# Does **vanishing self-wedge iff form is decomposable** hold in all generality?

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.

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.

• 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 $\mathbb{R}^4$. May 24 '20 at 9:42
• @DIdier_: This is a non-example :) May 24 '20 at 16:33
• @TedShifrin You are totally right. I gave another counter example (a true one!) as an answer to apologize. May 24 '20 at 16:39

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.