3
$\begingroup$

In complex analysis, by Poincare-Lelong theorem, we have $$ \frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z|^2)=T_{z=0} $$ as currents, where $$ T_{z=0}(\eta)=\int_{z=0}\eta. $$ Now suppose we have two variables $z_1$, $z_2$. We have $$ \frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z_1|^2)=T_{z_1=0} $$ and $$ \frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z_2|^2)=T_{z_2=0}. $$

On the other hand, we have the general principle that wedge products of differential forms correspond to intersections of subvarieties.

My question is: can we define the wedge product $\frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z_1|^2)\wedge \frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z_2|^2)$ so that it equals to $T_{z_1=0,z_2=0}$?

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, since this corresponds to a proper intersection, this kind of product is very robust and can be defined for a couple of different reasons:

  1. Since the unbounded loci of $\log |z_1|^2$ and $\log |z_2|^2$ are $\{ z_1 = 0 \}$ and $\{ z_2 = 0 \}$, and these intersect in a set of codimension $2$, the product can be defined by Demailly's extension of the Bedford-Taylor Monge-Ampère product through regularizing each function by a decreasing limit of smooth psh functions, see for example Demailly, Complex Analytic and Differential Geometry, Theorem III.4.5.

  2. Since the functions depend on different variables, the tensor product of the currents is indeed defined, cf., i.e., Hörmander, The Analysis of Linear Partial Differential Operators I, Theorem V.5.1.1. Since the action on a decomposable test form $\varphi_1(z_1) \varphi_2(z_2)$ equals the integration current along $\{ z_1 = z_2 = 0 \}$, it follows that the tensor product indeed is indeed equal to this integration current.

  3. Since $T_{z_1=0}$ and $T_{z_2=0}$ are integration currents along analytic sets that intersect properly, one may form the product by regularizing each term and taking a limit, and the limit equals the integration current along the intersection product (counted with multiplicity), cf., i.e., Chirka, Complex Analytic Sets, §16.2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.