Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbf Z)$. Let $\mathfrak H^*=\mathfrak H\cup\mathbf Q\cup\{\infty\}$. Shimura in his book Introduction to Arithmetic Theory of Automorphic Functions, section 6.7, considers the notion of a model for the compact Riemann surfac $\Gamma\backslash\mathfrak H^*$: let $\varphi$ be a modular function invariant under $\Gamma$ and let $V$ be a projective non-singular algebraic curve isomorphic to $\Gamma\backslash\mathfrak H^*$; then the pair $(\varphi,V)$ is called a model for $\Gamma\backslash\mathfrak H^*$ if $\varphi$ provides an isomorphism between $\Gamma\backslash\mathfrak H^*$ and $V$.

Let $\Gamma'$ be another congruence subgroup and suppose that $\alpha\Gamma\alpha^{-1}\subset \Gamma'$, where $\alpha \in \operatorname{GL}_2^+(\mathbf Q)$. Let $(\varphi',V')$ be a corresponding model. Then, according to Shimura, we have the following commutative diagram $\require{AMScd}$ \begin{CD} \mathfrak H^* @>\alpha>> \mathfrak H^*\\ @V \varphi V V @VV \varphi' V\\ V @>>T> V' \end{CD} Here $T$ is a rational map. For example if $\alpha=1$ and $\Gamma$ is a genus zero congruence subgroup of $\Gamma'=\operatorname{SL}_2(\mathbf Z)$ then we get the expression for the $j$-invariant as a rational function of the uniformizer for $\Gamma$.

What is a concrete example of this, when $\alpha$ is not the identity matrix?

  • 2
    $\begingroup$ What type of examples would you like? Would you like the expression for the $j$-invariant as a rational function of the uniformizer? Would you like genus zero examples? Higher genus? Examples where $\Gamma=\Gamma'$ like the Atkin-Lehner involution? $\endgroup$ – Will Sawin Sep 9 '19 at 22:50
  • $\begingroup$ @WillSawin, I would like see a concrete example for genus zero with $\alpha\neq 1$. Also I am especially interested in minimal polynomials of $f\circ \alpha$ over $\mathbf C(f)$ (or $\mathbf Q(f)$), and the relation of these polynomials to the minimal poynomials of $f$ over $\mathbf C(j)$ (or $\mathbf Q(j)$). $\endgroup$ – Shimrod Sep 9 '19 at 23:14

Here's an example. Let's take $\Gamma = \Gamma_{0}(4)$, and $\Gamma' = \Gamma(2)$. We'll let $\alpha = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$, so $\Gamma' = \alpha \Gamma \alpha^{-1}$. The function $$ f(z) = \frac{\eta^{8}(z)}{\eta^{8}(4z)} = q^{-1} - 8 + 20q - 62q^{3} + \cdots $$ is a uniformizer for $\Gamma_{0}(4)$ (having a simple pole at $\infty$ and a simple zero at the cusp at zero). The minimal polynomial for $f$ over $\mathbb{Q}(j)$ is $j = \frac{(f^{2} + 256f + 4096)^{3}}{f^{5} + 16f^{4}}$. This is found via linear algebra, and knowing that $X_{0}(4) \to X(1)$ is a degree $6$ cover (so the minimal polynomial will have degree $6$).

The function $h = f \circ \alpha^{-1} = f(z/2)$ is a modular form for $\alpha \Gamma \alpha^{-1}$. The minimal polynomial for $h(z)$ over $\mathbb{Q}(j)$ is $j = \frac{(h^{2} + 16h + 256)^{3}}{h^{2} (h+16)^{2}}$. If $X$ is the modular curve corresponding to $\Gamma_{0}(4) \cap \Gamma(2)$, then $X \to X_{0}(4)$ is a degree $2$ cover, and in fact one finds that $$ h^{2} + 16h + \frac{f^{5}}{65536} + \frac{f^{4}}{65536} (-j + 752) + \frac{769f^{3}}{256} + \frac{4863}{16} f^{2} + 8192 f + 65536 = 0. $$ This was found by factoring the minimal polynomial of $h$ over $\mathbb{Q}(f)$.


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