# Does Lefschetz-type theorems imply ampleness?

Let $X$ be a smooth $n$-dimensional complex projective variety and $D \subset X$ a smooth (effective) divisor. Consider the following properties:

1. $D$ is ample.
2. (Positivity) For any $k$-dimensional subvariety $Z \subset X$ the intersection $D^{n-k} \cdot Z$ is strictly positive.
3. (Hard Lefschetz) For any $k$, $1 \le k \le n$ the map $$\cup [D]^k \colon H^{n-k}(X, \mathbb{C}) \to H^{n+k}(X, \mathbb{C})$$ is an isomorphism.

4. (Weak Lefschetz / Lefschetz Hyperplane) If $j \colon D \hookrightarrow X$ is the tautological embedding, the induced maps $$j_* \colon H_k(D, \mathbb{Z}) \to H_k(X, \mathbb{Z})$$ are isomorphisms for $k<n-1$ and surjective for $k=n-1$.

5. (Zariski-Lefschetz / Homotopic Lefschetz Hyperplane) The induced maps $$j_* \colon \pi_k(D) \to \pi_k(X)$$ are isomorphisms for $k<n-1$ and surjective for $k=n-1$.

Of course, the first property implies all the rest (moreover, I guess this is the reason why ample divisors play such important role in algebraic geometry). It is also well-known, that 2) imples 1) (Kleiman criterion).So my questions are the following:

• Does 3) implies ampleness?
• Does 4) implies ampleness?
• It seems that Hard Lefschetz implies Weak one, however it is not clear to me, if it is equivalent. I am also not really sure, if 3) holds when one replaces cohomology with coefficients in $\mathbb{C}$ by integral cohomology.
• It is absolutely non-clear, if Hard Lefschetz implies 5. It would be very interesting to know if there are examples, when 4) holds, but 5) does not.

(Few remarks: It would be also interesting to know the answers to this questions for arbitrary ground-field, as far as they make sense.

For me, the most intriguing question from the list is "Does weak Lefschetz implies ampleness?". Certainly, it needs some additional discussion. First of all, there are examples due to B. Totaro when two smooth divisors from the same cohomology class have different topology, which can not happen with smooth divisors. Secondly, I know few boring counter-examples such as $C \times \{0\} \subset C \times \mathbb{P}^1$, where $C$ is any non-rational curve. May be one needs to ask $n \ge 3$ and/or simply-connectedness of $X$ plus some extra conditions to escape Totaro's-type issues.)

• It is not true that (3) implies ampleness. For example, let $X = \operatorname{Bl}_p \mathbb P^2$, and let $D = E$ be the exceptional divisor. Since $H^1(X,\mathbb C) = 0$, the only condition is that $D^2 \neq 0$. But $D^2 = -1$, showing both that (3) holds and that $D$ is not ample. – R. van Dobben de Bruyn Apr 7 '18 at 17:18
• It is a theorem of de Cataldo and Migliorini that $D$ satisfies hard Lefschetz if and only if it is semi-ample and the corresponding morphism is semi-small (in the sense of perverse sheaves). In fact, this is one of their main insights in their new proof of the Decomposition Theorem for semi-small maps. I think they call such classes "lef". They also show that lef classes satisfy weak Lefschetz, thus neither 3) nor 4) imply ampleness. – Geordie Williamson Apr 7 '18 at 18:54
• One more stupid comment: 3) certainly doesn't hold for integral coefficients in general (e.g. $\mathcal{O}(n)$ for $n \ge 2$ on projective spaces). Deciding when hard Lefschetz holds over $\mathbb{Z}$ (or with some primes inverted) seems to be very hard. – Geordie Williamson Apr 7 '18 at 18:57
• 2) implies 1) is the Nakai-Moishezon criterion, not the Kleiman criterion. – ulrich Apr 8 '18 at 5:35
• Stupid comment: in Hard Lefschetz, you can replace D with -D... – Piotr Achinger Apr 8 '18 at 8:20