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Jul 28, 2018 at 14:18 vote accept john mangual
Jul 24, 2018 at 17:07 answer added Neil Hoffman timeline score: 2
May 26, 2018 at 21:46 comment added john mangual @YCor I'm guessing elementary operations are like $(x,y) \mapsto (x+r \times y,y)$ and $(x,y) \mapsto (x, x+r \times y)$. And it's sufficient to have $r \in \{ \pm 1, \pm i \}$. it's amazing how quickly that turns into K-theory.
May 26, 2018 at 17:09 comment added YCor Generating subsets of such congruence subgroups have possibly been addressed in generating algebraic K-theory by the way.
May 26, 2018 at 17:08 comment added YCor Over an arbitrary Euclidean ring, $SL_2$ is generated by elementary matrices $e_{12}(r)$, $e_{21}(r)$ where $r$ ranges over the ring. So $\{e_{12}(1),e_{12}(i),e_{21}(1),e_{21}(i)\}$ generates $SL_2(Z[i])$. Hence $\{e_{12}(1),e_{12}(i),s\}$ also generates, where $s=[[0,-1][1,0]]$.
May 26, 2018 at 16:09 comment added john mangual @YCor I don't even know what the "standard" generators are. Sage will rather mechanically produce generating sets for invertible $2 \times 2$ matrices over $\mathbb{Z}$. Not sure what they'll say about the Bianch group specified here.
May 26, 2018 at 16:06 history edited john mangual CC BY-SA 4.0
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May 26, 2018 at 15:56 comment added YCor "the generators"? you probably mean "some generating family". It's a finite index subgroup in $SL_2(Z[i])$, so using standard generators from the latter (this does not use any algebraic group theory), you can get generators of the congruence subgroup as well with standard methods. (It's the kernel of a homomorphism onto $SL_2(Z/5Z)$ which has order 120, so I expect it will yield some generating subset much larger than necessary.)
May 26, 2018 at 15:55 history edited john mangual CC BY-SA 4.0
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May 26, 2018 at 15:47 history asked john mangual CC BY-SA 4.0