Motivating Intersection Theory via Enumerative
Geometry
Will Traves
Department of Mathematics
United States Naval Academy∗
James Madison University
27 APR 2015
* Any views or opinions presented in this talk are solely those of the presenter
and do not necessarily represent those of the U.S. Government
Traves
JMU
27 APR 2015
Questions
Question
Two distinct lines meet in how many points?
Question
Two distinct conics meet in how many points?
Question
How many lines in 3-space meet 4 randomly positioned lines?
The answer to the first question is obvious and ugly: 0 or 1.
Traves
JMU
27 APR 2015
Projective Space
Add points to R2 so that all parallel lines of slope m meet at a point at
infinity.
Question: What does the set of points at infinity look like?
Traves
JMU
27 APR 2015
Identification of points at infinity
The lines meet at infinity on each side of the lines.
Since parallel lines meet at only 1 point, antipodal points on the
boundary are identified.
Traves
JMU
27 APR 2015
The curve at infinity
Points at infinity form a curve wrapped twice around the boundary of
the R2 disk.
Build projective 2-space: P2 = R2 ∪ R1 ∪ R0
Traves
JMU
27 APR 2015
The Projective Plane: Thicken Line at Infinity
Question: What do we get if we thicken the curve at infinity to a band?
Traves
JMU
27 APR 2015
The linking numbers of triangles 124 and 356 is 0 because they are not linked;
apart from each other without being caught like a chain link.
A Möbius Band
Thicken line at infinity:
P2 = disk ∪ Möbius band
The linking numbers of triangles 246 and 135 is 1 because they are linked toge
P2 can’t be embedded in R3
(Conway, Gordon, Sachs (1983):
linked triangles in K6 )
If you add all of the linking numbers of all sets of triangles in this partic
find the sum to be 1 (246 and 135 are the only linked triangles). This is consist
Gordon and Sachs who claimed that as long as we are working in R3 then we w
linking number.
Traves
JMU
27 APR 2015
K6 embedded in P2
No linked triangles
Traves
JMU
27 APR 2015
Projective Coordinates: Möbius’s model of P2
(x, y) ∈ R2 ↔ (x, y, 1) ∈ R3
(x, y) ∈ R2 ↔ line through (0, 0, 0) and (x, y, 1)
Suggests: P2 = 1-dimensional subspaces of R3
Which points in P2 correspond to lines in the (x, y)-plane?
Traves
JMU
27 APR 2015
Lines parallel to the (x, y )-plane
Line through (x, y , z) represented by equiv. class (x : y : z) with
(x : y : z) (λx : λy : λz) for λ 6= 0.
If z 6= 0 then (x : y : z) = (x/z : y/z : 1) a point on our R2
but (x : y : 0) corresponds to lines parallel to z = 1.
The (x : y : 0) = (1 : y/x : 0) classes each correspond to a point at
infinity. Points at infinity characterized by z = 0.
Traves
JMU
27 APR 2015
Parallel Lines
Question: In what sense do parallel lines meet at infinity?
Line x = 0:
{(0, y)} ∼ {(0 : y : 1)} ∼ {(0 : 1 : 1/y )} −→ (0 : 1 : 0)
Line x = 1:
{(1, y)} ∼ {(1 : y : 1)} ∼ {(1/y : 1 : 1/y )} −→ (0 : 1 : 0)
In P2 the line x = 0 contains all points of the form (0 : y : z).
The line x = 1 contains all points of the form (z : y : z).
Traves
JMU
27 APR 2015
Implicit Equations
Implicit equations must be homogeneous: (1 : 2 : 1) satisfies y = x + 1
but (2 : 4 : 2) does not.
Homogeneous of degree d: F (λx, λy , λz) = λd F (x, y, z)
Implicit equations of lines are linear polynomials.
Traves
JMU
27 APR 2015
Homogenization
The parabola y = x 2 in R2 is not given by a homogeneous equation.
The parabola is a piece of the curve given by homogeneous equation
yz = x 2
Similarly, the circle (x − a)2 + (y − b)2 = r 2 can be homogenized to
(x − az)2 + (y − bz)2 = r 2 z 2 ,
which meets z = 0 at the two points satisfying x 2 + y 2 = 0:
(1 : i : 0) and (1 : −i : 0).
Traves
JMU
27 APR 2015
Bézout’s Theorem
Compactness of P2 allows us to count solutions:
Theorem (Bézout)
Any two curves, without common components, defined by the
vanishing of homogeneous polynomials of degrees d1 and d2 meet in
d1 d2 points in P2 , suitably interpreted.
Traves
JMU
27 APR 2015
Intersecting two conics
Question
How many points lie on two distinct conics?
Each conic is a degree 2 curve in projective space so they intersect in
2 · 2 = 4 points.
Traves
JMU
27 APR 2015
Intersection Ring: Elements
Ring elements are deformation classes [X ] of sub-objects in P2 .
[X ] = [Y ] if X and Y are fibers of a nice deformation.
Traves
JMU
27 APR 2015
Intersection Ring: Operations
[V ∩ W ] = [V ] ∗ [W ] if V and W
intersect transversely.
[V ∪ W ] = [V ] + [W ].
[degree d curve] = d [line]
[P2 ] is the identity element of the intersection ring for P2 .
Traves
JMU
27 APR 2015
Intersection Ring: Cell Decomposition
The cell decomposition of P2
P2 = R2 ∪ R1 ∪ R0
leads to an intersection ring
H(P2 , Z) = Z[R2 ] ⊕ Z[R1 ] ⊕ Z[R0 ].
The identity element of H is [R2 ] and ω = [R1 ] = [line] generates the
ring with ω 2 = [R0 ] = [point].
H(P2 , Z) = Z[ω]/(ω 3 ).
Traves
JMU
27 APR 2015
Computation in the Intersection Ring
[conic1 ∩ conic2 ] =
=
=
=
=
=
[conic1 ] · [conic2 ]
0
[line ∪ line0 ] · [line
∪ line ]
0
[line] + [line ] · [line] + [line0 ]
(2ω) · (2ω)
4ω 2
4[point]
The intersection of two conics consists of 4 points.
Traves
JMU
27 APR 2015
G(1, 3): Sketch
Space of lines in 3-space is a 4-dimensional object
RREF’s of 2x4 matrices give a cell decomposition of G(1, 3)
This leads to an intersection ring of G(1, 3).
Pieri Rules: How to intersect classes in G(1, 3)
Obtain results via Poncelet’s Principle
How many lines meet 4 general lines in 3-space?
Traves
JMU
27 APR 2015
Numerical Algebraic Geometry
Poncelet’s principle inspired
development of Numerical
Algebraic Geometry.
Deform equations to a system we
can solve.
Track the solutions as we move the
deformation back to the original
system.
Need to consider multiplicity and
solutions that go off to infinity.
Traves
JMU
27 APR 2015
Student Work: Andrew Bashelor
dent Project
Andy, Amy Ksir and me
Our
workto
grew
out of Bashelor’s Trident
(right
left)
project, a full-year undergraduate research
project focused on enumerative algebraic
geometry.
Traves
How many conics are tangent to
five conics in general position?
JMU
27 APR 2015
Student Work: Tom Paul
Tom Paul worked with me and
Max Wakefield
Traves
There are 16,695 braid
arrangements through 8 general
points.
JMU
27 APR 2015