Timeline for What is the normal closure of $GL_2(\mathbb{Z})$ inside $GL_2(\mathbb{Z}_\ell)$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2015 at 18:54 | comment | added | Will Chen | Yes sorry thats what I meant. | |
| Jul 23, 2015 at 17:12 | comment | added | YCor | You mean: under the abelianization map Out$(\hat{F_2})\to GL_2(\mathbf{Z}_\ell)$. | |
| Jul 23, 2015 at 17:10 | comment | added | YCor | No, I don't know much about $\mathrm{Out}(\hat{F_2})$, maybe just ask an independent question. | |
| Jul 23, 2015 at 17:05 | comment | added | Will Chen | @YCor Would you happen to have any idea about the normal closure of $GL_2(\mathbb{Z}) = Out(F_2)$ inside $Out(\widehat{F_2})$? (Here $\widehat{F_2}$ is the pro-$\ell$ completion of the free group of rank 2). Actually I'm only interested in the $SL_2(\mathbb{Z})$-case anyway, so the question is - is the normal closure of $SL_2(\mathbb{Z})$ inside $Out(\widehat{F_2})$ the full preimage of $SL_2(\mathbb{Z}_\ell)$ under the abelianization map $Out(\widehat{F_2})\rightarrow SL_2(\mathbb{Z}_\ell)$? | |
| Jul 23, 2015 at 16:45 | vote | accept | Will Chen | ||
| Jul 22, 2015 at 18:52 | history | edited | YCor | CC BY-SA 3.0 |
Correction after grghxy's comment.
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| Jul 22, 2015 at 18:47 | comment | added | YCor | (But it's true that both $R$ local or $R$ Euclidean imply $\mathrm{SL}_2(R)$ generated by elementary matrices, and the argument when $R$ is local is the simpler one.) | |
| Jul 22, 2015 at 17:09 | comment | added | YCor | The Euclidean argument works; yes I'm aware there are plenty of variants and generalizations, and since the question was about $\mathrm{SL}_2$ I chose the most elementary argument I had in mind. | |
| Jul 22, 2015 at 16:54 | comment | added | grghxy | Since ${\rm{diag}}(t,1/t)=u^+(t)u^{-}(-1/t)u^+(t-1)u^{-}(1)u^+(-1)$ for units $t$ and standard parameterizations $u^{\pm}$ of standard root groups of ${\rm{SL}}_2$, for any ring $R$ and simply connected Chevalley $R$-group $G$ the subgroup of $G(R)$ generated by $U_a(R)$'s for roots $a$ contains $\Omega(R)$ for an open cell $\Omega$. If $R$ is local then $\ker(G(R)\rightarrow G(k))\subset \Omega(R)$ (as $\Omega$ is an open subscheme of $G$), and translates of $\Omega$ by Weyl-representatives cover $G$ (enough on geometric fibers!), so $U_a(R)$'s generate $G(R)$ (no Euclidean stuff). | |
| Jul 22, 2015 at 16:39 | comment | added | grghxy | Also, a softer way to handle the squares (which works with more general integer rings in place of $\mathbf{Z}$) is to note that since the subset of squares in $\mathbf{Z}_{\ell}$ is open, the $\mathbf{Z}$-submodule they generate inside $\mathbf{Z}_{\ell}$ is open and hence closed, so it is a $\mathbf{Z}_{\ell}$-submodule. But it contains 1, so that $\mathbf{Z}$-submodule coincides with $\mathbf{Z}_{\ell}$. | |
| Jul 22, 2015 at 16:34 | comment | added | grghxy | One has to be careful about the "torus" quotient. More specifically, this argument shows that one gets at least ${\rm{SL}}_2(\mathbf{Z}_{\ell})$, so the answer is actually the image of ${\rm{GL}}_2(\mathbf{Z})$ in the abelian ${\rm{GL}}_2(\mathbf{Z}_{\ell})/{\rm{SL}}_2(\mathbf{Z}_{\ell}) = \mathbf{Z}_{\ell}^{\times}$ (via det), and this image is $\mathbf{Z}^{\times}$. So the normal closure is generated by ${\rm{SL}}_2(\mathbf{Z}_{\ell})$ and diag($u, 1$) for $u \in \mathbf{Z}^{\times}$, with quotient $\mathbf{Z}_{\ell}^{\times}/\mathbf{Z}^{\times}$. | |
| Jul 22, 2015 at 16:28 | history | answered | YCor | CC BY-SA 3.0 |