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.
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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.
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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.