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

- Is there an intuitive reason why $\gamma_2(\tau)$ has degree $h(D)$?
- What is the connection between Cartan subgroups of $ GL_2(\mathbb Z/N\mathbb Z)$ and this problem?
- Can someone please explain to me what a non-split Cartan subgroup is?

References

Jeremy Booher, "Modular curves and the class number one problem", Theorem $36$ and Definition $45$.

Serre: Lectures on the Mordell-Weil Theorem page 196.

Serre, "Proprietes galoisiennes des points d'ordre fini des courbes elliptiques", 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.

conjugateof $\mathbb F_{p^2}^\times$; you had to choose a basis of $\mathbb F_{p^2}$ as an $\mathbb F_p$-vector space, and varying the choice of basis covers all the conjugates anyway.) $\endgroup$ – LSpice May 29 '19 at 1:39