# Hard Lefschetz theorem for non-Kähler manifolds

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.

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.

• What do you mean by $T^3$? do you mean $T^3=T\wedge T\wedge T$?
– Tom
Aug 16 '21 at 2:10
• 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 $T^3=\int_X\tau^3$. Aug 16 '21 at 8:01
• Is the $\tau$ representing $T$ a non-degenerate form?
– Tom
Aug 16 '21 at 15:59
• 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. Aug 16 '21 at 21:49