Tutorial: The Birch and Swinnerton-Dyer Conjecture system:sage

Tutorial: The Birch and Swinnerton-Dyer Conjecture


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Nonsingular Plane Curves

A nonsingular plane algebraic curve is the set of solutions to a (nonsingular) polynomial:

$$F(X,Y) = 0$$

A rational point is $(x,y) \in \mathbf{Q}\times\mathbf{Q}$ such that $F(x,y) = 0$.

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Rational Points on Plane Curves

We find rational points on the curve you type in.

{{{id=63| @interact def f(F = ('0 = F(x,y) = ', 'x^2 + y^2 - 1'), search_bound=selector([1..10], buttons=True), box=(1..20), square=False) : R. = QQ[] try: F = R(F.lower()) except: print "Enter a polynomial in x, y with rational coefficients."; return C = Curve(F) P = C.rational_points(bound=search_bound) show(tuple(P)) eps = 0.1 xmax = max([box]+[p[0] for p in P])+eps; ymax = max([box]+[p[1] for p in P])+eps xmin = min([-box]+[p[0] for p in P])-eps; ymin = min([-box]+[p[1] for p in P])-eps g = implicit_plot(F, (x,xmin,xmax), (y,ymin,ymax), plot_points=300) if len(P) > 0: g += points(P,pointsize=40) if square: show(g, aspect_ratio=1) else: show(g) /// }}}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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The Congruent Number Problem

Definition: An integer $n$ is a congruent number if $n$ is the area of a right triangle with rational side lengths.

Open Problem: Give an algorithm to decide whether or not an integer $n$ is a congruent number.

This is a 1000-year old open problem, perhaps the oldest open problem in mathematics.

{{{id=39| T = line([(0,0), (3,0), (3,4), (0,0)],rgbcolor='black',thickness=2) lbl = text("3",(1.5,-.5),fontsize=28) + text("4",(3.2,1.5),fontsize=28) lbl += text("5",(1.5,2.5),fontsize=28) lbl += text("Area $n = 6$", (2.1,1.2), fontsize=28, rgbcolor='red') show(T+lbl, axes=False) /// }}} {{{id=40| /// }}} {{{id=3| /// }}}

Congruent Numbers and the BSD Conjecture

Theorem: A proof of the Birch and Swinnerton-Dyer Conjecture would also solve the congruent number problem.

Proof: Suppose $n$ is a positive integer.  Consider the cubic curve $y^2 = x^3 - n^2 x$.  Using algebra (see next slide), one sees that this cubic curve has infinitely many rational points if and only if there are rationals $a,b,c$ such that $n=ab/2$ and $a^2 + b^2 = c^2$.  The Birch and Swinnerton-Dyer conjecture gives an algorithm to decide whether or not any cubic curve has infinitely many solutions.

 

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Explicit Bijection

In fact, there is a bijection between

$$
 A = \left\{(a,b,c) \in \mathbf{Q}^3 \,:\, \frac{ab}{2} = n,\, a^2 + b^2 = c^2\right\}
$$
and
$$
 B = \left\{(x,y) \in \mathbf{Q}^2 \,:\, y^2 = x^3 - n^2 x, \,\,\text{with } y \neq 0\right\}
$$
given by the maps
$$
  f(a,b,c) = \left(-\frac{nb}{a+c},\,\, \frac{2n^2}{a+c}\right)
$$
and
$$
  g(x,y) = \left(\frac{n^2-x^2}{y},\,\,-\frac{2xn}{y},\,\, \frac{n^2+x^2}{y}\right).
$$

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5 is a Congruent Number

{{{id=5| n = 5; x,y = var('x,y') C = EllipticCurve(y^2 == x^3 - n^2 * x); C /// }}} {{{id=60| show(C.plot(), figsize=4) /// }}} {{{id=8| P = C.gens()[0] print P print "order of P = ", P.order() /// }}} {{{id=41| (-62279/1728)^2 == (1681/144)^3 - 25*(1681/144) /// }}} {{{id=86| /// }}} {{{id=43| /// }}}

1 is Not a Congruent Number

{{{id=45| n = 1 x,y = var('x,y') C = EllipticCurve(y^2 == x^3 - n^2 * x) C /// }}} {{{id=44| C.gens() /// }}} {{{id=47| /// }}}

Problem: Which positive integers $n\leq 10$ are congruent numbers?

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Finding Explicit Rational Right Triangles

{{{id=68| @interact def _(n=6, triangles=(1..10)): x,y = var('x,y') C = EllipticCurve(y^2 == x^3 - n^2*x) G = C.gens() html("rank = %s\n\n"%len(G)) if len(G) == 0: print "%s is not a congruent number"%n; return def g(x,y,n): return ((n^2-x^2)/y, -2*x*n/y, (n^2+x^2)/y) P = G[0] for i in [1..triangles]: a,b,c = g((i*P)[0], (i*P)[1], n) html("a=%s, b=%s, c=%s\n"%(a,b,c)) /// }}} {{{id=99| /// }}}

Problem: Find a rational right triangle with area $2009$.

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The $L$-function

Let $C$ be a cubic curve (+ a technical condition I'm not mentioning). For each prime number $p$, let $N_p$ be the number of solutions to the cubic modulo $p$.

Definition: For any cubic curve $C$, let $a_p = p-N_p$.

Theorem (Hasse): $|a_p| < 2\sqrt{p}$.

Theorem (Wiles et al.): The function $$L(C,s) = \prod_p\left(\frac{1}{1-a_p p^{-s} + p^{1-2s}}\right)$ extends to an entire complex-analytic function on $\mathbf{C}$.

{{{id=83| L = EllipticCurve([-2009^2,0])._pari_().elllseries show(line([(i,L(i)) for i in [0,0.03,..,2]]), figsize=[7,1.5], ymax=10) /// }}} {{{id=80| /// }}}

Problem: Use the following interact to plot the $L$-series for many small values of $n$ (e.g., 1,2,3,4,5,6,7,8,9,10).  For which $n\leq 10$ does $L(1)=0$?

{{{id=81| @interact def example(n=6): L = EllipticCurve([-n^2,0])._pari_().elllseries show(line([(i,L(i)) for i in [0,0.03,..,2]]), figsize=[7,1.5]) /// }}} {{{id=84| /// }}} {{{id=49| /// }}} {{{id=27| /// }}}

The Birch and Swinnerton-Dyer Conjecture

Heuristic Observation: If $C$ has infinitely many rational points, then the numbers $N_p$ will tend to be "large".  Since $L(C,1) "=" \prod_p \frac{p}{N_p}$, the number $L(C,1)$ will tend to be small.

Theorem (Mordell): There is a finite set $P_1, \ldots, P_r$ of rational points on $C$ so that all (non-torsion) rational points can be generated from these using a simple geometric process (chords and tangents). 

We call the smallest $r$ in Mordell's theorem the rank of $C$.

Conjecture (Birch and Swinnerton-Dyer): $$\text{ord}_{s=1} L(C,s) = \text{rank}(C)$$

This problem, exactly as stated, is the Clay Math Institute Million Dollar prize problem in algebraic number theory.  We proved above that its solution would also resolve the 1000-year old congruent number problem.

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Examples of the BSD Conjecture

{{{id=32| @interact def _(n=6): x,y = var('x,y') C = EllipticCurve(y^2 == x^3 - n^2*x) show(C) print "rank = ", C.rank(), "\n" L = C.lseries() print "L-series = ", L.taylor_series(1,53, 4) /// }}} {{{id=85| /// }}} {{{id=36| /// }}} {{{id=22| /// }}}

The Kolyvagin -- Gross-Zagier Theorem

Theorem: If $\text{ord}_{s=1} L(C,s) \leq 1$ then the Birch and Swinnerton-Dyer conjecture is true for $C$.


The proof involves Heegner points, modular curves, Euler systems and Galois cohomology.



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My Current Research

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