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If $C$ is the curve in $\mathbb{C}\mathbb{P}^{2}$ defined by the zero set of the polynomial $P^{\lambda}(x,y,z) = y^{2}z - x(x-z)(x-\lambda z)$, for $\lambda$ not $0$ or $1$. Then we know that $C$ is homeomorphic to a torus. Tori are homeomorphic to $\mathbb{C} / \Lambda $ for $\Lambda$ a lattice in $\mathbb{C}$. My question is given a specific value of $\lambda$, how would one approach calculating the lattice $\Lambda$?


Edit - $\textbf{Motivating example:}$

My motivation for asking this question came from question 5 on this past exam paper: https://www.dropbox.com/s/hfet5ki0nv3bm0d/2011.pdf?dl=0

In part a) we're asked to calculate a projective map $\sigma$ of $\mathbb{C}\mathbb{P}^{2}$ that sends the curve $C^{\lambda}$ (defined by the zero set of the polynomial $P^{\lambda}$), to the curve $C^{1 / \lambda}$ whilst fixing the point $[0,1,0]$.

Then we specialise to the case $\lambda = -1$ and you are then asked to show that $\sigma$ has order 4. Then we're told that holomorphic maps from $\mathbb{C} / \Lambda $ to $\mathbb{C} / \Lambda $ are of the form $ z \mapsto az + b$ for some $a,b \in \mathbb{C}$ and $a \cdot \Lambda \subseteq \Lambda$, and also that for every $\lambda \neq 0,1$ we have that, for some lattice $\Lambda$, $C^{\lambda} \cong \mathbb{C} / \Lambda$. Then we're asked to calculate the lattice $\Lambda_{-1}$ such that $C^{-1} \cong \mathbb{C} / \Lambda_{-1}$.


Others have shown me that I can use the Modular Lambda function to calculate $\Lambda$ for any $\lambda$, but I don't think this is what I'm meant to do, since we haven't been introduced to this function before. Would I be able to get some hints on how to tackle this? This is what I've done so far:


$\textbf{Progress:}$

If we let $\sigma$ be defined by the matrix $M(\lambda)$ where

$$ M(\lambda) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 / \sqrt{\lambda} & 0 \\ 0 & 0 & \lambda \end{bmatrix} $$

Then I think $\sigma$ maps $C^{\lambda}$ to $C^{1/\lambda}$ and fixes $[0,1,0]$. Then it is easy to show when $\lambda = -1$ that $M(\lambda)^{k}$ is not a scalar multiple of the identity, unless $k$ is a multiple of 4. Hence $\sigma$ has order 4.

Then suppose $\psi : \mathbb{C} / {\Lambda}_{-1} \rightarrow C^{-1}$ is an isomorphism. Then $\sigma' = \psi^{-1} \circ \sigma \circ \psi $ is a holomorphic map from $\mathbb{C} / {\Lambda}_{-1}$ to $\mathbb{C} / {\Lambda}_{-1}$, and so there exists $a, b \in \mathbb{C}$ with $a \cdot \Lambda_{-1} \subseteq \Lambda_{-1}$ such that $\sigma'(z) = az + b$. Then $\sigma'$ has order 4, and since $\sigma'^{4}(z) = a^4z + b(a^3 + a^2 + a + 1)$, we see that the map $z \mapsto a^4z + b(a^3 + a^2 + a + 1)$ must be the identity on $ \mathbb{C} / {\Lambda}_{-1}$.

Then $a^4z + b(a^3 + a^2 + a + 1) - z = (a^3 + a^2 + a + 1)((a-1)z + b) \in \Lambda_{-1}$ for every $z \in \mathbb{C}$.

But then I do not know how to proceed from here.


From wolfram alpha, I can see that the lattice must be $\Lambda_{-1} = \left< {1, \tau} \right>_{\mathbb{Z}}$ where $\tau = \frac{log(\operatorname{q}(-1))}{\pi \cdot i} = 1 + i $, where $\operatorname{q}$ is the "Elliptic Nome" function. But I have no idea where this comes from.

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    $\begingroup$ You want to invert this function: en.wikipedia.org/wiki/Modular_lambda_function $\endgroup$ – Qiaochu Yuan May 23 '18 at 22:59
  • $\begingroup$ @QiaochuYuan would you be able to explain in a bit more detail? I don’t really understand what the Modular lambda function is, nor how one would go about inverting it $\endgroup$ – Adam Higgins May 23 '18 at 23:09
  • $\begingroup$ Well you could find the bases of $\Lambda$ with the associated elliptic integrals but I am not sure those are easy to calculate. In order to find such a lattice I guess you would have to calculate the j-invariant of your elliptic curve and then find the associated lattice to that somehow. I haven't seen a constructive proof or an algorithm for that however but please post anything you find because I am really interested! $\endgroup$ – Μάρκος Καραμέρης May 23 '18 at 23:25
  • $\begingroup$ @ΜάρκοςΚαραμέρης if I do find anything I’ll make sure to post it. The question was motivated my a previous exam question that walked you through calculating it for a specific case. But the process was very specialised to that specific value of $\lambda$ which got me thinking about how one would approach doing it in general. The specific case In the exam question was $\lambda = -1$, and it asked you to construct an order 4 projective map $ \sigma : \mathbb{C}\mathbb{P}^{2} $ such that $C’ = \sigma (C)$ is another cubic curve, then proceed from there. $\endgroup$ – Adam Higgins May 23 '18 at 23:30
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    $\begingroup$ Well, $\tau+n$ is just as good as $\tau$ to generate a lattice. In fact, there is the entrie modular group which enters in this. The code returns one generator but does not mean it is the smallest one. $\endgroup$ – Somos May 25 '18 at 0:20
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The Wikipedia article Modular lambda function has a lot of information but you need only a few bits of it. For exmaple, in the context of Legendre and Jacobi elliptic functions, $\, m = \lambda(\tau) \,$ is called the parameter and $\, q = e^{\pi i \tau} \,$ is called the nome. Thus, in the Wolfram language, the code for $\,\tau\,$ is $\texttt{tau = Log[ EllipticNomeQ[ lambda ]] / (Pi I)} $ and the lattice $\,\Lambda\,$ is generated by $\,[1,\tau].\,$ For example, if $\,\lambda = 0.9,\,$ then $\,\tau = i\;0.6254...\,.$

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  • $\begingroup$ Thanks so much for this! Would you be able to give me some context for why this works? And what exactly this nome function does? $\endgroup$ – Adam Higgins May 24 '18 at 7:48
  • $\begingroup$ It works by definition $\,e^{\pi i \tau}=q(m)=q(\lambda).\,$ Read the Wikipedia link for nome q. $\endgroup$ – Somos May 24 '18 at 10:34

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