# Examples of models for modular curves

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?

• 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? – Will Sawin Sep 9 '19 at 22:50
• @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 $\mathbf Q(j)$). – 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 $$\mathbb{Q}(f)$$.