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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?

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Are $v'$ and $v''$ 1-forms? –  Mercy Sep 27 '12 at 12:51
    
@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 –  Nimza Sep 27 '12 at 13:22
    
What is then the meaning of the notation $v'\wedge dz_i$? –  Mercy Sep 27 '12 at 14:22
    
@Mercy If $v'$ is a function, then $v' \wedge dz_i = v' dz_i$. –  Nimza Sep 27 '12 at 14:41
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@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. –  Nimza Sep 27 '12 at 14:52

1 Answer 1

If $u$ is $C^2$, then $$(dd^c u)^n = n! \, 4^n \, \det \left(\frac{\partial^2u}{\partial z_j\partial\bar z_k}\right)\,dV. $$

So, (ignoring constants that are usually irrelevant), $(dd^c u)^n$ is the determinant of the complex Hessian. The operator $(dd^c \cdot)^n$ is called the complex Monge-Ampère operator and plays a crucial role in pluripotential theory, the systematic study of plurisubharmonic (psh) functions. It is possible to extend the domain of definition of $(dd^c \cdot)^n$ to some classes of non-smooth psh functions, but not to all psh functions. Finding the optimal domain of definition for the complex Monge-Ampère operator is a problem with a long complicated history and was only recently completely settled. (See Cegrell, The General Definition of the Complex Monge-Ampère Operator, Ann. de l'Inst. Fourier, 54, (2004), p.159-179 and Blocki, The domain of definition of the complex Monge-Ampère operator, Amer. J. Math. 128 (2006), 519-530.)

Added for the follow-up question I think the easiest way to see that $(dd^c u)^n$ is a determinant, is to note that $$\omega_1 \wedge \omega_2 \wedge \cdots \wedge \omega_n$$ is an alternating multilinear form, almost by definition of the wedge product. The only alternating $n$-linear form on an $n$-dimensional vector space is (up to a constant multiple) the determinant. If you really need the constant, you have to plug in some cleverly chosen vectors to see what it is.

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I know it. I only want to show this equality with determinant without direct calculations (I'm able to show it with direct calculations. My question is similar to question: is there an easy way to calculate laplacian in spherical coordinates - it is easy to do it but in general we have to write a lot). –  Nimza Sep 30 '12 at 7:26

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