2016-03-07-bsd-part2.sagews

Birch and Swinnerton-Dyer (part 2)

William Stein

Monday, March 7, 2016

Recall: BSD for elliptic curves over $\QQ$

BSD Rank: $\text{ord}_{s=1} L(E,s) = \text{rank}(E(\QQ))$

BSD Formula:

$$\frac{L^{(r)}(E,1)}{r!} = \frac{\Omega_E \cdot \prod c_p \cdot \text{Reg}_E \cdot \#\text{Sha}_E}{\#E(\QQ)_{\rm tor}^2}.$$

Exercise: Choose a curve of rank $2$ and compute what you can about the BSD formula for it.

Hint: use elliptic_curves.rank(2) to get at some curves of algebraic rank $2$.

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**Exercise:** Choose a curve of rank $4$ and compute what you can about the BSD formula for it.

Hint: use `elliptic_curves.rank(4)` to get at some curves of algebraic rank $4$.

Exercise: Choose a curve of rank $4$ and compute what you can about the BSD formula for it.

Hint: use elliptic_curves.rank(4) to get at some curves of algebraic rank $4$.

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## A Deep Theorem

**Theorem (Birch, Gross, Zagier):**
If $r_{\rm an} = \text{ord}_{s=1} L(E,s) \leq 3$, then there is an algorithm to compute $r_{\rm an}$.

Why?

A Deep Theorem

Theorem (Birch, Gross, Zagier): If $r_{\rm an} = \text{ord}_{s=1} L(E,s) \leq 3$, then there is an algorithm to compute $r_{\rm an}$.

Why?

Birch contribution: Connect $L(E,1)$ to modular symbosls.

See ((2.8.7) of [Cremona]) $L(E,1) = -2\pi i \cdot \int_{0}^{\infty} f_E(z)dz = - \langle \{0,\infty\}, f\rangle \in \RR$

The modular symbol $e = \{0,\infty\}$ is called the winding element (e is the first letter of "winding" in French).

Let $\Lambda_E = \langle H_1(X_0(N),\ZZ), f\rangle \subset \CC$, so that $E(\CC) = \CC/\Lambda_E$.

For $p\nmid N$, we have $(T_p - (p+1))(e) \in H_1(X_0(N),\ZZ)$. Using this (and the Hasse bound that $|a_p|< 2\sqrt{p}$), one can show that $\langle e, f\rangle \in \QQ\Lambda_E$.

More precisely, we get that for each $p\nmid N$, $$((p+1) - a_p )L(E,1) \in \Lambda_E$$ Note that $a_p = p+1 - \#E(\FF_p)$, so $(p+1) - a_p = \#E(\FF_p)$.

$\Omega_E$ is the least real element of $\Lambda_E$ (or twice it) and $\#E(\FF_p)$ is related to $\# E(\QQ)_{\rm tor}$.

The argument above shows that $$\frac{L(E,1)}{\Omega_E} \in \QQ.$$ With a little more work, it even shows that $$\# E(\QQ)_{\rm tor} \cdot L(E,1)/\Omega_E \in \ZZ,$$ which is evidence for the BSD formula.

Computing the image of $L(E,1)$ in $\QQ \Lambda$ is something one can do very algebraically. Just use linear algebra and Hecke operators to find a map $$H_1(X_0(N),\QQ,\{\text{cusps\}}) \to \QQ$$ with the same kernel as $x\mapsto \langle x, f\rangle$. Then compute the image of $e$ under this map. That exactly computes $L(E,1)/\Omega_E \in \QQ$.

(NOTE: I've brushed over issues of the "Manin constant"...)

Gross-Zagier contribution: Connect $L'(E,1)$ to rational points.

Elliptic curves are modular, so there is surjective morphism $\varphi: X_0(N)\to E$.

$X_0(N)$ is (coarse) moduli space of isomorphism classes of pairs $(F,C)$, where $F$ is an elliptic curve and $C$ is a cyclic subgroup of order $N$.

If $K$ is a quadratic imaginary field in which all primes dividing $N$ split, then the theory of CM elliptic curves produces an element $z \in X_0(N)(\CC)$, defined over some class field of $K$.

Heegner point: The trace of $\varphi(z)$ is a naturally constructed element of $y_K\in E(K)$!

Theorem (Gross-Zagier): $L'(E/K,1) = \text{(explicit nonzero factor)} \cdot h(y_K)$, where $h(y_K)$ is the canonical height of $y_K$.

Here $L(E/K,s) = L(E/\QQ,s) \cdot L(E^K/QQ,s)$, where $E^K$ is the quadratic twist of $E$ over $K$.

Theorem (Bump-Friedberg-Hoffstein, or Murty-Murty, or Waldspurger): There is a $K$ such that $\text{ord}_{s=1} L(E^K,s) \leq 1$ (and has opposite parity to $r_{\rm an}$).

Now suppose $r_{\rm an} =1$. Using Birch we see that $L(E,1)=0$. Then computing to some precision we find that $L'(E,1)\neq 0$ and we're done. (Aside: to prove BSD rank in this case -- Choose $K$ so that $L'(E/K,1)\neq 0$. Then the Gross-Zagier formula above implies that $y_K$ has infinite order and that $E(\QQ)$ has rank at least $1$. To get rank at most 1, one has to use an ingenous argument of Kolyvagin that involves a lot more Heegner points and Galois cohomology, and which led to one of the most central ideas in modern number theory ("Euler systems"). In this case, the formula above is so explicit that we also know that $L'(E/\QQ,1)/(\Omega_E\text{Reg}_E) \in \QQ$ as predicted by BSD.)

If $r_{\rm an} = 2$, then we can use Birch to prove that $L(E,1)=0$. Using the functional equation we automatically get that $L'(E,1) =0$. We then can compute $L''(E,1)$ to a few digits to prove that it is nonzero. choose a $K$ so that $L'(E^K,1)\neq 0$

If $r_{\rm an} = 3$, then we use Birch to check that $L(E,1)=0$ and from the functional equation we know that the order of vanishing is odd. Find a $K$ so that $L(E^K,1)\neq 0$. Then $L'(E/K,1) = \text{(explicit nonzero)} h(y_K)$. We then explicitly compute $y_K$ -- it's a concrete point after all -- and compute its height. We find that it is $0$, hence conclude that $0 = L'(E/K,1) = L'(E/\QQ,1) \cdot L(E^K,1)$. Since $L(E^K,1)\neq 0$, we conclude that $L'(E/\QQ, 1)=0$. We automatically get $L''(E/\QQ,1)=0$ by the functional equation. We then compute $L'''(E/\QQ,1)$ to some precision and get that it is nonzero.

If $r_{\rm an} = 4$, we can show that $L(E/\QQ,1)=0$ using Birch, and that $L'(E/\QQ,1)=0$ from parity. We can compute all day and night and see that $L''(E/\QQ,1)=0.00000000\ldots$. But there is no known (not even crazy wild conjecture!) for some relationship like $$L''(E/\QQ,1) = \text{height of something...}.$$ So there's no point $y_K$ to compute explicitly.