# Weierstrass elliptic function $\wp$ and $g_2 = 0$

Let $$\wp$$ be the Weierstrass ellptic function. One can show that $$\wp''(z) = 6\wp(z)^2 - \frac{g_2}{2}$$ where $$g_2 = \sum_{0\neq \omega \in L} \frac{60}{\omega^4}$$. Since I want to investigate the number of zeroes of $$\wp''$$, knowing that $$\wp$$ has degree 2 (when looked as a map from the usual lattice $$L = \{m\omega_1 + n\omega_2\}$$ to $$\mathbb{C}\cup \infty$$) it really depends on whether $$g_2 = 0$$ or not.

So can we describe the pairs $$(\omega_1, \omega_2)$$ for which $$g_2=0$$? I am not even sure whether there are such pairs. I have no idea how to approach this.

$$j$$ has exactly one zero $$\tau=e^{2i\pi /3}$$ in the fundamental domain $$\{ |z| \ge 1,\Im(z) >0,\Re(z)\in [-1/2,1/2)\}$$ of $$SL_2(\Bbb{Z}) \setminus \mathcal{H}$$ so $$j(z) =0$$ iff $$z = \frac{a\tau+b}{c\tau+d}$$ for some $$a,b,c,d\in \Bbb{Z},ad-bc=1$$ and $$g_2( u\Bbb{Z}+v\Bbb{Z})=0$$ iff $$(u,v) = (s(a\tau+b),s(c\tau+d))$$ for some $$a,b,c,d\in \Bbb{Z},ad-bc=1,s \in \Bbb{C}^*$$
For any $$L$$, $$\wp_L''(z) = 6\wp_L(z)^2 - \frac{g_2(L)}{2}$$ has $$4$$ zeros counted with multiplicity, if $$g_2(L) = 0$$ or $$6\wp_L(\omega/2)^2- \frac{g_2(L)}{2}=0$$ it is $$2$$ double zeros, otherwise it is $$4$$ simple zeros
• Sorry, I am still new to this $\wp$ function - how do we know that $\tau = e^{2\pi i/3}$ is the unique zero in the fundamental parallelogram? – DesmondMiles Apr 28 at 2:51
• As $g_2$, $j$ is a modular function, not an elliptic function en.wikipedia.org/wiki/J-invariant as $\wp$ it is meromorphic on some compact Riemann surface, we know it has one simple pole (at $i\infty$), so it has one simple zero – reuns Apr 28 at 2:52