# Grauert's criterion of ampleness

*Posted on October 1, 2015 by Dima*

Here’s a writeup of a proof of Grauert’s criterion for ampleness (here is the link to the original paper). Since often proving that a variety is algebraic is not far from proving that it is projective, this criterion can be useful in judging algebraicity of a variety.

We start with some observations on cohomology of invertible sheaves which are powers of the invertible sheaf associated to hyperplane section.

Recall that if \(D\) is a (Cartier) divisor on a variety \(X\) then it gives rise to a sheaf \[
{\cal O}(D) := \{ f \in k(X) \mid (f) + D \geq 0 \}
\] and a section \(s \in H^0(X, {\cal O}(D))\) such that \((s) = D\). Let \(V = H^0(X, L)\). Then a (generally speaking, partial) map \(\iota: X \dashrightarrow \mathbb{P}(V^\vee)\) is defined: \(x \mapsto (f \mapsto f(x))\). The value on the right, before projectivization, depends on trivialization, hence is only well-defined up to a constant, but since we projectivize, the map is well-defined. A line bundle (or a divisor \(D\)) is called *very ample* if \(\iota\) is a closed embedding, *ample* if some multiple of it is ample.

If \(D\) is ample, then \(H:=mD\) is a hyperplane section for some \(m > 0\). Then \(D_H^r\) is the degree of \(X\), hence positive.

**Theorem**. Let \(X=\mathbb{P}^d\)

- \(H^i(X, {\cal O}(n)) = 0\), if \(0 < i < d\) or \(i < 0\)
- \(H^i(X, {\cal O}(n)) \cong H^{n-i}(X, {\cal O}(-d-n-1))^\vee\)

A paranthesis on ampleness.

**Lemma**. Let \(X\) be a scheme covered by finitely many affine schemes \(X_i\) such that \(X_i\) is the locus of points \(x\) such that \(s_i\) generates \({\operatorname{\mathcal{O}}}_{X,x}\) for an \(s_i \in H^0(X, {\operatorname{\mathcal{L}}})\). Let \(s_{ij} \in H^0(X_i, {\operatorname{\mathcal{L}}})\) be sections such that \(s_{ij}/s_i\) is a base in \(H^0(X_i, {\operatorname{\mathcal{O}}}_X)\) for each \(i\). (Note that \(s_i\) generate \({\operatorname{\mathcal{L}}}\). ) Then the morphism \(X \to {\operatorname{Proj}}[s_i, s_{ij}]\) is an embedding.

**Theorem**. Let \({\operatorname{\mathcal{L}}}\) be a sheaf such that for any finitely generated quasi-coherent \({\operatorname{\mathbb{F}}}\) there exists an \(n_0\) such that \({\operatorname{\mathbb{F}}}\otimes {\operatorname{\mathcal{L}}}^n\) is generated by global sections for \(n \geq n_0\). Then there exists an \(m\) such that \({\operatorname{\mathcal{L}}}^m\) is very ample (defines an embedding into \(P^n\)).

*Proof* ([Liu, 5.1.34, p.169]). Let \(U\) be an affine neighbourhood of \(x\) such that \({\operatorname{\mathcal{L}}}|_U\) is free, and let \({\operatorname{\mathcal{I}}}\) be the sheaf of ideals that cuts out the complement of \(U\). There exists an \(n_0\) such that \({\operatorname{\mathcal{I}}}{\operatorname{\mathcal{L}}}^n\) is globally generated, so there is a section \(s \in H^0(X, {\operatorname{\mathcal{I}}}{\operatorname{\mathcal{L}}}^n) \subset H^0(X, {\operatorname{\mathcal{L}}}^n)\) that doesn’t vanish at \(x\). Since sectios of \({\operatorname{\mathcal{I}}}{\operatorname{\mathcal{L}}}^n\) that vanish on the complement to \(U\), \(X_s\) is contained in \(U\).

Now by compactness \(X\) is covered by finitely many affines of the form \(X_{s_i}\), \(s_i \in H^0(X, {\operatorname{\mathcal{L}}}^n)\) with \(H^0(X_{s_i}, {\operatorname{\mathcal{O}}})\) finitely generated. We can apply the previous lemma to conclude.

**Theorem**. Let \({\operatorname{\mathcal{L}}}\) be a line bundle such that for any sheaf of ideals \({\operatorname{\mathcal{I}}}\) there exists a number \(n\) such that \(H^1(X, {\operatorname{\mathcal{I}}}\otimes {\operatorname{\mathcal{L}}}^n)\) vanishes. Then \({\operatorname{\mathcal{L}}}\) is ample, i.e. \({\operatorname{\mathbb{F}}}\otimes {\operatorname{\mathcal{L}}}^m\) is globally generated for big enough \(m\).

*Proof*. ([Liu, 5.3.6, p.196])

As in the proof the previous theorem, let \(U\) be an affine neighbourhood of a point \(x\), \(I\) the sheaf of functios vanishing on the compement, and \(M\) the ideal sheaf of \(X\). Consider the sequence \[ 0 \to MI \to I \to I/MI \to 0 \] and tensor it with \({\operatorname{\mathcal{L}}}^n\) (exactness is kept, as \({\operatorname{\mathcal{L}}}^n\) is locally free and so flat) \[ 0 \to MI\otimes {\operatorname{\mathcal{L}}}^n \to I\otimes {\operatorname{\mathcal{L}}}^n \to I/MI \otimes {\operatorname{\mathcal{L}}}^n\to 0 \] and take the long exact sequence \[ \ldots H^0(X, I \otimes {\operatorname{\mathcal{L}}}^n) \to {\operatorname{\mathcal{L}}}^n \otimes k(x) \to H^1(X, MI{\operatorname{\mathcal{L}}}^n)=0 \] and so \({\operatorname{\mathcal{L}}}^n\) is globaly generated at \(x\), and we are done.

**Theorem** (Serre vanishing theorem). Let \(X\) be a compact complex manifold, and \(F\) be a sheaf. Suppose \(L\) is a line bundle that admits a hermitian metric with positive curvature (or is ample). Then there exists \(m\) such that \[
H^i(X, F \otimes L^n) =0, \textrm{ for all }i > 0, n \geq m
\]

Now to the main result.

**Theorem**. A line bundle \({\cal L}\) on a complex variety is ample if and only if for any subvariety \(Y\) of positive dimension there exists a section of \({\cal L}^{\otimes n}|_Y\) which vanishes at some point but does not vanish identically.

(references: Kleiman. Towards a numerical theory of ampleness *or* Cartier. Diviseurs amples. I have found them in this Mathoverflow discussion)

Note that, as opposed to Nakai-Moishezon or Kleiman criteria, Grauert’s criterion does not require that \(X\) be known to be projective.

*Proof*. The interesting direction is from right to left.

Proceed by induction on dimension of \(X\). When \(\dim X=1\), the statement reduces to the fact that a divisor is ample iff it is positive.

Since by assumption \({\cal L}^{\otimes n}\) admits a global section, \({\cal L}^{\otimes n} \cong {\cal O}(D)\) for some effective divisor \(D\). Wlog, assume \(n=1\). Then, again by assumption, \({\cal L}|_D\) has a non-zero section that vanishes at some point, and by induction hypothesis is ample.

Consider the short exact sequence \[ 0 \to {\cal O} \to {\cal O}(D) \to {\cal O}(D)|_D \to 0 \] Taking into account that \({\cal L} \cong {\cal O}(D)\) and tensoring with \({\cal L}\) several times we find that for all \(m\) the following sequence is exact \[ 0 \to {\cal L}^{\otimes (m-1)} \to {\cal L}^{\otimes m} \to {\cal L}^{\otimes m}|_D \to 0 \]

Note that \(H^i(X, i_*i^* {\cal L}^{\otimes m}) = H^i(D, i^*{\cal L}^{\otimes m})\) where \(i: D \to X\) is a closed embedding, because closed embeddings are affine (I guess I should say just ``acyclic’’ in complex analytic setting?). As \({\cal L}^{\otimes m}|_D\) is ample, the degree \(> 1\) cohomology of sufficiently big its tensor power vanishes by Serre vanishing and GAGA. Therefore \(\dim H^1(X, {\cal L}^{\otimes m})\) decreases as \(m\) increases, and becomes stationary after certain \(m_0\). Then in the sequence \[ \begin{array}{l} \ldots H^0(X, {\cal L}^{\otimes m}) \to H^0(X,{\cal L}^{\otimes m}|_D) \to H^1(X, {\cal L}^{\otimes (m-1)} \to \\ \to H^1(X, {\cal L}^{\otimes m}) \to H^1(X, {\cal L}^{\otimes m}|_D) \to \ldots \\ \end{array} \] we can replace three last terms by 0. Then the first map is surjective which implies that \({\cal L}^{\otimes m}\) is generated by global sections.

Here’s why. Suppose \(x \notin D\). Then the stalk of sections of \(\cal L\) at \(x\) is generated by the global section of \(\cal L\) that defines \(mD\). If \(x \in D\) then any generator of the stalk of \({\cal L}^{\otimes m}|_D\) at \(x\) is generated by a section of \({\cal L}^{\otimes m}\) on \(D\), by assumption. Because the map \(H^0(X, {\cal L}^{\otimes m}) \to H^0(X,{\cal L}^{\otimes}|_D)\) is surjective this generator can be extended to \(X\).

Let \(f: X \to \mathbb{P}^k\) be the map corresponding to \({\cal L}\). Since \({\cal L}\) is not trivial, the map is not constant. Let \(Y = f^{-1}(z_0)\) for some \(z_0 \in Im f\). Then \(\dim Y < \dim X\) and by induction, \({\cal L}_Y\) is ample. On the other hand, \({\cal L}_Y\) is trivial by construction of \(f\). Therefore \(f\) is a finite map, therefore some power of \({\cal L}\) is very ample.