Lyness Cycles, Elliptic Curves, and Hikorski Triples Jonny Griffiths, Maths Dept Paston Sixth Form College Open University, June 2012

MSc by Research, UEA, (Two years part-time) Supervisors: Professor Tom Ward Professor Graham Everest Professor Shaun Stevens

Mathematics

The Structure of this Talk 1. Lyness cycles (periodic recurrence relations) 2. An introduction to elliptic curves 3. The link between Lyness cycles and elliptic curves 4. Hikorski triples 5. Cross-ratio-type functions 6. Conclusions

Part 1: Lyness Cycles

u n+1 = u n + u n-1, The Fibonacci sequence z = x + y u 0 = 0, u 1 = 1 x, y, x + y, x + 2y, 2x + 3y, 3x + 5y,... Can this be periodic? x = 2x + 3y, y = 3x + 5y x = 0, y = 0. x y Order-2, periodic for these starting values, (locally periodic).

Can we have a recurrence relation that is periodic for (almost) all starting values? Globally periodic Globally periodic behaviour is very atypical of difference equations, and accordingly only a very restrictive class of functions f(x 1, x 2,...) exhibit this behaviour. Mestel. x 1, x 2,..., x n, f(x 1,...,x n ), f(x 2,...,f(x 1,...x n ))..., x 1, x 2... Order-n, period-m x m+1, x m+2 x n+1, x n

Robert Cranston Lyness,

Mathematical Gazette, 1942

Globally periodic for x and y non-zero, order-2, period-5.

Imagine you have series of numbers such that if you add 1 to any number, you get the product of its left and right neighbours. Then this series will repeat itself at every fifth step! The difference between a mathematician and a non- mathematician is not just being able to discover something like this, but to care about it and to be curious why it's true, what it means and what other things it might be connected with. In this particular case, the statement itself turns out to be connected with a myriad of deep topics in advanced mathematics: hyperbolic geometry, algebraic K-theory, and the Schrodinger equation of quantum mechanics. I find this kind of connection between very elementary and very deep mathematics overwhelmingly beautiful. Zagier

Regular Lyness Cycles Order 1 Order 2 Order 3

Order-1 regular Lyness cycles: what periods are possible?

For what values of n does have solutions for a, b, c, d and k in Q? Related question: what are the finite subgroups of GL 2 (Q)/N, Where N = { }? Answer: n = 1, 2, 3, 4, 6.

But that is it for rational coefficients...

..... so  is a root of  K (x), AND a rational quadratic equation. Since  K (x) is irreducible,  (K) = 1 or 2 (where  is the totient function).  (1) = 1,  (2) = 1,  (3) = 2,  (4) =2,  (5) = 4,  (6) =2,  (n) > 2 for n > 6. Proof: Cull, Flahive, Robson

Regular Lyness cycles order-2: what periods are possible? Possible periods: 2, 3, 4, 5, 6, 8, 12 All coefficients rational.

Can add constants easily to these Pseudo-cycle

But that is it for rational coefficients... Symmetric QRT maps (Quispel, Roberts and Thompson). Tsuda has given a theorem that restricts the periods for periodic symmetric QRT maps to 2, 3, 4, 5, 6.

Note: x, y, ky - x,... can have any period if you are choosing k from R. k = 1/ , period 5, k = √2, period 8, k = , period 10. x, y, |y| - x,... is period 9.

Part 2: An Introduction to Elliptic Curves

ax + by + c = 0 Straight line ax 2 + bxy + cy 2 + dx + ey + f = 0 Conics Circle, ellipse, parabola, hyperbola, pair of straight lines

ax 3 + bx 2 y + cxy 2 + dy 3 + ex 2 + fxy + gy 2 + hx + iy + j = 0 Elliptic curves UNLESS The curve has singularities; a cusp or a loop or it factorises into straight lines... y 2 = x 4 + ax 3 + bx 2 + cx + d can be elliptic too...

Any elliptic curve can be transformed into Weierstrass Normal Form Y 2 = X 3 + aX + b using a birational map; that is, you can get from the original curve to this normal form and back again using rational maps; The curves are said to be ISOMORPHIC.

a = -2.5 b = 1 a = 2.5 b = 1 Y 2 = X 3 + aX + b

For example... Transforming to Normal Form

This becomes...

Where does a straight line cross our elliptic curve in normal form? We are solving simultaneously y = mx + c, y 2 = x 3 + ax + b which gives x 3 - m 2 x 2 + x(a - 2cm) + b - c 2 = 0

This is a cubic equation with at most three real roots. Note; if it has two real roots, it must have a third real root. So if we pick two points on the curve, the line joining them MUST cut the curve in a third point.

P+Q+R=0 P+Q=-R

We can form multiples of a point by taking the tangent at that point.

Sometimes we find that kP = 0. In this case we say that P is a torsion point. y 2 =x P=0 P is of order 6

Amazing fact... The set of points on the curve together with this addition operation form a group. Closed – certainly. We want P and –P to be inverses. So P + -P = 0, and we define 0, the identity here, as the point at infinity.

Associativity? Geogebra demonstration demonstration

Notice also that if a, b are rational, then the set of rational points on the curve form a group. Closed – certainly. x 3 - m 2 x 2 + x(a - 2cm) + b - c 2 = 0 Inverses and identity as before If two roots are rational, the third must be. y = mx + c connects two rational points, so m and c must be rational.

Mordell Theorem (1922) Let E be an elliptic curve defined over Q. Then E(Q) is a finitely generated Abelian group. (Mordell-Weil Theorem [1928] generalises this.)

Siegel’s Theorem (1929) If a, b and c are rational, (and if x 3 + ax 2 + bx + c = 0 has no repeated solutions), then there are finitely many integer points on y 2 = x 3 + ax 2 + bx + c.

Mordell’s Theorem implies that E(Q) is isomorphic to E torsion (Q)  Z r The number r is called the RANK of the elliptic curve. How big can the rank be? Nobody knows.

y 2 + xy = x 3 − x Largest rank so far found; 18 by Elkies (2006) Curves of rank at least 28 exist.

Mazur’s Theorem (1977) The torsion subgroup of E(Q) is isomorphic to Z/nZ for some n in {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12} or to Z/2nZ  Z/2Z for some n in {1, 2, 3, 4}.

Part 3: The link between Lyness cycles and elliptic curves

x = 3, k = -14 Other roots are -7, -2, -1/3.

x = 3, k = -19/3 Other roots are -7, -2, -1/3.

An elliptic curve !

Additional note: simplifies to exactly the same set of curves. This curve has 5-torsion Note:

Choose A = (a, b) to be on the curve, which gives k. X = (0,-1) is a torsion point of order 5. What happens if we repeatedly add X to A? We expect to get back to A, and we do. But we get this sequence of points... The recurrence that built the curve is linked geometrically to adding a torsion point on it.

Does this work for other periods?

Note: This doesn’t work for period 8 and 12 pseudocycles.

Mazur’s Theorem link? Wouldn’t we expect the possibilities in Mazur’s Theorem to link with the possible periods for Lyness cycles? Mazur: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12 Lyness order-2 cycles: 2, 3, 4, 5, 6, 8, 12

Part 4: Hikorski Triples

What does mean to you?

What do mean to you?

Adding Speeds Relativistically Suppose we say the speed of light is 1. How do we add two parallel speeds?

Try the recurrence relation: x, y, (xy+1)/(x+y)… Not much to report... Try the recurrence relation: x, y, (x+y)/(xy+1)… still nothing to report...

But try the recurrence relation: x, y, -(xy+1)/(x+y)… Periodic, period 3

Now try the recurrence relation: x, y, -(x+y)/(xy+1)… Also periodic, period 3 Are both periodic, period 6

GCSE Resit Worksheet, 2002 How many different equations can you make by putting the numbers into the circles? Solve them!

Suppose a, b, c, and d are in the bag. If ax + b = cx + d, then the solution to this equation is x = There are 24 possible equations, but they occur in pairs, for example: ax + b = cx + d and cx + d = ax + b will have the same solution. So there are a maximum of twelve distinct solutions.

This maximum is possible: for example, if 7, -2, 3 and 4 are in the bag, then the solutions are:

If x is a solution, then –x, 1/x and -1/x will also be solutions. ax + b = cx + d a + b(1/x) = c + d(1/x) c(-x) + b = a(-x) + d a + d(-1/x) = c + b(-1/x)

The solutions in general will be: {p, -p, 1/p, -1/p} {q, -q, 1/q, -1/q} and {r, -r, 1/r, -1/r} where p, q and r are all ≥ 1

It is possible for p, q and r to be positive integers. For example, 1, 2, 3 and 8 in the bag give (p, q, r) = (7, 5, 3). In this case, they form a Hikorski Triple (or HT).

Are (7, 5, 3) linked in any way? Will this always work?

a, b, c, d in the bag gives the same as a + k, b + k, c + k, d + k in the bag. Translation Law

a, b, c, d in the bag gives the same as ka, kb, kc, kd in the bag. Dilation Law

So we can start with 0, 1, a and b (a, b rational numbers with 0 < 1 < a < b) in the bag without loss of generality.

a, b, c, d in the bag gives the same as -a, -b, -c, -d in the bag. Reflection Law

Suppose we have 0, 1, a, b in the bag, with 0 < 1 < a < b and with b – a < 1 then this gives the same as –b, – a, – 1, 0 which gives the same as 0, b – a, b – 1, b which gives the same as 0, 1, (b – 1)/(b – a), b/(b – a) Now b/(b – a) – (b – 1)/(b – a) = 1/(b – a) > 1

If the four numbers in the bag are given as {0, 1, a, b} with 1 1, then we can say the bag is in Standard Form. So our four-numbers-in-a-bag situation obeys three laws: the Translation Law, the Reflection Law and the Dilation Law.

Given a bag of numbers in Standard Form, where might the whole numbers for our HT come from?

The only possible whole numbers here are:

(b – 1)/a must be the smallest here.

Twelve solutions to bag problem are:

Pythagorean Triples Parametrisation? (2rmn, r(m 2 - n 2 ), r(m 2 + n 2 )) Hikorski Triples Parametrisation?

How many HTs are there? Plenty...

H(n) = # of HTs in which n appears = d(n 2 -1). All n > 2 feature in an least 4 HTs. If n separates a pair of primes, then d(n 2 -1) = 4. For how many n does n feature in exactly 4 HTs?

Is abc unique for each HT? The Uniqueness Conjecture If (a, b, c) and (p, q, r) are non-trivial HTs with abc = pqr, then (a, b, c) = (p, q, r).

Part 5: Cross-ratio-type functions

The Cross-ratio

Takes six values as A, B and C permute: Form a group isomorphic to S 3 under composition

So the cross-ratio and these cross-ratio-type functions all obey the three laws: the Translation Law, the Reflection Law and the Dilation Law. Cross-ratio-type functions

and Lyness Cycles x y ? ? = (x+y-1)/(x-1) Find a in terms of x and e in terms of y and then substitute...

x y

What if we try the same trick here? x y z ?

And here? x y ? So this works with the other cross-ratio type functions too...

Note: all the Lyness cycles generated by the various cross-ratio methods are regular.

Part 6: Conclusions

Consider the example we had earlier: is a periodic recurrence relation.

has a torsion point of order 3 when X = 0. X

If we take a point (p, q) on map to our cubic, add X and then map back, we get: (p, q) maps to (q, -(pq+1)/(p+q)) The recurrence that built the curve is linked geometrically to adding a torsion point on it.

is an elliptic curve

has integral points (5,3), (3,-2), (-2,5) If the uniqueness conjecture is true... and (30,1), (1,-1), (-1,30) and (1,30), (30,-1), (-1,1) and (3,5), (-2,3), (5,-2)

Why the name?

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