Let $$Q(x,y)=ax^2+bxy+cy^2$$ be a positive definite quadratic form with $a>0$ and $D=b^2-4ac<0$. Let $$\zeta_Q(s)=\sideset{}{'}\sum_{m,n}Q(m,n)^{-s},$$ the accent indicating that $(0,0)$ is omitted from the sum. According to the Kronecker limit formula we have $$\lim_{s\to1 +}\left(\lvert D\rvert^{s/2}\zeta_Q(s)-\frac{2\pi }{s-1}\right)=2\pi(2C-\log 4 +H(z_Q))$$ where $C$ is a constant, $z_Q$ is a point in the upper half-plane $\mathfrak H$ such that $Q(z_Q,1)=0,$ and

$$H(z)=-\log(\Im(z)\lvert \eta(z)\rvert^4)$$ $$\eta(z)=q^{1/24}\prod_{n=1}^\infty(1-q^n),\qquad q = e^{2\pi i z}.$$

Let $K=\mathbb Q(\sqrt D)$ and let $$L_K(s,\chi)=\sum_{\mathfrak a}\chi(\mathfrak a)N(\mathfrak a)^{-s}\quad\text{and}\quad \zeta(s,\mathcal A)=\sum_{\mathfrak a\in \mathcal A}N(\mathfrak a)^{-s}.$$ Here $\chi$ is a character of the class group $\operatorname{Cl}_K$. We have $$L_K(s,\chi)=\sum_\mathcal A\chi(\mathcal A)\zeta(s,\mathcal A).$$ Some work shows that actually $\zeta_Q(s)=\zeta(s,\mathcal A)$ if the class of the quadratic form $Q$ corresponds to the ideal class $\mathcal A$. Multiplying the limit formula by $\chi(\mathcal A)$, summing over the ideal classes, and letting $s\to1+$ therefore gives $$\frac {1}{2\pi}\lvert D \rvert^{1/2} L_K(1,\chi)=\sum_\mathcal A \chi (\mathcal A) H(z_\mathcal A).\label{f}\tag{1}$$ Here $z_\mathcal A$ is the CM point corresponding to the ideal class $\mathcal A$. Now we specialize $\chi$ to the genus character which is defined as follows in the case when $D$ is a fundamental discriminant, that is, $D$ is the discriminant of $K$. Let $D=D'D''$ be a decomposition of $D$ into fundamental discriminants. Let $\mathfrak p$ be a prime ideal of $K$. Then we define $$ \chi(\mathfrak p)= \begin{cases} \chi_{D'}(N(\mathfrak p))\quad \text{if } (D',\mathfrak p)=1\\ \chi_{D''}(N(\mathfrak p))\quad\text{if } (D'',\mathfrak p)=1 \end{cases} $$ Here $\chi_D$ is the Kronecker symbol modulo $\lvert D\rvert$ (the unique real primitive character modulo $\lvert D\rvert$). Each such decomposition corresponds to a decomposition of $L_K(s,\chi)$. Let $L_D(s)=\sum_{n=1}^\infty\chi_D(n)n^{-s}$. The decomposition is given by $$L_{D'}(s)L_{D''}(s)=L_K(s,\chi)\label{dec}\tag{2}$$. Now the Dirichlet class number formula says that $$L_{D}(1)=\begin{cases}\pi \lvert D\rvert^{1/2}h_D\quad \text{if } D<0\\2\lvert D\rvert^{1/2}h_D\log \varepsilon_D \quad \text{if } D>0\end{cases}\label{dir}\tag{3}$$ where $\varepsilon_D$ is the fundamental unit. Combining \eqref{f},\eqref{dec}, and \eqref{dir} we obtain $$\varepsilon_D^{h_{D'}h_{D''}}=\prod_{\mathcal A}\lvert\Im(z_\mathcal A)^{1/2}\eta(z_\mathcal A)^2\rvert^{-\chi(\mathcal A)}.\label{s}\tag{4}$$

We proved this formula for fundamental discriminants. How to prove it for arbitrary discriminants?

The case of non-fundamental discriminants arises in Siegel's solution of the class number one problem for imaginary quadratic fields.

Suppose that $K$ is an imaginary quadratic field with $h_K=1$. We may write $D_K=-p$, where $p>163$, $p\equiv \pm 2 \text{ ( mod 5)}$ is a prime.

Let $D=-5^2p$ and consider the decomposition $D=TG$, where $T=5,G=-5p$. Since $G$ is divisible by two distinct primes, we have $h_T=2m_p$ for some integer $m_p$. Further we have $h_G=1$, and $ \varepsilon = \varepsilon_g=(1/2)(1+\sqrt 5)$.

Siegel defines two modular functions $$\phi(z)=-\sqrt 5\frac{\eta\left(\frac {z}{5}\right)\eta\left(\frac {z+1}{5}\right)\eta\left(\frac {z-1}{5}\right)}{\eta\left(5z\right)\eta\left(\frac {z+2}{5}\right)\eta\left(\frac {z-2}{5}\right)}$$ $$\psi(z)=-\sqrt 5\frac{\eta\left(\frac {z}{5}\right)\eta\left(\frac {z+2}{5}\right)\eta\left(\frac {z-2}{5}\right)}{\eta\left(5z\right)\eta\left(\frac {z+1}{5}\right)\eta\left(\frac {z-1}{5}\right)}$$

Therefore for a suitable CM point $\tau\in \mathfrak H$ corresponding to the one ideal class of $K$ we have from \eqref{s} $$\varepsilon ^{2m_p} =\phi(\tau)$$ if $p\equiv 3 ~(5)$ and $$\varepsilon ^{2m_p} =\psi(\tau)$$ if $p\equiv 2~(5)$.

Next Siegel shows that the functions $\psi$ and $\phi$ are uniformizers for a certain modular curve of genus zero. He obtains relations of the form $j=R_1(\phi)$ and $j=R_2(\psi)$ where $R_1,R_2$ are rational function with coefficients in $\mathbb Q(\sqrt 5)$. We know that the $j$-invariant is a cube of a rational integer at $\tau$. We get therefore equations of the form $\gamma^3=R_i(\varepsilon^{2{m_p}} )$. Siegel is able to solve these, proving thereby that there is no tenth imaginary quadratic field with class number equal to one.

Imin Chen in his paper interprets Siegel's solution im more modern language. He shows that the Siegel modular functions are in fact uniformizers for the nonsplit Cartan modular curve $X_{ns}^+(5)$.

Can we always express the uniformizer for $X_{ns}^+(N)$ as an eta-quotient, similarly as Siegel? Which modular curves have uniformizers of this form? What are the canonical references for the Cartan modular curves?

  • 1
    $\begingroup$ What do you mean when you say "uniformizer" for a modular curve? I imagine most mathematicians would interpret a "uniformizer" to be a modular function $f$ so that the function field $K(X)$ is equal to $K(f)$ (where $K$ is some base field, either $\mathbb{Q}$, a number field, or $\mathbb{C}$). This requires the modular curve $X/K$ to be isomorphic to $\mathbb{P}^{1}$ and in particular to have genus zero (which Siegel's modular curve that is "uniformized" by $\psi$ and $\phi$ is not). $\endgroup$ – Jeremy Rouse Jul 27 '19 at 0:33
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    $\begingroup$ For any congruence subgroup the generators of the field are polynomial functions of $j,j(\frac{az+ b}{d})$ thus eta quotients. Hence you need to make clear that your two functions generate not only the field of a modular curve but also the coordinate ring of some associated affine curve (defined at first in term of isomorphism classes of elliptic curves with some $N$-th level data) @JeremyRouse so that for each non-singular point there is an uniformizer which is a polynomial in those two functions. $\endgroup$ – reuns Jul 27 '19 at 1:01
  • $\begingroup$ @JeremyRouse, Thanks, the Siegel's modular curve has genus zero, I have erroneously written one. $\endgroup$ – Shimrod Jul 27 '19 at 8:38

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