Week 6 The Kleiman Criterion, the Mori Cone Theorem and the MMP for surfaces RR on a nonsingular projective surface For curves, the important invariant is the genus g(C). It has several interpretations: -- in topology, C is a sphere with g handles -- the number in RR l(D) >= 1-g + deg D, with equality if deg D >> 0 -- h^1(OC) = g -- h^0(KC) = g -- deg KC = 2 - 2*g -- C deforms in a moduli space of complex dimension 3*g-3 Surfaces are more complicated. I give a long-winded digression on chi(OS) and its interpretation below. Once chi(OS) is out of the way, RR theorem for a surface says chi(OS(D)) = h^0(D) - h^1(D) + h^2(D) = chi(OS) + 1/2*(D^2-D*KS). Serre duality says h^i(D) = h^{n-i}(KS-D). Assume S is a nonsingular projective surface. Choose an embedding S in PP^N, and write H for the hyperplane section. Sometimes work with n*H for n >> 0. Note chi(OS(D)) is an inhomogeneous quadratic function of D. This has coarse consequences: ==== Corollary 1: if D^2 > 0 then either h^0(nD) or h^0(-nD) grows like n^2/2 * D^2. The two cases are distinguished by D.H > 0, or D.H < 0. To prove the corollary: chi(nD) = n^2*1/2*D^2 + l.o.t. Claim. If h^2(n*D) = h^0(KS-nD) grows like n^2, so does h^0(-nD). This is a coarse argument that we use repeatedly. Write the divisor KS \lineq A - B with A,B both effective divisors. If h^0(A-B-nD) grows quadratically then so does h^0(A-nD). Now A is a sum of effective curves. Consider the restriction to A 0 -> OS(-nD) -> OS(A-nD) -> OA(A-nD) -> 0. Clearly H^0(OA(A-nD)) grows at most linearly. If you want to analyse this further, let A = sum Ai with irreducible Ai. Then restrict to each Ai inductively (this is called "dévissage"). There are finitely many steps, and at each step OAi(A'-nD) is a given irreducible curve, with given A' and the only varying piece is -nD, that grows at most linearly because it is RR space on an irreducible curve. QED Corollary 2 (Algebraic index theorem) (I) If H is an ample divisor then any divisor D with H.D = 0 has D^2 <= 0. (II) Moreover, H.D = D^2 = 0 implies that D \numeq 0. The next clause assumes the theorem that the Néron-Severi group NS(X) (divisors modulo algebraic equivalence) is finitely generated. (III) On Div / numeq, the intersection pairing D1.D2 has index (1, -(rho-1)). In other words, if I diagonalise it by the Gram--Schmidt process, it is given in some basis by diag(1,-1,..-1). (I) follows directly from Corollary 1. If D^2 > 0 then we have seen that H.D > 0 or H.D < 0. (II) Suppose that H.D = D^2 = 0 but there is some other divisor E with D.E <> 0. Then playing around gives a contradiction. [Some E' = E+a*H is orthogonal to H but has the same intersection with D. Now with D.E' <> 0 implies that (D+b*E')^2 > 0 for some very small b of the right sign.] QED Remark. We can treat intersection numbers D1.D2 as the symmetric bilinear form obtained by polarising the inhomogeneous quadratic form given by the RR formula: D1.D2 = chi(D1+D2) - chi(D1) - chi(D2) + chi(OS). The above corollaries relate directly to this polarisation. Kleiman's paper does the same kind of thing with the Snapper polynomial: chi(n1D1+.. ndDd) is a polynomial of degree d = dim. ===== Lecture 2 The story so far: D nef and H ample implies that D + ep*H is an ample QQ-divisor for every ep in QQ, ep > 0. Simple to state without QQ-divisors as N*D + H ample for every N > 0. This only used the Nakai-Moishezon criterion for ample: D^2 > 0 and D.C > 0 for every C. (This follows by Serre vanishing. RTP: for every coherent sheaf F on S, there is a N such that H^i(F tensor OS(ND)) = 0 for i = 1 and 2. In this comes from a devissage of F: F can be written as an extension of associated submodules (supported on a single irreducible component). The devissage consists of saying that for each associated submodules Fj, there exists an n such that H^i(Fj tensor OX(-nD)) = 0 for i = 1,2. Then pass to n big enough to do that for all j at the sdame time.) Now the Kleiman cone: write NE S in N_1(S)_RR, which is finite-dim RR^rho. Give N_1(S)_RR any norm (they are all uniformly equivalent). Define NEbar(S) to be the real closure of NE(S). Theorem (Kleiman criterion) TFAE 1. D ample 2. D.C > ep*||C|| for every C in NE(S) 3. D.C > 0 for every C in NE(S) \ 0. 4. The section D.z = 1 intersect NEbar(S) is compact. Then D ample ==> D.C > ep*||C|| for every C in NE(S) for some ep > 0. This implies in particular D > 0 on NEbar \ 0. In fact since all norms are equivalent, for the given D we can take the norm to be diag (1,-1,..-1) with D^2 = 1 and the remaining basis elements orthogonal to D. For C in NE(S) write C = A + B with A = al*D and B.D = 0. Then D.C = al, and ||C||^2 = al^2 + B^2 > al^2. Therefore D.C > ||C||^2. Changing norm introduces an ep > 0 (this is what uniformly equivalent means). Conversely D > 0 on NEbar \ 0 gives that the intersection D.z = 1 intersect NEbar is compact. In fact, consider the projective space PP^{\rho-1} = PP(N_1(S)_RR). D.z is a linear form on N_1(S)_RR, so D.z intersect NEbar maps homeomorphically to a closed subspace of a compact space. D is positive, so is bounded below. Suppose that H is some ample. Then H.z = 1 is a compact slice of NEbar(S), and D is positive on it. Then D.z >= ep*(H.z) for every z in NEbar(S) Therefore D-ep*H is nef, and D = (D-ep*H) + ep*H is ample by above result. ==== Lecture 3. The Mori cone So far, we have only used the quadratic leading term in RR chi(OS(D)) = chi(OS) + 1/2*(D^2-KS.D). For the Mori cone theorem, we need also the linear term -KS.D. Approaching NEbar from the Kz < 0 side has a particular flavour to it. View H + la*K as defining a pencil of hyperplane sections of NEbar. When la = 0, the Kleiman criterion means that H.z = 0 only for z=0. Assume from now on that K is not nef. Then K.z < 0 for some z, so that as la increases, the hyperplane section must cut into NEbar. Define the nef threshold t = sup { la in RR | H + la*K is nef }. It exists by the completeness of the reals. The point about this is that just before t, la < t so that H+la*K is ample. Just after t, la > t so that H+la*K is not nef, so there exists an effective 1-cycle z in NE(S) with (H+la*K).z < 0. Lemma (Rationality lemma) The threshold t is rational, with denominator < 3. The idea is to take the round-down s = [t] (or integral part). If t is not an integer, s < t so that H + s*K is ample, and at the same time H + (s+1)*K is not nef. However, this divisor is of the form (H + s*K) + K, so is ample plus the canonical class. In characteristic zero, Kodaira vanishing would give H^i(H + (s+1)*K) = 0 for all i > 0, so chi(OS(H + (s+1)*K)) is given by the RR theorem, that is, H^0(OS(H + (s+1)*K) = chi(OS(H + (s+1)*K) = chi(OS) + 1/2 (H+(s+1)*K)*(H+s*K) (*) = chi(OS) + 1/2 (H+s*K)^2 + 1/2 (H+s*K)*K. In fact, we don't need Kodaira vanishing. H^2 = 0 is true by the same argument used in the proof of the index theorem, so that H^0 > RHS of (*). We can eventually adjust this argument to give H (a multiple of the original) and an effective divisor D \lineq H + (s+1)*K. Then D is effective so D = sum ni*Ga_i, but not nef. So one of its components has D.Ga < 0. Then the threshold value t is exactly when (H + tK).Ga_i = 0 for some Ga_i. In other words t = min H.Ga_i/(-K.Ga_i) min taken over Ga_i with K.Ga_i < 0. This implies t is rational. In other words the threshold value t is not just when the hyperplane (H + tK)^perp touches NEbar(S), it is when (H+tK)Ga = 0 for some curve. This is Mori's extremal curve. More formally, assume that t is irrational. We apply the same argument to multiples nH and s_n = [n*t]. On the RHS of (*) we can't use anything about chi(OS), but (H+s*K)^2 >= 0. If (H+t*K)^2 > 0 for n >> 0 the (H+s*K)^2 in the second summand follows it closely, and grows quadratically. Failing that (H+t*K)^2 = 0 and H+s*K is ample. If H+t*K is not \numeq 0 then (H+t*K)*K < 0 The fractional part {n*t} = n*t - [n*t] runs through real values in (0,1) that are uniformly distributed, so for some n we can get it close to 0 or to 1. If t is rational with denominator >= 4, the same argument proves {n*t} takes values <= 1/4 or >= 1/4 infinitely often, which is just as good for current purposes. At that point, if D^2 > 0 we get KGa <0 and Ga^2 < 0, so it is a -1-curve D0 = H+t*K the threshold value. Then when we pass to mH and do the same argument, the threshold is mH+mt*K. Then [mt] is n < mt < n+1. Write mt = n+al. H^0(mH + (n+1)K) >= RR formula = chi(OS) + 1/2*(m^2 D0^2 + m(1-2*al) D0*K - al(1-2*a)K^2) If D0^2 > 0 the first term goes to infty quadratically If D0^2 = 0 but D0 not numeq 0 then we must have D0*K < 0. We can choose m large with (1-2*al) > c > 0, so RHS -> infty If D0 numeq 0 then KS numeq -(1/t)H is ample. So t is rational in any case. In cases: D0^2 > 0 D_0L=0, L is a -1-curve D0^2 = 0 but D0 not numeq 0. H+tK is not effective for any t > t0. If 2t not integral then again we can choose m >> 0 with (1-2*al) < c < 0 . ===== Discussion of chi(OS) and Hodge symmetry chi(OS) = h^0(OS) - h^1(OS) + h^2(OS) = 1 - q + pg In good cases pg = geometric genus, q = irregularity. Serre duality gives h^2(OS) = h^0(KS) and h^1(OS) = h^1(KS). There is a small ambiguity here that we need to navigate. A main case is nonsingular projective surfaces over a field characteristic zero (or over CC). The complex vector space H^n(S,CC) (= H^n(S,QQ) x CC) only depends on the topology of S as a 4-dimensional real manifold. In this case the coherent cohomology is controlled by Hodge theory, with Hodge symmetry: write H^{p,q}(S) = H^q(Om^pS). Then H^n(S,CC) = directsum_{p+q=n} H^{p,q}, and H^{q,p} = complex conjugate of H^{p,q}, so in particular h^{p,q} = h^{q,p}. When Hodge symmetry holds, (1) h^1(OS) = h^0(Om^1) = q and (2) pq = h^2(OS) = h^0(Om^2S) (1) says first cohomology of OS has dimension equal to h^0(Om^1) the dimension of the space of global regular 1-forms. This then equals the dimension of the Picard scheme and the Albanese variety. (2) say second cohomology of OS has dimension equal to h^0(Om^2) = h^0(KS) the dimension of the space of canonical forms or global regular 2-forms. In this case, the table of h^{p,q} "Hodge diamond" is h^{0,2} = pg h^{1,1} = q h^{2,2} = 1 h^{0,1} = q h^{1,1} h^{2,1} = q h^{0,0} = 1 h^{1,0} = q h^{2,0} = pg depending on the 3 numbers pg, q, and h^{1,1}. The topological Euler number is e(S) = 2 + 4q + 2pg + h^{1,1} and the coherent chi(OS) = 1 - q + pg. On the other hand, we can construct nonsingular projective surfaces in characteristic p for which Hodge symmetry fails, and with h^1(OS) > h^0(Om^1) and h^2(OS) > h^0(Om^2). I will give examples later if time permits. In these cases it is not safe to say "irregularity" or "geometric genus" without qualification. Max Noether's formula chi(OS) = 1/12*(c1^2 + c2) = 1/12*(KS^2 + e(S)) gives an interpretation of chi(OS) in terms of Chern numbers.