direct image of currents

Question

I'm studing currents from Demailly's Complex Geometry, and the author defines the direct image of a current by a $C^{\infty}$ map and also for the case of a submersion. My question is about the compatibility of the direct image of the wedge product of smooth forms, in particular if $F:X\to Y$ is a submersion and $\omega,\tau$ are smooth $(p,p)$ forms, is it true that $F_{*}(\omega\wedge\tau)=F_{*}(\omega)\wedge F_{*}(\tau)$? This is true on the locus where $F$ is not singular, but it could exist a set where $F$ could not be defined, I think for example to a birational map, for example a blow-up or something like this.

Edit: I'll try to explain better. (I'm sorry but this argument is new to me and very difficult too.) If $\omega$ and $\tau$ are smooth forms of bidegree $p$, I can always define the wedge product as smooth forms, but I can see it also as a wedge product of currents in particular (as I understand it from Demailly). So, according to Demailly's book, I can define the direct image of the current $\omega\wedge\tau$; is this correct? Now, I have no idea when the wedge product of current is defined, but let's suppose that $F_{*}(\omega)\wedge F_{*}(\tau)$ is well defined. Now considering $F_{*}(\omega)$ and $F_{*}(\tau)$ as currents, is it true that $F_{*}(\omega)\wedge F_{*}(\tau)=F_{*}(\omega\wedge\tau)$?

Answer 1

Here is an example to show that your question has a negative answer, even for a simple case as blowup in 2 dimension. Just say you have a map $f:X\rightarrow Y$ and two closed smooth forms $u$ and $v$ and assume you have $f_*(u)\wedge f_*(v)=f_*(u\wedge v).$

Not talking about how to do wedge product of two general currents, no one really knows how to do so. Just say that it must be compatible with the pushforward on cohomology level. Then you should then have $f_*[u]\wedge f_*[v]=f_*[u\wedge v]$. Here I used $[.]$ for the cohomology class of a smooth closed form.

Now let $f \colon X\rightarrow \mathbb{P}^2$ be the blowup a point, and let $E$ be the exceptional divisor. Let $u=v$ be a closed smooth form representing the cohomology class of $E$. Then $f_*(u)$ and $f_*(v)$ must have the cohomology class of $f_*(E)$ which is zero, while $f_*(u\wedge v)$ has the cohomology class of $f_*(E.E)$ which is minus of a point.

Answer 2

The pushforward $f_*\Omega$ of a smooth nondegenerate volume form $\Omega$ on $X$ with respect to the holomorphic map $f:X\to Y$ is defined as follows: From definition of pushforward of a current by duality, for any continuous function $\psi$ on $Y$, we have

$$\int_{X_{can}}\psi f_*\Omega=\int_{X}(f^*\psi)\Omega=\int_{y\in Y}\int_{f^{-1}(y)}(f^*\psi)\Omega$$ and hence on regular part of $Y$ we have

$$f_*\Omega=\int_{f^{-1}(y)}\Omega$$

Hence by this formula $f_{*}(\omega\wedge\tau)=f_{*}(\omega)\wedge f_{*}(\tau) $ is not correct in general.

Moreover, Demailly showed that

$$\omega=f^*f_*\omega+\lambda{E}$$ where $E$ is the exceptional divisor and $\lambda\geq -v(\omega,Z)$ where $v(\omega,Z)=\inf_{x\in Z}v(\omega,x)$ and $v(\omega,x)$ is the Lelong number. Hence $f_{*}(\omega\wedge\tau)=f_{*}(\omega)\wedge f_{*}(\tau) $ is not correct.

Definition of Lelong number: Let $W\subset \mathbb C^n$ be a domain, and $\Theta$ a positive current of degree $(q,q)$ on $W$. For a point $p\in W$ one defines $$\mathfrak v(\Theta,p,r)=\frac{1}{r^{2(n-q)}}\int_{|z-p|<r}\Theta(z)\wedge (dd^c|z|^2)^{n-q}$$ The Lelong number of $\Theta$ at $p$ is defined as

$$\mathfrak v(\Theta,p)=\lim_{r \to 0}\mathfrak v(\Theta,p,r)$$