Understanding arbitrary currents and currents of integration.

Question

I've been doing some readings, and there are some notational issues with currents that I do not understand. Let $\Omega\subset \mathbb C^n$ be open. A current $T$ of bidegree $(p,q)$ is an element of the dual space $(D^{(n-p,n-q)}(\Omega))'$ where $D^{(n-p,n-q)}(\Omega)$ is the space of test forms (differential forms of bidegree $(p,q)$ with smooth coefficients of compact support). One can define similary define a current of order zero as an element of $(D_0^{(n-p,n-q)}(\Omega))'$ where now the coefficients of the forms are simply continuous with compact support. Alternatively, we can think of $T$ as a differential form with distributional coefficents $T_{I,J}$: $$ T=\sum_{I,J}T_{I,J}dz^I\wedge d\bar z^J $$ If $T$ is a current of order $0$, then each $T_{I,J}$ is a Radon measure (a result of Riesz's Theorem). Now, we can define $T_f$ as the current of integration given by $$ T_f(\phi)=\int_{\Omega}f\wedge \phi$$ where $f$ is a $(p,q)$ form and $\phi$ is an $(n-p,n-q)$ form. In this specific case we can write $$ f=\sum_{I,J}f_{I,J} dz^I\wedge d\bar z^J$$ where $f_{I,J}$ are smooth functions (or continuous functions if $T_f$ is of order $0$). This is where my confusion starts. Some texts I read say that given a current $T$, there is the duality pairing given by $$ \langle T,\phi\rangle = \int_{\Omega} T\wedge \phi$$ If $T$ is a current like $T_f$, then I understand this pairing. Otherwise I don't know how to integrate a differential form WITH distributional (or measure) coefficents. If I can write $T$ as a differential forms with smooth or continuous coeffients, then this pairing makes sense. Another example is given a current $T$ of order zero and a compact set $K\subset\Omega$, we can define $$ ||T||_K=\int_{K}\sum_{I,J}|T_{I,J}|$$ Given that $T_{I,J}$ are measures, I understand the norm above. One can show that if $T\geq 0$, then there are positive constants $C_1,C_2$ such that $$ C_1||T||_K\leq \int_K T\wedge \beta^p\leq C_2||T||_K$$ where $\beta$ is the Kahler form. I want to try to prove this statement, but I don't know how to interpret the integral in the middle of the inequality above. Can all currents be written an a current of integration?

Answer 1

After talking to someone far more knowledgeable then myself, there is an interpretation when the current $T$ has measure coefficients. I summarize below. Let $I'$ and $J'$ be the increasing complements of $I$ and $J$ respectively and let $$ \phi=\sum_{I',J'}\phi_{I',J'}dz^{I'}\wedge d\bar z^{J'}$$ be a $(n-p,n-q)$ form.

1. $T_f$ is a current of integration, i.e. f is a $(p,q)$ form. In this case the distributional coefficients $T_{I,J}$ are $$ T_{I,J}(\psi)=c_{I,J}\int_{\Omega}f_{I,J}\psi d\lambda$$ where $d\lambda$ is Lebesgue measure, $\psi$ a smooth function of compact support, and $c_{I,J}$ is a constant chosen so that $$c_{I,J} dz^I\wedge d\bar z^J\wedge dz^{I'}\wedge dz^{J'}=d\lambda$$ This of course agrees with the definition of a current of integration: $$ \int_{\Omega}f\wedge \phi =\sum_{I,J}\int_{\Omega} \big(f_{I,J}\phi_{I',J'}\big)dz^I\wedge d\bar z^J\wedge dz^{I'}\wedge d\bar z^{J'}$$

2. $T$ is a current of order zero. Then it's coefficients $T_{I,J}$ are measures, so we can interpret $$ \int_{\Omega}T\wedge \phi=\sum_{I,J}\int_{\Omega} \phi_{I',J'}T_{I,J}$$

3. If the coefficients of $T$ are not measures, then $$ \int_{\Omega}T\wedge \phi$$ is strictly a notation for $T(\phi)$.