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Let $\Gamma = SL_2(\mathbb{Z})$. Let $\mathcal{H} = \{z \in \mathbb{C}\ |\ Im(z) > 0\}$. Let $\sigma = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $\Gamma$. Let $z \in \mathcal{H}$. We write $$\sigma z = \frac{pz + q}{rz + s}$$ It is easy to see that $\sigma z \in \mathcal{H}$ and $\Gamma$ acts on $\mathcal{H}$. Let $z, w \in \mathcal{H}$. We say $z$ and $w$ are equivalent if there exists $\sigma \in \Gamma$ such that $w = \sigma z$. It is well known the stabilizer of a point of $\mathcal{H}$ is trivial except it is equivalent to $i$ or $\rho = (-1 + i\sqrt 3)/2$(see here). Note that $\rho$ is a primitive cube root of unity.

On the other hand, it is well known that the unit group of an imaginary quadratic number field is trivial except it is $\mathbb{Q}(i)$ or $\mathbb{Q}(\sqrt{-3}) = \mathbb{Q}(\rho)$.

My question Are these two facts related or mere coincidence? If they are related, how?

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Makoto Kato
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1 Answers1

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Yes, these facts are related. If you view $Y=\mathcal H/SL_2(\mathbf Z)$ as the moduli space of complex elliptic curves, the orbits of $i$ and $\rho$ correspond to elliptic curves which have extra automorphisms. Namely, the curve $\mathbf C/\mathbf Z[i]$ has an automorphism of order $4$ induced by multiplication by $i$, and similarily the curve $\mathbf C/\mathbf Z[\rho]$ has an automorphism of order $3$ given by multiplication by $\rho$. As a consequence, $Y$ fails to be a fine moduli space, and this is reflected in the presence of these "conical" points. (However, they are not really singular points of $Y$, which is in fact smooth. Some people call them "orbifold cusps". What distinguishes them from the other points is really the behavior of the map $\mathcal H \to \mathcal H/SL_2(\mathbf Z)$ over them.)

Bruno Joyal
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