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Math Help - 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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  2. #2
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    Quote Originally Posted by housi View Post
    Hy. Im looking for a proof that \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 x\in\mathbb{Z}, denote by \bar{x} its image in the quotient ring \mathbb{Z}_m = \mathbb{Z}/m\mathbb{Z}. I assume that \textsf{SL}_n(\mathbb{Z}_m) denotes the set of all matrices with entries in \mathbb{Z}_m with determinant \bar{1}, and that the map from \textsf{SL}_n(\mathbb{Z}) to \textsf{SL}_n(\mathbb{Z}_m) is the natural quotient map given by (a_{ij})\mapsto(\bar{a}_{ij}).

    Now take m = 7 and let 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 \det(A)=\bar{1}, but any lifting of A to a matrix in \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 \begin{bmatrix}2&0&7\\-7&2&0\\0&7&86\end{bmatrix} is a lifting of that matrix A to an element of \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 04:38 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 \widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\mathrm{SL  }_n(\widehat{\mathbb{Z}}) (pro-finite completion) in the case n\geq3 (this is actually what i want to prove). With the solution of the congruence subgroup problem one can get \widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\lim_{m>0}  \mathrm{SL}_n(\mathbb{Z})/\Gamma(m) (projective limes) where \Gamma(m) denotes the kernel of \mathrm{SL}_n(\mathbb{Z})\to\mathrm{SL}_n(\mathbb{  Z}/ m \mathbb{Z}). Because it shouldn't be difficult to prove \mathrm{SL}_n(\widehat{\mathbb{Z}})\cong\lim_{m>0}  \mathrm{SL}_n(\mathbb{Z}/ m\mathbb{Z}) it remains to prove \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 \mathrm{id}_n+E_{ij} where E_{ij} has 1 at the place (i,j), i\neq j and 0 else. But have no idea how to show...
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