Motivating 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

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 : Thicken Line at Infinity

Question: What do we get if we thicken the curve at infinity to a band?

Traves JMU 27 APR 2015

Projective space is also the union of a disc in R2 and a Möbius strip, and is equivalent to the sphere S2 with a blow up at one point.

2 3 The Relationship between RP and R .

Earlier we claimed that RP2 is not a subset of R3, that it does not “fit” into R3. For this proof we call upon Conway, Gordon and Sachs’ 1983 result (ams.org) that 6 points all cannot be linked to one another such that the total linking number of all triangles formed is even, in R3. We will give an example of 6 points linked in RP2 , K6, such that the linking number of the set is even.

First, to demonstrate Conway, Gordon and Sachs’ proof, an example of K6, 6 points linked in R3:

Figure 16

The linking numbers of triangles 124 and 356 is 0 because they are not linked; they could be pulled AMobius¨ apart Band from each other without being caught like a chain link.

Thicken line at infinity: 2 P = disk ∪ Mobius¨ band The linking numbers of triangles 246 and 135 is 1 because they are linked together.

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 particular linking, you will find the sum to be 1 (246 and 135 are the only linked triangles). This is consistent with Conway, Gordon and Sachs who claimed that as long as we are working in R3 then we will have an odd total linking number. Traves JMU 27 APR 2015 ! 20 ! 2 K6 embedded in P

No linked triangles Traves JMU 27 APR 2015 Projective Coordinates: Mobius’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 = x2 in R2 is not given by a homogeneous equation.

The parabola is a piece of the curve given by homogeneous equation

yz = x2

Similarly, the circle (x − a)2 + (y − b)2 = r 2 can be homogenized to

(x − az)2 + (y − bz)2 = r 2z2,

which meets z = 0 at the two points satisfying x2 + y 2 = 0:

(1 : i : 0) and (1 : −i : 0).

Traves JMU 27 APR 2015 Bezout’s´ Theorem

Compactness of P2 allows us to count solutions: Theorem (Bezout)´ Any two curves, without common components, defined by the vanishing of homogeneous polynomials of degrees d1 and d2 meet in 2 d1d2 points in P , 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

2 2 1 0 P = R ∪ R ∪ R

leads to an intersection ring

2 H(P , Z) = Z[R2] ⊕ Z[R1] ⊕ Z[R0].

The identity element of H is [R2] and ω = [R1] = [line] generates the ring with ω2 = [R0] = [point].

H(P2, Z) = Z[ω]/(ω3).

Traves JMU 27 APR 2015 Computation in the Intersection Ring

[conic1 ∩ conic2] = [conic1] · [conic2] = [line ∪ line0] · [line ∪ line0] = [line] + [line0] · [line] + [line0] = (2ω) · (2ω) = 4ω2 = 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

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 USNA Trident Project

This talk reports on joint work with MIDN Andrew Bashelor and my colleague at USNA, Amy Ksir.

Andy, Amy Ksir and me Our(right work togrew left) out of Bashelor’s Trident How many conics are tangent to project, a full-year undergraduate research project focused on enumerative algebraic five conics in ? geometry.

Traves JMU 27 APR 2015 Student Work: Tom Paul

There are 16,695 braid Tom Paul worked with me and arrangements through 8 general Max Wakefield points.

Traves JMU 27 APR 2015