# Can the cohomology vanish on $X$ but on no restriction to hyperplane sections?

I would need a result asserting that if, say, a locally free sheaf $\mathcal{F}$ on a projective $X$ has non-vanishing cohomology when restricted to any smooth hyperplane section, then $H^1(X, \mathcal{F}) \neq 0$, too. I fear that this is too naive, so I will be more specific in two steps, to contextualize. Also, to be fair, I should substitute "complex of sheaves" to "sheaf" (and "hypercohomology" to "cohomology").

Let $X$ be a projective complex surface, and $F^\bullet = F^1 \xrightarrow{\theta} F^2 \xrightarrow{\theta} F^3$ a 3-steps complex of locally free sheaves. Suppose that we know that for every smooth hyperplane section $i \colon C \subset X$, the hypercohomology $\mathbb{H}^1(C, i^*F^\bullet)$ of the restriction $i^*F^\bullet = i^*F^1 \xrightarrow{i^*\theta} i^*F^2$ is not zero (in my situation, I actually know that this has just 2 steps). Is it possible that $\mathbb{H}^1(X, F^\bullet) = 0$?

Some more background (explaining the, probably useless, condition on lengths of complexes). I actually have a flat bundle $\mathcal{V}$ on $X$, with a decomposition $\mathcal{V} = \bigoplus_k U_k$ and maps $\theta \colon U_k \to \mathcal{A}^{1,0}(U_{k+1})$, where $\mathcal{A}^{1,0}$ denotes $(1,0)$-forms, such that $\theta \wedge \theta = 0$, i.e., for every $k$ a complex $$F_{X,k}^\bullet = U_k \xrightarrow{\theta} \mathcal{A}^{1,0}(U_{k+1}) \xrightarrow{\theta} \mathcal{A}^{2,0}(U_{k+2}).$$ By Lefschetz hyperplane theorem, since $\pi_1(C) \twoheadrightarrow \pi_1(X)$, we get an injection $H^1(X, \mathcal{V}) \hookrightarrow H^1(C, i^*\mathcal{V})$. The above decomposition gives a decomposition $$H^1(X, \mathcal{V}) \cong \mathbb{H}^1\big(\mathcal{V} \xrightarrow{\theta}\mathcal{A^{1,0}(V)} \xrightarrow{\theta} \mathcal{A^{2,0}(V)}\big) = \bigoplus_k \mathbb{H}^1(F_{X,k}^\bullet),$$ and we actually have an analogous decomposition on $C$ (of course, with only 2 steps complexes, since on curves $\mathcal{A}^{2,0} = 0$). The natural injective map, then, sends $\mathbb{H}^1(F_{X,k}^\bullet)$ to $\mathbb{H}^1(F_{C,k}^\bullet)$, and every restriction must be injective. I would like to deduce that if the former vanishes for some $k$, then the latter does, too, for the same $k$ and at least one smooth hyperplane section $C$ (a fortiori, in my situation this will imply that it vanishes for every such $C$, but some other arguments are needed).

I apologize in advance for the long trivial-but-disguised-as-serious question, no matter how, basic algebraic geometry always escapes me...

• Unless I'm mistaken, it seems as though you could use the Lefschetz hyperplane theorem. It doesn't apply to hypercohomology, on the nose, but it seems like using the second standard spectral sequence you could obtain the result. – Alex Youcis Apr 30 '14 at 20:14
• If $X$ has dimension $1$, then $H^1$ vanishes for the restriction of $F$ to any hyperplane section, but does not vanish in general for $F$ itself. – user143488 May 1 '14 at 8:19
• @cant_log: I know, I have specified that $X$ has to have dimension 2. The point is that for higher dimension (using what I have written in the third paragraph) Lefschetz theorem gives an isomorphism, for dimension 1 it is clearly false, for dimension 2 I would need the weaker statement that there exists at least one hyperplane section which determines the cohomology being 0 or not (not much hope to have an isomorphism in general, in dimension 2, if $F$ is a constant sheaf it should be false already) – Spi May 1 '14 at 10:18
• @AlexYoucis : I fear I do not understand your idea. If I get it right, Lefschetz hyperplane theorem gives just an immersion in dimension 2, and only if $F$ is the constant sheaf $\mathbb{C}$ (or maybe for harmonic bundles, but I don't have that). I must admit I didn't work out the details of the spectral sequences, I will do that and let you know! – Spi May 1 '14 at 10:22
• By hyperplane sections, do you means those defined by a given embedding in a projective space or do you allow taking varying embeddings (therefore allowing hypersurface sections). – user143488 May 1 '14 at 13:16

## 1 Answer

I misread the question (in the opposite direction). The answer is yes, this can happen. Let $n\ge 1$ be an integer and let $F=O_X(-n)$. Let $C$ be any hyperplane section in $X$. We have $$H^1(X, O_X(-n)|_C)=H^1(C, O_X(-n)|_C)\simeq H^0(C, \omega_C(n)).$$ The last term is not zero for $n$ big enough by Riemann-Roch. And the bound on $n$ depends only on the arithmetic genus of $C$, equal to $1-\chi(O_X)+\chi(O_X(-1))$. So for $n$ big enough, $H^1(X, O_X(-n)|_C)\ne 0$ for any hyperplane section $C$ (even singular, reducible etc).

On the other hand, at least when $X$ is Cohen-Macaulay (e.g. smooth), we have $$H^1(X, O_X(-n))\simeq H^1(X, \omega_{X/k}(n))=0$$ for $n$ big enough by Serre vanishing.

• Ok, of course you are very right... I should have checked more carefully before asking! This puts a stop to my most naive hopes, but it's good to have such a definitive answer, thanks :-) – Spi May 2 '14 at 17:08