# When does the vanishing wedge product of two forms require one form to be zero?

Let $\alpha$ and $\beta$ be two complex $(1,1)$ forms defined as:

$\alpha = \alpha_{ij} dx^i \wedge d\bar x^j$

$\beta= \beta_{ij} dx^i \wedge d\bar x^j$

Let's say, I know the following:

1) $\alpha \wedge \beta = 0$

2) $\beta \neq 0$

I want to somehow show that the only way to achieve (1) is by forcing $\alpha = 0$. Are there general known conditions on the $\beta_{ij}$ for this to happen?

The only condition I could think of is if all the $\beta_{ij}$ are the same. However, this is a bit too restrictive. I'm also interested in the above problem when $\beta$ is a $(2,2)$ form.

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note that $dx^i\wedge dx^i = 0$ – user20266 Jun 23 '12 at 12:37
If you find an expression for the $\gamma$s (in terms of the $\alpha_{ij}$, $\beta_{ij}$ of cause) in e.g. $\alpha\wedge\beta = \sum_{i<k,j<l} \gamma_{ijkl} dx^i\wedge dx^k\wedge d\overline{x}^j \wedge d\overline{x}^l$, you will have a condition for the product to vanish, namely all the $\gamma_{ijkl}$ must vanish, by the uniqueness of this representation. – Ben Jun 23 '12 at 15:03
I believe my question might have been misunderstood due to the title (which I chose for shortness). I have rephrased it to reflect the actual question – FMN Jun 24 '12 at 13:25

\begin{eqnarray} \alpha\wedge\beta &=&\sum_{i,j,k,l}\alpha_{ij}\beta_{kl}dx^i\wedge d\bar{x}^j\wedge dx^k\wedge d\bar{x}^l\cr &=&(\sum_{i<k,j<l}+\sum_{i<k,l<j}+\sum_{k<i,j<l}+\sum_{k<i,l<j})\alpha_{ij}\beta_{kl}dx^i\wedge d\bar{x}^j\wedge dx^k\wedge d\bar{x}^l\cr &=&\sum_{i<k,j<l}(-\alpha_{ij}\beta_{kl}+\alpha_{kj}\beta_{il}+\alpha_{il}\beta_{kj}- \alpha_{kl}\beta_{ij})dx^i\wedge dx^k\wedge d\bar{x}^j\wedge d\bar{x}^l. \end{eqnarray} Thus $$\alpha\wedge\beta=0 \iff \alpha_{ij}\beta_{kl}+ \alpha_{kl}\beta_{ij}=\alpha_{kj}\beta_{il}+\alpha_{il}\beta_{kj} \quad \forall \ i<k,j<l.$$
How about $(dx^1\wedge d\overline{x}^1 - dx^2\wedge d\overline{x}^2)\wedge(dx^1\wedge d\overline{x}^1 + dx^2\wedge d\overline{x}^2) = 0$? – Ben Jun 23 '12 at 14:52
What do you mean by "not linear in the $\beta$"? – Mercy King Jun 23 '12 at 21:25
@Mercy In the final expression, the $\beta_{kj}$ belongs to the $\alpha_{il}$ at the end. – Ben Jun 24 '12 at 9:30