Problems and Progress on Points in P2
Speaker: B. Harbourne
Affiliation: Department of Mathematics and Statistics,
University of Nebraska-Lincoln, Lincoln, NE 68588-0323, USA
Zero-Dimensional Schemes and Applications
This abstract has been revised and updated to include
results known to the author as of March 14, 2000.
- R = k[P2] = homogeneous
coordinate ring of P2 = polynomial
ring in three indeterminates over algebraically closed field k
- p1, ... , pn are general points
- m = (m1, ... , mn)
is a vector of nonnegative integer multiplicities
- Ipi is the ideal in R
generated by all forms vanishing at pi
- Z(m) denotes the fat points subscheme
m1p1 + ... + mnpn
- Z(n,m) denotes the fat points subscheme
m(p1 + ... +pn)
- The defining ideal for Z(m)
is the intersection IZ(m)
in R of the ideals
Ip1m1, ... ,
is the mith power of the ideal
- D(x) = the greatest integer less than or equal to x (i.e., round x Down)
- U(x) = the least integer greater than or equal to x (i.e., round x Up)
- a(m) = the least degree t such that
the homogeneous component
is nontrivial in degree t
- Alternatively, if m and n are positive integers, then a(n,m) will denote
a(m), where m = (m, ... , m)
has n entries
is the Hilbert function of IZ(m); i.e.,
- 0 -> F1(Z(m))
-> F0(Z(m)) ->
IZ(m) -> 0 denotes a
minimal free graded resolution of
IZ(m) over R
- [P1] Determine hIZ(m).
- In what degrees t is (IZ(m))t=0?
- In what degrees t does Z(m) impose independent conditions
on forms of degree t?
- [P2] Determine F0(Z(m)) as a graded R-module.
- What would be a reasonable conjecture for the structure of
F0(Z(m)) as a graded R-module?
For t sufficiently small, we know hIZ(m)(t) = 0,
while for t sufficiently large we know
[(t+2)(t+1) - m1(m1+1) - ... - mn(mn+1)]/2.
max(0, [(t+2)(t+1) - m1(m1+1) - ... - mn(mn+1)]/2)
as the ``expected'' value of hIZ(m)(t),
although there are cases for which this expected value is known
not to be the actual value; when
[(t+2)(t+1) - m1(m1+1) - ... - mn(mn+1)]/2
we say that Z(m) imposes independent
conditions on forms of degree t.
Short of determining
hIZ(m)(t) for every t, it is
also of interest
to bound a(m) below and to bound tind(m) above
(where a(m) is defined above and tind(m) is the least degree t such that
Z(m) imposes independent
conditions on forms of degree t). As with a(m) and a(n,m),
we will also write
tind(n,m) for tind(m)
when m1 = ... = mn.
For n <= 9, hIZ(m) is known
(see M. Nagata, On rational surfaces, II. Mem. Coll. Sci. Kyoto (A) 33 (1960), 271-293,
or, for a more modern and more general
approach, see B. Harbourne, Anticanonical Rational Surfaces, Trans. Amer.
Math. Soc. 349 (1997), 1191-1208).
Thus interest has centered on the case that n > 9.
[P1] The Segre-Harbourne-Gimigliano-Hirschowitz
Several people have put forward conjectures regarding the value
These conjectures all turn out to be equivalent, and allow one
to explicitly compute all values of hIZ(m)
(the equivalence of Segre's conjecture to the others was only recently recognized,
by Ciliberto, Clemens and Miranda). Here are some additional references:
- B. Segre: Alcune questioni su insiemi finiti di punti in Geometria Algebrica,
Atti del Convegno Internaz. di Geom. Alg., Torino (1961).
- B. Harbourne: The geometry of rational surfaces and Hilbert
functions of points in the plane,
Can. Math. Soc. Conf. Proc. 6 (1986), 95-111.
- A. Gimigliano: On linear systems of plane curves. Ph.D. thesis,
Queen's University (1987).
- A. Hirschowitz, Une conjecture pour la cohomologie
des diviseurs sur les surfaces rationelles génériques,
Journ. Reine Angew. Math. 397 (1989), 208-213.
- B. Harbourne, Points in Good Position in
Zero-dimensional schemes, Proceedings of the
International Conference held in Ravello, Italy, June 8-13, 1992,
De Gruyter, 1994.
- R. Miranda, Linear Systems of Plane Curves, Notices Amer. Math. Soc.,
Rather than discuss the general form of the SHGH conjecture,
we will consider a special case. Say that m
and by extension Z(m) is quasi-uniform
if n > 9 and m1 = ... = m9 >=
m10 >= ... >= mn >= 0,
and that Z(m) is uniform
if n > 9 and m1 = ... = mn. Then the
conjectures above imply:
(Quasi-)Uniform SHGH Conjecture: If
Z(m) is (quasi-)uniform, then
has its expected value in every degree t >= 0.
Some results are known:
- The Uniform SHGH Conjecture holds when:
- (in characteristic 0) n is a square not divisible
by a prime bigger than 5. See L. Evain's paper
``La fonction de Hilbert de la réunion
de 4h gros points génériques
de P2 de même multiplicité'',
J. Alg. Geom. 8 (1999), 787--796.
- (in characteristic 0) m <= 12: see
C. Ciliberto/R. Miranda, Linear Systems of plane curves with base points
of equal multiplicity, preprint (1998), to appear, Trans. Amer. Math. Soc.
- n is sufficiently large compared with m: see
and A. Hirschowitz,
"An Asympotic Vanishing Theorem for Generic Unions of Multiple Points", preprint (1997).
- Various results when no multiplicity is more than 4
have been obtained for the full SHGH Conjecture.
Here is a web form for computing actual and expected values
of the Hilbert function.
There is also a lot of work on bounding a(n,m).
Easy Fact: If n >= 0 and m >= 0, then
a(n,m) <= D([1/4 + n(m2 + m)]1/2 - 1/2).
(This follows from Riemann-Roch.
Moreover, for n > 9, the SHGH Conjecture implies this is an equality.)
Nagata's Conjecture (M. Nagata, On the 14th
Problem of Hilbert, Amer. J. Math. 33 (1959), 766-772): If n > 9 with
m > 0, then a(n,m) > mn1/2.
Nagata proved this Conjecture for any square n > 9. The strict
inequality is a bit tricky; an easy argument specializing the n points to
a curve of degree n1/2 gives a(n,m) >= mn1/2
when n is a square. More generally, specializing to a curve of
degree D(n1/2) or, resp., degree U(n1/2)
gives the bounds:
The best general bound (applicable in all characteristics,
almost always better and never worse than mD(n1/2)
seems to be (see B. Harbourne, On Nagata's Conjecture,
- a(n,m) >= mD(n1/2)
- a(n,m) >= mn/U(n1/2)
In general, it does not seem easy to tell which d and r give
the best result, but taking d = D(n1/2) and
r = U(D(n1/2)n1/2) is quite good and gives:
- a(n,m) >= mnd/r, whenever (r/d)2 >= n >= r
- For the best choice of r and d we always
have n1/2 - 1/(2n3/2) > nd/r >=
n1/2 - 1/n1/2
- For n = x2+2, the best possible choices of r and d
are r = x2+1, and d = x, which puts
nd/r quite close to the bound n1/2 - 1/(2n3/2).
However, J. Roé (1998)
defines a quantity R(n,m) for n > 2 and any m such that a(n,m) >= R(n,m).
For m not bigger than
about n1/2, R(n,m) seems overall to be the best lower bound
constructed so far since, for m < n1/2,
R(n,m) seems (although this does not seem easy to prove)
to exceed Nagata's conjectured bound; here is a link to a
more complete description of how R(n,m) is
defined and computed.
See B. Harbourne, On Nagata's Conjecture,
preprint (1999), for discussion
and references, or link here
for a more complete discussion. (Also worth
mentioning are bounds given by G. Xu for integral curves in characteristic 0;
see ``Curves in P2 and symplectic packings,'' Math. Ann. 299 (1994), 609-613.)
- a(n,m) >= U(mnD(n1/2)/U(D(n1/2)n1/2))
- Note that nD(n1/2)/U(D(n1/2)n1/2)
>= n1/2 - 1/n1/2, but that when n - (D(n1/2))2
is even, then nD(n1/2)/U(D(n1/2)n1/2)
>= n1/2 - 1/(2n1/2).
Here is a web form for computing some of the bounds on a(n,m)
A connection between bounding tind(n,m) and a(n,m)
has been noted by Ziv Ran:
if a(n,m) >= cm for some constant c, then
tind <= max(n1/2, n/c)(m+1) - 2. (If c < n1/2,
this simplifies to tind <= U(n(m+1)/c) - 3.)
Thus the bounds above on a(m,n) give bounds on tind(n,m).
gives what seem to be the best general bound for large m:
- a(n,m) >= mnd/r, whenever (r/d)2 >= n >= r
When n is a square, one can do better, however:
-  tind(n,m) <= U((m+1)r/d) - 3, whenever (r/d)2 >= n >= r
- ignoring all terms but that involving m, this bound
(taking best choices for r and d) satisfies
m(n1/2 + 1/(2n3/2)) < mr/d <=
m(n1/2 + 1/n1/2)
-  Also see the bound d1(m,n)
given by J. Roé (Linear systems of plane curves
with imposed multiple points, (2000) preprint), which,
although hard to analyze, seems to be the best
bound currently available when m is not too large (always at least as good
as (m+1)((n + 2.5)1/2+Pi/8)-1).
-  tind(n,m) <= mU(n1/2) + U((U(n1/2) - 3)/2)
- See Lemma 5.3 of Harbourne/Holay/Fitchett
- When n > 1 is a square and m is sufficiently large,
it is easy to check that tind(n,m) cannot be less than
mU(n1/2) + U((U(n1/2) - 3)/2),
so in fact
tind(n,m) = mU(n1/2) + U((U(n1/2) - 3)/2)
for m >> 0 when n > 1 is a square.
- When n > 9 is a square and m is sufficiently large,
it is interesting to note that
mU(n1/2) + U((U(n1/2) - 3)/2)
is equal to the expected value
D([1/4 + n(m2 + m)]1/2 - 1/2) of a(n,m).
Here are some previous bounds for tind(n,m):
When n is a square and m is large enough,  is always the best bound,
since it is exact. When m is small
(m up to about n1/2 when n is not a square),
numerical evidence suggests  d1(m,n)
is the best bound, although  is sometimes close.
When m is larger than about n1/2 and n is not
a square,  seems to be the best bound.
- A. Gimigliano (Regularity of Linear Systems of Plane Curves, J. Alg 124 (1989), 447 - 460)
and A. Hirschowitz, (Une conjecture pour la cohomologie
des diviseurs sur les surfaces rationelles
Journ. Reine Angew. Math. 397 (1989), 208-213)
give upper bounds of approximately m(2n)1/2 for tind(n,m)
- G. Xu, Ample line bundles on smooth surfaces, Journ. Reine Angew. Math. 469 (1995), 199 - 209
gives an upper bound of approximately m(10n/9)1/2 for tind(n,m)
-  E. Ballico (Curves of minimal degree with prescribed singularities,
(1997), preprint) gives an upper bound of approximately m((n+1)1/2 + 1)
Here is a web form
for computing some of the bounds on tind(n,m)
[P2] Determine F0(Z(m)) as a
Question: What should one
expect for the structure of F0(Z(m))
for arbitrary m (i.e., when the multiplicities
are not all equal)?
- F0(Z(m)) and
together determine F1(Z(m)).
- F0(Z(m)) is now known
for any m with n <= 8 (in preparation,
Fitchett, Harbourne, and Holay).
- (Quasi-)Uniform Resolution Conjecture:
If n > 9 and m is (quasi-)uniform, then
the minimal free resolution of IZ(m) is
0 -> R[-a-2]a+1-h x R[-a-1]c
-> R[-a-1]b x R[-a]h ->
IZ(m) -> 0,
a = a(m),
h = hIZ(m)(a),
b = max(0,a+2-2h), and
c = max(0,2h-a-2).
- Most of the evidence for this conjecture is for the uniform case:
- The Uniform Resolution Conjecture holds for m=1 (A. V. Geramita,
D. Gregory, L. Roberts, Minimal ideals and points in
projective space, J. Pure Appl. Alg. 40 (1986), 33-62).
- The Uniform Resolution Conjecture holds for m=2 (M. Idà,
The minimal free resolution for the first infinitesimal
neighborhoods of n general points in the plane,
J. Alg. 216 (1999), 741-753).
The SHGH Conjecture implies the Uniform Resolution Conjecture
for (HHF 1999):
- infinitely many m for each n > 9,
- for all m >= (n1/2 - 2)/4 if n is an even square, and
- for all m >= (n - 9)/8 if n is an odd square.
- In fact, if n > 9 is an even square and m >= (n1/2 - 2)/4,
then the expected resolution given above holds whenever
a(n,m) has its expected value of mn1/2 + (n1/2-2)/2
(see HHF 1999).
Thus, for even squares, proving the Uniform Resolution Conjecture
(for m sufficiently large) amounts to proving that a(m,n) has its expected value.
- Therefore the Uniform Resolution Conjecture
holds for n = 16 (note that SHGH holds for n = 16 uniform
multiplicities since a(16,m) >= 4m + 1 by Nagata and
tind <= 4m + 1 by a bound given above), and, in characteristic 0,
(applying Evain's paper mentioned above) for m >= (n1/2 - 2)/4
if n > 9 is an even square not divisible by a prime bigger than 5.
- No general conjecture for F0(Z(m))
has been given for n > 8.
- We know F0(Z(m)) is the
direct sum over all degrees t of R[-t]ct,
where ct is the dimension of the cokernel
of the multiplication map
R1 x (IZ(m))t-1 ->
(IZ(m))t. Let X be the surface
obtained by blowing up the points p1, ... , pn,
with L a general line on X and E1, ... , En
the exceptional divisors of the blow ups.
Let Ft(m) =
(tL - m1E1 - ... - mnEn),
where m = (m1, ... , mn).
Then to compute ct for all t for every
m = (m1, ... , mn),
it turns out to be enough to be able in general
to compute ct for those t such that
Ft(m) meets every exceptional
curve on X nonnegatively. Assuming the SHGH Conjecture, it
is in turn enough in general just
to compute ctind. There are two nontrivial cases:
- a(m) < tind(m), and
- a(m) = tind(m).
- For the case a(m) < tind(m),
it is enough (again assuming SHGH, as shown
in S. Fitchett's 1997 thesis)
to consider m and t such that
Ft(m) = L + sE for some
exceptional curve E on X, with 1 <= s <= L.E and also with
u < s < U, where u is the minimum of t and t - m,
U is the maximum of t and t - m, and
m is the maximum of mi over all i.
- It is easy to generate test cases, especially with n = 9.
The exceptional curves are precisely the divisors of
the form E = E9 + A + bKX
So for example, we can take A = a(E1 - E2).
Then with E = E9 + A + bKX
and with a2 + a < s < 2a2 - a,
this gives Ft(m) = L + sE
satisfying u < s < U but for which it is not known
what value to expect for ct.
- KX = 30L - E1 - ... - E9,
- A = a0L + a1E1
+ ... + a8E8,
- a0, ... , a8 are any integers such that
3a0 - a1 - ... - a8 = 0 (i.e., A.KX = 0), and
- b = (a02 - a12 -
... - a82)/2 (i.e., b = A2/2).
is a link to a web based computation of the
minimal free resolution of IZ(m)
for n <= 7 and n = 8.
is a link to a web based computation of the resolution (or at least the
for the Uniform Resolution Conjecture.