I am trying to prove the following from a book I am reading through.

Thm: If $\omega$ is a 2-form on $\mathbb{R}^4$ and $\omega \wedge \omega = 0$, then $\omega$ is decomposable. Note decomposable means can be written of the form $\omega = x \wedge y$.

I know that I can write my 2-form as $\omega = a_1 e_1 \wedge e_2 + a_2 e_1 \wedge e_3 + a_3 e_1 \wedge e_4 + a_4 e_2 \wedge e_3 + a_5 e_2 \wedge e_4 + a_6 e_3 \wedge e_4$, where $e_1, e_2, e_3, e_4$ are basis elements. Then I have $0 = \omega \wedge \omega = (a_1a_6-a_2a_5+a_3a_4)(e_1 \wedge e_2 \wedge e_3 \wedge e_4)$. Thus $(a_1a_6-a_2a_5+a_3a_4)=0$.

I don't know what to do from here. Thanks for the help!

  • $\begingroup$ The theorem that $\omega\wedge\omega=0$ implies $\omega=x\wedge y$ is proved using induction by Nigel Hitchin in Projective geometry (chapter 3) see pages 48-49. $\endgroup$ – Andrey Sokolov Jun 17 '13 at 5:17

You can prove this by proving the contrapositive. Suppose $\omega$ is not decomposable.

That means that $\omega$ necessarily must be the sum of two (and only two) decomposable terms, like $\omega = x \wedge y + z \wedge w$. Consider what happens if you add a term like $y \wedge z$. You should realize that you can lump that into one of the two terms and just create a new basis vector to maintain the same form.

Now, look at the wedge product of $\omega$ with itself. You should be able to argue that (1) as long as $x, y, z, w$ are linearly independent, $\omega \wedge \omega \neq 0$, and (2) if they are linearly dependent, then the original supposition that $\omega$ is not decomposable has been violated.

If you can do that, then you have proven then the following statement: If $\omega$ is not decomposable, then $\omega \wedge \omega \neq 0$. The contrapositive of this statement is also true as a result, and it is that contrapositive that you're interested in.


One problem with your approach is that your basis is too generic, which is inconvenient for computations. Try some special basis: e.g. Combine $a_1e_1 \wedge e_2 + a_2 e_1 \wedge e_3 + a_3 e_1\wedge e_4$ as $e_1 \wedge (a_1e_2 + a_2e_3 + a_3e_4)$. Then let $e_2' = a_1e_2 + a_2e_3 + a_3e_4$ we can replace the sum of first three terms as $e_1 \wedge e_2'$. Eliminate $e_2$ by substituting $e_2 = \frac{1}{a_1} (e_2' - a_2e_3 - a_3e_4)$. There are degenerate cases ($a_1 = 0$) that can be handled case by case - I'll write more details here if you need it.

Using this idea you can show that $\omega$ can be written as $$a e_1 \wedge e_2 + b e_2 \wedge e_3 + c e_3 \wedge e_4$$ for a suitable basis $\{e_1,\cdots,e_4\}$. Now can you proceed?

  • $\begingroup$ If I could write it of your form wouldn't that mean that $\Lambda^k(V*)$ had dimension 3 when I know it has dimension 6? $\endgroup$ – Leo Spencer Apr 21 '13 at 1:40
  • $\begingroup$ Also, where are you suggesting I replace $e_2$ with $e'_2$ $\endgroup$ – Leo Spencer Apr 21 '13 at 1:47
  • $\begingroup$ @Leo Spencer, answer updated. For your question, notice I am claiming is that every 2 tensor can be written in that form for SOME basis (that depends on the 2 tensor in hand), rather than all 2 tensors can be written in that form for a fixed basis. $\endgroup$ – user27126 Apr 21 '13 at 6:10

This can be proved using induction on the dimension of V

$\omega=\sum_{i}a_{i}e_{1}\wedge e_{i}+\alpha=e_{1}\wedge \beta+\alpha$, where $\alpha$ is a 2-form which does not involve $e_{1}$, and $\beta=\sum_{i}a_{i}e_{i}$ does not involve $e_{1}$. Since $\omega\wedge \omega=0$, $\alpha \wedge \beta=0$, and $\alpha \wedge \alpha=0$.

By induction hypothesis, $\alpha$ can be written as $\alpha_{1}\wedge \alpha_{2}$.

Since $\alpha_{1} \wedge \alpha_{2}\wedge \beta=0$, $\alpha_{1}, \alpha_{2}, \beta$ are linear dependent. One of them can be written as linear combaination of the other two.

Substitute to the expression of $\omega=\alpha_{1}\wedge \alpha_{2}+e_{1}\wedge\beta$, you can find it is decomposable.


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