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Strictly Nef Divisors but Not Ample 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): x2C\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 2 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) 2 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)jIjP (−xI ): I⊂{1;:::;rg 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) = (D1jDr : ··· :Dr−1jDr ). 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.
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