Regarding the derivative of the $j$-invariant

Question

Is anyone aware of a formula for the derivative of the $j$-invariant $j(\tau)$ with respect to $\tau$? Here, $\tau$ is in the upper half-plane.

I would image there are probably quite a few formulae for $j'(\tau)$, but they are not well-known. I looked through a few books but I could not find a single formula.

Any help would be greatly appreciated.

Answer 1

Yes, $\frac{j'(\tau)}{j(\tau)}=-\frac{E_6(\tau)}{E_4(\tau)}$ (here $\ '=\frac{d}{2 \pi i d \tau}$), where $E_4$ and $E_6$ are Eisenstein series.

Here's a straightforward proof: Use $j(\tau)=\frac{E_4^3(\tau)}{\eta(\tau)^{24}}$, known fact that $\vartheta_{k}(f):=f'-\frac{k}{12}E_2(\tau) f$ maps modular forms of weight $k$ to modular forms of weight $k+2$, and ${\rm dim}(M_{14}(SL(2,\mathbb{Z}))=1$, spanned by $E_4^2(\tau) E_6(\tau)$. This allows you to conclude that $\frac{j'(\tau)}{j(\tau)}$ is proportional to $\frac{E_6(\tau)}{E_4(\tau)}$.

Answer 2

One may recursively prove the formula, for any $\gamma\in\mathrm{SL}_2(\mathbb R)$, $$\frac{d^k}{dz^k}(f|_{\gamma,1-k})=\big(\frac{d^k}{dz^k}f\big)|_{\gamma,k+1}$$ where if $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $$f_{\gamma,k}(z):=(cz+d)^{-k}f(\gamma z).$$ The formula shows $\frac{d^k}{dz^k}$ of a modular function of weight $1-k$ is a modular function of weight $k+1$. In particular, since $j$ is nearly holomorphic of weight $2$ (i.e., is holomorphic everywhere except for $\infty$), $j'(z):=\frac1{2\pi i}\frac d{dz}j(z)=q\frac d{dq}j(z)$ is nearly holomorphic of weight $2$.

Moreover, $j(z)=q^{-1}+O(1)$, so $j'(z)=-q^{-1}+O(1)$. Thus, $\Delta\cdot j'$ is a modular form of weight $12+2$ (since $\Delta=\eta^{24}$ is cuspidal). However, the space of modular forms is the graded algebra $\mathbb C[E_4,E_6]$, so $M_{14}=\mathbb C E_4^2E_6$. Thus, $j'(z)\Delta(z)=c E_4^2E_6$ for some $c\in\mathbb C$. Comparing Fourier coefficients tells us $c=-1$.