- Prove that \(\wp ^ { \prime \prime }\) is a quadratic polynomial in \(\wp\)
Solution
The Weierstrass function is elliptic, so \(\wp''\) and \(\wp^2\) are elliptic. Near a lattice point, by translation it suffices to work near \(0\). From the defining series, \[ \wp(z)=\frac1{z^2}+c_2z^2+O(z^4) \] for some constant \(c_2\); there are no odd powers because \(\wp\) is even. Consequently \[ \wp''(z)=\frac6{z^4}+2c_2+O(z^2), \qquad 6\wp(z)^2=\frac6{z^4}+12c_2+O(z^2). \] Thus \[ H(z):=\wp''(z)-6\wp(z)^2 \] has a removable singularity at every lattice point. Hence \(H\) extends to an entire elliptic function. It is bounded on a fundamental parallelogram and therefore bounded on \(\mathbb C\); Liouville’s theorem gives \(H\equiv C\) for some constant \(C\). Therefore \[ \boxed{\wp''(z)=6\wp(z)^2+C}, \] which is a quadratic polynomial in \(\wp\).
More explicitly, if \[ g_2=60\sum_{\omega\in\Lambda\setminus\{0\}}\omega^{-4}, \] then the Laurent expansion gives \(c_2=g_2/20\), hence \(C=-g_2/2\) and \[ \wp''=6\wp^2-\frac{g_2}{2}. \]