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I am currently reading through the book An Introduction to Teichmüller Spaces by Imayoshi and Taniguchi. In Section 1.2, we see that $M_1$, the moduli space of tori, can be identified with $\mathbb{H}/PSL(2,\mathbb{Z})$. This is clear, since two tori $R_\tau$ and $R_{\tau'}$ - generated by normalized lattices having sides $1, \tau$ and $1, \tau'$, respectively - are biholomorphically equivalent if and only if $\tau=\tau'$ (where $\tau \in \mathbb{H}$), and in particular we can identify this with $\mathbb{C}$ using the $j$-invariant of an elliptic curve (associated to a torus generated by the normalized lattice with sides $1, \tau$).

We also know that for a given cross ratio $\{z_1, z_2, z_3, z_4\} = \lambda$, permutation of the $z_i$ by an element of $S_4$ results in the cross ratio being one of the following: $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda-1}{\lambda}, \frac{\lambda}{\lambda-1}$, and we can put the 4-tuple having cross ratio $\lambda$ in "canonical" form by setting it to be $\{0, 1, \lambda, \infty\}$. Next, a torus $S_\lambda$ is given by the following equation (depending on $\lambda$): $w^2 = z(z-1)(z-\lambda)$. Then two tori $S_\lambda$ and $S_{\lambda'}$ are biholomorphically equivalent if and only if there is a linear fractional transformation taking $\{0, 1, \lambda, \infty\}$ to $\{0, 1, \lambda', \infty\}$, so then $\lambda'$ can only be one of the values after a permutation as above.

Let $G$ be the group (which is actually just $S_3$) generated by the two functions $\lambda \mapsto \frac{1}{\lambda}$ and $\lambda \mapsto 1-\lambda$ (note these functions are analytic automorphisms of $D = \mathbb{C}-\{0, 1\}$). In particular we have the $S_4/V \approx S_3$, where $V$ is the Klein 4-group (and is in fact the permutations that fix the cross ratio). The book goes on to say that this shows $M_1 \approx D/G$ and there is a biholomorphic map $F: D/G \to \mathbb{C}$ given by $F([\lambda]) = f(\lambda) = \frac{{(\lambda^2-\lambda+1)}^3}{\lambda^2{(\lambda-1)}^2}$.

I do not understand where exactly $f(\lambda)$ comes from or how the identification is clear, but the setup makes sense to me. Does this function have a special name? Does it appear anywhere else (in a significant way)?

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  • $\begingroup$ I don't see what you mean with cross ratio $\{z_1, z_2, z_3, z_4\} = \lambda$. Then $j,\lambda$ are complicated functions of $\tau$. It is more convenient to look at $f(\lambda) = j(E_\lambda)$ where $E_\lambda : w^2 = z(z-1)(z-\lambda)$ and $j(E)=j(E')$ iff $E \cong E'$. And $G$ is the group of biholomorphic functions $g:D \to D$ such that $f(g(\lambda)) = f(\lambda)$ thus $D/G \cong f(D)$, to identify it with $\mathbb{H}/PSL(2,\mathbb{Z})$ you need to make clear the isomorphism between isomorphism class of complex tori and isomorphism class of elliptic curves. $\endgroup$ – reuns Apr 15 at 23:05
  • $\begingroup$ I don't really see how this answers the question. For the time being, I am just following the book, which presents the material in this manner. It's primarily focused on geometry, so the $j$-invariant isn't really crucial, or even explicitly mentioned - just that it gives a way to identify $\mathbb{H}/PSL(2,\mathbb{Z})$ and $\mathbb{C}$. $\endgroup$ – nilradical1 Apr 15 at 23:07
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    $\begingroup$ Consider the field of invariants of $\mathbb{C}(\lambda)$ under the action of $G$. I believe that you will find that it is generated by $f(\lambda)$. As to why $f(\lambda)$ and not another generator, look at what the points $\{0,1,\infty\}$ have to go to. $\endgroup$ – Kapil Apr 16 at 1:58

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