Let $\tau$ be a CM point of discriminant $D$. Assume that $D$ is not divisible by $3$. Then $j(\tau)$ is an algebraic integer of degree equal to the class number $h(D)$. Let $ \gamma\_2(\tau)=j(\tau)^{1/3}$, the cube root being chosen in such a way that $\gamma\_2(\tau)$ is positive on the imaginary axis.


Weber had shown that the degree of $ \gamma\_2(\tau) $ is $h(D)$ instead of the expected $3h(D)$. In other words, $\mathbb Q(\gamma\_2(\tau))$=$\mathbb Q(j(\tau))$.


The modular function $\gamma\_2$ has level $3$, and the exact subgroup under which it is invariant is
$$\Gamma(\gamma\_2)=\bigg \lbrace \begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL\_2(\mathbb Z) :\begin{pmatrix}a&b\\c&d\end{pmatrix}\equiv\begin{pmatrix}0&\*\\\*&0\end{pmatrix}\text{or}\begin{pmatrix}\*&b\\b&\*\end{pmatrix}\text{mod } 3\bigg \rbrace.$$


Questions


1. Is there an intuitive reason why $\gamma\_2(\tau)$ has degree $h(D)$?
2. What is the connection between Cartan subgroups of $ GL\_2(\mathbb Z/N\mathbb Z)$ and this problem?
3. Can someone please explain to me what a non-split Cartan subgroup is?


References


[Jeremy Booher, "Modular curves and the class number one problem"](https://pdfs.semanticscholar.org/cebc/9b8eb74d114a1896ab1a5590a25846df92d6.pdf), Theorem $36$ and Definition $45$.


[Serre: Lectures on the Mordell-Weil Theorem](https://www.springer.com/us/book/9783663106340) page 196.


[Serre, "Proprietes galoisiennes des points d'ordre fini
des courbes elliptiques"](https://www.wstein.org/sage_summer/bsd_comp/Serre-properties_galoisiennes_des_points_dordre_fini_des_courbes_elliptiques.pdf), Section 5.3.b


I would be especially interested in some comments on the third referenced article (in connection to the problem at hand). Serre writes that this is an elementary proof that $j$ is a cube.



Intuitive reason why the $j$-invariant is a cube?
There is a map from the $\mathbb P^1$ with coordinate $\gamma\_2$ to the $\mathbb P^1$ with coordinate $j$ given by $j= \gamma\_2^3$. We want to check that the fiber over $j$ has a $\mathbb Q(j)$ rational point. Because this is cubic covering, it can't gain a rational point over a quadratic extension if it didn't have one already, so for simplicity we can check that the fiber has a $\mathbb Q( j, \tau , \sqrt{-3})$-rational point.


Over $\mathbb Q(\mu\_3)$, the $3$-torsion points define an $SL\_2(\mathbb Z/3)$-covering of the modular curve, and this map is simply the map associated by the Galois correspondence to the subgroup $\Gamma(\gamma\_2)$ of $SL\_2(\mathbb Z/3)$. So by the Galois correspondence, the fiber has a rational point if and only if the action of the absolute Galois group of $\mathbb Q(j, \tau, \sqrt{-3})$ on the fiber factors through $\Gamma(\gamma\_2)$.


To check this, we split into two cases depending on whether $3$ is split or inert in $\tau$. In either case, there is a natural action of the ring of integers of $\mathbb Q(\tau)$ on the $3$-torsion points which the Galois group commutes with. The action necessarily factors through the ring of integers mod $3$. If $3$ is split, the ring of integers mod $3$ is $\mathbb F\_3 \times \mathbb F\_3$, acting diagonalizably, and so the Galois group consists of elements that commute with it, which must lie in $\mathbb F\_3^\times \times \mathbb F\_3^\times$ (or the determinant $1$ elements of it). If $3$ is inert, the ring is $\mathbb F\_9$, and so the Galois group must lie in $\mathbb F\_9^\times$ (or the determinant $1$ elements of it). No matter how one of these two subgroups is conjugated, they must lie in the subgroup you have written down (for instance because it contains all the elements of order a power of $2$).


The group $\mathbb F\_3^\times \times \mathbb F\_3^\times$ here is a split Cartan, as are all its conjugates, and $\mathbb F\_9^\times$ and all its conjugates are non-split Cartans.


