Enumerative Geometry of Hyperplane
Arrangements
Will Traves
Department of Mathematics
United States Naval Academy
University of Nebraska, Lincoln
6 DEC 2013
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Warm-up Question
Joint work with Max Wakefield and Tom Paul
Question
How many generic arrangements of k lines pass through d points in
general position in P2 ?
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Warm-up Question
Joint work with Max Wakefield and Tom Paul
Question
How many generic arrangements of k lines pass through d points in
general position in P2 ?
If d is too small there are infinitely many arrangements.
If d is too large there are no arrangements.
If d = 2k then there is a finite number of arrangements through the d
points.
Traves (USNA)
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6 DEC 2013
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Warm-up Question
Joint work with Max Wakefield and Tom Paul
Question
How many generic arrangements of k lines pass through d points in
general position in P2 ?
If d is too small there are infinitely many arrangements.
If d is too large there are no arrangements.
If d = 2k then there is a finite number of arrangements through the d
points.
No line can pass through 3 points (by general position assumption);
PHP shows that each line goes through 2 points.
2k
2k − 2
2
Count =
···
/k! = (2k − 1)!!
2
2
2
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The Intersection Lattice of A
Problem
Given a hyperplane arrangement A in Pn , count the number of
hyperplane arrangements with intersection lattice isomorphic to L(A)
that pass through d points in general position.
The intersection lattice (or poset) of A encodes how the hyperplanes
intersect:
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The Moduli Space MA
Definition
If A is an arrangement of k hyperplanes in Pn then set
MA := {arrangements with intersection lattice ∼
= L(A)} ⊂ (Pn∗ )k /Sk
with dimension d = dA .
Hyperplane H : a0 x0 + · · · + an xn = 0 corr. to (a0 : . . . : an ) ∈ Pn∗
A passes through P ⇐⇒ (H1 · · · Hk )(P) = 0 – a codim 1 condition
dimension d = # point conditions giving a finite counting problem
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Degree NA of MA
The degree NA of MA is the number of arrangements in MA through
d points in general position.
The map
(Pn∗ )k −→ P(C[x0 , . . . , xn ]k )
(H1 , . . . , Hk ) 7−→ H1 · · · Hk
factors through (Pn∗ )k /Sk .
The condition that A passes through a point, (H1 · · · Hk )(P) = 0,
is a linear condition on P(C[x0 , . . . , xn ]k ) and pulls back to a
multilinear condition on (Pn∗ )k .
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Finding d the Easy Way...
Dim MA = # degrees of freedom you have in drawing A and
represent each degree by a point constraint:
Virtual dimension:
vdim(MA ) = 2k −
X
(np − 2)
with np =| {H ∈ A : p ∈ H} |
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Finding d the Easy Way...Fails!
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An (Intractible?) Open Problem
Problem
Find a combinatorial algorithm to compute dim MA from L(A).
Mn̈ev’s Universality Theorem ⇒ MA arbitrarily complicated
Need to “see” syzygies among defining equations for MA from L(A).
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Cohomological Approach
Goal: count arrangements without precise location of constraint points.
Move points −→ arrangements vary −→ number remains constant
Codim-d constraint −→ degree-d class ∈ (graded) cohomology ring.
Deformed constraints have the same class.
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Cohomology ring structure
The structure of the cohomology ring depends on the structure of the
underlying moduli space X .
The cells of a cell-decomposition of X determine a Z-module basis for
H ∗ (X , Z).
Cohomology class with degree d
represented by dual homology
class (via Poincaré duality).
[V ∩ W ] = [V ] ∗ [W ] if V and W
intersect transversely.
[V ∪ W ] = [V ] + [W ].
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Deformation
Deformed constraints have the same class.
Example: Line ax + by + cz = 0 in P2 represented by [a : b : c].
Line through P[1 : 0 : 0] ⇔ a = 0.
Line through Q[0 : 1 : 0] ⇔ b = 0.
Deformation F (t): a(1 − t) + bt = 0 so [P-pencil] = [Q-pencil].
Q[0 : 1 : 0]
P [1 : 0 : 0]
t=0
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Example: Pn and Bézout’s Theorem
Pn has one cell in each dimension d = 0, . . . , n
H ∗ (Pn , Z) = Zc0 ⊕ Zc1 ⊕ · · · ⊕ Zcn w/ cd codim-d class
c0 = [Pn ] = Identity element
c12 = c2 = [codim-2 plane]
c1 = [hyperplane]
c1n = cn = [point]
H ∗ (Pn , Z) = Z[h]/(hn+1 ) w/ h = c1 = [hyperplane].
Deformation: [degree d hypersurface] = dh.
Bézout: [d1 -hyp ∩ d2 -hyp ∩ · · · ∩ dn -hyp] = (d1 h) · · · (dn h) = d1 · · · dn hn .
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Example: Generic Arrangements in Pn
Kunneth formula: H ∗ ((Pn )k , Z) = Z[h1 , h2 , . . . , hk ]/(h1n+1 , . . . , hkn+1 ).
Question
How many generic arrangements of k hyperplanes go through d = kn
points?
A goes through P ⇔ H1 (P)H2 (P) · · · Hk (P) = 0.
[H1 (P)H2 (P) · · · Hk (P) = 0] = h1 + · · · + hk
[{arrs satisfying kn point conditions
}]
kn
kn
= (h1 + · · · + hk ) = n,n,...,n (h1 h2 · · · hk )n =
kn
so answer is n,n,...,n
/k !.
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kn
n,n,...,n
points/arrs
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Characteristic Numbers
Definition
Given a family of plane curves, the characteristic number N(p, `)
counts the number of such curves that pass through p points and are
tangent to ` lines in general position.
Let Nk (p, 2k − p) be the nontrivial characteristic numbers for the family
of generic arrangements of k lines in P2 .
We can compute these for k = 3 and k = 4 and use them to solve
more general enumerative problems.
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The Heinz Problem
Question
How many generic arrangements of 3 lines pass through 3 points and
are tangent to 3 lines?
M = {(L1 , L2 , L3 , P12 , P13 , P23 ) : Pij ∈ Li ∩ Lj } ⊂ (P2∗ )3 × (P2 )3
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A Comhology Calculation
M = {(L1 , L2 , L3 , P12 , P13 , P23 ) : Pij ∈ Li ∩ Lj } ⊂ (P2∗ )3 × (P2 )3
3 )
H ∗ ((P2∗ )3 × (P2 )3 , Z) = Z[x1 , x2 , x3 , y12 , y13 , y23 ]/(x13 , . . . , y23
xi = [hyperplane (point) condition on Li ]
yij = [hyperplane (line) condition on Pij ]
[Pij ∈ Li ] = [{ax0 + by0 + cz0 = 0}] = [{ax0 }] = xi + yij
[M] = (x1 + y12 )(x2 + y12 )(x1 + y13 )(x3 + y13 )(x2 + y23 )(x3 + y23 )
Labeled Count = [M](x1 + x2 + x3 )3 (y12 + y13 + y23 )3 = 342[pt]
Count = N3 (3, 3) = 342/3! = 57.
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Why is this called the Heinz Problem?
57 Varieties!
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Characteristic Numbers for k = 3
Theorem (Paul, Wakefield, T-)
The characteristic numbers 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? → Duality!
Why is the table unimodal?
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4 Lines
Consider N4 (0, 8). Then 82 62 quadruple lines
each count as 6! labeled arrangements in
M = MA ∪ M 0 .
N4 (0, 8) = ([M](y12 + · · · + y34 )8 −
= 16, 695.
8
2
6
2
6!)/4!
By duality there are
16,695 braid
arrangements
through 8 points.
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Table of Characteristic Numbers
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? → Excess Intersection
– higher dimensional collections of arrangements that contribute a
finite number to the degree computation.
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d-coned Generic Arrangements
We consider arrangements where each hyperplane contains a
common linear space, but the hyperplanes are otherwise generic.
d-coned generic arrangement A in Pn :
H∼
= Pn−(d+1) containing generic arr. B
Ω linear space of dim. d, Ω ∩ H = ∅
each H in A is span of hyp in B with Ω.
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Running Example
Question
How many 1-coned arrangements in P5 of 6 hyperplanes pass through
D points in general position?
Each hyperplane is determined by its normal vector modulo line Ω, i.e.
by a point in P(C6 /Ω) ∼
= P3 .
D = dim MA = dim G(1, 5) × (P3 )6 = 2 × 4 + 6 × 3 = 26.
G(1, 5) = Grassmannian of lines in P5 .
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Cohomology of the Grassmannian
Each line in P5 is the span of 2 points in C6 .
Cellular decomposition of G(1, 5): cells determined by position of
leadingones in RREF of spanning
matrix: 1 ∗ ∗ 0 ∗ ∗
σ20 = M ∈ M2×6 : M ∼
→ codimσ20 = 2
0 0 0 1 ∗ ∗
Schubert calculus tells us how to multiply cohomology classes
Giambelli formula → special classes
Pieri formula → multiplication of special classes
Duality: σα σα0 = [single Ω]
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A cohomology class
Z := {(Ω, H) ∈ G(1, 5) × P5 : Ω ⊂ H}.
H must contain 2 independent points on Ω: so codim Z = 2.
[Z ] = aσ20 + bσ11 + cσ10 h + dσ00 h2
Use duality to find coefficients:
[Z ] = σ11 + σ10 h + σ00 h2
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# 1-coned arr. of 6 hyperplanes through 26 points in P5
Pull-back to G(1, 5) × (P5 )6 :
[Zi ] = σ11 + σ10 hi + σ00 hi2
ANS = (h1 + h2 + h3 + h4 + h5 + h6 )26
6
Y
[Zi ]
i=1
Schubert Calculus gives:
2 4
2 3
σ00
σ11 = σ00 σ10
σ11 = 1σ44 = 1[arr]
h
6
26
ANS = 1 2,4
3,3,5,5,5,5 + 1
= 10270301132391300
Traves (USNA)
6
1,2,3
4 2
σ10
σ11 = 2[arr].
i
6
26
+ 2 4,2
/6!
4,4,4,4,5,5
Catalan numbers
26
3,4,4,5,5,5
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General Result
We have a general expression for the dimension D of MA for d-coned
arrangements of k hyperplanes in Pn .
We can also express the number of such arrangements passing
through D points using Schubert Calculus.
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Parting thoughts
1
Need a better sense of how the intersection lattice of A affects the
dimension of MA and the enumerative counts.
2
Is there a recursive relationship between the enumerative counts
for d-coned arrangements of k hyperplanes in Pn passing through
D = dim MA points?
3
Is there a role for quantum cohomology here? If so, are the
cohomology rings associatitve and can we leverage this to get
nice generating functions for some class of enumerative problems
(c.f. moduli of curves)?
4
This project is in its infancy and there’s room for many others to
contribute.
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