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Consider the congruence subgroup

$$\Gamma(N) = \left\{\left(\begin{array}{cc} a & b \\ c & d\end{array} \right) \in SL_2(\mathbb{Z})\ ;\ \left(\begin{array}{cc} a & b \\ c & d\end{array} \right) \equiv \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array} \right)\mod N\right\}$$

acting on the upper half-plane $\mathfrak{h} = \{\tau \in \mathbb{C} \ ;\ \Im(\tau)>0\}$ by fractional linear transformations:

$$ \left(\begin{array}{cc} a & b \\ c & d\end{array} \right) \cdot \tau = \frac{a\tau + b}{c\tau + d} $$

It is known that the quotient $Y(N) = \Gamma(N)\backslash\mathfrak{h}$ is an algebraic curve for every $N$. For instance,

$$Y(2) = \mathbb{P}^1 \setminus \{0,1,\infty\}$$

which can be proved by considering modular function $\lambda : \mathfrak{h} \to \mathbb{C}$. It is also well known that the compactification $X(7)$ of $Y(7)$ is isomorphic to Klein's quartic in $\mathbb{P}^2$.

What about the other small values of $N$? How do I compute $Y(N)$ for $N=3,4,5,6$? I know I can look at the genus of the compactification $X(N)$, but is there a way of getting the precise form of $Y(N)$ as in the example for $N=2$?

Bonus: What about the universal elliptic curve over $Y(N)$, is it also computable in these small cases?

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    Is there a reason you want to understand the full level structure? If you want to undersatnd $\Gamma_0(N)$-level structure, this has been worked on by may people (e.g. see Galbraith's thesis: https://www.math.auckland.ac.nz/~sgal018/thesis.pdf). The case of $X(N)$ is somehow less interesting to people, I would guess, than $X_0(N)$ since the latter has a clear connection to modularity whereas the former doesn't. – Alex Youcis Aug 25 '20 at 18:59
  • I just think this is a very natural question. It's quite strange that I cannot find the answer in the literature so easily. I know other congruence subgroups are perhaps more interesting in arithmetic questions, but full level structure does play an important role in the general theory, and is in some sense more basic than other level structures... –  Aug 25 '20 at 19:14
  • Note: if you're looking for a model over $\mathbb{Q}$ which represents the corresponding moduli problem you will need to require $N \geq 3$ (for e.g., $Y(1)$ it doesn't quite work out since there are non-isomorphic elliptic curves with the same $j$-invariant) – Mummy the turkey Aug 26 '20 at 09:41
  • There are several ways, and this has been studied extensively in the literature, though usually the groups of interest are non-split Cartan, and not the full modular subgroup. The idea is the same as in Galbraith's thesis cited above. Simply compute a basis for the space of modular forms, and look for polynomial relations. Note: this modular curve has a model only over the cyclotomic field. – assaferan Oct 04 '20 at 03:18

1 Answers1

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Let me try to give a complete answer.

First, for $ N = 3,4,5 $ the genus of the compactification $X(N)$ is $0$, therefore it is isomorphic to $\mathbb{P}^1$. Since $X(N) = Y(N) \cup C(N)$, where $C(N)$ is the finite set of cusps of $\Gamma(N)$, namely $C(N) = \Gamma(N) \backslash \mathbb{P}^1(\mathbb{Q})$, we see that $Y(N) = \mathbb{P}^1 \setminus C(N)$, where $C(N)$ is a finite set of points. Specifically, computing the number of cusps, we see that $|C(2)| = 3, |C(3)| = 4, |C(4)| = 6$ and $|C(5)| = 12$.

Moreover, we know that the isomorphism $X(1) \rightarrow \mathbb{P}^1$ is given by the $j$-map. We can compute the natural covering map $X(N) \rightarrow X(1)$, and by composition obtain the natural corresponding isomorphism. (This is the modular function $\lambda : \mathfrak{h} \rightarrow \mathbb{C}$ you have described above). This map also allows you to represent the universal elliptic curve, as it describes $j$ as a rational function in $\lambda$, and one may simply substitute it in the description of $E_j$.

This was done for several cases, using towers of prime-power level subgroups, basically using two methods.

In the paper https://projecteuclid.org/euclid.ant/1513090725 [Sutherland, Andrew; Zywina, David. Modular curves of prime-power level with infinitely many rational points. Algebra Number Theory 11 (2017), no. 5, 1199--1229. doi:10.2140/ant.2017.11.1199] the authors used Siegel functions to describe the $j$-map for $X(2), X(3), X(4), X(5)$. The tables in the end of the paper show the covering maps to intermediate covers, but composing them all together one obtains the result.

In the paper https://link.springer.com/article/10.1007/s40993-015-0013-7 [Rouse, J., Zureick-Brown, D. Elliptic curves over ℚ and 2-adic images of Galois. Res. Number Theory 1, 12 (2015). https://doi.org/10.1007/s40993-015-0013-7] the authors used Eisenstein series to describe the $j$-map for $X(2)$ and $X(4)$, again by maps to intermediate covers.

Finally, for $N = 6$, the genus of $X(6)$ is $1$, and since it has rational points, we know it is an elliptic curve. As $\Gamma(6)$ has $12$ cusps, $Y(6)$ is this elliptic curve without $12$ points. Generalizing and adapting the methods of Rouse and Zureick-Brown, I calculated that $X(6)$ is the Elliptic Curve $y^2 = x^3 + 1$, and the cusps are mapped to the $12$ points $$ S = \left \{ \infty, (-1,0), (2, \pm 3), (0,\pm 1), (\zeta_6^{\pm1},0 ), \left(-2 + \sqrt{-3}, \pm (3 + \sqrt{-3}) \right), \left(-2 - \sqrt{-3}, \pm (-3 + \sqrt{-3}) \right) \right \}$$ so that $$ Y(6) = E \setminus S $$

I also ran it for the other curves above to see where the cusps map to. One obtains the following models.

$$ Y(3) = \mathbb{P}^1 \setminus \left \{ 0, \frac{1}{3}, \frac{1}{2} \left( 1 \pm \frac{\sqrt{-3}}{3} \right) \right \} $$

$$ Y(4) = \mathbb{P}^1 \setminus \left \{ 0, \pm 1, \infty, \pm i \right \} $$

$$ Y(5) = \mathbb{P}^1 \setminus \left \{ 0, \infty, \phi \zeta_5^j, -\phi^{-1} \zeta_5^j \right \} $$

where $\phi = \frac{1 + \sqrt{5}}{2}$, and $\zeta_5$ is a $5$-th root of unity.

As for the universal elliptic curve, it's slightly tedious to explicitly write it down here, but it is simply substituting the $j$-map in each of the cases.

assaferan
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