It seems when people talk about modular forms they tend to forget that they are very related to families of elliptic curves. Here I want to explain some simple way to understand the connection. We will consider modular forms for the full modular group $SL(2, \mathbf Z)$.

So consider the simplest family of elliptic curves, the Weierstrass family:

$y^2=x^3+ax+b.$

Here $a$ and $b$ are meant to be some formal parameters. This indeed defines and elliptic curve over the ring $k[a,b,\Delta^{-1}]$, where $k$ is the base field, which is supposed to be of characteristic ${0}$, and $\Delta$ is the discriminant:

$\Delta=-16(4 a^3 + 27 b^2).$

When we write $y^2=x^3+ax+b$ we in fact mean the corresponding projective variety $E$ over $Spec\; k[a,b]$ with equation

$\bar y^2 \bar z = \bar x^3 + a \bar x^2 \bar z + b \bar z^3.$

Let us denote the affine chart with coordinate functions $x, y$ by $U = E\setminus \{0\}$ and the point at infinity by ${0}$ since it is the zero point for the addition on the curve.

Now we are going to compute some Laurent series expansions at ${0}$. First we choose local parameter $t=-x/y$. Indeed, $x$ has pole of order $2$ and $y$ has pole of order $3$ at ${0}$, therefore $t$ has simple zero there. To find expansion of $x$ we solve the following equation in Laurent series:

$\frac{x^2}{t^2} = x^3 + a x + b.$

Rewriting it as

$(x t^2)^3 - (x t^2)^2 + a t^4 (x t^2) + b t^6=0$

we obtain a polynomial equation in $x t^2$ which can be solved by Newton’s method starting with $x t^2 = 1 + O(t)$. We obtain

$x = t^{-2}-a t^2-b t^4-a^2t^6 - 3 a b t^8+O(t^{10}),$

$y= -t^{-3}+a t + b t^3 + a^2 t^5 + 3 a b t^7+O(t^9).$

Let us compute the expansion of the invariant differential $\omega = \frac{dx}{2y}$:

$\frac{dx}{2y} = (1+2a t^4+3b t^6+6 a^2 t^8+20 a b t^{10}+O(t^{12})) dt.$

We see that it is possible to integrate this series formally and make it the new local parameter:

$z = \int \frac{dx}{2y} = t+ \frac{2a}5 t^5+\frac{3b}7 t^7+\frac{2 a^2}3 t^9+\frac{20 a b}{11} t^{11}+O(t^{13}).$

Then the expansions of $x$ and $y$ with respect to the new local parameter are:

$x = z^{-2}-\frac{a}5 z^2-\frac{b}7 z^4+\frac{a^2}{75}z^6 + \frac{3 ab}{385}z^8+O(z^{10}),$

$y = \frac{\partial}{2\partial z}x = -z^{-3}-\frac{a}5 z -\frac{2b}7 z^3 + \frac{a^2}{25} z^5 + \frac{12 a b}{385} z^7+O(z^9).$

We also consider the formal integral of $-x dz$:

$v_0:=-\int x dz = z^{-1} +\frac{a}{15}z^3 +\frac{b}{35} z^5 - \frac{a^2}{525}z^7 - \frac{ab}{1155} z^9 + O(z^11).$

Consider the power series

$\frac{1}{e^{z}-1} + \frac{1}{2}-\frac{z}{12}= z^{-1} + \sum_{k\geq 2}\frac{B_{2k}}{(2k)!}z^{2k-1}.$

If we substitute this power series in place of $v_0$ and find

$x = -\frac{\partial v_0}{\partial z} = \frac{1}{(e^z-1)^2} + \frac{1}{e^z-1} + \frac{1}{12},$

$y = \frac{\partial x}{2 \partial z} = -\frac{1}{(e^z-1)^3} - \frac{3}{2(e^z-1)^2} - \frac{1}{2(e^z-1)},$

then we can easily verify that

$y^2=x^3 - \frac{x}{48} + \frac{1}{864},$

i.e. we have found a solution for $a=-\frac{1}{48}$, $b=\frac{1}{864}$.

This explains that we should in general put

$a = -\frac{E_4}{48} \qquad b=\frac{E_6}{864}$

and define $E_{2k}$ in such a way that

$v_0 = z^{-1} + \sum_{k\geq 2}\frac{B_{2k} E_{2k}}{(2k)!}z^{2k-1}.$

In this way we obtain $E_{2k}$ as a polynomial of $E_4$, $E_6$, but in fact it is true that this polynomial is the same polynomial that expresses the Eisenstein series of weight $2k$ in terms of the Eisenstein series of weights $4$ and $6$. So for us modular forms will be homogeneous polynomials of $a$ and $b$ where weight of $a$ is $4$ and weight of $b$ is $6$.

To define the weight more geometrically let us consider the action of the multiplicative group on $E$:

$(a,b,x,y)\longrightarrow (\lambda^4 a, \lambda^6 b, \lambda^2 x, \lambda^3 y), \qquad (\lambda\in k^\times).$

Then a modular form $f$ of weight $k$ is a function of $a, b$ which transforms like

$f \longrightarrow \lambda^k f.$

If we consider not only functions of $a, b$, but functions of $a, b, x, y$ then we obtain Jacobi forms of index ${0}$.

Derivatives of modular forms

We want to apply this language to understand some natural operations on modular forms. The first operation is the Euler derivative $\delta_e$. This simply takes a modular form $f$ of weight $k$ and sends it to $kf$. It is easy to see that this is exactly the action of the Lie algebra of the multiplicative group. Next we want to reconstruct the Serre derivative.

Suppose we have a derivation $\partial$ on $k[a,b]$. Let us try to lift it to obtain a derivation of the ring of functions on $U$ (which is generated by $a,b,x,y$). We would have $\partial^* x$, $\partial^* y$ satisfying a relation

$2 y \partial^* y = (3 x^2 + a) \partial^* x + x \partial a + \partial b.$

But note that we could simply apply $\partial$ to the Laurent series expansions of $x,y$ term by term (denote it by $\partial x, \partial y$) and get a solution to the relation above. Therefore the difference must satisfy

$2y (\partial y-\partial^* y) = (3 x^2 + a) (\partial x - \partial^* x).$

But we also have a solution to the equation above! Namely it is the operator $\frac{d}{dz}$ which will be denoted simply by $'$. Therefore we must have a Laurent series $\alpha$ which satisfies

$\partial y = \partial^* y + \alpha y',\qquad \partial x = \partial^* x+ \alpha x'.$

Using the fact that $', \partial$ commute it is easy to obtain

$\alpha' x'=2\partial^* y - (\partial^* x)'.$

We expect $\partial^* x, \partial^* y$ to be regular functions on $U$. Clearly one can assume $\partial^* x$ to contain only even powers of $z$ and $\partial^* y$ to contain only odd powers of $z$ – this corresponds to $\partial^*$ being invariant under the involution $(x,y)\longrightarrow (x,-y)$. We see that the right hand side is a regular function on $U$ which contains only odd powers of $z$. Therefore it is a product of $y$ and a polynomial in $x, a, b$. So we write

$\alpha' x' = 2 y P(x).$

Noting that $x' = 2y$ gives

$\alpha' = P(x)$.

Next observation is that for any polynomial in $x$ we can express it as a derivative of an expression of the form

$y Q(x) + A z + B v_0 \qquad (Q\in k[a,b][x], \; A\in k[a,b],\; B\in k[a,b].)$

In fact $z$ is the formal integral of $\omega=dz$ and $v_0$ is the formal integral of $-\eta = -\frac{x dx}{2y} = -x dz$ and these forms generate the first cohomology of $E$. So,

$\alpha = y Q(x) + A z + B v_0.$

It implies that

$\partial x = R(x) + (A z + B v_0) x' \qquad (R\in k[a,b][x]).$

But we know that

$\partial x = -\frac{\partial a}{5} z^2 - \frac{\partial b}{7} z^4 + O(z^6).$

Looking at the power series expansions we conclude that

$\partial x = A( z x'+ 2x) + B(v_0 x' + 2 x^2 + \frac {4a}3).$

So it is natural to consider a derivation for which $(A,B)=(1,0)$ and a derivation for which $(A,B) = (0,1)$. In the former case we obtain

$\alpha = z,\; \partial a = 4 a, \; \partial b = 6b,\; \partial^*x = 2x,\; \partial^* y=3y.$

It is easy to see that we have got the Euler operator. In the latter case we obtain

$\alpha = v_0,\; \partial a = 6 b,\; \partial b = - \frac{4 a^2}{3},\; \partial^* x = 2 x^2 + \frac{4a}3,\; \partial^* y = 3xy.$

Using our convention $a = -\frac{E_4}{48} \; b=\frac{E_6}{864}$ one can see that this is the Serre derivative $\delta_s$:

$\delta_s E_4 = -\frac{E_6}{3},\qquad \delta_s E_6 = -\frac{E_4^2}{2}.$

It is important that we did not only obtain $\delta_s$ as a certain canonical derivation which lifts to a derivation on $k[U]$, but we also computed $\delta_s x$ which can be interpreted as a formula which gives the Serre derivatives of all the Eisenstein series.

In the end I would like to mention that using this approach and studying the Gauss-Manin connection one can explain some other things which appear in the theory of modular and quasi-modular forms and seem mysterious, like Bol’s identity and Rankin-Cohen brackets.

The main idea is: “the ring of modular forms, or the ring of quasi-modular forms come naturally equipped with an elliptic curve over it.

Also here is a useful formula for values of modular forms. If $f$ is a modular form of weight $k$ and a curve $y^2 = x^3 + a_0 x + b_0$ has periods $\omega_1, \omega_2$, then

$f(a_0,b_0) = f(\frac{\omega_1}{\omega_2}) \left(\frac{\omega_2}{2\pi i}\right)^{-k}$.

On the left we have the values of $f$ as a polynomial of $a, b$ and on the right we have its value as a function on the upper half plane. There is a corresponding formula relating values of quasi-modular forms and periods of differentials of second kind.