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Generalizing Pascal’s Mystic Hexagon Theorem

Will Traves

Department of MathematicsUnited States Naval Academy

Mathematics and Statistics ColloquiumDalhousie University

Halifax

18 APR 2013

Traves (USNA) Generalizing Pascal’s Theorem Halifax, 18 APR 2013 1 / 24

Ulterior Motives

Really in Halifax to honor my father as he nears the end of his term asPresident at Dalhousie.

Tom Traves


Start at the Beginning

Line Arrangement due to Pappus of Alexandria (Synagogue; c. 340)

Richter-Gebert: 9 proofs in Perspectives on Projective Geometry, 2011


Pascal’s Mystic Hexagon Theorem

Pascal: placed the 6 intersection points on a conic (1639)

Converse: Braikenridge-Maclaurin Theorem


Why Mystic?


The Projective Plane

P2 is a compactification of R2 obtained by attaching a line at infinity

Thicken line at infinity:P2 = disk ∪ Mobius band

P2 can’t be embedded in R3

(Conway, Gordon, Sachs (1983):linked triangles in K6)

! 20 !

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.

The Relationship between RP2 and R3.

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 apart from each other without being caught like a chain link.

The linking numbers of triangles 246 and 135 is 1 because they are linked together.

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.


Bezout’s Theorem

Compactness of P2 allows us to count solutions:

Theorem (Bezout)Any two curves, without common components, defined by thevanishing of polynomials of degrees d1 and d2 meet in d1d2 points inP2, suitably interpreted.Lines meeting an Ellipse

y=0

4x2+9y2=36

y=1 y=2 y=3 Line meets curve in two

points (possibly imaginary).

Double point: tangency when y=2 4x2 + 9(4) = 36 so 4x2 = 0

It seems that such a curve of degree 1 always meets such a curve of degree 2 in (2)(1) points, if we count them properly.


The 8⇒ 9 Theorem

Theorem (Cayley-Bacharach-Chasles)If two cubics C1 and C2 meet in 9 distinct points, then any other cubicC through 8 of these points goes through the ninth too.

Elementary proof on Terry Tao’s blog (July 15, 2011)


Proof of Pascal’s Theorem

C1: deg 3 curve consisting of red lines

C2: deg 3 curve consisting of blue lines

C: deg 3 curve consisting of conic andline through 1 and 2

Since C goes through 8 points of C1 ∩ C2, C goes through the ninthpoint too (by CBC Theorem).


From Pascal to Mobius

Theorem (Mobius)Inscribe a polygon with 4k + 2 sides into a conic and consider the2k + 1 points where opposite edges meet. If 2k of these points arecollinear, then so is the last one. Inscribe two 2k-gons in a conic andassociate edges cyclicly. The associated edges meet in 2k points andif 2k − 1 of these point are collinear, then so is the last one.


Folklore

TheoremSuppose that k red lines meet k blue lines in a set Γ of k2 distinctpoints. If S = 0 is an irreducible curve of degree d through kd points ofΓ then the remaining points lie on a unique curve C of degreet = k − d.

Proof (Existence):R: deg k poly defining red lines.B: deg k poly defining blue lines.Pick P ∈ S \ Γ. Choose

Fa,b = aR + bBto vanish at P. Then S = 0 is acomponent of Fa,b = 0 (by Bezoutand S irred) and Fa,b/S defines C.�


d-Constructible Curves

DefinitionA curve C of degree t is d-constructible if there exist k = d + t redlines and k blue lines so that:(1) red and blue lines meet in a set Γ of k2 distinct points(2) dk points in Γ lie on a degree-d curve S(3) the remaining tk points of Γ lie on C.


Density: definition

DefinitionA curve C of degree t is d-constructible if there exist k = d + t redlines and k blue lines so that:(1) red and blue lines meet in a set Γ of k2 distinct points(2) dk points in Γ lie on a degree-d curve S(3) the remaining tk points of Γ lie on C.

DefinitionThe d-construction is dense in degree t if almost every degree-tcurve is d-constructible (i.e. there is a nonempty Zariski-open setU ⊂ P(C[x , y , z]t ) so that every degree-t curve in U is d-constructible).


First Example

ExampleFor any d , all lines are d-constructible.

Arrange d + 1 points on the line and red and blue lines through thesepoints (with distinct intersection points).

Folklore Theorem⇒ ∃ complementary curve of degree d .


Second Example

ExampleFor any d , all irreducible conics are d-constructible.

Arrange 2(d + 2) points, joined cyclicly by alternating red and bluelines, ensuring distinct intersection points.

Folklore Theorem⇒ ∃ complementary curve of degree d .


Counting Dimensions

Theorem (T-)If d ≥ 3, then the d-construction is not dense in degrees d + 4 orhigher. The 2-construction is not dense in degrees five or higher. The1-construction is not dense in degrees six or higher.


Elliptic Curves

Elliptic curves are genus 1 smooth curves (isomorphic to a real torus).The points on the curve form a group.

Any elliptic curve can beembedded in P2 as a smooth cubicE with 0E a point of inflection atinfinity.

Group law:A + B + C = 0 ⇐⇒A, B, C are collinear pts of E .

Can use CBC Theorem to check that this group law is associative.

The smooth cubics form a dense set of all cubics in P2.


Elliptic Curves are d-constructible

Theorem (T-)All elliptic curves are d-constructible so the d-construction is dense indegree 3.

Need to arrange 3(d + 3) points on E with a blue and red line througheach of them; construction has d + 4 parameters.

Case d even:


Elliptic Curves are d-constructible: Case d odd

Theorem (T-)All elliptic curves are d-constructible so the d-construction is dense indegree 3.

Case d odd (c.f. Mobius):


Density


1-Constructible Curves

Theorem (Roth and T-)The 1-construction is dense in degrees 4 and 5.

deg 4 pf: Find f = (f1, . . . , f15) : C17 → {1-constr deg-4 curves}.

Thm of generic fiber: dim(imf ) + dim(gen fiber) = dim(domain).

Suppose P ∈ fiber f−1(c); for unit tangent vector v to fiber,

Dvf (P) =

(∂fixj

) ∣∣∣∣∣x=P

v = Jac(f )(P)v = 0.

Rank Jac(f )(P0) = 15⇒ dim(fiber) = 2⇒ dim(gen fiber) ≤ 2.Thus dim(im f) ≥ 15 = dim(deg 4 curves).

Chevalley’s Thm⇒ im(f) ⊃ dense open set of 1-constr deg-4 curves.


Density

ConjectureThe d-construction is dense in degree 4 for all d.


Strongly Inscribed Polygons

Call a polygon P strongly inscribed in a curve C if every edge,extended to a line, meets C only in points p with exactly two edge linespassing through p.

The d-construction is dense in degree t if and only if a dense set ofdegree-t curves admits a strongly inscribed polygon with 2(d + t)edges.


Strongly Inscribed Polygon Results

Theorem (T-)Every elliptic curve admits a strongly inscribed polygon with 2k ≥ 8edges.

Easy to see that if degree C is odd then only polygons with an evennumber of edges can be strongly inscribed.

QuestionWhich degree 4 curves admit a stronglyinscribed polygon with an odd numberof sides?


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