L-function of Elliptic Curves, Complex Multiplication and Analytic Continuation

Consider an elliptic curve over ${\mathbb Z}$ in Weierstrass form given by

$\displaystyle Y^2=X^3+AX+B$

The discriminant ${\Delta}$ is defined as

$\displaystyle \Delta=-16\left(4 A^{3}+27 B^{2}\right)$

which is essentially the discriminant of the polynomial ${X^3+AX+B}$ .

Also define the invariant ${c_6}$ (coming from the coefficients of Weierstrass function)

$\displaystyle c_{6}=-864 B.$

The curve ${\bmod}$ ${p}$ is non-singular precisely when ${p \nmid \Delta}$ .

Let ${N_p}$ be the number of solutions to ${Y^2=X^3+AX+B}$ mod ${p}$ . We define the quantities

$\displaystyle a_p = p-N_p.$

At prime dividing the discriminant the points ${E(\mathbb F_p)}$ for either an additive group ${\mathbb F_p}$ or a multiplicative group isomorphic to ${\mathbb F_p^{*}}$ or a torus of norm one elements inside ${\mathbb F_{p^2}^{*}}$ This corresponds to the singular point being a cusp, node with tangent over ${\mathbb F_p}$ and a node with tangent over ${\mathbb F_p^2}$ . We call these cases additve reduction, split multiplicative reduction and non-split multiplicative reduction. Therefore ${a_p =0, 1, -1}$ in these cases.

Concretely we have

$\displaystyle \begin{aligned} &E \text { is non-singular } \Leftrightarrow \Delta \neq 0\\ &E \text { has node (mutiplicative, split/non-split} \Leftrightarrow \Delta=0, A \neq 0\\ &E \text { has cusp (additive reduction)} \Leftrightarrow \Delta=0, A=0. \end{aligned}$

Further

$\displaystyle \begin{aligned} &E \text { has split mutiplicative reduction} \Leftrightarrow \Delta=0, A \neq 0, -c_6 =\square \\ &E \text { has non-split multiplication reduction} \Leftrightarrow \Delta=0, A \neq 0, -c_6 \neq \square \end{aligned}$

For instance ${\Delta=0, A=0}$ implies that the curve looks like ${Y^2=X^3 \mod p}$ and hence ${N_p=p.}$ (The solutions are ${(X, Y)=(t^2, t^3), t\in \mathbb F_p}$ ) The multiplicative case is slightly harder.

Remark: These properties of reduction don’t depend on the choice of coordinates, but if we extend the scalar say from rationals to a bigger field, the reduction might change. For instance it’s possible that over a prime lying over our prime of additive reduction, the curve become non-singular!

Now form the L-function

$\displaystyle L(E, s) = \prod_{p \mid \Delta}\left(1-a(p) p^{-s}\right)^{-1} \prod_{p \nmid \Delta}\left(1-a(p) p^{-s}+p^{1-2 s}\right)^{-1}$

We have the estimate due to Hasse (RH for elliptic curves over ${\mathbb F_p}$ )

$\displaystyle |a(p)|<2 \sqrt{p}$

This implies that

$\displaystyle L(E,s)=\sum_{1}^{\infty} \frac{a(n)}{ n^{s}}$

converges for ${Re(s)> 3/2.}$

The series is defined by the Euler product and hence the analytic properties like continuation, functional equation are very hard to establish. In fact they are in general proved by proving the modularity of the elliptic curves. That is by showing that the same L-function comes from L-function for a cusp form on ${\Gamma_0(N)}$ .

The functional equation looks

$\displaystyle \left(\frac{\sqrt{N}}{2 \pi}\right)^{s} \Gamma(s) L(E, s)=\eta\left(\frac{\sqrt{N}}{2 \pi}\right)^{2-s} \Gamma(2-s) L(E,2-s)$

where ${N}$ is the conductor of the elliptic curve and ${\eta = \pm 1}$ the root number.

Conductor of ${E}$ is an important invariant that measure the “ramification”. Root numer is another important invariantt defined in terms of local root numbers. Both these quantities are hard to compute and depend a lot on the arithmetic properties of ${E}$ .

The conductor is made of the prime dividing the discriminant and we have

$\displaystyle N=\prod_{p}p^{f_p}$

$\displaystyle \begin{aligned} &E \text { has good reduction at } p \implies f_p=0\\ &E \text { has multiplication reduction at } p \implies f_p=1 \\ &E \text { has multiplication reduction at } p\ge 5 \implies f_p=2 \\ &E \text { has multiplication reduction at } p=2, 3 \implies f_p=2+\delta \end{aligned}$

The definition of ${\delta}$ is complicated.

The root number is defined as

$\displaystyle \eta=\prod_{p \leq \infty} W_{p}(E)$

$\displaystyle \begin{aligned} W_{\infty}=-1 \\ E \text { has good reduction at }p \implies W_{p}=1 \\ E \text { has split multiplicative reduction at} p \implies W_{p}=-1 \\ E \text { has nonsplit multiplicative reduction at} p \implies W_{p}=1\\ \end{aligned}$

The definitions for additive reduction for complicated.

But for special curves with complex multiplication, it’s easy to show that the L-functions matches with L-functions defined over the field of complex multiplication.

We are going to prove the analytic continuation and functional equation for the following family of elliptic curves.

$\displaystyle Y^2=X^3-k^2X$

The discriminant is ${\Delta =64k^6}$ , ${c_6 =0}$ .

So at prime dividing ${2k}$ , we have bad reduction. At primes dividing ${2k}$ , we have ${a_p =p}$ which implies all of them have additive reduction.

Detecting the square in terms of Legendre symbol, we get

$\displaystyle N_p = \sum_{(\bmod p)}\left(1+\left(\frac{x^3-k^2x}{p}\right)\right)$

$\displaystyle \implies a_p =-\sum_{x(\bmod p)}\left(\frac{x^{3}-k^{2} x}{p}\right)=-\left(\frac{k}{p}\right) \sum_{x(\bmod p)}\left(\frac{x^{3}-x}{p}\right)$

Therefore

$\displaystyle a_k(n) = \left(\frac{k}{n}\right) a_1(n)$

where ${a_k(n)}$ are the Dirichlet coefficients for ${y^2=x^3-k^2x}$ , ${a_1(n)}$ for ${y^2=x^3-x}$ .

Note that this formula is even true for primes dividing ${2k}$ because ${a(p) = p-N_p=0}$ in this cases.

So the L-function for parameter ${k}$ is a “twist” of the L-function for the curve ${y^2=x^3-x}$ by the quadratic character ${\left(\frac{k}{n}\right) }$

The curve ${y^2=x^3-x}$ has complex multiplication, that we have a map ${(x, y) \rightarrow (iy, -x)}$ that takes one solution to another.

We compute the coefficient ${a_1(n)}$ for this curve.

We already saw ${a_1(2)=0}$ .

$\displaystyle a_1(p)=-\sum_{x(\bmod p)}\left(\frac{x^{3}-x}{p}\right)$

Changing ${x}$ by ${(-x)}$ , we get

$\displaystyle a_1(p)=-\left(\frac{-1}{p}\right) \sum_{x(\bmod p)}\left(\frac{x^{3}-x}{p}\right)=\sum_{x(\bmod p)}\left(\frac{x^{3}-x}{p}\right)$

So for ${p =3 \mod 4}$ , ${\left(\frac{-1}{p}\right) =-1 \implies a_1(p)=0.}$

Let’s fix a prime ${p=1 \mod 4.}$ We have

$\displaystyle p=a^2+b^2=\pi \bar{\pi}$

where ${\pi =a+ib}$ is chosen so that ${\pi = 1 \mod 2(1+i)}$ . That is ${a +b \equiv1 \mod 4}$ .

Now the above exponential sum can be computed in terms of Jacobi sums and we get

$\displaystyle a_1(p) =2a.$

For every ${ \alpha}$ in ${\mathbb Z[i]}$ coprime to ${ (2(1+i))}$ , there exist a unit ${u(\alpha)}$ such that ${u(\alpha) \alpha = 1 \mod (2(1+i))}$

Now consider the following character on ${ \left(\mathbb Z[i]/(2(1+i)) \right)^{\times}}$ defined by

$\displaystyle \chi((\alpha)) = u(\alpha) \alpha$

So we we have ${(\alpha) = (a+ib)}$ with ${a+b\equiv 1 \mod 4}$ , and we set ${\chi((\alpha)) =a+ib}$ .

Now it has an attached L-function

$\displaystyle L(s, \chi) =\sum_{\mathfrak a} \frac{\chi(\mathfrak a)}{N(\mathfrak a)^s} = \prod_{\mathfrak p}\left(1-\chi(\mathfrak{p})(N \mathfrak p)^{-s}\right)^{-1}$

with coefficients

$\displaystyle \lambda(n) =\sum_{N\mathfrak a =n}\chi(\mathfrak a).$

The previous computations on ${a_1(p)}$ can used to check that

$\displaystyle \lambda(n) = a_1(n).$

${\lambda(2)}$ is zero because ${ 2= (1+i) (1-i) \equiv 0 \mod 2(1+i)}$
${\lambda(p)}$ for ${p =3 \mod 4}$ is zero because ${p}$ cannot be written as sums of squares ie as ${N(\mathfrak a)}$ . For ${p=1 \mod 4}$ ,

$\displaystyle \lambda(p) = \lambda(\mathfrak p) + \lambda(\bar{ \mathfrak p}) = a+ib + a-ib =2a = a_1(p).$

Thus the problem is reduced to showing analytic continuation of this L-function. This is now reduced to harmonic analysis on ${\mathbb X/\mathbb Z[i]}$ , that is how ${\mathbb Z[i]}$ sits inside ${\mathbb C}$ as a lattice. Just like ${\mathbb Z \hookrightarrow \mathbb R}$ , and poisson summation gives the analytic properties of ${\zeta(s)}$ and Dirichlet L-functions, Poisson summation over ${\mathbb Z[i]}$ gives analytic continuation for the above L-function.

In fact, the theta series

$\displaystyle f(z)=\frac{1}{4} \sum_{\alpha \in \mathbb{Z}[i]} u(\alpha) \alpha e\left(z|\alpha|^{2}\right)$

is the object of interest. It’s a cusp form of weight ${2}$ that is attached to the elliptic curve. The L-function of ${f}$ matches with the L-function ${L(s, \chi)}$ and hence the L-function of the elliptic-curve.