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Nov 13, 2015 at 11:35 vote accept Puzzled
Jun 27, 2014 at 18:18 answer added Steffen Kionke timeline score: 3
S Jun 25, 2014 at 21:59 history bounty ended CommunityBot
S Jun 25, 2014 at 21:59 history notice removed CommunityBot
Jun 18, 2014 at 6:31 comment added Puzzled $A\equiv 0\: mod(D)$ means that $A = D\cdot B$ for some $B\in M_2(\mathbb{Z})$.
Jun 18, 2014 at 6:31 history edited Puzzled CC BY-SA 3.0
added 89 characters in body
Jun 18, 2014 at 2:48 comment added Keerthi Madapusi What does mod (D) mean?
Jun 18, 2014 at 2:15 history edited Stanley Yao Xiao CC BY-SA 3.0
added 7 characters in body
Jun 17, 2014 at 22:07 comment added few_reps (Now that I've seen the definition, I'm less sure ... but at least it should be a good point to start)
Jun 17, 2014 at 22:01 comment added few_reps The strong approximation theorem shows that $Sp_{2n}(\mathbf Z)$ projects onto all quotients $Sp_{2n}(\mathbf Z/m)$. This implies your mod 2 reduction is onto (if I understand the question).
Jun 17, 2014 at 21:57 comment added Puzzled I added the precise definition of $\Gamma_2(1,d)$ in the question.
Jun 17, 2014 at 21:53 history edited Puzzled CC BY-SA 3.0
added 411 characters in body
Jun 17, 2014 at 21:00 comment added Keerthi Madapusi What is the definition of $\Gamma_g(1,d)$?
S Jun 17, 2014 at 20:23 history bounty started IMeasy
S Jun 17, 2014 at 20:23 history notice added IMeasy Draw attention
Jun 6, 2014 at 9:50 history asked Puzzled CC BY-SA 3.0