Let $X$ be a compact complex manifold in Fujiki class $\mathcal C$, that is bimeromorphic to a compact Kähler manifold, let $T$ be a Kähler current of $X$, then we have the De Rham class $[T]\in H^{1,1}(X,\mathbb R)$, pick a smooth form $\tau$ in the same class as $[T]$, then does the wedge map $\tau^q\wedge :H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ induces a surjective map?

This theorem is already known for the Kähler case, for a compact Kähler manifold $X$, replace here $\tau$ by a Kähler form $\omega$, then the wedge map $\omega^q\wedge:H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ induces a surjective map. What if we generalize the Kähler case to Fujiki class $\mathcal C$? do we have a similar conclusion as stated above? Does anyone knows any reference about this problem?

Added:from Ang14, page7 theorem 0.10, we know

For a compact manifold $X$ endowed with a symplectic structure $\omega$, $X$ satisfies the hard Lefschetz conditon and $X$ satisfies the $dd^{\wedge}$-lemma (namely, every $d$-exact $d^{\wedge}$-closed form being $dd^{\wedge}$-exact) are equivalent.

see the same page for the defintion of $d^{\wedge}$. So this provides a special case of our question, that is if we further assume $\tau$ being a symplectic form and the $dd^{\wedge}$-lemma is satisfied, the map $\tau^q\wedge :H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ is a surjective map. But our original problem remained open.


I think that the answer to your question is no.

In order to construct a counter-example, the idea is the following: let $T$ be a Kahler current, and $E$ a $d$-closed positive current on a compact complex $3$-fold $X$ such that $T^3>0$ and $E^3<0$. Then, for $t>0$, the current $S_t:=T+tE$ is a Kahler current. Under the above conditions, you can find $t>0$ such that $S_t^3=0$. Indeed, $S_t^3=T^3+3tT^2E+3t^2TE^2+t^3E^3$ and for $t=0$, $S_0^3=T^3>0$ while for $t\to \infty$, $S_t^3<0$, so there is a $t>0$ so that $S_t^3=0$.

Now, for this $t$, denote by $\tau_t$ a smooth representative of the class of $S_t$. Then $\tau_t^3=0$. Now for instance, for $q=n=3$, the map $\tau_t^3\wedge:H^0(X,\Omega^0)\to H^3(X,\Omega^3)$ is the zero map, while $H^0(X,\Omega^0)$ and $H^3(X, \Omega ^3)$ are $1$-dimensional.

In order to fulfill these conditions, you can take $Y$ to be Hironaka's example, it is a modification of $CP^3$, so there is a map $p:Y\to CP^3$, and denote by $\omega$ the FS metric on $CP^3$

Then the $X$ mentioned above is the blow up of $Y$ at a point $x$, denote by $\pi:X\to Y$ the blow-up. If $E$ is the exceptional divisor, then $[E]$ is a positive current, and $E^3=-1<0$, and $T$ is $\pi^*p^*\omega+\varepsilon S$, where $S$ is some Kahler current on $X$ ans $\varepsilon$ is such that $T^3>0$. Then, as mentiond above, you can find $t>0$ such that $T+t[E]$ has zero self-intersection.

  • $\begingroup$ What do you mean by $T^3$? do you mean $T^3=T\wedge T\wedge T$? $\endgroup$
    – Tom
    Aug 16 '21 at 2:10
  • $\begingroup$ No, $T\wedge T\wedge T$ cannot be defined, by $T^3$ I mean its self intersection, take $\tau$ a smooth representative of $T$, and then $T^3=\int_X\tau^3$. $\endgroup$
    – user48958
    Aug 16 '21 at 8:01
  • $\begingroup$ Is the $\tau$ representing $T$ a non-degenerate form? $\endgroup$
    – Tom
    Aug 16 '21 at 15:59
  • $\begingroup$ Of course not. Why should $\tau$, which is chosen arbitrarily, be non-degenerate? You add a boundary to $\tau$, and all the properties that $\tau$ had are all gone. $\endgroup$
    – user48958
    Aug 16 '21 at 21:49

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