# Hypersurfaces without variable cohomology

Let $X$ be a smooth projective variety over $\mathbb C$ of dimension $n+1$. If $Y$ is a smooth very ample hypersurface, we know that except $H^n(X;\mathbb Q)\rightarrow H^n(Y;\mathbb Q)$, the restriction of cohomology (rational coefficients) at other degrees are surjective. My question is the following.

What are the restrictions we can put on $X$, so that there exists a smooth very ample hypersurface $Y$ such that the restriction $H^*(X;\mathbb Q)\rightarrow H^*(Y;\mathbb Q)$ is surjective on all degrees. For example, I would love to know whether it is true if $X$ is a smooth projective toric variety (or with more restrictions). We can replace very ample by ample if it's better. Boundary divisors(torus invariant divisors) satisfies the cohomology requirements, but they might not even be ample. Other possible conditions for $X$ are welcomed as well.

(Should this question be made into community wiki?)

p.s.

My motivation is that in my project, I want to find an ample hypersurface all of whose cohomology classes can be lifted to the ambient space. But uniqueness is not needed. I realized such a hypersurface may not be found for just any variety (e.g., hypersurface in $\mathbb P^n$ if the degree is sufficiently large). So I want to put a reasonable condition on $X$. I have tried to decompose $H^n(Y;\mathbb Q)$ into $H^n(Y;\mathbb Q)=H^n_{fix}(Y;\mathbb Q)\oplus H^n_{var}(Y;\mathbb Q)$ where $H^n_{fix}(Y;\mathbb Q)$ is the image of the restriction from $X$, and $H^n_{var}(Y;\mathbb Q)$ is generated by the vanishing cycles in a Lefschetz pencil. I kind of feel that requiring the monodromy to be trivial does not simplify the problem. But I also realized that if the monodromy $H^n_{var}(Y;\mathbb Q)$ is "very big", I will be equally happy in my project. But what I want for "very big" is quite a messy condition. Any discussion along this line is also welcomed in the comment because there are definitely nice results about monodromy that I'm not aware about.

• What do you mean by "boundary divisors"? For the smooth projective toric variety $X=\mathbb{P}^1\times \mathbb{P}^2$ with the standard action of $\mathbb{G}_m\times (\mathbb{G}_m)^2$, one of the irreducible components of the complement of the open orbit is $Y=\{0\}\times \mathbb{P}^2$. The restriction map $H^2(X;\mathbb{Q})\to H^2(Y;\mathbb{Q})$ is not injective. – Jason Starr Jan 28 '16 at 10:25
• @JasonStarr Right. But I only need the restriction to be surjective. I think in your example it's still ok. Probably it's better for me to change boundary divisors into torus invariant divisors. – Honglu Jan 28 '16 at 16:01
• This will be pretty rare--the morphism in question is a map of Hodge structures, so it is surjective (hence an isomorphism by Lefschetz) iff it is an isomorphism on every Hodge component. The obstruction to surjectivity on $H^n(X, \mathcal{O}_X)\to H^n(Y, \mathcal{O}_Y)$ is the kernel of the map $H^{n+1}(X, \mathcal{O}_X(-Y))\to H^{n+1}(X, \mathcal{O}_X)$. But for $Y$ sufficiently ample, this kernel will be huge. – Daniel Litt Jan 28 '16 at 16:27

## 1 Answer

Suppose $X$ is a regular surface, i.e. $H^1(X)=0$, e.g. a toric surface. Then $Y$ is a smooth rational curve. So $X$ has a very ample smooth rational curve. This immediately requires $X$ to be rational by taking a pencil of smooth rational curves. Does every rational surface have such a curve? It's ok for projective space (lines) and Hirzebruch surfaces (high degree sections).

Not for $\mathbb P^2$ blown up at $k$ points for $k \geq 2$. Let $D = n H - \sum_{i=1}^k m_i E_i$. Then $D \cdot D = n^2 - \sum_{i=1}^k m_i^2$ and $K = -3 H + \sum_{i=1}^k E_i$ so $D \cdot K = -3n + \sum_{i=1}^k m_i$

By adjunction we have

$$-2= D \cdot D + D \cdot K = n^2 - \sum_{i=1}^k m_i^2 -3n + \sum_{i=1}^k m_i$$

$$= \left(n- \sum_{i=1}^k m_i\right) \left(n + \sum_{i=1}^k m_i\right) +\sum_{1 \leq i< j \leq k} m_i m_j -3n + \sum_{i=1}^k m_i$$

$$= \left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j -2$$

Thus

$$\left(n- \sum_{i=1}^k m_i-1\right) \left(n + \sum_{i=1}^k m_i-2\right) +\sum_{1 \leq i< j \leq k} m_i m_j=0$$

But in fact it is positive:

$\left(n- \sum_{i=1}^k m_i-1\right)>0$ by ampleness because it is the intersection number with the line through the points.

$m_i\geq 1$ by ampleness because it's the intersection number with $E_i$, so $\left(n + \sum_{i=1}^k m_i-2\right) >0$ and $m_im_j>0$

This is a contradiction so this is impossible. When $k=2$ this variety is toric, showing that being toric is not a sufficient condition.

• Thanks. So the condition in the question is really strong even for surfaces. I'm also curious about whether it's true for rational surface. But it looks subtle even for blow-ups of $\mathbb P^2$. – Honglu Jan 28 '16 at 20:15
• @Honglu My suspicion is yes because the equation $D \cdot D + D \cdot K = -2$ becomes easier to solve when the Picard rank gets larger. But I don't know how to check which solutions are very ample. – Will Sawin Jan 28 '16 at 20:22
• I think it's not true for blow-ups of $\mathbb P^2$. Let's blow up n points $p_1,…,p_n$ that are collinear in $X:=\mathbb P^2$ to get $\tilde X$. Let $E_1,\dotsc,E_n$ be the corresponding exceptional curves. Let $C\subset \tilde X$ be a smooth very ample rational curve. Its image in $X$ (denoted by $\overline C$) is a rational curve passing through $\{pi\}$. Each component near $\{p_i\}$ has multiplicity at most $2$. Since $C$ is very ample, $C$ must intersect each $E_i$ (say their intersection number is $e_i$). – Honglu Jan 28 '16 at 22:41
• Notice the intersection of $\overline C$ and the line is at least $\sum e_i$. So the arithmetic genus of $\overline C$ is at least $(\sum e_i-1)(\sum e_i-2)/2$. But the geometric genus of it is $0$. However, the difference between arithmetic genus and geometric genus can be calculated at the singularities $\{p_i\}$. Each of them should contribute a multiple of $e_i$ (forgot the formula). The point is, the difference between arithmetic/geometric genus should be linear in terms of $\sum e_i$. If $n$ is sufficiently large, it may not be possible. @Will Sawin – Honglu Jan 28 '16 at 22:41
• @Honglu I don't understand your multiplicity bound - I think $y^n- x^{n+1}$ becomes smooth when you blow up - but I see a way to do it just with intersection theory. Think it works already for blowing up 2 points, which I guess would be toric. I'll write that up soon. – Will Sawin Jan 28 '16 at 23:11