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Let $a_p$ denote the matrix $[[1,0],[0,p]]$, where $p$ is prime.

Then $SL_2(\mathbb{Z}[1/p])$ can be presented as the amalgamated product

$$SL_2(\mathbb{Z})*_{\Gamma_0(p)} a_pSL_2(\mathbb{Z})a_p^{-1}$$ where $\Gamma_0(p)$ are the matrices which are upper-triangular mod $p$, and the inclusions into the two factors are the natural ones.

I'd like to construct the group $GL_2(\mathbb{Z}[1/p])$ (in the software package GAP) as a semidirect product of $SL_2(\mathbb{Z}[1/p])$ and $p^\mathbb{Z}$.

For this, I need to tell GAP the action of conjugation of $a_p$ on $SL_2(\mathbb{Z}[1/p])$ in terms of the generators of the amalgamated product.

I'm sure this must have been done somewhere. Does anyone have a reference? (Or perhaps can someone give an explicit presentation of this conjugation action?)

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  • $\begingroup$ Alas this seems not to be possible at the moment, starting with the difficulty of actually defining $SL_2(\Z[1/p])$ or the lack of methods to construct infinite semidirect products for a genertic action. $\endgroup$ – ahulpke Jun 10 '19 at 19:00
  • $\begingroup$ @ahulpke I've actually already constructed the group $SL_2(\mathbb{Z}[1/p])$ as an FpGroup using the amalgamated product description above. Surely if there is an explicit description of the conjugation action in terms of generators of the amalgam, I can just explicitly give the semidirect product using generators and relations? $\endgroup$ – user355183 Jun 10 '19 at 19:13
  • $\begingroup$ OK, so your question is more about writing down a presentation rather than using the SemidirectProduct constructor in GAP (which I was talking about). $\endgroup$ – ahulpke Jun 12 '19 at 0:48
  • $\begingroup$ @ahulpke Yes, though unfortunately doing even simple calculations with this group seems to be prohibitively inefficient. The presentation I'm working with has 3 generators and 8 relators of total length 85. Are such fp groups just too complicated to work with? Specifically I'm trying to intersect two 2-generator subgroups of this group, and computing the index of the intersection inside the two 2-generator subgroups. $\endgroup$ – user355183 Jun 12 '19 at 17:31
  • $\begingroup$ Unless the subgroups have finite index in the large group, I doubt that there is an algorithm that could calculate the intersection. $\endgroup$ – ahulpke Jun 12 '19 at 21:34
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Well, this is an exercise, what had you tried?

Let $u=e_{12}(1)$, $v=e_{21}(1)$, $w=e_{12}(p^{-1})$, $x=e_{21}(p)$ be the given generators (actually $u=w^p$ and $x=v^p$ are redundant).

Then for $a=a_p$, we have $aua^{-1}=w$, $ava^{-1}=x$, $axa^{-1}=x^p$. It remains to express $awa^{-1}=e_{12}(p^{-2})$ in terms of the generators.

We have $e_{12}(t^{-1})e_{21}(-t)e_{12}(t^{-1})=s(t):=\begin{pmatrix}0 &t^{-1}\\ -t & 0\end{pmatrix}$, and $s(t)s(-1)=d(t):=\begin{pmatrix}t^{-1} & 0\\ 0 & t\end{pmatrix}$. We have $d(t)e_{12}(1)d(t)^{-1}=e_{12}(t^{-2})$.

So, we have $$awa^{-1}=d(p)ud(p)^{-1}=s(p)s(-1)us(-1)^{-1}s(p)^{-1}=wx^{-1}wu^{-1}vuv^{-1}uw^{-1}xw^{-1}.$$

(If no stupid mistake in the computation.)

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  • $\begingroup$ I'm a bit confused. Would you mind explaining your notation a bit? As for what I've tried, it's hard to explain, but beyond the fact that the conjugation action on the generators on the $SL_2(\mathbb{Z})$-factor on the left is easy, but I had no systematic way of understanding the action on the generators of the factor on the right. In my experience brute force working with matrices isn't especially enlightening or effective without having an understanding of what you're really doing, which is why I hoped someone who actually understands what they're doing might be able to help. $\endgroup$ – user355183 Jun 10 '19 at 21:54
  • $\begingroup$ Ah okay I assume you mean $u$ is the upper triangular parabolic $[[1,1],[0,1]]$ and $v$ is its conjugate. Okay I see. $\endgroup$ – user355183 Jun 10 '19 at 21:59
  • $\begingroup$ $(E_{ij})$ being the standard basis, the notation is $e_{ij}(t)=I+tE_{ij}$. $\endgroup$ – YCor Jun 10 '19 at 22:05
  • $\begingroup$ Fantastic! I was using the two torsion generators of orders 4 and 6, which may be why I got nowhere. You seem to have done this very quickly - would you mind saying a few words about how you were led to consider these calculations? Is this just the first thing you happened to try or were you guided by some Lie theoretic intuition? $\endgroup$ – user355183 Jun 10 '19 at 22:26
  • $\begingroup$ For the only tricky part, I sort of remembered that $s(t)$ was obtained with a computation involving $e_{12/21}(\pm t^{-1})$ and got the formula by trying. $\endgroup$ – YCor Jun 10 '19 at 22:28

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