Let $\Gamma_n$ be the $n$-th congruence subgroup of $GL(2,\mathbb{Z}_p)$. So $\Gamma_n$ consists of matrices in $GL(2,\mathbb{Z}_p)$ which are congruent to the identity matrix modulo $p^n$. Let $Z(\Gamma_n)$ the the center of $\Gamma_n$.

My question is to compute the index $[\Gamma_n/Z(\Gamma_n):\Gamma_{n+1}/Z(\Gamma_{n+1})]$

I know that the index of $[\Gamma_n:\Gamma_{n+1}]=p^4$. I also know that if $G_n$ is the $n$-th congruence subgroup of $SL(2,\mathbb{Z}_p)$, then the index of $[G_n:G_{n+1}]$ is probably $p^3$

(because the topological generators of $G_n$ are $1+p^nE_{2,1}, 1+p^nE_{1,2} $ and $ (1+p^n)E_{1,1}+(1+p^n)^{-1}E_{2,2}$).

Here $E_{i,j}$ is the elementary matrix having $1$ at $(i,j)$-th place and $0$ elsewhere.

I can't figure out the the index $[\Gamma_n/Z(\Gamma_n):\Gamma_{n+1}/Z(\Gamma_{n+1})]$, that is the index of the congruence subgroups of $PGL_2(\mathbb{Z}_p)$.

Any help is welcome. Thanks in advance.

  • 2
    $\begingroup$ (Better asked on Math Stack Exchange...) $\endgroup$ – paul garrett Jan 22 at 20:59
  • 3
    $\begingroup$ Garrett: I asked on Math Stack Exchange on Oct 13, without any reply. Here is the link math.stackexchange.com/questions/2953359/… $\endgroup$ – MathStudent Jan 22 at 21:21
  • 3
    $\begingroup$ The centre of a finite index subgroup $\Gamma $ of $GL(2,{\mathbb Z}_p)$ is the intersection of $\Gamma$ with the group of scalar matrices since $\Gamma$ is Zariski dense in $GL(2)$ ). If $p$ is an odd prime, then $\Gamma _n$ is the product of scalars in $\Gamma _n$ and the intersection $\Delta_n$ of $\Gamma _n$ with $SL(2,{\mathbb Z}_p)$. This will reduce the computation to the index $[\Delta _n:\Delta _{n+1}]$ and that is $p^3$. For odd primes, $[\Gamma _n:\Gamma _{n+1}]$ should be $p^4$. $\endgroup$ – Venkataramana Jan 23 at 1:43
  • 1
    $\begingroup$ It's true, but not obvious, that $\Gamma_n/\operatorname Z(\Gamma_n)$ is a subgroup of $\operatorname{PGL}_2(\mathbb Z_p)$. (A priori it need not embed in any quotient of $\operatorname{GL}_2(\mathbb Z_p)$.) Also true is that $\operatorname Z(\Gamma_n) \cap \Gamma_{n + 1}$ equals $\operatorname Z(\Gamma_{n + 1})$, so that you're just computing $[\Gamma_n : \Gamma_{n + 1}]\cdot[\operatorname Z(\Gamma_n) : \operatorname Z(\Gamma_{n + 1})]^{-1}$. $\endgroup$ – LSpice Jan 24 at 20:25
  • 1
    $\begingroup$ It may help to realise that most of these computations can be reduced to the Lie algebra; $X \mapsto 1 + X$ induces an isomorphism of $\operatorname{Lie}(\Gamma_n)/\operatorname{Lie}(\Gamma_{n + 1})$ with $\Gamma_n/\Gamma_{n + 1}$, with the hopefully obvious meaning of the notation. $\endgroup$ – LSpice Jan 24 at 20:26

Your Answer

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

Browse other questions tagged or ask your own question.