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This may be a naïve question, but I have been thinking about the structure of the the matrix group $SL_2(\mathbb{Z}[i]) \subset SL_2(\mathbb{C})$. One thing that has been on my mind is whether or not $SL_2(\mathbb{Z}[i])$ is finitely generated. We know that $SL_2(\mathbb{Z}) \subset SL_2(\mathbb{R})$ is finitely generated, namely by $$S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}; \quad T =\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$

Since, $S$ and $T$ are in $SL_2(\mathbb{Z}[i])$ is it possible to come up with a generating set? My thought would be potentially having $S, T$ and maybe $i I_2$. If anyone has any insight, please share.

J. W. Tanner
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An Isomorphic Teen
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    $iI_2\notin \mathsf{SL}_2(\mathbb{Z}[i])$, since the determinant is $-1$, not $1$. You do have $\left(\begin{array}{cc}0&i\i&0\end{array}\right)$, and the diagonal matrix with diagonal entries $i$ and $-i$. – Arturo Magidin Feb 22 '23 at 22:21
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    Won't the same argument as for $SL_2(\Bbb{Z})$ work? Using $S$ and $T_1,T_i$ to apply the gcd algorithm to the bottom row. – reuns Feb 22 '23 at 22:25
  • Good point regarding $i I_2$ – An Isomorphic Teen Feb 22 '23 at 22:28
  • @reuns are you suggesting that its possible to have a generating set of the matrices? Also, im not sure I have seen the argument regarding the gcd algorithm. I was shown a proof using the linear fractional transformation as a group action. – An Isomorphic Teen Feb 22 '23 at 22:31
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    Yes. Show us your non $\gcd$ algorithm proof – reuns Feb 22 '23 at 22:32
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    The idea I think would be to do something very close to the Smith normal form calculation, except using only row/column operations of determinant 1 (so in particular, in place of swapping rows/columns you would use an operation like $S$, and similarly instead of multiplying a row by a unit you would need to for example multiply row 1 by the unit and row 2 by its inverse). Then at the end, you should get a power of $\begin{bmatrix} i & 0 \ 0 & -i \end{bmatrix}$. – Daniel Schepler Feb 22 '23 at 22:33
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    Bonjour reuns ! @AnIsomorphicTeen : a way of proving such things is to reason using elementary operations on lines and columns. Given a matrix, it is sometimes possible to make it become the identity matrix by applying elementary transformations on it. These elementary transformations act in the same way as multiplying on the left or the right your matrix by elementary matrices. Do you need to divide? Can you switch lines and columns? Check "Smith normal form". – Plop Feb 22 '23 at 22:36
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    Of course it is even finitely presented, since it is an arithmetic group. – Moishe Kohan Feb 22 '23 at 22:37
  • This is fantastic news! – An Isomorphic Teen Feb 22 '23 at 22:38

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From a relatively elementary viewpoint (noting @MoisheKohan's observation in comments), it is indeed possible to prove the finite generation, by explicit matrices, in effect by explicit proof of unicuspidality of the quotient of hyperbolic three-space by this group. Namely, translations $\pmatrix{1&1\cr 0&1}$ and $\pmatrix{1& i\cr 0&1}$ generate all translations, and together with inversion $\pmatrix{0 & -1\cr 1& 0}$ (and maybe $\pmatrix{\pm i& 0\cr 0&\mp i}$ and $\pmatrix{-1&0\cr 0&-1}$, depending how we want to pose things), do succeed in moving every element in hyperbolic three-space into an explicit Siegel set... with good further details. This is done in some detail in chapter one of my CUP book(s) "Modern Analysis of Automorphic Forms, I, II"... also available (legally) on-line at www.math.umn.edu/~garrett/m/v/current_version.pdf

paul garrett
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