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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$2012-06-23
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    If you find an expression for the $\gamma$s (in terms of the $\alpha_{ij}$, $\beta_{ij}$ of cause) in e.g. $\alpha\wedge\beta = \sum_{i$\gamma_{ijkl}$ must vanish, by the uniqueness of this representation. – 2012-06-23
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    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 question2012-06-24

1 Answers 1

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\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

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    How about $(dx^1\wedge d\overline{x}^1 - dx^2\wedge d\overline{x}^2)\wedge(dx^1\wedge d\overline{x}^1 + dx^2\wedge d\overline{x}^2) = 0$?2012-06-23
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    @BenA.You're right. I just modified my previous statement.2012-06-23
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    There is something wrong in the last equation (it is not linear in the betas)2012-06-23
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    What do you mean by "not linear in the $\beta$"?2012-06-23
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    @Mercy In the final expression, the $\beta_{kj}$ belongs to the $\alpha_{il}$ at the end.2012-06-24