Let $u$ be a $C^2$ function from $\mathbb{C}^n$ to $\mathbb{C}$. Define $$ \partial u = \sum\limits_{i=1}^n \frac{\partial u}{\partial z_i}dz_i, \\ \overline{\partial} u = \sum\limits_{i=1}^n \frac{\partial u}{\partial \overline{z}_i}d\overline{z}_i $$ If $v = v' \wedge dz_i$, we can define $\partial v = \partial v' \wedge dz_i$, $\overline{\partial} v = \overline{\partial} v' \wedge dz_i$ and by analogy we can define such operators for $v = v'' \wedge d \overline{z}_i$. Next we define operators $$ d = \partial + \overline{\partial}, \;\;\; d^c = i(\overline{\partial} - \partial) $$ We have $dd^c u = 2i \partial \overline{\partial} u$. I want to compute $n$-th exterior power of $dd^c u$. Is it possible to do it without direct calculations that involve change of summation variables?
Easy way to calculate $(dd^c u)^n$
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complex-analysis
several-complex-variables
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0Are $v'$ and $v''$ 1-forms? – 2012-09-27
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0@Mercy, $v'$ and $v''$ are functions for which operators $\partial$ and $\overline{\partial}$ are already defined. Then such operators are defined on $1$-forms and you can think of $v'$ and $v''$ as of $1$-forms to define it on $2$-forms – 2012-09-27
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0What is then the meaning of the notation $v'\wedge dz_i$? – 2012-09-27
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0@Mercy If $v'$ is a function, then $v' \wedge dz_i = v' dz_i$. – 2012-09-27
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0Not a common notation. I think you should just remove it! Just to make sure, by $n$-th exterior power of $\omega=dd^cu$ you mean $\omega\wedge\ldots\wedge\omega$ ($n$ times) right? – 2012-09-27
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1@Mercy Notation is from Kiselman's "Plurisubharmonic functions and potential theory in several complex variables", 1998, page 3. Furthermore, this notation is well used in differential geometry in MSU. I don't know why you don't like it. And $\omega^n = \omega \wedge \ldots \wedge \omega$ ($n$ times), you're right. – 2012-09-27