4
$\begingroup$

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?

$\endgroup$
2
  • 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
7
$\begingroup$

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 $\mathbb{Q}(f)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.