Pole order of $\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2}\left(=j(\tau)\right)$

Question

Concerning the relation $$j=\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2},$$ I understand, that the RHS is an element of $\mathbb{C}(j)$, and thus the LHS can be expressed as a polynomial in $\lambda$. I also understand, that if the order of the pole of the RHS at infinity is equal to 1, the highest degree of the polynomial can only be 1.

But no matter how I approach the RHS, the poleorder always seems to be two. For instance: If $\tau \to i\infty$, then $\lambda(\tau)\to 0$, thus we have a $0^2$ in the denominator. Thus the poleorder should be 2.

I suspect, that if I can express $\lambda$ as a 1 periodic Fourie series, that the problem may resolve, but my skills seem to be failing me.

Any help would be greatly appreciated.

Answer 1

No $j(\tau)$ is not a polynomial in $\lambda(\tau)$. You already have a rational function expression in $\lambda(\tau)$, it is unique.

Answer 2

The relation defines the modular lambda function, which is a Hauptmodul (generator of the function field) for the congruence subgroup $$\Gamma(2)=\left\{M\in SL(2,\mathbb Z): M\equiv \mathbb 1\mod 2\right\}.$$ The modular curve $\Gamma(2)\backslash \overline{\mathbb H}$ has three cusps $\tau=i \infty$, $\tau=0$ and $\tau=1$, each of width $2$. This corresponds to the rational function $j(\lambda)$ having second order poles at $\lambda\in\{0,1,\infty\}$.