Let \(\sigma\) be the canonical Weierstrass sigma function of a lattice. Prove that \[ \wp'(z)=-\frac{\sigma(2z)}{\sigma(z)^4}. \]
Solution
Set \[ G(z)=-\frac{\sigma(2z)}{\sigma(z)^4}. \] We prove that \(G\) and \(\wp'\) are the same elliptic function.
1. The function \(G\) is \(\Lambda\)-periodic.
For a basis period \(\lambda_i\), E-HM213A17-140 gives \[ \sigma(z+\lambda_i) =-e^{\eta_i(z+\lambda_i/2)}\sigma(z). \] Applying this identity twice, with an arbitrary variable \(w\), gives \[ \begin{aligned} \sigma(w+2\lambda_i) &=-e^{\eta_i(w+3\lambda_i/2)}\sigma(w+\lambda_i)\\ &=e^{2\eta_i(w+\lambda_i)}\sigma(w). \end{aligned} \] Taking \(w=2z\) yields \[ \sigma(2z+2\lambda_i) =e^{4\eta_i z+2\eta_i\lambda_i}\sigma(2z). \] On the other hand, \[ \sigma(z+\lambda_i)^4 =e^{4\eta_i z+2\eta_i\lambda_i}\sigma(z)^4. \] The factors cancel, so \[ G(z+\lambda_i)=G(z) \qquad(i=1,2). \] Hence \(G\) is elliptic for \(\Lambda\).
2. Modulo \(\Lambda\), \(G\) has a triple pole at \(0\) and simple zeros at the three nonzero points of order two.
The sigma function has simple zeros precisely at the lattice points.
At \(z=0\), both \(\sigma(2z)\) and \(\sigma(z)\) vanish simply. Thus the quotient \[ \frac{\sigma(2z)}{\sigma(z)^4} \] has pole order \(4-1=3\).
If \(c\notin\Lambda\) and \(2c\in\Lambda\), then \(\sigma(2c)=0\) while \(\sigma(c)\ne0\). Since the zero of \(\sigma\) at \(2c\) is simple and multiplication by \(2\) is locally biholomorphic, \(\sigma(2z)\) has a simple zero at \(c\). Thus \(G\) has a simple zero at each of the three nonzero classes in \[ E[2]=\tfrac12\Lambda/\Lambda. \] There are no other zeros or poles modulo \(\Lambda\).
3. The elliptic function \(\wp'\) has the same divisor.
Near a lattice point, periodicity reduces to the expansion at \(0\): \[ \wp(z)=z^{-2}+O(z^2), \qquad \wp'(z)=-2z^{-3}+O(z). \] Thus \(\wp'\) has a triple pole at \(0\) modulo \(\Lambda\). If \(c\notin\Lambda\) and \(2c\in\Lambda\), then oddness and periodicity give \(\wp'(c)=\wp'(c-2c)=-\wp'(c)\), so \(\wp'(c)=0\). Since \(\wp'\) has a single triple pole modulo \(\Lambda\), it has exactly three zeros counted with multiplicity; hence its zeros are exactly the three nonzero points of order two, each simple. Hence \(G\) and \(\wp'\) have identical zeros and poles with identical multiplicities.
4. Their quotient is constant, and the normalized principal term makes that constant \(1\).
By <1>1–<1>3, \[ H(z)=\frac{G(z)}{\wp'(z)} \] is an elliptic function with no zeros or poles. It therefore descends to a holomorphic function on the compact torus \(\mathbb C/\Lambda\) and is constant.
From the canonical product, \[ \sigma(z)=z+o(z) \qquad(z\to0). \] Consequently \[ G(z) =-\frac{2z+o(z)}{(z+o(z))^4} \sim -\frac2{z^3}. \] Since also \[ \wp'(z)\sim-\frac2{z^3}, \] we have \[ \lim_{z\to0}\frac{G(z)}{\wp'(z)}=1. \] Thus \(H\equiv1\), and therefore \[ \boxed{\wp'(z)=-\frac{\sigma(2z)}{\sigma(z)^4}.} \]