5
$\begingroup$

$\DeclareMathOperator\SL{SL}$For an odd prime $p$, let $$\Gamma_1(p)=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \SL_2(\mathbb{Z}):\begin{pmatrix}a & b \\ c & d\end{pmatrix}\equiv\begin{pmatrix}1 & * \\ 0 & 1\end{pmatrix} \mod (p)\right\}\subset \SL_2(\mathbb{Z})$$ be the congruence subgroup of level $p$. Let $e$ be the matrix $$\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.$$

Question: What is the normal closure of $\{e\}$ in $\Gamma_1(p)$?

For $$A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in\Gamma_1(p),$$ we can see that $$AeA^{-1}=\begin{pmatrix}1-ac & a^2 \\ -c^2 & 1+ac\end{pmatrix}$$ lies in the subgroup $$\Gamma=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \SL_2(\mathbb{Z}):a\equiv d\equiv 1 \mod (p),\quad c\equiv 0 \mod \left(p^2\right)\right\}.$$ My guess is that the normal closure is exactly this subgroup $\Gamma$.

I have checked it for $p=3$, in which case $\Gamma$ is generated by $$\left\{\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix},\ \begin{pmatrix}4 & -1 \\ 9 & -2\end{pmatrix},\ \begin{pmatrix}7 & -4 \\ 9 & -5\end{pmatrix}\right\},$$ where $$\begin{pmatrix}4 & -1 \\ 9 & -2\end{pmatrix}=\begin{pmatrix}1 & 1 \\ 3 & 4\end{pmatrix}\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}4 & -1 \\ -3 & 1\end{pmatrix}$$ and $$\begin{pmatrix}7 & -4 \\ 9 & -5\end{pmatrix}=\begin{pmatrix}-2 & 1 \\ -3 & 1\end{pmatrix}\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & -1 \\ 3 & -2\end{pmatrix}$$ are conjugates of $e^{-1}$.

For $p=5$, I used Magma to obtain a set of generators of $\Gamma$, while not all of which are conjugates of $e^n$. However, it could still be possible that they are products of conjugates of $e^n$. I have no idea how it would be proven for general $p$.

$\endgroup$
5
  • $\begingroup$ A successive-approximation argument at least shows that the analogous normal closure with the $p$-adic integers $\mathbb Z_p$ in place of $\mathbb Z$ is what you'd like. Do you happen to know that the normal closure contains some congruence subgroup? $\endgroup$
    – LSpice
    Commented Apr 11 at 1:24
  • 4
    $\begingroup$ My guess is that the normal closure of this matrix is an infinite-index subgroup. The point here is that your subgroup of SL(2,Z) is the fundamental group of a finite cover of the modular curve, and I believe that once your prime p is large enough that cover will have positive genus. Conjugates of your matrix e will correspond to loops around cusps since e is parabolic, so at best your normal closure is the subgroup generated by the cusps. (sorry for the poor typesetting and explanation — I am writing this on a phone) $\endgroup$ Commented Apr 11 at 1:44
  • $\begingroup$ @AndyPutman You're absolutely correct! It turns out that the genus of the modular curve for $\Gamma$ is already nonzero when $p=7$ (it has genus 3). Thanks! $\endgroup$
    – Max
    Commented Apr 11 at 4:40
  • $\begingroup$ @LSpice I checked that the normal closure and $\Gamma$ do have the same image in $SL_2(\mathbb{Z}/p^n)$. It seems that the normal closure cannot be a congruence subgroup by Andy's argument when $p\ge7$. $\endgroup$
    – Max
    Commented Apr 11 at 4:45
  • $\begingroup$ @Max: Happy to help! $\endgroup$ Commented Apr 11 at 12:01

1 Answer 1

5
$\begingroup$

$\DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \newcommand\bbH{\mathbb{H}}$ I'm at my office now, so I'll write a properly typeset version of my comment so this question will leave the "unanswered" queue.

Recall that the modular curve $Y$ is $\bbH^2/\SL_2(\Z)$. Topologically, $Y$ is a once-punctured $2$-sphere with two orbifold points. Any finite-index subgroup $\Gamma$ of $\SL_2(\Z)$ gives a finite orbifold cover $Y(\Gamma)$ of $Y$. For your subgroup $\Gamma_1(p)$, write $Y_1(p)$ for $Y(\Gamma_1(p))$.

The space $Y_1(p)$ is topologically a punctured surface of finite type. The punctures are the cusps. Your element $e$ is a parabolic element of $\SL_2(\Z)$, so considered as an element of the fundamental group of $Y_1(p)$ it corresponds to a loop that is homotopic into one of the cusps. The same is true for all conjugates of $e$. It follows that the normal closure of $e$ in $\Gamma_1(p)$ is contained in the subgroup of the fundamental group of $Y_1(p)$ generated by loops around the cusps. This is an infinite-index proper subgroup if $Y_1(p)$ has positive genus, which it does for $p \geq 7$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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