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Let $\mathcal M_m$ be the set of $2$-by-$2$ primitive (relatively prime entries) matrices with determinant $m$. Let $\alpha \in \mathcal M_m$ and let $\Gamma\subset \operatorname{SL}_2(\mathbb Z)$. Define $$R_{\alpha,\Gamma}=\lbrace M\in \Gamma\backslash\mathcal M_m : M \equiv \alpha \text{ in } \Gamma\backslash\mathcal M_m/\Gamma\rbrace.$$ Thus $R_{\alpha,\Gamma}$ is a set of representatives $M$ for the left action of $\Gamma$ on $\mathcal M_m$ such that there exist matrices $A,B\in \Gamma$ with the property $M=A\alpha B$.

Let $F_{N,\mathbb C}$ be the field of modular functions of level $N.$ Suppose that $f\in F_{N,\mathbb C}$, $\mathbb C(j)\subset \mathbb C(f)$, and $\alpha \in \mathcal M_m$. The Galois group of $F_{N,\mathbb C}$ over $\mathbb C(j)$ is isomorphic to $\operatorname{SL}_2(\mathbb Z/N\mathbb Z)/\{\pm 1\}$. The subgroup corresponding to the subfield $\mathbb C(f)$ is $$\Gamma(f)=\lbrace A\in\operatorname{SL}_2(\mathbb Z/N\mathbb Z): f\circ A = f\rbrace.$$

It is easy to see that $f\circ \alpha \in F_{mN,\mathbb C}$, and that the minimal polynomial of the function $f\circ \alpha$ over $\mathbb C(f)$ is $$\prod_{M\in R_{\alpha, \Gamma(f)}}X-f\circ M.\label{m}\tag{1}$$

Suppose that in addition $f$ has rational Fourier coefficients. Is the polynomial above the also the minimal polynomial of $f\circ\alpha$ over the field $\mathbb Q(f)$?

Update

The result is true when $\alpha = \begin{pmatrix}m & 0 \\0& 1\end{pmatrix}$. In this case the function $f\circ \alpha$ has rational Fourier coefficients. It is therefore invariant under the action of $(\mathbb Z/mN\mathbb Z)^\times$ on $F_{N}$.

To show that the polynomial $\eqref{m}$ in question is also the minimal polynomial of $f\circ\alpha$ over $\mathbb Q(f)$, we must show that the set of roots of $\eqref{m}$ is stable under all automorphisms fixing $\mathbb Q(f)$. Each such automorphism can be represented as

$$\begin{pmatrix}d & 0 \\0& 1\end{pmatrix}\gamma,$$

where $\gamma \in \Gamma(f)$ and $d\in (\mathbb Z/mN\mathbb Z)^\times$. This is because $\operatorname{Gal}(F_{mN}/\mathbb Q(j)) \cong \operatorname{PGL}_2(\mathbb Z/ mN\mathbb Z)$. Since $f\circ\alpha(z)=f(mz)=\sum_ka_kq^{mk/N}$ has rational coefficients, the matrix $ \begin{pmatrix}d & 0 \\0& 1\end{pmatrix}$ acts trivially on $f\circ\alpha$. The set $R_{\alpha,\Gamma(f)}$ is stable under the action of $\gamma$ by definition.

Update II

We prove that the result is true when both $f$ and $f\circ\alpha$ have rational Fourier coefficients. Let $k=mN$. First note that if $d\in(\mathbb Z/k\mathbb Z)^\times$ and $\gamma \in \Gamma(f)$ then any lift to $\operatorname{SL}_2(\mathbb Z)$ of the matrix $$\begin{pmatrix}1 & 0\\0& d\end{pmatrix}\gamma\begin{pmatrix}1 & 0 \\0& d\end{pmatrix}^{-1}$$ also lies in $\Gamma(f)$ because it has determinant equal to $1$ and the automorphism induced by it fixes $f$ (here we use that $f$ has rational coefficients).

We know that $F_k=\mathbb Q(j,h^{(r,s)}:(r,s)\in \mathbb Z^2, \not \in k\mathbb Z^2)$, where $$h^{(r,s)}(\tau)=\frac{g_2(\tau)}{g_3(\tau)}\wp_\tau\left(\frac{r\tau+s}{k}\right)$$ are the Fricke functions. They satisfy $h^{(r,s)\gamma}=h^{(r,s)}\circ\gamma$ for $\gamma\in \operatorname{SL}_2(\mathbb Z)$, and $(h^{(r,s)})^{\sigma_d}=h^{(r,sd)}$ where $\sigma_d$ is the automorphism induced by $d\in(\mathbb Z/k\mathbb Z)^\times$.

Let $M\in R_{\alpha,\Gamma(f)}$. We must show that for each $d\in(\mathbb Z/k\mathbb Z)^\times$ we have $(f\circ M)^{\sigma_d}=f\circ M'$ for some $M'\in R_{\alpha,\Gamma(f)}$.

Write $f\circ \alpha = Q(h^{(r,s)})$ where $Q$ is a rational function with rational coefficients. Here $(r,s)$ ranges over some finite set of representatives.

We have $A\alpha B=M$ for some $A,B\in \Gamma(f)$; thus $$f\circ M=f\circ \alpha \circ B = Q(h^{(r,s)B}).$$

Let $D = \begin{pmatrix}1 & 0\\0& d\end{pmatrix}$. Then $$(f\circ M)^{\sigma_d}= Q(h^{(r,s)BD})=Q(h^{(r,s)DB'}),$$ where $B'=D^{-1}BD$ lies in $\Gamma(f)$ by above discussion. Therefore $$(f\circ M)^{\sigma_d}=Q(h^{(r,s)D}\circ B').$$ On the other hand, $(f\circ \alpha)^{\sigma_d}= Q(h^{(r,s)D})$, so $(f\circ M)^{\sigma_d}=(f\circ \alpha)^{\sigma_d}\circ B'$. Now we use the fact that $f\circ\alpha$ has rational Fourier coefficients to deduce that $(f\circ M)^{\sigma_d}=f\circ \alpha\circ B'$. Thus we can take $M' = \alpha B'$.

For the Fricke functions see Shimura: An introduction to the theory of automorphic functions, Section 6.2.

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  • $\begingroup$ I don't understand your definition of $R_{\alpha,\Gamma}$: if $\Gamma=\mathrm{SL}_2(\mathbb{Z})$ then $\Gamma \begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix} \Gamma$ is the disjoint union of $p+1$ cosets $\Gamma M$ (they correspond to the sublattices of $\mathbb{Z}^2$ with index $p$). Furthermore $\Gamma \backslash \mathcal{M}_m / \Gamma$ is in bijection with the pairs $(a,b)$ with $a|b$ and $ab=m$, so if $m$ is composite there is more than one representative. $\endgroup$ – François Brunault Apr 8 '19 at 13:19
  • $\begingroup$ @FrançoisBrunault, Thank you for your comment. I have forgotten the assumption that the matrices in the set $\mathcal M_m$ must be primitive. In this case there is only one representative (see for example this question(mathoverflow.net/questions/273919/…)). However when $\Gamma$ is smaller that the modular group, there may be more representatives. The set $R_{\alpha, \Gamma}$ is a set of representatives congruent to $\alpha$ from the right but not congruent from the left. $\endgroup$ – Shimrod Apr 8 '19 at 13:43
  • $\begingroup$ I see, then the second setence of second paragraph should be before the definition of $R_{\alpha,\Gamma}$ (as it stands, it seems to say $R_{\alpha,\Gamma}$ has cardinality 1). $\endgroup$ – François Brunault Apr 8 '19 at 13:51
  • $\begingroup$ @FrançoisBrunault, It says that $R_{\alpha, \Gamma}$ has cardinality $1$ when $\Gamma$ is the full modular group. $\endgroup$ – Shimrod Apr 8 '19 at 13:59
  • $\begingroup$ But e.g. $\begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix}$ are distinct elements in $R_{\alpha,\Gamma}$ since the condition $M \equiv \alpha$ is vacuous in the case $\Gamma=\mathrm{SL}_2(\mathbb{Z})$. $\endgroup$ – François Brunault Apr 8 '19 at 14:17
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EDIT. The answer is yes as soon as $f$ and $f \circ \alpha$ both have rational Fourier coefficients.

To see this, we recall Shimura's theorem: for any modular form $f$ of level $N$, any $g \in \mathrm{SL}_2(\mathbb{Z})$ and any $\sigma \in \mathrm{Aut}(\mathbb{C})$, we have $(f | g)^\sigma = f^\sigma | g_\lambda$ where $g_\lambda \in \mathrm{SL}_2(\mathbb{Z})$ is any lift of the matrix $\begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}^{-1} g \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$, and $\lambda$ is defined by $\sigma(e^{2\pi i/N})=e^{2\pi i\lambda/N}$. See the comments below for the precise reference.

Note that $g \to g_\lambda$ defines an action of $(\mathbb{Z}/N\mathbb{Z})^\times$ on $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$. Shimura's theorem implies immediately the following.

Lemma. Let $f$ be a modular function (or modular form) of level $\Gamma(N)$ with rational Fourier coefficients. Then the stabilizer $\Gamma(f)$ of $f$ in $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ is stable under the action of $(\mathbb{Z}/N\mathbb{Z})^\times$ defined above.

Now, let $\alpha \in M_2(\mathbb{Z})$ be any matrix with positive determinant, and assume that $f \circ \alpha$ has rational Fourier coefficients. The function $f \circ \alpha$ is modular of some level $N'$ divisible by $N$. We want to show that the polynomial $P_{f,\alpha}$ defined by (1) has coefficients in $\mathbb{Q}(f)$. For this, it is enough to show that the set of roots of $P_{f,\alpha}$ is stable by $\mathrm{Aut}(\mathbb{C})$. So let $M$ be any matrix in $\Gamma(f) \alpha \Gamma(f)$, which we write $M=\gamma' \alpha \gamma$. Then \begin{equation*} (f|M)^\sigma = (f | \alpha \gamma)^\sigma = (f \circ \alpha)^\sigma | \gamma_{\lambda'} = f | \alpha \gamma_{\lambda'} \end{equation*} for some $\lambda' \in (\mathbb{Z}/N'\mathbb{Z})^\times$. By the lemma above applied in level $N'$, the matrix $\gamma_{\lambda'}$ belongs to $\Gamma(f)$, which finishes the proof.


In general however, this is not always true. For example, take a modular function $f$ of level $N$ with rational Fourier coefficients and $\Gamma(f)=\{\pm I\}$. Such an $f$ exists: the modular curve $Y(N)(\mathbb{C}) = \Gamma(N) \backslash \mathcal{H}$ can be defined over $\mathbb{Q}$, so we have an extension of function fields $\mathbb{Q}(Y(N))/\mathbb{Q}(j)$ with Galois group $\mathrm{PSL}_2(\mathbb{Z}/N\mathbb{Z})$, then we can take any $f$ generating this extension.

Take $m=1$ and let $\alpha$ be any matrix in $\mathrm{PSL}_2(\mathbb{Z}/N\mathbb{Z})$. Note that $R_{\alpha,\Gamma_f} = \Gamma(N) \alpha$, so we simply have $P_{f,\alpha}(X)=X-f \circ \alpha$.

In general $f \circ \alpha$ will not have rational Fourier coefficients. In fact, a theorem of Shimura tells us that for any $\sigma \in \mathrm{Aut}(\mathbb{C})$, we have $(f \circ \alpha)^\sigma = f^\sigma \circ \alpha_\sigma$ where $\alpha_\sigma = \begin{pmatrix} 1 & 0 \\ 0 & \chi(\sigma) \end{pmatrix}^{-1} \alpha \begin{pmatrix} 1 & 0 \\ 0 & \chi(\sigma) \end{pmatrix}$ and $\chi(\sigma) \in (\mathbb{Z}/N\mathbb{Z})^\times$ is the cyclotomic character, defined by $\sigma(e^{2\pi i/N})=e^{2\pi i \chi(\sigma)/N}$. See for example this MO answer.

In general however, it seems that you are asking whether the Hecke operator $T_\alpha =\Gamma(f) \alpha \Gamma(f)$ is defined over $\mathbb{Q}$. Maybe there will still be a relation of the form $T_\alpha \circ \sigma = \sigma \circ T_{\alpha'}$ at least when $m$ is prime to $N$ (this is just a guess, I have to check). In any case, this kind of property is more naturally stated using the adelic language: you take a compact open subgroup $K$ of $\mathrm{GL}_2(\hat{\mathbb{Z}})$ and consider the Hecke correspondence $T_\alpha = K\alpha K$ where $\alpha$ is any matrix in $\mathrm{GL}_2(\mathbb{A}_f)$, here $\mathbb{A}_f = \hat{\mathbb{Z}} \otimes \mathbb{Q}$ are the finite adeles of $\mathbb{Q}$. Then the Galois action will essentially correspond to the determinant of $\alpha$ via abelian class field theory.

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  • $\begingroup$ Thank you for your answer, please see the updated question. Could you please give me the reference? What is a good place to start thinking in terms of modular curves, rather than modular functions? $\endgroup$ – Shimrod Apr 10 '19 at 15:04
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    $\begingroup$ @Shimrod The reference is Shimura's book "Arithmeticity in the theory of automorphic forms", volume 82 of Mathematical Surveys and Monographs. AMS, Providence, RI, 2000. See also arXiv 1807.00391 for at least one other proof. $\endgroup$ – François Brunault Apr 10 '19 at 15:20
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    $\begingroup$ For modular curves I would suggest Diamond-Shurman's book A first course in modular forms and Diamond-Im's article Modular forms and modular curves. Rohrlich's article (Chapter 3 in Cornell, Silverman & Stevens, Modular Forms and Fermat's Last Theorem) also discusses the function field of $Y(N)$. For adelic modular forms and adelic modular curves you could try to start here: warwick.ac.uk/fac/sci/maths/people/staff/david_loeffler/… (see Lecture 4). $\endgroup$ – François Brunault Apr 10 '19 at 15:21
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    $\begingroup$ @Shimrod Sorry, I forgot to write this is Lemma 10.5 in Shimura's book. Also, I don't see why the result is true for $\alpha=\begin{pmatrix} m & 0 \\ 0 & 1 \end{pmatrix}$. $\endgroup$ – François Brunault Apr 10 '19 at 17:18
  • $\begingroup$ I updated the question, is it clear now? $\endgroup$ – Shimrod Apr 10 '19 at 19:28

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