Periods and $\mathscr{P}(w)$ of the elliptic curve $y^2=4(x-1)(x-2)(x-3)$

Question

I changed the curve to Weierstrass form as suggested: $y^2=4(x+1)x(x-1)$ We have $e_1=-1$, $e_2=0$, $e_3=1$ and $e_1<e_2<e_3$. Firstly, I will find the half periods $w_1,w_2,w_3$.

$\begin{array}. w_3=\int_1^\infty\frac{1}{\sqrt{4(x+1)x(x-1)}}dx\overset{x=1+\tan^2\theta}=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{2-sin^2\theta}}d\theta=\frac{1}{\sqrt{2}}K(\frac{1}{\sqrt{2}}).\end{array}$

$\begin{array}.w_2=w_3+\int_0^1\frac{1}{\sqrt{4(x+1)x(x-1)}}dx\overset{x=\tan^2\theta}=w_3+\int_0^{\frac{\pi}{4}}\frac{\pm i}{\sqrt{1-2sin^2\theta}}d\theta=\frac{1}{\sqrt{2}}K(\frac{1}{\sqrt{2}})\pm i\frac{1}{\sqrt{2}}K(\frac{1}{\sqrt{2}}).\end{array}$

$\begin{array}.w_1=w_2+\int_{-1}^0\frac{1}{\sqrt{4(x+1)x(x-1)}}dx\overset{x=-1+\tan^2\theta}=w_2+\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{(1-2\cos^2\theta)(1-3\cos^2\theta)}}d\theta=\frac{1}{\sqrt{2}}K(\frac{1}{\sqrt{2}})\pm i\frac{1}{\sqrt{2}}K(\frac{1}{2})\pm \frac{1}{\sqrt{2}}K(\frac{1}{\sqrt{2}}).\end{array}$

So, $w_3=\frac{1}{{\sqrt{2}}}K(\frac{1}{\sqrt{2}})$ and $w_1=i\frac{1}{{\sqrt{2}}}K(\frac{1}{\sqrt{2}})$.

Also, its associated Weierstrass $\mathscr{P}$-function is: $$\mathscr{P}(w)=e_3+\frac{e_1-e_3}{sn^2(\sqrt{e_1-e_3}w)}=3-\frac{2}{sn^2(i\sqrt{2}w)}$$ where $k=\frac{\sqrt{e_2-e_3}}{\sqrt{e_1-e_3}}=\frac{1}{\sqrt{2}}$.

My questions:

1. Are my computations correct? I am confused about the standart formulas/definitions. I just took them from a book. Especially those $\pm$s confused me.

2. I don't know much about the Weierstrass's $\mathscr{P}$-function. What is the use of it for the elliptic curve?

3. Solve $\mathscr{P}(w)=0$.

Thanks for any helpfull comments.

Answer 1

You are a bit confused and almost correct. Your question is about the elliptic curve $$ y^2=4(x-1)(x-2)(x-3). $$ Note that this is not in the standard Weierstrass form because the coefficient of $\;x^2\;$ is not zero. Thus, let $\;x = X+2\;$ and now the equation is $$ y^2=4(X+1)X(X-1)=4X^3-4X=4X^3 -g_2X-g_3 . $$ where $\;g_2 = 4,\; g_3 = 0.\;$ Now the roots of the cubic are $$e_1=1,\quad e_2=-1,\quad e_3=0.$$ The corresponding real half period $\; w_1 \approx 1.3110\;$ and the imaginary half period $$ w_2=iw_1 \quad \text{ while }\quad w_3 = -w_1-w_2 = -(1+i)w_1.$$ Verify that $$ \wp(w_1) = \wp(-w_1) = e_1,\; \wp(w_2) = \wp(-w_2) = e_2,\; \wp(w_3) = \wp(-w_3) = e_3. $$

You wrote:

I am confused about the standard formulas/definitions. I just took them from a book.

but did not mention which book. I recommend the DLMF Chapter 23 about Weierstrass elliptic functions. It specifically mentions the different conventions about notations for these elliptic functions which can be confusing.

To make the connection of your elliptic curve to Weierstrass functions, let $\;x(z) = \wp(z)+2,\; y(z) = \wp'(z).\;$ Then verify that $\; y(w_1) = y(w_2) = y(w_3) = 0\;$ and that $\; x(w_1) = 3,\; x(w_2) = 1,\; x(w_3) = 2.\;$

I used the Mathematica code

g23 = {N[4, 20], 0};
{w1, w2} = WeierstrassHalfPeriods[g23];

wp[z_, g23_:g23] := WeierstrassP[z, g23];
wpp[z_, g23_:g23] := WeierstrassPPrime[z, g23];
x[z_] := wp[z] + 2;
y[z_] := wpp[z];
xy[z_] := {x[z], y[z]};
chop[v_] := Rationalize[v, 10^20];

w3 = -w1-w2; (* or w3 = w1+w2 *)
vw = {w1, w2, w3};
ve = wp[vw];
vxy = xy /@ vw;

Print["e1..e3 = ", chop[ve]];
Print["xy1..e3 = ", chop[vxy]];
(*
e1..e3 = {1, -1, 0}
xy1..e3 = {{3, 0}, {0, 0}, {2, 0}}
*)

and you can use the code yourself in the Wolfram Cloud.

Alternatively, you can use this PARI/GP code in your browser.

[g2, g3] = [4, 0];
E = ellinit(-[g2, g3]/4);
[w1, w2] = ellperiods(E)/2;
w3 = -w1-w2; /* orw3 =  w1+w2 */
vw = [w1, w2, w3];
ve = [ellwp(E, z) | z <- vw];
vxy = [xy(z) | z <- vw];
xy(z) = ellwp(E, z, 1) + [2, 0];
chop(v) = [bestappr(z, 10^20) | z <- v];

print("e1..3 = ", chop(ve));
print("xy(w1..3) = ", chop(vxy))
/* 
e1..3 = [1, -1, 0]
xy(w1..3) = [[3, 0], [1, 0], [2, 0]]
 */