Quasi-Modular Forms, E_2

Consider holomorphic modular forms and take their derivatives. The resulting functions are not modular.

In fact, let

$\displaystyle D=\frac{1}{2\pi i} \frac{d}{dz} = q\frac{d}{dq}.$

Take any weight ${k}$ modular form ${f}$ , that is a holomorphic function satisfying

$\displaystyle (c z+d)^{-k} f\left(\frac{a z+b}{c z+d}\right) =f(z)$

and ${f}$ is holomorphic at infinity, hence looks like

$\displaystyle f(z) =\sum_{n=0}^{\infty}a_n e(z).$

We have the following formula for it’s derivatives

$\displaystyle \begin{aligned} &(c z+d)^{-(k+2 m)} {D}^{m} f\left(\frac{a z+b}{c z+d}\right) \\ &\quad=\sum_{j=0}^{m}\left(\begin{array}{c} m \\ j \end{array}\right) \frac{(k+m-1) !}{(k+m-j-1) !}\left(\frac{1}{2 i \pi}\right)^{j} \mathrm{D}^{m-j} f(z)\left(\frac{c}{c z+d}\right)^{j}. \end{aligned}$

So instead of just the term ${D^{m} f}$ , we have extra terms involving the factors ${\left(\frac{c}{c z+d}\right)^{j}.}$

In general, a function that transforms similar to the above expression is called a quasi-modular function. More precisely, if

$\displaystyle (c z+d)^{-k} f\left(\frac{a z+b}{c z+d}\right)=\sum_{j=0}^{s} f_{j}(z)\left(\frac{c}{c z+d}\right)^{j}$

we call ${f}$ a quasi-modular of depth ${s}$ . Here ${f_i}$ are holomorphic functions. (We will see that they have to quasi-modular)

Note that the the first function ${f_0}$ should be equal to ${f}$ because if we take the identity matrix ${\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)}$ , the transformation gives ${f(z) = f_0(z).}$

The functions ${f_j}$ are quasi-modular and satisfy the equation

$\displaystyle (c z+d)^{-k+2j}f_j =\sum_{k=0}^{s-j}\left(\begin{array}{c}k+j \\ k\end{array}\right) f_{j+k} \left(\frac{c}{c z+d}\right)^{k}$

So they have weight ${k-2j}$ and depth ${s-j}$ .

The proof is easy, just compare the transformation formula for a product of matrices ${\gamma_1}$ and ${\gamma_2}$ to get some relations which can be inverted to the above formulae.

Quasi-Modular forms: If the function ${f_s}$ is a modular form, we call the function ${f}$ a quasi-modular form.

The ${f_i}$ ‘s have fourier expansions that have no negative terms, that is

$\displaystyle f_j =\sum_{n=0}^{\infty}a_j(n) e(nz).$

The derivatives ${Df}$ are quasi-modular form of weight ${k+2}$ and depth ${s+1}$ if ${f}$ has weigh ${k}$ and depth ${s}$ .

Let

$\displaystyle \Delta=q \prod_{n \geq 1}\left(1-q^{n}\right)^{24}=\eta(z)^{24}.$

Then the function

$\displaystyle E_2= \frac{D\Delta}{\Delta} =1-24 \sum_{n=1}^{+\infty} \sigma_{1}(n) e^{2 i \pi n z}$

is a quasi-modular form of weight ${2}$ and depth ${1}$ .

We have

$\displaystyle (c z+d)^{-2} E_2\left(\frac{a z+b}{c z+d}\right) = E_2(z) + \frac{6}{\pi i} \left(\frac{c}{c z+d}\right).$

Note that we can define ${E_2}$ using the usual formula

$\displaystyle G_{2}(z)=\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} \frac{1}{(m z+n)^{2}}$

but the sum is not absolutely convergent, that’s reason the proof of modularity fails in this cases, and is in fact reflected by the fact that we have an extra term in the transformation formula above.

We have a nice property. If ${f}$ is a modular form of weight ${k}$

$\displaystyle Df -\frac{k}{12}fE_2$

is a modular form of weight ${k+2}$ , although ${Df}$ is just quasi-modular.

We have the examples of derivatives of modular forms, $E_2$ , do we have other examples of quasi-modular forms?

No, all the quasi-modular forms (for ${SL_2(\mathbb Z)}$ are generated by ${E_2}$ and the ring of modular forms. That is ${\mathbb C[E_2, E_4, E_6]}$ is the algebra of quasi-modular forms. To see this note that ${f- \left(\frac{i \pi}{6}\right)^{s}f_s E_{2}^{s}}$ is of strictly smaller depth than ${f}$ .

So quasi-modular forms are modular forms and the almost modular ${E_2}$ and is closed under their derivatives.

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