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Let $M$ be an $m$ manifold, $T^{*}_p(M) \equiv\Omega_p^1$ the cotangent space at some point $p \in M$, and $\Omega^r_p$ the space or $r$-forms at $p$, then the exterior algebra for $T^{*}_p(M)$ is the direct sum: $\Omega_p=\Omega_p^0\oplus\Omega_p^1\oplus\Omega_p^2\oplus \cdots\oplus\Omega_p^m$.

For a two form it is clear how the correspondence holds between a vanishing wedge and being decomposable. My understanding is that a decomposable $r$-form is synonymous to being simple, i.e. given $ \omega \in \Omega_p^r$ then $\omega$ simple (decomposable) iff $\omega=v_1\wedge v_2\wedge\cdots\wedge v_r$, $\{v_i\}\in \Omega_p^1$. Please correct this if wrong.

However, given any simple $w$ then self wedge should vanish by virtue of having:

$w \wedge w = (\text{some sign})v_1 \wedge v_1\wedge v_2 \wedge v_2\wedge\cdots\wedge v_r \wedge v_r =0$. Hence for any decomposable $r$-form its self wedge should vanish. Is the reverse not true for anything other than a 2-form? Please let me know if there is a mistake in my reasoning.

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It is not true that any form whose exterior product with itself is zero must be decomposable. In fact, $\omega\wedge\omega = 0$ for all odd-degree forms $\omega$. This is because of the alternating nature of the exterior product: for any $r$-form $\omega$, we have $\omega\wedge\omega = (-1)^r\omega\wedge\omega$, which implies both sides are zero if $r$ is odd.

The claim is not true for even $r\neq 2$, either. For example, if $m \geq 7$, then

$\eta = v_1\wedge v_2 \wedge v_3\wedge v_4 + v_1\wedge v_5 \wedge v_6\wedge v_7$,

where $v_1,\dots ,v_7$ form a linearly independent set, is not decomposable, and yet $\eta \wedge\eta = 0$. This can be extended to other even forms. Also note in the above example with $m=7$, all 4-forms satisfy $\omega\wedge\omega = 0$, as there are no 8-forms.

In other words, the reverse of what you're saying is indeed not true for anything other than a $2$-form.

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    $\begingroup$ Another shorter example is $\omega=\mathrm{d}x^1 \wedge \mathrm{d}x^2 + \mathrm{d}x^3 \wedge \mathrm{d}x^4$ on $\mathbb{R}^4$. Here $\omega\wedge \omega$ is twice the standard volume form on $\mathbb{R}^4$. $\endgroup$
    – Didier
    May 24 '20 at 9:42
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    $\begingroup$ @DIdier_: This is a non-example :) $\endgroup$ May 24 '20 at 16:33
  • $\begingroup$ @TedShifrin You are totally right. I gave another counter example (a true one!) as an answer to apologize. $\endgroup$
    – Didier
    May 24 '20 at 16:39
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I would say that the statement is false because of a simple fact if $n \geqslant 3$. Say $V$ is of dimension $n$. Take any $(n-1)$-form $\alpha$. Then $\alpha \wedge \alpha = 0$ because $\Lambda^{2(n-1)}V = 0$ as $2(n-1) > n$. Thus any non decomposable $(n-1)$ form is a counter example.

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