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Suppose $M$ is a complex $n$-dimensioanl compact Kähler manifold and $\omega$ a Kähler class. Suppose $\alpha\in H^{1,1}(M,\mathbb{R})$ is a nef class belonging to the boundary of the Kähler cone of $M$. If for some $1\leq k\leq n-1$ we have

$$\int_M\alpha^k\omega^{n-k}=0,$$

can we conclude that $\alpha^k=0\in H^{k,k}(M,\mathbb{R})?$

If the answer is generally no, any counterexample?

Thanks in advances!

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    $\begingroup$ This seems to be (almost directly) the statement of the hard Lefschetz theorem. $\endgroup$ – Alex Degtyarev Sep 26 '16 at 16:23
  • $\begingroup$ Ample classes are nef. I presume you want to consider classes which are in the boundary of the Kähler cone. $\endgroup$ – ACL Sep 26 '16 at 16:25
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    $\begingroup$ @GunnarÞórMagnússon For $k=1$ the answer is yes, and that is why I feel it should hold for $2\leq k<n$. But at this moment I cannot prove it.The reason why k=1 is correct: The assumption tells us that $\alpha$ is primitive. Then Teissier inequality tells us that $\int_M\alpha^2\omega^{n-2}=0$. Then the conclusion follows from the Hodge-Riemann bilinear relation. $\endgroup$ – Kevin Sep 26 '16 at 20:49
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    $\begingroup$ @GunnarÞórMagnússon: how could there be a torsion class in $H^{1,1}(M,\mathbf R)$? That is a real vector space. $\endgroup$ – Lazzaro Campeotti Sep 27 '16 at 9:51
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    $\begingroup$ @GunnarÞórMagnússon: I am only too familiar with this phenomenon. $\endgroup$ – Lazzaro Campeotti Sep 27 '16 at 13:41
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@Kevin I think the answer is 'yes'. Here is a proof: if $\alpha$ is nef, then, for every $\varepsilon >0$ the class $\alpha+\varepsilon \omega$ is Kahler, and in particular the class $(\alpha+\varepsilon\omega)^k$ contains a positive $(k,k)$-current. We let $\varepsilon$ go to $0$, and obtain in the class $\alpha^k$ a positive $(k,k)$-current $T$. Since $\int_X\alpha^k\wedge\omega^{n-k}=\int_XT\wedge \omega^{n-k}=0$, it follows that $T\wedge\omega^{n-k}=0$, hence the trace measure of $T$ is $0$, hence $T$ has to be zero.

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Just for fun, here is an answer in the purely algebraic setting.

So, suppose that $X$ is irreducible projective algebraic of dimension $n$, $\alpha=c_1(\mathcal O_X(D))$ is the class of a nef divisor $D$, and $\omega=c_1(\mathcal O_X(A))$ is the class of an ample divisor.

What we want to show is that if there exists an integer $1\le k \le n$ such that $D^k\cdot A^{n-k}=0$, then for every irreducible subvariety $Z\subseteq X$ of dimension $k$, one has $D^k\cdot Z=0$.

By induction on $n$, the case $n=1$ being clear.

If $k=n$, then there is nothing to prove. Otherwise, fix a $k$-dimensional irreducible subvariety $Z\subseteq X$ and, by taking a large multiple $mA$ of $A$, choose a divisor $H\in|mA|$ containing $Z$, say $$ H=m_1 H_1+\cdots+m_N H_N, $$ where the $H_j$'s are the reduced irreducible components of $H$, $m_j>0$, and $Z\subseteq H_1$.

Now, $$ \begin{aligned} 0 & =D^k\cdot A^{n-k}=\frac 1m\, D^k\cdot H\cdot A^{(n-1)-k} \\ & =\frac 1m\sum m_j\, D^k\cdot H_j\cdot A^{(n-1)-k} \\ & =\frac 1m\sum m_j\, D_{H_j}^k\cdot A_{H_j}^{(n-1)-k} \end{aligned} $$ where the $D_{H_j}$'s and $A_{H_j}$'s are respectively the restrictions of $D$ and $A$ to $H_j$.

Since $D_{H_j}$ is nef and $A_{H_j}$ is ample, we obtain that each $D_{H_j}^k\cdot A_{H_j}^{(n-1)-k}$ must be zero. In particular, $D_{H_1}^k\cdot A_{H_1}^{(n-1)-k}=0$.

Since $\dim H_1=n-1$, by induction hypothesis we have that $D_{H_1}^k$ intersects in zero each subvariety of $H_1$ of dimension $k$. But then, $0=D_{H_1}^k\cdot Z=D^k\cdot Z$.

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  • $\begingroup$ Thanks @divierietti! Just to have fun a bit longer: since numerical and homological equivalence are not known to coincide in general, the numerical triviality of $D^k$ seems to be everything you can obtain algebraically, is it? $\endgroup$ – ACL Oct 12 '16 at 19:32

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