# Cohomology of twisted holomorphic forms on Fano threefolds

Given a Fano threefold $X$, its index $ind(X)$ is the largest integer $r$ such that there exists a divisor $H$ such that $rH \cong -K_X$. Let $\mathcal{L}$ be the associated (ample) line bundle and define the twisted forms $\Omega^q(k) = \Omega^q \otimes \mathcal{L}^k$. When do the twisted cohomology groups $H^p(X, \Omega^q(k))$ vanish? The Kodaira-Nakano vanishing theorem states that the cohomology groups vanish for $p+q > 3.$ For simple examples such as $\mathbb{CP}^3$ and the quadric hypersurface, many more of the twisted cohomology groups vanish. What results are known? Can anything stronger be said if $X$ has a Kahler-Einstein metric?

Added: There are few scattered results in the literature. For example:

The only two threefolds with non-vanishing $H^0(X, \Omega^1_X (1))$ are the Mukai–Umemura threefold $V_{22}$ and $V_{18}$ [math.AG/0310390].

-

## 1 Answer

The Hodge groups $H^1(X,\Omega^2_X)$ and $H^2(X,\Omega^1_X)$ are usually not zero. These are the groups used to construct the Griffiths intermediate Jacobian of $X$. For many types of Fano threefolds, e.g., smooth cubic hypersurfaces in $\mathbb{C}P^4$, the Torelli theorem holds -- one can uniquely reconstruct $X$ from its polarized intermediate Jacobian (or, equivalently, polarized weight 3 Hodge structure).

$\textbf{Update}$. The OP asks for examples where $k$ is positive. Let $X$ be a smooth, cubic hypersurface in $\mathbb{P}^4$. I claim that $h^1(X,\Omega^2_X(1))$ is nonzero. Let $S$ be a general hyperplane section of $X$ so that $\mathcal{O}_X(S)$ equals $\mathcal{O}_{\mathbb{P}^4}(1)|_X$. Then I even claim that the induced map of cohomology groups, $$H^1(X,\Omega^2_X) \to H^1(X,\Omega^2_X(S)),$$ is injective. Using the long exact sequence of cohomology, to prove this, it suffices to prove that $h^0(S,\Omega^2_X(S)|_S)$ equals $0$.

Let $C\cong \mathbb{P}^1$ be a smooth conic in the smooth cubic surface $S$. Then $T_X|C$ is isomorphic to $T_C\oplus (N_{C/X})$, which is $\mathcal{O}_{\mathbb{P}^1}(2)\oplus ( \mathcal{O}_{\mathbb{P}^1}(1) )^{\oplus 2}$. Thus $\Omega_X|_C$ is isomorphic to $\Omega_C\oplus T_C^{\dagger}$, i.e., $\mathcal{O}_{\mathbb{P}^1}(-2) \oplus (\mathcal{O}_{\mathbb{P}^1}(-1))^{\oplus 2}$, where $T_C^\dagger$ denotes the annihilator of $T_C$. Taking the second exterior power, $\Omega^2_X|_C$ is isomorphic to $(\Omega_C\otimes T_C^\dagger)\oplus \bigwedge^2 T_C^\dagger$, i.e., $(\mathcal{O}_{\mathbb{P}^1}(-3))^{\oplus 2} \oplus \mathcal{O}_{\mathbb{P}^1}(-2)$. Finally, $\Omega^2_X(+1)|_C$ is isomorphic to $(\mathcal{O}_{\mathbb{P}^1}(-1))^{\oplus 2} \oplus \mathcal{O}_{\mathbb{P}^1}$. Therefore, $H^0(C,\Omega^2_X(S)|_C)$ is $1$-dimensional, with the unique global section (up to scaling) in the "direction of" $\bigwedge^2 T_C^\dagger$. One way of thinking of this is, if you contract this $2$-form with a tangent vector in $T_C$ at a point $p$ of $C$, then you get the zero $1$-form.

However, for a general point $p$ of $S$, there are $27$ different conics passing through this point with $27$ different tangent spaces that span the entire tangent space of $S$ at $p$. For a global section in $H^0(S,\Omega^2_X(S)|_S)$, its restriction to each of these $27$ conics $C$ containing $p$, the contraction of the $2$-form with $T_{C,p}$ gives zero, and the spaces $T_{C,p}$ span $T_{S,p}$. Thus this $2$-form must contract to zero with every element in $T_{S,p}$. Since $T_{S,p}$ is a codimension $1$ linear subspace of $T_{X,p}$, the only $2$-form that contracts to zero with all of $T_{S,p}$ is the zero $2$-form (there are nonzero $1$-forms that contract to zero with all of $T_{S,p}$). Therefore, every global section of $\Omega^2_X(S)|_S$ must vanish at every sufficiently general point $p$ of $S$. Since $\Omega^2_X(S)|_S$ is a locally free sheaf on an integral scheme $S$, this means that every global section is zero.

$\textbf{Second Update}$. I want to add one word about why $N_{C/X}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)\oplus \mathcal{O}_{\mathbb{P}^1}(1)$, as opposed to $\mathcal{O}_{\mathbb{P}^1}(1)\oplus \mathcal{O}_{\mathbb{P}^1}(1)$. This is equivalent to saying that the induced map $H^0(C,N_{C/X}) \to H^0(C,N_{C/X}|_{0,\infty})$ is surjective, i.e., the "evaluation map" from the parameter space of $2$-pointed conics in $X$ to $X\times X$ is submersive. By Sard's theorem / generic smoothness, it suffices to prove the map is dominant (since we are in characteristic $0$). Given general points $x_0$, $x_\infty$ of $X$, the line $L$ spanned by $x_0$ and $x_\infty$ intersects $X$ in a third point $x_1$, which is also a general point of $X$ (since $X$ is a cubic hypersurface and the points are general). Thus there are $6$ lines $M$ in $X$ containing $x_1$. For each such line, for the plan $\Pi = \text{span}(L,M)$, the intersection of $\Pi$ with $X$ equals $M\cup C$, where $C$ is a conic containing $x_0$ and $x_\infty$. Also, in all likelihood, this computation of $H^1(X,\Omega^2_X(1))$ appears somewhere in the publications of Markushevich, Tikhomirov and Iliev, since this comes up in their analysis of the Abel-Jacobi maps associated to cubic threefolds.

-
The Hodge groups are for $k=0$. Can anything be said about $k > 0$? For the quadric, the groups $H^1(X, \Omega_X^2(k))$ vanish for all $k > 1.$ –  Richard Eager May 8 '13 at 0:40