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## Question on Kähler/ample cone, cone of curves…

Assume $X$ is smooth "simply connected" complex projective variety and $Y\subset X$ a smooth hyperplane section. ( $Y= X\cap H$, $H\subset \mathbb{P}^n$).

Let's $NE(X)$ be the cone of effective 1-cycles modulo numerical-equivalence. Let's $\mathfrak{K}_X$ be the Kahler cone.

I have couple of question on these cones.

1- If two curves $C$ and $C'$ are numerically equivalent in $X$, then are they the same in $H_2(X)$.

2- If the answer to previous question be yes then we can look at the image of $NE(X)$ in $H_2(X,\mathbb{R})$. Is this cone open, i.e. of maximal dimension $=\dim H_2(X)$?

3- If two divisors $D$ and $D'$ be numerically equivalent, then are they the same in $H^2(X)$?

In general when numerically equivalent $\rightarrow$ equivalence of homology classes.

4- Assume $\dim X=4$. Then by Lefschetz hyperplane theorem $\dim H^2(X,\mathbb{Z})= \dim H^2(Y,\mathbb{Z})$ and we obviousely have $\mathfrak{K}_X \subset \mathfrak{K}_Y$. Is it possible for them to be not equal? i.e. is it possible to have a line bundle on $X$ which is ample on $Y$ but not on $X$? (For this part you may assume $X$ is Fano)

I may add to this list later:) . P.M. I know there are two more good discussions on Kähler cone ... on mathoverflow. i.e. http://mathoverflow.net/questions/30926/structure-of-kahler-cone and http://mathoverflow.net/questions/27249/what-does-the-ample-cone-look-like

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Dear Mohammad, a very minor point: it is probably a little confusing to use K_X to denote the K\"ahler cone, since that is already standard notation for the canonical bundle. – Artie Prendergast-Smith Feb 16 2011 at 8:13
Yes, I'll change the notation. – Mohammad F.Tehrani Feb 16 2011 at 14:53
Mohammad, I added a concrete example for Q4 to my original answer. – Sándor Kovács Feb 21 2011 at 3:45

EDIT Added a concrete example for the answer for Q4.

By a result of Kleiman (somewhere in SGA6 and also in Lazarsfeld's book around 1.1.20) a numerically trivial line bundle has a power that's in ${\rm Pic}^\circ X$ which is trivial if $X$ is simply connected, so on such an $X$ a numerically trivial line bundle is necessarily torsion, but that corresponds to a finite étale cover, which is again trivial by the simple connectedness assumption, so there are no numerically trivial line bundles.

The long exact cohomology sequence of the exponential sequence $$0\to \mathbb Z \to \mathscr O_X \to \mathscr O_X^* \to 0$$ Shows that the kernel of the map ${\rm Pic} X\to H^2(X,\mathbb Z)$ is ${\rm Pic}^\circ X$. By the above this means that ${\rm Pic} X\to H^2(X,\mathbb Z)$ is injective and since there are no torsion line bundles, it remains injective after tensoring with $\mathbb R$.

Then the answers are:

1) Yes, if $X$ is a surface since curves are divisors. In general, without the simple connectedness assumption the answer is certainly no, but that was not what you asked.

2) Actually no. Mostly no. Many examples, perhaps the simplest one is any (smooth projective) curve with $b_2>1$ or any smooth projective surface: $NE(X)$ will land in $H^{1,1}$ but $H^{2,0}\neq 0$. But even greate\r difference is possible. Take a general K3 surface: $NE(X)$ is $1$-dimensional and $H^2$ is $22$.

3) Yes, by the first two paragraphs.

4) No. Or yes, it is possible to have a line bundle like that. Let $\pi: X\to Z$ be a flat morphism from a smooth projective $4$-fold $X$ to a smooth projective $3$-fold $Z$, for instance a $\mathbb P^1$-bundle. Let $H\subset X$ be the pull-back of an ample line bundle from $Z$. Now if $Y$ is a general hyperplane section of $X$ that does not contain any fibre of $\pi$, then the induced map $Y\to Z$ is finite and hence $H\cap Y$ is an ample divisor.

Here is a concrete example: Let $Z$ be an arbitrary smooth projective threefold and let $\mathscr E$ be a rank $2$ vector bundle on $Z$ such that there exists a surjective map $\mathscr E\to \mathscr L$ onto a line bundle on $Z$. Consider $\pi: X=\mathbb P(\mathscr E)\to Z$ and observe that after twisting by the pull-back of a sufficiently ample line bundle on $Z$ we may assume that $\mathscr O_{\mathbb P(\mathscr E)}(1)$ is very ample. The surjective morphism $\mathscr E\to \mathscr L$ produces a section of $\pi$ and let us denote the image of that with $Y$. Just for kicks, notice that $Y\simeq Z$ so it can be any smooth projective variety and $Y\cap H$ corresponds to the divisor that pulls back to $H$ so it is indeed ample on $Y$. Also notice that $Y$ is a divisor on $X$ representing $\mathscr O_{\mathbb P(\mathscr E)}(1)$, so by the above assumption it is a hyperplane section of $X$. Choosing $\mathscr E\to \mathscr L$ generally makes $Y$ general but this is actually not necessary for the construction. Since $Y$ is a section, it intersects every fiber of $\pi$ in exactly $1$ point, so it cannot contain a fibre.

3.5) (inspired by Dave's answer): An interesting related problem is the connection between algebraic and homological equivalence. Again algebraic implies homological equivalence, but by a result of Griffiths (On the periods of certain rational integrals. I, II. Ann. of Math. (2) 90 (1969), 460-495; (2) 90 1969 496–541) they are not the same and in fact the difference can be quite large as proved by Clemens (Homological equivalence, modulo algebraic equivalence, is not finitely generated. Inst. Hautes Études Sci. Publ. Math. No. 58 (1983), 19–38 (1984)).

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 About Q2: Can you say why $NE(X)$ is one dimensional for a generic projective $K3$. What if $X$ is a C.Y 3-fold? – Mohammad F.Tehrani Feb 16 2011 at 1:56 Dear S\'andor, about your answer to 1): what is "that map"? – Artie Prendergast-Smith Feb 16 2011 at 8:11 Artie, thanks for pointing that out, I mixed up things about curves and divisors. I'll try to clean this up.... – Sándor Kovács Feb 16 2011 at 16:40 The Picard number (the rank of the Picard group as an abelian group) of a general K3 or more generally CY manifold is 1. The rank of $NE(X)$ is equal the Picard number. For explicit examples consider hypersurfaces a general hypersurface $X\subset \mathbb P^n$ of degree $n+1$ is a CY manifold and by the Lefschetz hyperplane theorem (for n=3 Noether's theorem) its Picard group is isomorphic to the Picard group of the ambient space, which is $\mathbb Z$. – Sándor Kovács Feb 16 2011 at 16:42 Two more questions: 1-Consider the example of C.Y 3-fold $X$ in $\mathbb{P}^1 \times \mathbb{P}^3$ which is a double cover over $\P^3$ and $K3$ fibration over $\P^1$. In this case, can you tell me what is $NE(X)$ or $\bar{NE}(X)$ ? I can see two curves in it. One the generator of cone of a $K3$ fiber and the other one, the pull-back of a line from $K3$. ($X$ has the properties you mentioned in answering Q2) 2- In simply connected situation as before, is Kahler cone the dual of $NE(X)$ .i.e. is $\bar\mathfrak{K}_X=$ set of $w\in H^2(X)$ such that $w(C)\geq 0$ for any $C\in \bar{NE}(X)$? – Mohammad F.Tehrani Feb 16 2011 at 21:50

To augment Sándor's answer, especially for "Question 3.5": On a smooth variety $X$, for cycles of any codimension, homological equivalence always implies numerical equivalence, so you're asking when the two are the same. They certainly differ by torsion, so let's use ${\Bbb Q}$ coefficients and ask when $$Z^k/Hom^k \to Z^k/Num^k$$ is an isomorphism. This question is part of Grothendieck's "standard conjectures" and is unresolved in general; however it is always true for $k=1$, and apparently also for $k=2$, as well as for any $k$ on abelian varieties; see Fulton's Intersection Theory, 19.3. (In case the notation isn't self-explanatory, $Z^k$ is codimension $k$ cycles, $Hom^k$ is the kernel of the map to $H^{2k}$, and $Num^k$ is cycles numerically equivalent to zero.)

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This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1) I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this paper. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2) Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.

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 For Q1: Kahler cone is 2-dimensional by Lefschetz Hyperplane theorem, and cone of curves is at least 2 dimensional since I can give you two curves with different homology class and so it is two dimensional because $dim H_2(X)=2$. But I don't know what are the generators of these 2-dimensional cones. For example for Kahler cone, two generators of campactified Kahler cone are $h$ and $H-ah$ where $h$ and $H$ are pullbacks of $O(1)$ from $\P^1$ and $P^3$ respectively; but I can't find $a$. It should be negative and not less than $-1/6$ but I can't find the actual value. – Mohammad F.Tehrani Feb 17 2011 at 15:05 How do you know that the Kähler cone is $2$-dimensional? Are you saying that $X$ is an ample divisor on $\mathbb P^1\times \mathbb P^3$? In that case this is OK. But I don't see why $a$ would not be $0$? Both $h$ and $H$ are nef and then by Kleiman's criterion should be in the closure of the Kähler cone. However, they cannot be in the interior, because they are not ample. The rank is $2$ so there can be at most two classes on the boundary and hence these two classes generate the Kähler cone. – Sándor Kovács Feb 18 2011 at 5:23 I first thought $X\rightarrow \mathbb{P}^3$ is finite (2:1) cover and so is of the type you mentioned. But I was wrong, there are 64 points in $\mathbb{P}^3$ over which the fiber is whole $\mathbb{P}^1$, rather than two points. I am still looking for an example of your argument in answering fourth question. – Mohammad F.Tehrani Feb 18 2011 at 15:32 Mohammad, I added a concrete example to my original answer. – Sándor Kovács Feb 21 2011 at 3:45

There is an interesting special case when the answer to your Question 4 is positive. Assume that $X$ is a smooth fourfold, and $Y$ an ample anticanonical hypersurface (in particular then $X$ is Fano and $Y$ is Calabi-Yau). In this case, the Kahler cones of $X$ and $Y$ coincide. This is proved by Kollar in an appendix to Borcea, Homogeneous vector bundles and families of Calabi-Yau threefolds II, in: Proc. Symp. Pure Math. 52. Part II.

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 Yes you are right, I asked Kollar and he told me about that. – Mohammad F.Tehrani Feb 24 2011 at 0:08