Strictly Nef Divisors but not Ample

Yen-An Chen October 19, 2016

Abstract In this talk, I will introduce some notions of positivity of line bundles, which are nefness, strict nefness, and ampleness. Moreover, I will give examples of line bundles which are strictly nef but not ample.

1 Preliminaries

1.1 Intersection Definition 1.1. Let C and D be two distinct irreducible curves on a surface S. Let x be a point of C ∩ D. We define the intersection multiplicity of C and D at x to be

mx(C ∩ D) = dimk Ox/(f, g) where f and g are generators of C and D at x, respectively. Then we define X (C.D) = mx(C ∩ D). x∈C∩D

0 Remark 1.2. If we consider the scheme-theoretical intersection C∩D, then (C.D) = h (S, OC∩D). Theorem 1.3.

(C.D) = χ(S, −C − D) − χ(S, −C) − χ(S, −D) + χ(S, OS).

Proof. Note that we have the following exact sequence.

s (t,−s) t 0 / OS(−C − D) / OS(−C) ⊕ OS(−D) / OS / OC∩D / 0 where s and t are sections of OS(C) and OS(D), respectively. Remark 1.4. By this theorem, we may define the intersection of two invertible sheaves. Also, this definition depends only on the linear equivalence class of C and D. Moreover, it defines a bilinear pairing on Pic(X). Now we want to generalized the intersection to higher dimension.

1 Theorem 1.5. Let D1,...,Dr be Cartier divisors on a projective scheme over k. The function

(m1, . . . , mr) 7→ χ(X, m1D1 + ... + mrDr)

r takes the same values on Z as a polynomial f ∈ Q[x1, . . . , xr] of total degree ≤ dim X. Lemma 1.6. For a fixed integer d, let f : Zr → Z be a map such that the map

m 7→ f(n1, . . . , ni−1, m, ni+1, . . . , nr)

r−1 is a rational polynomial of degree ≤ d for each (n1, . . . , ni−1, ni+1, . . . , nr) ∈ Z . Then f takes the same values as a rational polynomial in r indeterminates. Proof. We proceed by induction on r. When r = 1, we are done. Assume r > 1, there exist r−1 functions f0, . . . , fd : Z → Q such that

d X j f(m1, . . . , mr) = fj(m1, . . . , mr−1)mr. j=0

Now we pick distinct integers c0, . . . , cd. By induction hypothesis, we have polynomials Pi with rational coefficients such that

d X j f(m1, . . . , mr−1, ci) = fj(m1, . . . , mr−1)ci = Pi(m1, . . . , mr−1). j=0

j Note that the matrix (ci ) is invertible and its inverse has only rational entries. Thus, every fj is a linear combination of P0,...,Pr. Proof (Theorem 1.5). We proceed by induction on r first then on dim X. When r = 1 and 0 dim X = 0, χ(X,D) = h (X, OX ) for all D, we are done. When r = 1 and dim X ≥ 1, we write D ∼ E − F where E and F are effective. Then we consider the following exact sequences.

0 / OX (mD − E) / OX (mD) / OE(mD) / 0

0 / OX ((m − 1)D − F ) / OX ((m − 1)D) / OF ((m − 1)D) / 0

Thus, we have χ(X, mD) − χ(X, (m − 1)D) = χ(E, mD) − χ(F, (m − 1)D). By induction hypothesis, the right hand side is a rational polynomial in m with degree < dim X. Then χ(mD) is a rational polynomial in m with degree ≤ dim X. Note that for any divisor D0 on X, the function m 7→ χ(X,D0 + mD) is also a rational polynomial in m with degree ≤ dim X. Thus, by lemma 1.6, we have that

χ(X, m1D1 + ... + mrDr) = P (m1, . . . , mr) is a rational polynomial. Let d be its total degree. Also let c1, . . . , cr be integers such that the degree of Q(M) = P (c1m, . . . , crm) is still d. Since Q(M) = χ(X,M(c1D1 + ... + crDr)), we have d ≤ dim X by the case of r = 1.

2 Definition 1.7. Let D1,...,Dr be Cartier divisors on a projective scheme over k with r ≥ dim X. We define the intersection number (D1. ··· .Dr) to be the coefficient of m1 . . . mr in f(x1, . . . , xr)

Remark 1.8. • It only depends on the linear equivalence of Di.

• For any polynomial P (x1, . . . , xr) of total degree ≤ r, the coefficient of x1 . . . xr in P is X (−1)|I|P (−xI ). I⊂{1,...,r} Thus, this number is an integer.

• When r > dim X, then this number vanishes.

• The map (D1,...,Dr) 7→ (D1. ··· .Dr) is multi-linear.

• If Dr is effective, then (D1. ··· .Dr) = (D1|Dr . ··· .Dr−1|Dr ).

1.2 Positivity Definition 1.9. An invertible sheaf L on a noetherian scheme X is said to be ample if for every coherent sheaf F on X, there is an integer n0 > 0 depending on F such that for all n n ≥ n0, the sheaf F ⊗ L is generated by global sections. Theorem 1.10 (Nakai-Moishezon). An invertible sheaf L on X is ample if and only if L dim V .V > 0 for all positive dimensional irreducible varieties V . Definition 1.11. An invertible sheaf L on a noetherian scheme X is said to be nef (numeri- cally effective) if L dim V .V ≥ 0 for all positive dimensional irreducible varieties V . Theorem 1.12 (Kleiman). L is nef if and only if L .C ≥ 0 for all irreducible curves C. Proof. We proceed by induction on n = dim X. When X is a curve, we are done. Thus, we may assume that Lm.V ≥ 0 for all irreducible varieties V with dimension m ≤ n − 1. So it suffices to show Ln ≥ 0. Now we consider P (t) = (L + tH)n. Note that Ln ≥ 0 is equivalent to P (0) ≥ 0. Assume that P (0) < 0. Note that (Lj.Hn−j) ≥ 0 since Hn−j is represented by an effective j-cycle by ampleness of H. So the coefficient of tn−j is non-negative for j < n. It follows that P (t) has only one root for t ≥ 0, say t0.

Claim. L + tH is ample for t > t0. Indeed, we will show that (L + tH)m.V > 0 for all irreducible variety V of dimension m for t > t0. For V = X, it is just P (t). Note that P (t) > P (t0) = 0. For V ( X, we notice that L.V ≥ 0 by induction hypothesis and that Lj.Hm−j.V ≥ 0 by ampleness of H. Thus, all coefficients are non-negative and the leading coefficient Hm.V is positive. So (L+tH)m.V > 0. Now we write P (t) = Q(t) + R(t) where

Q(t) = L.(L + tH)n−1 and R(t) = tH.(L + tH)n−1

3 Note that Q(t) ≥ 0 since L is nef and L+tH is ample for t > t0. By continuity, Q(t0) ≥ 0. For R(t), we know that all coefficient of R(t) is non-negative and the leading coefficient is positive. So R(t0) > 0. Thus, we get a contradiction. Question : Is there such Kleiman’s type theorem for ampleness? No!

Definition 1.13. L is strictly nef if L.C > 0 for all irreducible curves C.

Before giving an example of a strictly nef divisor which is not ample, I need the following preparation.

1.3 Ruled Surfaces

Let C be a smooth projective curve over C and E be a vector bundle of rank 2 on C. We could consider the ruled surface π : X = P(E) → C. Let D be the divisor corresponding to OX (1). For any effective curve Y on X, we denote by m(Y ) the degree of Y over C. By linearity, we have the map m : Pic X → Z. Proposition 1.14. We have the following exact sequence.

π∗ m 0 / Pic C / Pic X / Z / 0. Proposition 1.15. For any m > 0, there is a one-to-one correspondence between

• effective curves Y on X (possibly reducible with multiple components) without any fibers as components and of degree m over C, and

• sub-line bundle M of Sm(E).

More explicitly, the correspondence is given by Y 7→ π∗OX (m − Y ) and M 7→ subscheme of X defined by the homogeneous ideal generated by M. Furthermore, we have D.Y = md − deg M.

Proof. For D.Y , we consider the exact sequences on X,

0 → OX (m − Y ) → OX (m) → OY (m) → 0

0 → OX (m − 1 − Y ) → OX (m − 1) → OY (m − 1) → 0. 1 Since OX (m − Y ) is trivial along the fibers, R π∗OX (m − Y ) = 0 by Grauert theorem. Also 1 π∗OX (m − 1 − Y ) = R π∗OX (m − 1 − Y ) = 0. Therefore, we have the following exact sequence.

m 0 → M → S (E) → π∗OY (m) → 0

m−1 0 → S (E) → π∗OY (m − 1) → 0. Claim. Let π : Y → C be a finite morphism of curves and N be an invertible sheaf on Y . Then

deg π∗N = deg π∗OY + deg N.

4 Assuming claim, we have

m deg S (E) = deg M + deg π∗OY + m(D.Y )

m−1 deg S (E) = deg π∗OY + (m − 1)(D.Y ) Note that 1 deg Sm(E) = (m2 + m)d 2 where d = deg E. Hence, we get md = deg M + D.Y . Proof (Claim). Note that, for any line bundle L on C,

∗ deg π∗(N ⊗ π L) = deg π∗N ⊗ L = deg π∗N + deg π deg L and deg N ⊗ π∗L = deg N + deg π deg L. Then we may replace N by N ⊗ π∗L. So we may assume that N is an ideal sheaf on Y defining a subscheme Z. Then we have

0 → N → OY → OZ → 0.

Since π is finite, we also have

0 → π∗N → π∗OY → π∗OZ → 0.

Then deg π∗N = deg π∗OY − length π∗OZ and length π∗OZ = length OZ = − deg N. Corollary 1.16. D2 = deg E.

Definition 1.17. A vector bundle E on a variety X is said to be stable (resp. semi-stable) if for every sub-bundle E0 ( E, we have deg E0 deg E < rank E0 rank E (resp. ≤).

2 Examples

Theorem 2.1 ([Har70]). Let C be a curve of genus g ≥ 2 over the complex numbers. Then

1. If E is stable, every symmetric power Sm(E) is semi-stable, and

2. For any r > 0 and d ∈ Z, there exists a stable bundle E of rank r and degree d such that all its symmetric powers Sm(E) are stable.

5 Proof. We will use the theorem [NS65] of Narashimhan and Seshadri which says that stable vector bundles E of rank n corresponds to irreducible unitary representations of a group π where π is generated by a1, b1, . . . , ag, bg, c with relations

−1 −1 −1 −1 a1b1a1 b1 ··· agbgag bg c = 1 cn = 1.

Note that the symmetric powers of a unitary representation are still unitary (possibly reducible), we have that if E is stable, then Sm(E) is a direct sum of the stable bundles, hence semi-stable. For the second statement, we want to construct an irreducible unitary representation of π all of whose symmetric powers are irreducible. For simplicity, we only prove for the case of rank 2. Note that if we have chosen any two unitary matrices A1,B1 ∈ U(2), then we can find further unitary matrices A2,B2,...,Ag,Bg,C(= id) satisfying the relations above. Thus, it suffices to find unitary matrices A, B such that for all m > 0, Sm(A) and Sm(B) form an irreducible pair. In other words, they have no common fixed subspace. Now let λ 0  A = 1 0 λ2 where |λi| = 1 and λ2/λ1 is not a root of unity. Then A is unitary and for any m > 0,

 m  λ1 0 0 0  0 λm−1λ 0 0  m  1 2  S (A) =  ..   0 0 . 0  m 0 0 0 λ2

m Note that all eigenvalues of S (A) are distinct because of our choice of λ1 and λ2. Thus, the only fixed subspace of Sm(A) are the subspaces spanned by subsets of the standard basis. Now we write µ µ  B = 11 12 µ21 µ22. Claim. If B is very general, then all the entries of all the matrices of Sm(B) are non-zero. Indeed, for each (i, j, m), the (i, j)-th entry of Sm(B) is given by a certain polynomial Φi,j,m(µ11, µ12, µ21, µ22), which is not identically zero. On the other hand, U(2) is a real 4- manifold in GL(2, C), which is a complex 4-manifold. Also, U(2) is not contained in any analytic hypersurface of GL(2, C). Hence, by Baire category theorem, there exist unitary matrices B = (µ) such that none of Φi,j,m(µ) is zero. Thus, Sm(A) and Sm(B) form an irreducible pair. Example 2.2. There is a complete smooth surface X and a divisor D such that D.Y > 0 for all effective curves Y , but D is not ample. Now we consider a smooth curve of genus g ≥ 2 over C. Then, by theorem 2.1, there exists a stable bundle E of rank 2 and degree 0 such that all its symmetric powers Sm(E) are stable. Let X = P(E) and D be the divisor corresponding to OX (1).

6 Claim. D is strictly nef.

If Y is a fiber, then D.Y = 1. Otherwise, if Y is an irreducible curve of degree m over C, then Y corresponds to a sub-line bundle M ⊂ Sm(E). Since Sm(E) is stable of degree 0, deg M < 0. Thus, D.Y = md − deg M > 0.

Claim. D is not ample.

Indeed, D2 = d = 0 by corollary 1.16.

References

[Har70] R. Hartshorne, Ample subvarieties of algebraic varieties.

[NS65] M. Narasimhan and C. Seshadri, Stable and unitary vector bundles on a compact riemann surface, Annals of Mathematics (1965), 540–567.

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