Differentiating the inverse Weierstrass P-function

Question

I will begin with some background:

The solutions $\theta$ of $$\cos \theta=x $$ constitute of two families, each of which is an arithmetic progression. Namely, if $\arccos x$ denotes any particular solution then the families are $$\left\{ \arccos x+2 \pi k: k \in \mathbb{Z} \right\} \cup \left\{ -\arccos x+2 \pi k: k \in \mathbb{Z} \right\}. $$ If, instead, we solve the system $$\cos \theta=x \\ \sin \theta=y $$ with $x,y$ satisfying the compatibility condition $$x^2+y^2=1 $$ then the solutions $\theta$ form a single arithmetic progression $$\{\theta^*+2 \pi k: k \in \mathbb{Z} \} $$ where $\theta^*$ is any particular solution. Since $\theta$ is then uniquely defined up to additive integer multiples of $2 \pi$, the differential $\mathrm{d} \theta$ is well-defined, and one can see that $$\mathrm{d} \theta=-y \mathrm{d} x+x \mathrm{d} y. $$

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Analogously, one can think of the equation $$\wp(\theta;g_2,g_3)=x $$ as being an underdetermined system, since it has the reflection symmetry $\theta \mapsto -\theta$, as well as the translations $\theta \mapsto \theta+2 \omega_i, \; i=1,2$, where $2\omega_1,2\omega_2$ are generators of the period lattice $$\Lambda=\{2m \omega_1+2n \omega_2:m,n \in \mathbb{Z} \}.$$ If, instead one looks at the system $$\wp(\theta;g_2,g_3)=x \\ \wp'(\theta;g_2,g_3)=y$$ with $x,y$ satisfying the compatibility condition $$y^2=4x^3-g_2 x-g_3 $$ then $\theta$ is uniquely defined only up to translations by points on the period lattice. In fact, one of the branches of $\theta$ is what Mathematica calls InverseWeierstrassP[{x,y},{g2,g3}] or $$\wp^{-1} (x,y;g_2,g_3).$$

My questions are:

• Is there a nice formula for $\mathrm{d} \theta$ in this case?
• Similarly, are there simple formulas for the partial derivatives below? $$\frac{\partial}{\partial x} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) \\ \frac{\partial}{\partial y} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) $$

Answer 1

Edit and correction (thanks to Nemo for pointing out):

First note that even though $d\theta$ is well defined, it is well defined as a differential of a map on $S^1$ and not on $\mathbb{R}^2$, which means that its representation as a linear sum of $dx$ and $dy$ is not unique. In fact, the tangent space (and the cotangent space) are one-dimensional. Therefore, the choice of map varies the differential (as a differential on $\mathbb{R}^2$) - e.g. the linear combination you have written can be obtained by using $\theta = \arctan(y/x)$.

In our case, the explicit description of this branch of the inverse Weierstrass function is $\wp^{-1}(x,y)=\int_{-\infty}^{x} \frac{1}{\sqrt{4t^3-g_2t-g_3}}dt$. This means that $$ d\theta = y^{-1}dx + (6x^2 - \frac{g_2}{2})^{-1}dy $$

This makes sense regarding the previous answer, where I had forgotten to invert the results in the end.

Old Answer:

Note that the differential $d\theta$ is simply the gradient of the path defined above. In the case of $\wp$ we see that the path is given by $\theta \mapsto (\wp(\theta), \wp'(\theta)) = (x,y)$, hence the gradient is given by $(\wp'(\theta), \wp''(\theta))$. Next, we derive the functional equation $$ (\wp'(\theta))^2 = 4(\wp(\theta))^3 - g_2 \wp(\theta) - g_3 $$ to obtain $$ 2 \wp'(\theta) \wp''(\theta) = 12 (\wp(\theta))^2 \wp'(\theta) - g_2 \wp'(\theta) $$ and hence $$ \wp''(\theta) = 6(\wp(\theta))^2 - \frac{g_2}{2} $$ Recalling that $\wp(\theta) = x$, we see that $\wp''(\theta) = 6x^2-\frac{g_2}{2}$, hence we may write $$ d\theta = ydx + (6x^2 - \frac{g_2}{2})dy $$

Hope that this fully answers your question.

Answer 2

For a fixed $g_2$, we will consider a change of coordinates from $(x,y)$ to $(g_3,\theta)$ defined by the equations

$$\begin{align}x&=\wp(\theta;g_2,g_3), \\ y&=\wp'(\theta;g_2,g_3). \end{align} $$

Taking the differential of both equations, one gets

$$\begin{align} \mathrm{d} x&=\wp'(\theta;g_2,g_3) \mathrm{d} \theta+\frac{\partial \wp}{\partial g_3}(\theta;g_2,g_3) \mathrm{d} g_3 \\ \mathrm{d} y&=\wp''(\theta;g_2,g_3) \mathrm{d} \theta+ \frac{\partial \wp'}{\partial g_3}(\theta;g_2,g_3) \mathrm{d} g_3 \end{align} $$

Simplifying, using the 2nd order ODE for $\wp$, and identity 18.6.19 in Abramowitz and Stegun, yields

$$\begin{align} \mathrm{d} x=&y \mathrm{d} \theta+\frac{y \left(6 g_2 \zeta \left(\theta ;g_2,g_3\right)-9 g_3 \theta \right)+12 g_2 x^2-18 g_3 x-2 g_2^2}{2 \left(g_2^3-27 g_3^2\right)} \mathrm{d} g_3 \\ \mathrm{d} y=&\left( 6x^2-\frac{1}{2} g_2 \right) \mathrm{d} \theta+\frac{6 g_2 \left(12 x^2-g_2\right) \zeta \left(\theta ;g_2,g_3\right)-54 g_3 \left(2 \theta x^2+y\right)+9 g_2 \left(g_3 \theta +4 x y\right)}{4 \left(g_2^3-27 g_3^2\right)} \mathrm{d} g_3 \end{align} $$

Now, inverting these relations gives $$\begin{align} d\theta=&\frac{ \left(6 g_2 \left(g_2-12 x^2\right) \zeta \left(\theta ;g_2,g_3\right)+54 g_3 \left(2 \theta x^2+y\right)-9 g_2 \left(g_3 \theta +4 x y\right)\right)}{2 \left(9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3\right)}\mathrm{d} x+\frac{\left(6 g_2 \left(y \zeta \left(\theta ;g_2,g_3\right)+2 x^2\right)-9 g_3 (2 x+\theta y)-2 g_2^2\right)}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3} \mathrm{d} y \\ dg_3=&\frac{ \left(g_2^3-27 g_3^2\right) \left(12 x^2-g_2\right)}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3}\mathrm{d} x-\frac{2 \left(g_2^3-27 g_3^2\right) y}{9 g_2 x \left(g_3+8 x^3-2 y^2\right)+27 g_3 \left(y^2-4 x^3\right)-18 g_2^2 x^2+g_2^3} \mathrm{d} y \end{align} $$

Lastly, since $g_3=4x^3-g_2 x-y^2$ the equations take the form

$$ \begin{align} \mathrm{d} \theta=&\frac{ \left(3 g_2 \left(-2 \left(g_2-12 x^2\right) \zeta \left(\theta ;g_2,4 x^3-g_2 x-y^2\right)-3 g_2 \theta x+48 \theta x^3+30 x y-3 \theta y^2\right)-54 \left(4 x^3-y^2\right) \left(2 \theta x^2+y\right)\right)\mathrm{d} x}{108 g_2 x \left(y^2-4 x^3\right)+54 g_2^2 x^2-2 g_2^3+54 \left(y^2-4 x^3\right)^2}+\frac{ \left(g_2 \left(-6 y \zeta \left(\theta ;g_2,4 x^3-g_2 x-y^2\right)+2 g_2-30 x^2-9 \theta x y\right)+9 \left(4 x^3-y^2\right) (2 x+\theta y)\right) \mathrm{d} y}{54 g_2 x \left(y^2-4 x^3\right)+27 g_2^2 x^2-g_2^3+27 \left(y^2-4 x^3\right)^2} \\ \mathrm{d}g_3=&(12x^2-g_2)\mathrm{d} x-2y \mathrm{d} y \end{align}.$$ This finally gives $\frac{\partial \theta}{\partial x},\frac{\partial \theta}{\partial y} .$