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Does anyone know how to prove that the following special value of the Modular Lambda Function is correct?

$$\lambda(\sqrt{2}i)=(\sqrt{2}-1)^2$$

I have a somewhat promising observation that might help us derive this special value, but it hasn't panned out for me so far. If we consider the elliptic function $\wp$ with $\tau=\sqrt{2}i$, the fundamental domains look like rectangles with side lengths of $1$ and $\sqrt{2}$, which exhibit a certain self-similarity: one of these rectangles can be dissected into two similar copies of the same rectangle, scaled down by a factor of $\sqrt{2}$. However, I haven't figured out how to use this fact to my advantage.

Another note: I did figure out how to prove the special value $\lambda(i)=1/2$: it follows trivially from the functional equation $\lambda(-1/\tau) =1-\lambda(\tau)$.

Any help is appreciated!

Franklin Pezzuti Dyer
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    There are tons of method to derive these values, telling us what you already know is helpful for us to provide the proof that best suits you. For example, if you know the relation between $j$ and $\lambda$, with $j(\sqrt{-2}) = 20^3$. Another way: use the modular equation of $\lambda$ with index $2$. Yet another is the action of Galois group on $\lambda$. Yet yet another is to use elliptic singular value: this last method has lots of variation, like Ramanujan invariants $g_n, G_n$ or Dedekind's $\eta$, all depends on what you know in advance. – pisco Jun 01 '21 at 19:09
  • @pisco Fair enough. All I know about $\lambda$ is what I've read in Ahlfors - namely, its definition and basic functional equations, as derived directly from its definition in terms of the $\wp$ function. It would be nice if there was a way to prove this special value using functional equations that can be derived from the properties of $\wp$, like the example I gave for $\lambda(i)$. – Franklin Pezzuti Dyer Jun 01 '21 at 19:25
  • So you're defining $\lambda$ as something like $(e_1-e_2)/(e_1-e_3)$, with $e_i$ values of $\wp$ at half-periods? – pisco Jun 01 '21 at 19:27
  • @pisco Yep, that's precisely the definition I'm using (from Ahlfors). – Franklin Pezzuti Dyer Jun 01 '21 at 19:29
  • Ahlfors gives a similar exercise in one of his books: https://math.stackexchange.com/q/349806/72031 – Paramanand Singh Jun 04 '21 at 07:27

2 Answers2

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As OP requested, here is a solution based on Weierstrass elliptic function, with $$\lambda = \frac{e_1 - e_2}{e_1-e_3}$$

Though I think such an approach is almost the worst one, among the multitude of other approaches I mentioned in comment.

Let $\wp(z)$ has period $\{1,\sqrt{-2}\}$. Then $\wp(\sqrt{-2}z)$ is also periodic under this lattice, and this is even function, so we have coprime polynomials $f,g$ such that $$\wp(\sqrt{-2}z) = \frac{f(\wp(z))}{g(\wp(z))}$$ by looking at order of poles at $z=0$, we see $\deg f = \deg g + 1$. Note that $\deg f \geq 2$. I claim the $\deg f$ is exactly $2$.

Note that $f(x)$ has distinct roots (because $\wp$ is surjective on $\mathbb{C}$ and every zero is simple). If $f(x)$ has more than 2 roots, then $\wp(\sqrt{-2}z)$ has more than 4 zeroes (modulo $\mathbb{Z}+\mathbb{Z}\sqrt{-2}$), contradiction, as this number should be exactly $4$.

So let $$\wp(\sqrt{-2}z) = a\wp(z) + b + \frac{1}{c\wp(z)+d}$$ write $\wp(z) = \frac{1}{z^2} + \frac{g_2}{20}z^2 + \frac{g_3}{28}z^4 + \frac{g_2^2}{1200}z^6 + \cdots$ then one sees immediately $b=0, a=-1/2$. So $$c\wp(z)+d = - \frac{40}{3g_2 z^2} -\frac{200g_3}{7g_2^2}+ \left(\frac{10}{9}-\frac{3000g3^2}{49g_2^3}\right)z^2 + o(z^2)$$ giving $98g_2^3 = 3375g_3^2$, so we can normalize $g_2 = 30, g_3 = 28$.

Hence $e_i$ are roots of $4x^3-g_2 x -g_3 = 0$, i.e. $\{-2,\frac{1}{2} \left(2-3 \sqrt{2}\right),\frac{1}{2} \left(3 \sqrt{2}+2\right)\}$, there are six combination of $(e_1-e_2)/(e_1-e_3)$, one of them will be $3-2\sqrt{2}$, as desired.

pisco
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The simpler approach is based on Ramanujan's class invariants and this is quite amenable to hand calculation.

However for the current question it is just enough to use the modular equation of degree $2$ (corresponding to $\tau =\sqrt{2}i$) namely $$l=\frac {2\sqrt{k}}{1+k}$$ and then put $l=k'=\sqrt {1-k^2}$.

On squaring we get $$1-k^2=\frac{4k}{(1+k)^2}$$ or $$k^2=\frac{(1-k)^2}{(1+k)^2}$$ ie $$k=\frac{1-k}{1+k}$$ or $$ k^2+2k-1=0$$ This gives us $$k=\frac{-2+\sqrt{8}}{2}=\sqrt{2}-1$$ And then the value of modular function $\lambda$ is given by $$\lambda=k^2=(\sqrt{2}-1)^2$$

We can use the same approach for $\tau=\sqrt{3}i$ based on a modular equation of degree $3$. But for larger values of $n$ with $\tau=i\sqrt{n} $ we need a combination of modular equations and Ramanujan class invariants.

Paramanand Singh
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