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Whittaker and Watson mention that when the invariants $g_2, g_3$ of Weierstrass $\wp$ function are real and such that $g_2^3 - 27g_3^2 > 0$, and if $2\omega_1$ and $2\omega_2$ are its periods then the function takes real values on the perimeter of the parallelogram with vertices $0,\omega_1,-\omega_2,-\omega_1-\omega_2$.

I am not able to understand why $\wp$ takes real values. Can someone help?

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  • $\begingroup$ That's §20.32, ex.2 which presumes from ex.1 that $g_2,g_3$ are real. $\endgroup$
    – ccorn
    Jun 18, 2015 at 20:35
  • $\begingroup$ @ccorn Indeed as I told you the invariants must be real, I am editing the question and adding the hypothesis. Also what is the reference of the book where the exercise is taken?? $\endgroup$ Jun 18, 2015 at 21:28
  • $\begingroup$ Whittaker & Watson: A Course of Modern Analysis. Merchant Books, 2nd edition 1915. They actually talk about a rectangle in §20.32 ex.2 (p. 437) after having requested the reader to establish purely real/imaginary periods as part of ex.1. $\endgroup$
    – ccorn
    Jun 18, 2015 at 23:06

1 Answer 1

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Late answer, but here it is why.

If $g_2, g_3 \in \mathbb{R}$ and $g_2^3-27g_3^2>0$ it can bee seen that the half periods are of the form $\omega_1=\alpha$ and $\omega_2=i \beta$ with $\alpha, \beta \in \mathbb{R}$.

This essentially does the trick since it is easily seen by a direct computation that if $\wp(z)=\wp(z|\alpha, i \beta)$ then \begin{equation} \overline{\wp(z)}=\wp\left(\overline{z}\right) \tag{1} \end{equation} Now take $z \in \{ iy, \ \alpha+iy, \ x, \ x+i \beta : x,y \in \mathbb{R}\}$.

i): If $z=iy$ with $y \in \mathbb{R}$, by (1) it follows that $\overline{\wp(z)}=\overline{\wp(iy)}=\wp\left(\overline{iy}\right)=\wp(-iy)$, but since $\wp$ is even then $\wp(-iy)= \wp(iy)$, thus indeed $$ \overline{\wp(z)}=\wp(iy)=\wp(z), \ \text{hence} \ \wp(z) \in \mathbb{R} $$ ii): If $z=\alpha + iy$ with $y \in \mathbb{R}$, again by (1), $\overline{\wp(z)}=\overline{\wp(\alpha+iy)}=\wp\left(\overline{\alpha + iy}\right)=\wp(\alpha-iy)$, but $2\alpha$ is a period of $\wp$, that gives $\wp(\alpha-iy)= \wp(\alpha-iy-2\alpha)=\wp(-(\alpha+iy))$, again being $\wp$ even we have $\wp(-(\alpha+iy))=\wp(\alpha+iy)$, therefore $$ \overline{\wp(z)}=\wp(\alpha+iy)=\wp(z), \ \text{hence } \ \wp(z) \in \mathbb{R} $$ iii): If $z=x$ with $x \in \mathbb{R}$, (1) gives that $\overline{\wp(z)}=\overline{\wp(x)}=\wp(\overline{x})=\wp(x)=\wp(z)$, then of course $\wp(z) \in \mathbb{R}$

iv): Finally when $z=x + i\beta$ with $x \in \mathbb{R}$, by (1), $\overline{\wp(z)}=\overline{\wp(x+i\beta)}=\wp\left(\overline{x + i\beta}\right)=\wp(x-i\beta)$, since $2i\beta$ is a period of $\wp$ then $\wp(x-i\beta)= \wp(x-i\beta+2i\beta)=\wp(x+i\beta)$, therefore again $$ \overline{\wp(z)}=\wp(x+i\beta)=\wp(z), \ \text{hence } \ \wp(z) \in \mathbb{R} $$

Thus indeed $\wp$ takes real values in all the parallelograms generated by the half periods $\omega_1$ and $\omega_2$, as wanted.

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  • $\begingroup$ @ccorn the case I am using is when te invariants $g_2$ and $g_3$ are real, this is the only case when $\wp$ is real in the whole boundary of the parallelogram generated by the half periods. I am making an edit to clarify this. $\endgroup$ Jun 18, 2015 at 20:37
  • $\begingroup$ I noticed that shortly after commenting here, which is why I switched to commenting the question instead. Sorry for the irritation. $\endgroup$
    – ccorn
    Jun 18, 2015 at 23:16
  • $\begingroup$ @ccorn no worries ! I'm glad you take interest in my answer $\endgroup$ Jun 19, 2015 at 0:01
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    $\begingroup$ Thank you Alonso for your help. This point is important for a problem at hand. I am wondering if you would be interested to collaborate on it. If you wish to explore, can you send me an email on dr.amey.joshi@gmail.com? $\endgroup$
    – Amey Joshi
    Jun 20, 2015 at 2:19

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