On the cusps of $\Gamma_0(3) \cap \Gamma(2)$

Question

I've been trying to compute the number of non-equivalent cusps of $\Gamma_0(3) \cap \Gamma(2)$. My approach so far has been the following: I believe that this is finite index subgroup of $\Gamma_0(3)$ so it should have the same cusps if I'm not mistaken.I wanted to work with these cusps and see when could they be equivalent by an element in the intersection. Cusps of $\Gamma_0(n)$ should be $0,\infty$ and $\cfrac{a}{dc}$ where $d$ is some divisor of $n$ and $(a,dc)=1$.I've been told that this subgroup has $6$ non-equivalent cusps in total which I am not sure how that is true as an upper bound of for the non-equivalent cusps of $\Gamma_0(3)$ is its index which is $4$ in this case. Maybe it would be easier to work with the cusps of $\Gamma(2)$?

Answer 1

I note several different things here.

Let $G := \Gamma(2) \cap \Gamma_0(3)$. The set of all cusps is the same (as noted by Shimura), and we know exactly what these are: $\mathbb{Q} \cup \{ \infty \}$. Implicitly this question is actually about the set of inequivalent cusps.

As $G$ is a subgroup of $\Gamma(2)$ and $\Gamma_0(3)$, it has fewer elements and hence fewer matrices that can relate different cusps. We should generically expect for there to be more inequivalent cusps in $G$ than in either $\Gamma(2)$ or $\Gamma_0(3)$.

(Explicitly here: $\Gamma(2)$ has three inequivalent cusps: $\{ 0, 1, \infty \}$; and $\Gamma_0(3)$ has two inequivalent cusps: $\{ 0, \infty \}$; and $G$ has $6$ inequivalent cusps: $\{ 0, 1, -\frac{1}{2}, -\frac{1}{3}, \frac{2}{3}, \infty \}$).

It is not true that any subgroup of finite index in $\Gamma_0(3)$ has the same set of inequivalent cusps, for example.

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To generate the list of inequivalent cusp representatives for $G$, I first computed coset representatives for $G$ in $\mathrm{SL}(2, \mathbb{Z})$. This gave a list $\{ g_1 = Id, \ldots, g_{24} \}$ of $\mathrm{SL}(2, \mathbb{Z})$ matrices. All inequivalent cusps will be of the form $g_{j} \infty$ for some coset representative $g_j$. (This is essentially a reparsing of the statement that the coset representatives can be used to construct a fundamental domain, and any fundamental domain will touch all inequivalent cusps).

For the set of coset representatives I chose, I got a list of $8$ cusps of the form $g_j \infty$. Now it remains to check which of these cusps are actually inequivalent under the action of $G$.

To do this, to each cusp $c = a/b$, I first solved $1 = sa + tb$. Then I made the matrices $\sigma_{c, j}$ given by

$$ \sigma_{c, j} = \begin{pmatrix} a & -(t + ja) \\ b & s - jb \end{pmatrix}. $$

These matrices satisfy $\sigma_{c, j} \infty = c$ for any $j$. I note that $\Gamma(6) \subset G$, as $\Gamma(6)$ is clearly in both $\Gamma(2)$ and $\Gamma_0(3)$. Thus every matrix taking $\infty$ to $c$ is $G$-equivalent to one of $\sigma_{c, j}$ for $0 \leq j \leq 5$.

Thus every matrix taking a cusp $c_1$ to another cusp $c_2$ is $G$-equivalent to $\sigma_{c_2, j_2} \sigma_{c_1, j_1}^{-1}$ for some $0 \leq j_1, j_2 \leq 5$. I looked at those products and determined if any were in $G$, and used this to see which cusps were identified.

This showed the final $6$ cusps.

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Alternatively, it would be possible to determine inequivalent cusps from the list of coset representatives for $G$ inside $\mathrm{SL}(2, \mathbb{Z})$ by examining the resulting fundamental domain and recognizing which edges match up. If two edges match up at a cusp, then the two cusps share the same cusp representative. But I didn't do this.