Numerical Evidence for the Birch andSwinnerton-Dyer Conjecture

Brendan CreutzUC Primer, May 23 2014

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A quote“The subject of this lecture is rather a special one. I want to describesome computations undertaken by myself and Swinnerton-Dyer onEDSAC, by which we have calculated the zeta-functions of certain

elliptic curves."

Bryan Birch, 1965

Figure : Bryan Birch, 2011Brendan Creutz Evidence for BSD UC Primer 2 / 19

The Electronic Delay Storage Automatic Calculator (Cambridge 1958)

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Rational points on algebraic curves

A polynomial relation F (x , y) = 0 defines an algebraic curve C.The points on C correspond to solutions of F (x , y) = 0.When the coefficients of F (x , y) are rational numbers, it isreasonable to consider rational points, i.e., points whosecoordinates are rational numbers.

QuestionsHow many rational points are there on a given curve?How can we find the rational points on a given curve?

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Equations of degree two

Example

Rational points on the curve C : x2 + y2 − 1 = 0 correspond toPythagorean triples.

Figure : Pilmpton Tablet 1700BC

Example

The curve C : x2 + y2 + 1 = 0 has no rational points.

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Equations of degree three

Example (Diophantus, ca 300AD)There are infinitely many rational points on the genus one curve

C : y2 = x3 − x + 9

Figure : Pilmpton Tablet 1700BC

(−1,3) ( 199 , 109

27

)

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Equations of degree three

Example (Fermat, 1637; Euler, 1770)There are only finitely many rational points on the genus one curve

C : x3 + y3 = 1

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Equations of degree three

Example (Stoll, 2002)

The curve C : y2 = x3 + 7823 has infinitely many rational points. Thesmallest has coordinates:

x =2263582143321421502100209233517777143560497706190989485475151904721

y =186398152584623305624837551485596770028144776655756

1720094998106353355821008525938727950159777043481

The moralSolving equations over Q is hard.

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An easier problem

Solving equations modulo an integer is much easier.

Example

The equation y2 ≡ x3 + 7823 hastwo solutions modulo 2: (0,1) and (1,0).five solutions modulo 5: (1,2), (1,3), (2,1), (2,4), and (3,0)two solutions modulo 7: (0,2) and (0,5).

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The Conjecture (informally)The Birch and Swinnerton-Dyer conjecture suggests that one candetermine the number of rational points on a cubic curve by countingthe number of solutions modulo p for enough prime numbers p.

Figure : Sir Peter Swinnerton-Dyer

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Elliptic Curves

DefinitionAn elliptic curve is a cubic curve with a rational point.

Every elliptic curve can be defined by an equation of the form

E : y2 = x3 + ax + b with a,b ∈ Q such that 4a3 − 27b2 6= 0.

Theorem (Mordell 1922)The set E(Q) of rational points on anelliptic curve forms a finitely generatedabelian group. Hence,

E(Q) ' Zr × (finite group)

QP

P + Q

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Solving equations over Q is hardThere is no known algorithm to finding generators of E(Q).There is no known algorithm for finding the rank of E(Q).

Try something easierFor each prime number p, E gives a curve over the finite field Fp

E/Q // E/Fpy2 = f (x) // y2 ≡ f (x) (mod p)

Then try and relate the twoSolutions over Q give solutions over Fp. So

E(Q) infinite ?=⇒ many solutions modulo p

E(Q) finite ?=⇒ few solutions modulo p

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p versus Np

The number Np of solutions modulo p is approximately p.

Birch and Swinnerton-Dyer considered∏

p≤M

Np

p.

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∏p≤M

Np

p?∼ log(M)r

• for an elliptic curve of rank 0• for an elliptic curve of rank 1

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∏p≤M

Np

p?∼ log(M)r

• for an elliptic curve of rank 0• for an elliptic curve of rank 1• for an elliptic curve or rank 2

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The L-series (zeta function)

DefinitionThe L-series associated to an elliptic curve E is the function of thecomplex variable s,

L(E , s) :=∏

p prime

(1−

ap

ps +1

p2s−1

)−1

,

where ap = p + 1− Np (for all but finitely many primes).

A formal (and completely unjustified) calculation gives

L(E ,1) =∏

p

pNp

.

So we expect: L(E ,1) = 0 ⇐⇒ E(Q) is infinite.

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Conjecture 1 (Birch and Swinnerton-Dyer, 1962)For any elliptic curve E , the Taylor expansion of L(E , s) at s = 1 hasthe form

L(E , s) = c(s − 1)r + higher order terms

where r is the rank of E(Q).

Millennium Problem: prove or disprove Conjecture 1.

Conjecture 2 (Birch and Swinnerton-Dyer, 1964?)The leading coefficient is given by the formula

c =#X(E) · Reg(E) ·

∏p Tp

(#E(Q)tors)2

Problem: prove that X(E) is finite.

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What is known

Theoretical EvidenceIf ords=1 L(E , s) ≤ 1, then r = ords=1 L(E , s).(Gross-Zagier, Kolyvagin 1980s,Wiles 1994, Breuil et al. 2001)The rank conjecture is true for a positive proportion of curves.(Bhargava-Shankar 2014)

Computational EvidenceConj. 1 holds for all curves of rank ≤ 3 and conductor < 350k.(2 million curves) (Cremona et al. 2000s)Conj. 2 holds for all curves of rank ≤ 1 and conductor < 5k.(35 thousand curves) (Grigorov et al 2009; Creutz-Miller 2012)

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Thank You!

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