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(I couldn't find many sources of the derivation on the internet, so I might as well show most of my work here)

I started by using the Argument sum formulas (see this post), which require specific values of sl and cl. To find them, I started by investigating the poles of sl and cl:

To find the poles of sl and cl, we consider the integral representation of arcsl and arccl:

$$\text{arcsl}(\infty)=\int_0^{\infty}\frac{1}{\sqrt{1-t^4}}\, dt=\int_0^{1}\frac{1}{\sqrt{1-t^4}}\, dt +\int_1^{\infty}\frac{1}{\sqrt{1-t^4}}\, dt$$

we let $t\mapsto \frac{1}{t}$ on the latter integral, $$=\int_0^{1}\frac{1}{\sqrt{1-t^4}}\, dt+\int_0^{1}\frac{1}{\sqrt{t^4-1}}\, dt=\int_0^{1}\frac{1}{\sqrt{1-t^4}}\, dt\pm i\int_0^{1}\frac{1}{\sqrt{1-t^4}}\, dt=\frac{1\pm i}{2}\varpi$$

Hence sl has poles on $\frac{1\pm i}{2}\varpi$

Remark: Notice if we replace $\infty$ with any direction of $\infty$, we still get the result, meaning that the poles of sl really go to $e^{i\theta}\infty$ for all $\theta$.

Using that sl$(-z)= -$sl$(z)$, we also have that $\text{sl}(-\frac{\varpi}{2}\pm \frac{i\varpi}{2})\longrightarrow \infty$.

Meanwhile using the formula $$\text{sl}(u+v)=\frac{\text{sl}(u)\sqrt{1-\text{sl}^4(v)}+\text{sl}(v)\sqrt{1-\text{sl}^4(u)}}{1+\text{sl}^2(v)\text{sl}^2(u)}$$

Notice that $\sqrt{1-\text{sl}^4(v)}\sim \text{sl}^2(v)$ as sl$(v)$ grows, let $u\longrightarrow \frac{\varpi}{2}+i\frac{\varpi}{2}$ and $-v\longrightarrow \frac{\varpi}{2}-i\frac{\varpi}{2}$

$$\text{sl}(i\varpi)+\text{sl}\Big(\frac{\varpi}{2}+i\frac{\varpi}{2}-(\frac{\varpi}{2}-i\frac{\varpi}{2})\Big)\sim \frac{1}{\text{sl}( u)}-\frac{1}{\text{sl}(v)}\longrightarrow 0$$

where we used the fact that sl$(\frac{\varpi}{2}+i\frac{\varpi}{2})$ and sl$(\frac{\varpi}{2}-i\frac{\varpi}{2})$ diverge to infinity.

We therefore have sl$(i\varpi)=0$

Now we consider the following integral $$\int_0^i \frac{1}{\sqrt{1-t^4}}\, dt$$ substituting $-it=w$ gives $i\frac{\varpi }{2}$. Hence $\text{sl}(i\frac{\varpi }{2})=i$

Using the argument sum formula for cl:

$$\text{cl}(u+v)=\frac{\text{cl}(u)\text{cl}(v)-\text{sl}(v)\text{sl}(u)}{1+\text{sl}(u)\text{sl}(v)\text{cl}(v)\text{cl}(u)}$$

letting $u=v\longrightarrow i\frac{\varpi}{2}$ gives

$$\text{cl}(i\varpi)=\frac{\text{cl}^2( i\frac{\varpi}{2})-\text{sl}^2( i\frac{\varpi}{2})}{1+\text{sl}^2( i\frac{\varpi}{2})\text{cl}^2( i\frac{\varpi}{2})}$$

But we know that cl$(i\frac{\varpi}{2})=$ sl$(\frac{\varpi}{2}-i\frac{\varpi}{2})$ diverges, since $\text{sl}(i\frac{\varpi }{2})=i$, we have

$$\text{cl}(i\varpi)=\frac{1-0}{0-1}=-1$$

Giving cl$(i\varpi)=-1$

We are now ready to prove double periodicity, we compute $\text{sl}((1\pm i)\varpi)$

$$\text{sl}((1\pm i)\varpi)=\frac{\text{sl}(\varpi)\text{cl}( i\varpi)\pm \text{sl}(i\varpi)\text{cl}(\varpi)}{1\mp \text{sl}(\varpi)\text{cl}( i\varpi)\text{sl}(i\varpi)\text{cl}(\varpi)}=0$$

while similarly using argument sum for cl, $\text{cl}((1\pm i)\varpi)=1$

$$\text{sl}(z+(1\pm i)\varpi)=\frac{\text{sl}(z)\text{cl}((1\pm i)\varpi)+\text{cl}(z)\text{sl}((1\pm i)\varpi)}{1-\text{sl}((1\pm i)\varpi)\text{cl}((1\pm i)\varpi)\text{cl}(z)\text{sl}(z)}$$

Note that $\text{sl}((1\pm i)\varpi)=-\text{sl}(\pm i\varpi)=0$

therefore, we have that $\text{sl}(z+(1\pm i)\varpi)=\text{sl}(z)$.

This proves that $\text{sl}$ has the double period $\{(1+i)\varpi,(1-i)\varpi\}$

Does this construct the fundamental domain?

To answer this question I have to show that there does not exist $0<\xi<1$ such that

$\text{sl}((1+i)\xi \varpi )=0$ (Because sl$(0)=0$). By applying the argument sum formula, it boils down to proving that there does not exist $0<\xi<1$ such that

$$\text{sl}(\varpi \xi)\text{cl}(i\xi \varpi )=\pm \text{cl}(\xi \varpi )\text{sl}(i\xi \varpi )$$

And now I am stuck.

Note that I cannot use the Mittag pole expansion to prove this, I used this to prove the pole expansion

Dqrksun
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1 Answers1

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Notice that all we have to do is to prove that there does not exist a $\xi \in(0,1)$ such that $$\text{sl}((i\pm 1) \varpi \xi)=0$$

To prove this, I proceeded to the following.

Starting with $$\text{arcsl}\Big(\frac{z\sqrt{2}}{\sqrt{1+z^4}}\Big)=\int_0^{\frac{z\sqrt{2}}{\sqrt{1+z^4}}}\frac{1}{\sqrt{1-t^4}}\, dt$$

Substitute $t=\frac{\sqrt{2} u}{\sqrt{1+u^4}}$ , $dt= \sqrt{2}\Big(\frac{1-u^4}{(1+u^4)^{3/2}}\Big)\, du$

$$\implies =\int_0^z \frac{1}{\sqrt{1-\frac{4u^4}{(1+u^4)^2}}}\, \sqrt{2}\Big(\frac{1-u^4}{(1+u^4)^{3/2}}\Big)\, du =\sqrt{2}\int_0^z \frac{du}{\sqrt{1+u^4}}$$

Let $ue^{\pm i\pi /4} =v$ , $du= e^{\mp i\pi /4} dv$

$$\sqrt{2}\int_0^{ze^{\pm i\pi /4}} \frac{1}{\sqrt{1-v^4}} \, e^{\mp i \pi/4}dv =(1\mp i)\text{arcsl}(z e^{\pm i\pi /4}) $$ $$\implies \text{arcsl}\Big(\frac{z\sqrt{2}}{\sqrt{1+z^4}}\Big)= (1\mp i)\text{arcsl}(z e^{\pm i\pi /4})$$ set $z\mapsto z e^{\mp i\pi /4}$ $$\implies \text{arcsl}\Big(\frac{z(1\mp i)}{\sqrt{1-z^4}}\Big)= (1\mp i)\text{arcsl}(z)$$ $z\mapsto \text{sl}(z)$ $$\implies \text{arcsl}\Big(\frac{\text{sl}(z)(1\mp i)}{\sqrt{1-\text{sl}^4z}}\Big)= (1\mp i)z$$ Giving the formula: $$\text{sl}((1\pm i)z)=(1\pm i) \frac{\text{sl}(z)}{\text{sl}'(z)}$$

let $\xi \in (0,1)$: $$\implies \text{sl}((1\pm i)\varpi \xi)=(1\pm i) \frac{\text{sl}(\varpi \xi)}{\text{sl}'(\varpi \xi)}$$

This is zero only if $\text{sl}(\varpi \xi)=0$ or $\text{sl}'(\varpi \xi)\longrightarrow \infty$. But we know that both are impossible for $\xi \in (0,1)$. This proves that $\{(1+i)\varpi,(1-i)\varpi\}$ indeed constructs the fundamental domain.

Dqrksun
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