The Heinz Problem
Will Traves
Department of Mathematics
United States Naval Academy
Commutative Algebra Meets Algebraic Combinatorics
Dalhousie University, Halifax
24 JAN 2014
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The Heinz Problem
Joint work with Max Wakefield and Tom Paul
Question
How many generic arrangements of 3 lines pass through 3 points
and are tangent to 3 lines (in general position)?
tangency: cubic ∩ line has a double (or higher) point
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Examples and Non-examples
Example
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Non-example (not generic arr.)
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Cohomological Approach
Goal: count arrangements without precise location of constraint points.
Move points −→ arrangements vary −→ number remains constant
Deformed constraints have the same cohomology class
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Constraints on a Single Line
L : ax + by + cz = 0 ∈ P2∗ through P(x0 : y0 : z0 ) ∈ P2
⇐⇒ ax0 + by0 + cz0 = 0 (linear constraint on a, b, c)
There is a 1-dimensional “line” in P2∗ of lines through P.
The constraint “goes through P” corresponds to the class
h ∈ H ∗ (P2∗ , Z). Each point P gives the same class h.
The cells of a cell-decomposition of P2∗ determine a Z-module basis
for H ∗ (P2∗ , Z):
H ∗ (P2∗ , Z) ∼
= Z · P2∗ ⊕ Z · h ⊕ Z · point.
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Cohomology ring structure
Cohomology class represented by
dual homology class (via Poincaré
duality).
[V ] ∗ [W ] = [V ∩ W ] if V and W
intersect transversely.
[V ] + [W ] = [V ∪ W ].
h2 = h ∗ h = [L1 ∩ L2 ] = [point]
[P 2∗ ] =identity of H ∗ (P2∗ , Z).
H ∗ (P2∗ , Z) = Z · P2∗ ⊕ Z · h ⊕ Z · point = Z · 1 ⊕ Z · h ⊕ Z · h2 = Z[h]/(h3 )
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Three lines
Kunneth formula:
H ∗ ((P2∗ )3 , Z) ∼
= H ∗ (P2∗ , Z) ⊗ H ∗ (P2∗ , Z) ⊗ H ∗ (P2∗ , Z)
∼
= Z[h1 , h2 , h3 ]/(h3 , h3 , h3 ).
1
2
3
P1 ∈ ∪ 3 lines ⇐⇒ P1 ∈ L1 or P1 ∈ L2 or P1 ∈ L3
constraint ↔ class h1 + h2 + h3
Q: How to deal with tangency without counting non-generic
arrangements?
A: Introduce intersection points Pij = Li ∩ Lj .
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Correspondence
Introduce intersection points Pij = Li ∩ Lj .
Require: each given L : a0 x + b0 y + c0 z = 0 passes through a
Pij (x0 : y0 : z0 ) ⇒ removes (most) double-line arrs from count
Pij ∈ L ⇐⇒ ax0 + by0 + cz0 = 0 ↔ class hij ∈ H 1 (P2 , Z).
L contains one of P12 , P13 , P23 ↔ h12 + h13 + h23 ∈ H ∗ ((P2 )3 , Z).
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All Constraints
Work on space X = (P2∗ )3 × (P2 )3 with
H ∗ (X , Z) =
Z[h1 , h2 , h3 , h12 , h13 , h23 ]
3 , h3 , h3 )
(h13 , h23 , h33 , h12
13 23
M = {(L1 , L2 , L3 , P12 , P13 , P23 ) : Pij ∈ Li ∩ Lj } ⊂ X
[Pij ∈ Li ] = [ax0 + by0 + cz0 = 0] = hi + hij
[M] = (h1 + h12 )(h2 + h12 )(h1 + h13 )(h3 + h13 )(h2 + h23 )(h3 + h23 )
Count = [M](h1 + h2 + h3 )3 (h12 + h13 + h23 )3 = 342(h1 h2 h3 h12 h13 h23 )2
Unlabeled Count = 342/3! = 57.
[Why Heinz?]
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Why is this called the Heinz Problem?
57 Varieties!
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Characteristic Numbers for 3 Generic Lines
The characteristic numbers for a family of curves are the number of
such curves passing through p points and tangent to ` lines.
Theorem (Paul, Wakefield, T-)
The characteristic numbers N3 (p, `) for the family of generic
arrangements with 3 lines are:
Points p
Lines `
N3 (p, `)
0
6
15
1
5
30
2
4
48
3
3
57
4
2
48
5
1
30
6
0
15
Why is the table symmetric?
Why is the table unimodal?
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Enumerative Problem
Problem
How many arrangements of 3 lines passing through 3 points are
tangent to 3 elliptic curves?
Class of a smooth curve of degree d is d(d − 1).
Fulton-MacPherson:
x 3 (6x + 3y )3 = 63 x 6
Count
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+62 3
3
2
x 5y
+6 · 32
3
2
x 4 y 2 +33 x 3 y 3
= 216N3 (6, 0) +324N3 (5, 1) +162N3 (4, 2)
= 216(15)
+324(30)
+162(48)
= 22, 275
Heinz Problem
+27N3 (3, 3)
+27(57)
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Characteristic Numbers for 4 Lines
With 4 lines there still are a finite number of quadruple lines that must
be removed from our correspondence count.
Theorem (Paul, Wakefield, T-)
The characteristic numbers for the family of generic arrangements with
4 lines are:
Points p
Lines `
N4 (p, `)
0
8
16695
1
7
17955
2
6
13185
3
5
8190
4
4
4410
5
3
2070
6
2
855
7
1
315
8
0
105
What happens with 5 lines? → Correspondence doesn’t remove
quintuple lines → Excess Intersection → higher dimensional
collections of arrangements that contribute a finite number to the
degree computation.
[Braid?]
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Some Details for Four Lines
Consider N4 (0, 8). Then 82 62 /2 quadruple lines
each count as 6! labeled arrangements in [M].
N4 (0, 8) = ([M](h12 + · · · + h34 )8 −
= 16, 695.
(82)(62)
2
6!)/4!
By duality there are
16,695 braid
arrangements
through 8 points.
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