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Thread: Special Linear Group

  1. #1
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    Special Linear Group

    Hy. Im looking for a proof that SL_n(IZ)->SL_n(IZ/mIZ) is onto. Can you help me?
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    Quote Originally Posted by housi View Post
    Hy. Im looking for a proof that $\displaystyle \textsf{SL}_n(\mathbb{Z})\to \textsf{SL}_n(\mathbb{Z}/m\mathbb{Z})$ is onto. Can you help me?
    Have I interpreted the question correctly? If so, then I think that the result is false.

    For $\displaystyle x\in\mathbb{Z}$, denote by $\displaystyle \bar{x}$ its image in the quotient ring $\displaystyle \mathbb{Z}_m = \mathbb{Z}/m\mathbb{Z}.$ I assume that $\displaystyle \textsf{SL}_n(\mathbb{Z}_m)$ denotes the set of all matrices with entries in $\displaystyle \mathbb{Z}_m$ with determinant $\displaystyle \bar{1},$ and that the map from $\displaystyle \textsf{SL}_n(\mathbb{Z})$ to $\displaystyle \textsf{SL}_n(\mathbb{Z}_m)$ is the natural quotient map given by $\displaystyle (a_{ij})\mapsto(\bar{a}_{ij}).$

    Now take m = 7 and let $\displaystyle A = \begin{bmatrix}\bar{2}&\bar{0}&\bar{0}\\ \bar{0}&\bar{2}&\bar{0}\\ \bar{0}&\bar{0}&\bar{2}\end{bmatrix}\in\textsf{SL} _3(\mathbb{Z}_7).$ Then $\displaystyle \det(A)=\bar{1},$ but any lifting of A to a matrix in $\displaystyle \textsf{GL}_3(\mathbb{Z})$ will have determinant of the form 8+28k (mod 49), which can never be equal to 1.

    Edit. Stupid mistake: 8 + 28k can be equal to 1 (mod 49). In fact, 8 + 12×28 = 7^3 + 1; and in fact $\displaystyle \begin{bmatrix}2&0&7\\-7&2&0\\0&7&86\end{bmatrix}$ is a lifting of that matrix A to an element of $\displaystyle \textsf{SL}_3(\mathbb{Z}).$ So it looks to me as though the result may be true after all. But it's not at all obvious!
    Last edited by Opalg; May 29th 2008 at 03:38 PM.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Have I interpreted the question correctly? If so, then I think that the result is false.
    Yes you have interpretet the question correctly! More precisely i think i need this result to prove $\displaystyle \widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\mathrm{SL }_n(\widehat{\mathbb{Z}})$ (pro-finite completion) in the case $\displaystyle n\geq3$ (this is actually what i want to prove). With the solution of the congruence subgroup problem one can get $\displaystyle \widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\lim_{m>0} \mathrm{SL}_n(\mathbb{Z})/\Gamma(m)$ (projective limes) where $\displaystyle \Gamma(m)$ denotes the kernel of $\displaystyle \mathrm{SL}_n(\mathbb{Z})\to\mathrm{SL}_n(\mathbb{ Z}/ m \mathbb{Z})$. Because it shouldn't be difficult to prove $\displaystyle \mathrm{SL}_n(\widehat{\mathbb{Z}})\cong\lim_{m>0} \mathrm{SL}_n(\mathbb{Z}/ m\mathbb{Z})$ it remains to prove $\displaystyle \mathrm{SL}_n(\mathbb{Z}/ m\mathbb{Z})\cong\mathrm{SL}_n(\mathbb{Z})/\Gamma(m)$. Thus i need to show that this natural projection is onto. Maybe one can solve this using matrices of the type $\displaystyle \mathrm{id}_n+E_{ij}$ where $\displaystyle E_{ij}$ has 1 at the place $\displaystyle (i,j), i\neq j$ and 0 else. But have no idea how to show...
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