Explicit Division and Torsion Points on Superelliptic Curves and - - PowerPoint PPT Presentation

Explicit Division and Torsion Points on Superelliptic Curves and Jacobians

Vishal Arul’s Thesis Defense

MIT

April 3, 2020

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 1

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Fruit problem

99.9999% of people cannot solve this! Solve if you are a genius! + + + + + = 16 Can you fjnd positive whole values for , , and ?

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 2

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Infjnite descent: irrationality of √ 2

Arithmetic Show that √ 2 is irrational, i.e, that one cannot fjnd two nonzero integers a, b such that √ 2 = a b .

1 If such a, b existed, they

would both be even

2 Divide a, b by 2 to get a

smaller solution a

2, b 2

• 3 Can’t divide by 2

forever! Geometry Show that there is no point (a, b) ∈ (Z \ {0})2 that lies

• n the (degenerate)

hyperbola x2 − 2y2 = 0. x y

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 3

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Rational points on curves

Informally, a curve is a 2-variable polynomial equation f(x, y) = 0 A rational point on this curve is a solution (a, b) where a, b ∈ Q. Example: x3 − y2 + 1 = 0 Some solutions: (−1, 0) (0, 1) (0, −1) (2, 3) (2, −3) x y

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 4

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Rational points on conics

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 Some examples: x2 + 2y2 − 1 = 0 2x2 + y − 2 = 0 x2 − 3y2 − 1 = 0

1 Are there any rational points? 2 If so, how do we fjnd one? 3 Is there a formula that gives all solutions?

See Abhinav Kumar’s Lecture 24 for MIT’s 18.781 for more information.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 5

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Example: x2 + y2 = 1

(0, 1) is one rational point. Take any line through (0, 1) of rational slope; it intersects the circle at exactly one other point, which must be a rational point! This procedure generates all the rational points. Example: Line is y = 3

2x + 1

Intersection point is

• − 12

13, − 5 13

• x

y

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 6

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Elliptic curves

Informally, elliptic curves are given by an equation f(x, y) = 0 where deg f = 3. After a change of coordinates, the equation is y2 = x3 + Ax + B. We can add points. The sum of two rational points is another rational point. x y P Q −(P + Q) P + Q y2 = x3 − 2x + 2

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 7

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Strategy for fjnding rational points on elliptic curves

By the Mordell-Weil theorem, there exist fjnitely many rational points that generate all of them.

1 Use a computer to fjnd small rational points. 2 Determine whether we have found them all using 2-descent.

(We return to what this means later.)

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 8

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Back to the fruit problem

Bremner and MacLeod solved this problem (and did more)1 x y + z + y x + z + z x + y = 16. Clearing denominators, this becomes a degree 3 equation – an elliptic curve! Computer fjnds P = (−1729, 1909, 2511) is a solution, but x(P) is negative. Try multiples of P!

P

x = −1729 y = 1909 z = 2511

2P

x = −59704795693360 y = 58440761954029 z = 60611523515451

3P

x = −2860839847498395711385911675329 y = 5502830308807460377711231185551 z = 3225095062622507332335161309589

4P

x = 5595948611224060224364017631176103062582886764573084679 y = 1736089310886316841024156986935534509488541328603612320 z = −5500750928993484313189193690198645161760377304528912439

5P

x = 29306973965939511385616058054230695096641981174315979687608373450722909732659496417749 y = 33410032195872509393087670433431148212674791764424544744567912881635223088249073886351 z = −26556547643917023101089714889680809129736599315550969939439263851901574038728121206849

1Andrew Bremner and Allan MacLeod. “An unusual cubic

representation problem.” Annales Mathematicae et Informaticae. 2014.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 9

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Solution to the fruit problem

Calculations are restricted to 120 seconds. Input is limited to 50000 bytes. Running Magma V2.25-4. Seed: 1519260268; Total time: 0.310 seconds; Total memory usage: 85.16MB. Found a solution after multiplying by 11! Number of apples is 75695883920707641508654826369959980969484451183645312281679\ 4437380752701088755812980091049260660373658301414260453983030970812870540605007\ 7357739096869256607369466849049598689320109753370254292760383586908218687016167\ 1591261838569615705225865940066075319029896125903861891981127258138299976866161\ 7652089849345328389884032389254792615170485647887842866692663123727097675015331\ 5515608939686715005617866255912952511 Number of tangerines is 7166369758780814676912316128994352747367541683001253028\ 3003079424699216676553002703222405496100897367085440114493618868063648054160182\ 0518276882452685812328425046020925426588717716068885887864369991064394215179763\ 5162781962169934058777034950490334867644810567745551982537193820787907197051080\ 8009030074990221125473470601943040943508285885974430400086910850816628931992243\ 9346786021192652347943472958586821777673871 Number of bananas is 3739353473639792037899660845586527588666096869066509865973\ 5784952220869157385809042423277508375964340579066162049915895424902178387079529\ 1682616970345665316405351511025345400253316500560354614645904106598045889574950\ 8195105532469600315226908892502589998301861315135515383031079377766503151960589\ 6705877276805455129511173175225722575347604765502284715832294577561038757745387\ 491222687245758103143877455222635105109

https://math.mit.edu/~varul/fruits

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 10

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2-descent on elliptic curves

For an elliptic curve E, we want to compute the group of rational points E(Q). (Assume we already found E(Q)tors.)

1 Use a computer to fjnd small rational points. 2 Determine whether we have found them all using 2-descent.

If we are lucky, then we will know when we found enough rational points to generate E(Q)/2E(Q).

1 Suppose that P is a rational point not expressible as a sum

• f our generators. By shifting P by a combination of our

generators, we can make it divisible by 2.

2 Divide P by 2 to get a smaller rational point (one which

• nly has about 25% the digits that P does!)

3 Repeat this until the height of P is small enough. Manually

search for points of small height we may have missed. In fact, Fermat’s infjnite descent for X4 − Y 4 = Z2 is equivalent to 2-descent for the elliptic curve y2 = x3 + 4x!

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 11

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The genus of a curve

We go back to the general case of a curve C given by some equation f(x, y) = 0. Topologically, the complex points of C look like a g-holed torus, where g is the genus of the curve.2 Conics have genus 0. Elliptic curves have genus 1. Faltings’ Theorem. When g ≥ 2, there are only fjnitely many rational points. Unfortunately, when g ≥ 2, there is no way to add points of C.

2Pictures taken from Wikipedia. Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 12

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The jacobian of a curve

We work in a bigger space J where it is possible to add points. There are two ways to think about elements of J . A point of J is an unordered collection of k ≤ g points of the curve C. This interpretation “talks to the curve” but addition (the group structure) is not

• bvious to write down.

A point of J is a point of Cg/Λ, where Λ ≃ Z2g is the period lattice of C. (As a group, the jacobian is (S1)2g.) This interpretation makes addition easy to write down (just add in Cg) but it does not “talk to the curve.” The Abel-Jacobi map goes from the left interpretation to the right interpretation, but the Abel-Jacobi map is not algebraic; it is described in terms of integrals of holomorphic 1-forms.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 13

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Odd-degree hyperelliptic curves and their jacobians

An odd-degree hyperelliptic curve C is one of the form y2 = f(x) where d := deg f is odd. Its genus is g = (d − 1)/2. Mumford representation is an effjcient way to represent k ≤ g unordered points of C, i.e, a point of its jacobian. Suppose the points are P1, . . . , Pk. The Mumford data is two polynomials (U(x), V (x)) such that

1 The roots of U(x) are x(P1), …, x(Pk). 2 For each i, we have y(Pi) = V (x(Pi)). Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 14

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Division by 2 on odd-degree hyperelliptic curves

Suppose that we want to “divide by 2” on an odd-degree hyperelliptic curve C. Each point P of C will have 22g halves, i.e, points D on the jacobian J such that 2D = P. Zarhin gives the Mumford representation of each such D. Example: the genus 2 hyperelliptic curve y2 = x(x + 1)(x + 4)(x + 9)(x + 16) and P = (0, 0). Then P has 22g = 16 halves in J . One of them has Mumford representation U(x) = x2 − 35x + 24 V (x) = 300x − 240 which in terms of points is 1 2(35 ± √ 1129), 5010 ± 150 √ 1129

• .

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 15

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Division by 1 − ζ on superelliptic curves

I generalize Zarhin’s work to the superelliptic situation yn = f(x) where d := deg f is coprime to n. Then C has genus (n − 1)(d − 1)/2. The automorphism ζ : (x, y) → (x, ζny) of C induces one on its jacobian J , so 1 − ζ will be an endomorphism

• f J . I provide a formula for dividing by 1 − ζ. In terms of

Cg/Λ, this divides by the complex number 1 − e2πi/n. There is no Mumford representation because there are n points

• f C with the same x-coordinate. Instead, my representation for

points of J is n equations whose simultaneous vanishing locus gives g points on C. When n = 2, this recovers Zarhin’s formula.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 16

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Example of division by 1 − ζ

Consider the point (0, 4) on the genus 3 superelliptic curve C: y3 = (x + 8)(x + 1)(x − 1)(x − 8). Applying the formula to divide (0, 4) by 1 − ζ3 gives the three polynomials G1,1 = x2 + 20 − 5ζ−1

3 y

G1,2 = −21x − ζ−2

3 xy

G1,3 = 16 + 5x2 + 4y + y2 Solving the equations G1,1 = G1,2 = G1,3 = 0 on C gives the points

• (0, 4ζ3), (
• −20 − 105ζ3, −21ζ2

3), (−

• −20 − 105ζ3, −21ζ2

3)

• .

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 17

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Connections to Anderson–Ihara theory

Anderson–Ihara theory gives a connection between the Galois action on πpro-ℓ

1

(P1 \ {s1, . . . , sr}, ⋆) and the Galois action on the “higher circular units” associated to {s1, . . . , sr}. Superelliptic curves are branched n-fold coverings of P1, so when n = ℓk, Anderson–Ihara theory applies. Using my formula for 1 − ζn, I hope to make this connection explicit (fjnd equations for the higher circular ℓ-units) for small quotients of this π1.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 18

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Torsion points on curves

A torsion point of C is one whose image in J is torsion; i.e, some nonzero multiple of it is zero. More precisely, P is a torsion point with respect to a basepoint B if the degree zero divisor class [P − B] is torsion. When C has genus at least 2, Raynaud’s theorem states that C has only fjnitely many torsion points over C (with respect to a fjxed basepoint). How can we fjnd them all?

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 19

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Torsion points on the superelliptic Catalan curve

The Catalan problem asks: what are all pairs of consecutive perfect powers? The Catalan curve Cn,d is yn = xd + 1. The following are torsion points of Cn,d: { ∞

• 1-torsion

, (0, ⋆)

d-torsion

, (⋆, 0)

n-torsion

} Any torsion point not in this list will be called an “exceptional torsion point.” Theorem (A.). Assume n, d ≥ 2 coprime with g = (n − 1)(d − 1)/2 ≥ 2. There are no exceptional torsion points unless n + d = 7.

1 C2,5. Exceptional torsion points = (ζi

5

5

√ 4, ± √ 5). (5-torsion)

2 C4,3. Exceptional torsion points = (2ζi

3, ζj 4

√ 3) (12-torsion)

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 20

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Strategy for showing no other torsion points exist

Suppose we had an exceptional torsion point P.

1 We can produce more exceptional torsion points by

applying GQ and Aut(Cn,d) to P.

2 If there are enough torsion points, there will be relations

between them.

3 If there are low-degree relations, we get a low degree map

to P1, which would contradict the Castelnuovo–Severi inequality: if C has a degree d1 map and a degree d2 map to P1 where d1 and d2 are coprime, then genus(C) ≤ (d1 − 1)(d2 − 1).

4 If Castelnuovo–Severi isn’t suffjcient, we have another

geometric tool: Weierstrass weights. The sum of all the Weierstrass weights on a genus g curve must be g3 − g. Torsion points of small order have nonzero weight, so if there are too many of them, we get a contradiction. (This is a sketch when there are no exceptional torsion points.)

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 21

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Galois action on torsion

Informally, GQ acts by permuting roots of irreducible

• polynomials. For example, it can move

√ 2 to − √ 2 but cannot move √ 2 to √ 3. By proving a new congruence for Jacobi sums, I explicitly computed the GQ-action on the p-torsion of Jp,q. Theorem (A.). The fjeld generated by the p-torsion of Jp,q is Q   ζpq,

p

• p−1
• t=1

(1 − ζt

pζi q)

• cyclotomic unit

(t

j) : 1 ≤ i ≤ q − 1, 0 ≤ j ≤ p − 3

   . This means that if were to try to write explicit equations for Jp,q, the coeffjcients would lie in this fjeld. The cyclotomic units that show up are related to Iwasawa theory, Vandiver’s conjecture…

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 22

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Torsion on the generic superelliptic curve

Now consider a “generic” superelliptic curve yn = (x − a1) · · · (x − ad) where n, d ≥ 2 are coprime and a1, . . . , ad are indeterminates. Then {∞, (ai, 0)} are torsion points; defjne an exceptional torsion point to be one outside this list. Assume that the genus g = (n − 1)(d − 1)/2 is at least 2. Theorem (Poonen-Stoll). When n = 2, there are no exceptional torsion points. Theorem (A.). When d ≥ 3, there are no exceptional torsion

• points. When d = 2 and n ≥ 7, there are no exceptional torsion
• points. There are exceptional torsion points when

(d, n) = (2, 5), and they were already classifjed by work of Coleman.

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 23

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Thank you

Thank you for attending my thesis defense!

Vishal Arul’s Thesis Defense Explicit Division and Torsion Points 24

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