oct 26 '17 at 20:29
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    $\begingroup$ @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. $\endgroup$ – rschwieb Oct 26 '17 at 23:20
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    $\begingroup$ @GuerlandoOCs if $a=v\wedge\cdots$ then $a\wedge a=(-1)^jv\wedge v\wedge\cdots=0\wedge\cdots=0$ after anticommuting things around. $\endgroup$ – rschwieb Oct 27 '17 at 0:04
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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.

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

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