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Let $g,n$ be positive integers, is there a reference that $\mathrm{Sp}(2g,\mathbb{Z})\to\mathrm{Sp}(2g,\mathbb{Z}/n\mathbb{Z})$ is surjection?

The only reference I could find is lemma 5.16 in Deligne–Mumford, but to be honest, I don't quite understand the argument: Why is the reduction map surjective on unipotent elements?

Deligne and Mumford - The irreducibility of the space of curves of given genus

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    $\begingroup$ Because unipotent elements in ${\rm Sp}(2g , {\mathbb Z}/n{\mathbb Z})$ lift in an obvious way. $\endgroup$ Commented Feb 8, 2018 at 21:09
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    $\begingroup$ @PaulBroussous: that is basically the argument given in Deligne-Mumford (just as tersely too). :) $\endgroup$
    – nfdc23
    Commented Feb 9, 2018 at 2:56

3 Answers 3

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This was originally proved in

M. Newman, J. R. Smart, Symplectic modulary groups, Acta Arith 9 (1964), 83-89.

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    $\begingroup$ It's hard to believe this was not known to people who cared before then... $\endgroup$
    – Igor Rivin
    Commented Feb 8, 2018 at 23:18
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    $\begingroup$ @IgorRivin: The result is basically equivalent to knowing generators for $\text{Sp}(2g,\mathbb{Z}/n)$. For $n$ prime, these were certainly known earlier (e.g. it was known since symplectic groups were first studied that symplectic groups over fields are generated by transvections). But I've search extensively, and I can't find much of anything in print before this about symplectic groups over rings like $\mathbb{Z}/n$ for $n$ not a prime (which are not even integral domains). I am morally certain that nothing equivalent to this appeared earlier in the written literature. $\endgroup$ Commented Feb 9, 2018 at 2:13
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    $\begingroup$ (even today it is nontrivial to find a good source for the fact that $\text{Sp}(2g,\mathbb{Z})$ is generated by transvections -- the proof that works for fields and appears in many places does not generalize in a simple way) $\endgroup$ Commented Feb 9, 2018 at 2:16
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    $\begingroup$ @IgorRivin: I don't have that book handy, but if I recall correctly he only considers the classical groups over fields (or, more generally, skew fields). It is not at all routine to generalize things to more general rings, even to $\mathbb{Z}$. For an elementary example, it is trivial to show that $\text{SL}(n,k)$ is generated by elementary matrices for $k$ a field. That proof actually gives a version of bounded generation. But for $k=\mathbb{Z}$, things are more complicated and you have to work harder (precisely because bounded generation is false for $n=2$). $\endgroup$ Commented Feb 9, 2018 at 3:54
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    $\begingroup$ Actually (and this is not relevant to the Newman reference), there IS a good reference - Hahn and O'Meara do it in their section 6.2, in fairly great generality. $\endgroup$
    – Igor Rivin
    Commented Feb 9, 2018 at 4:39
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According to Demazure, when he was a graduate student he asked Serre about the surjectivity of $G(\mathbf{Z}) \to G(\mathbf{Z}/(n))$ for all $n>0$ with $G = {\rm{Sp}}_{2g}$ or maybe ${\rm{SL}}_N$ (I can't remember which he said it was). Serre told him that he should ask Grothendieck for the "right" answer. Grothendieck told him that this was the "wrong question", and to approach such matters from the "right" point of view was the original goal of the SGA3 seminar on reductive group schemes over rings, etc. (killing a fly with a sledgehammer?).

That being said, a slick perspective which transforms the problem into something more robust over fields goes as follows. Let $\mathbf{A}_f$ be the ring of finite adeles for $\mathbf{Q}$ (this is a $\mathbf{Q}$-algebra, namely $\mathbf{Q} \otimes_{\mathbf{Z}} \widehat{\mathbf{Z}}$ in which $\widehat{\mathbf{Z}}$ is a compact open subring), so $K = G(\widehat{\mathbf{Z}})$ is a compact open subgroup of $G(\mathbf{A}_f)$ which meets $G(\mathbf{Q})$ in $G(\mathbf{Z})$. Since $G$ is $\mathbf{Z}$-smooth, the reduction map $K = G(\widehat{\mathbf{Z}}) \to G(\mathbf{Z}/(n))$ is surjective for Hensel's Lemma reasons and has open kernel. Thus, the original surjectivity question has an affirmative answer if $G(\mathbf{Q})$ is dense in $G(\mathbf{A}_f)$ (as this implies $G(\mathbf{Z})$ is dense in $G(\widehat{\mathbf{Z}})=K$, so $G(\mathbf{Z})$ maps onto $K/K'$ for any open normal subgroup $K'$ of $K$).

In other words, we can now forget about $\mathbf{Z}$-structures and instead attack a question entirely in terms of the $\mathbf{Q}$-group $H = G_{\mathbf{Q}}$: when is $H(\mathbf{Q})$ dense in $H(\mathbf{A}_f)$? We'd like an affirmative answer for at least $H$ equal to ${\rm{SL}}_N$ and ${\rm{Sp}}_{2g}$. In the case of $H=\mathbf{G}_{\rm{a}}$, such density holds and is just a special case of the classical "strong approximation" property for adele rings of global fields. This makes contact with Paul Broussous' comment about unipotent groups when $H$ is a split simply connected $\mathbf{Q}$-group due to the way in which such groups have an "open cell" expressed in terms of certain ${\rm{SL}}_2$-subgroups (especially the direct product structure for a split maximal torus in terms of a basis of simple coroots for the cocharacter lattice, one way of characterizing the simply connected case).

For an actual argument along these lines (i.e., bootstrapping from strong approximation for adele rings using the structure theory of split simply connected groups over fields), see http://math.stanford.edu/~conrad/248BPage/handouts/strongapprox.pdf (and note how it makes essential use of working over a field rather than over a ring of integers). This has a vast generalization to arbitrary simply connected groups over global fields relative to a suitable finite non-empty set of places (such as the place $\{\infty\}$ for $\mathbf{Q}$ when the group is split) via the "strong approximation theorem" for such groups, but that lies far deeper than the split case discussed above.

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The canonical reference is Morris Newman's Integral Matrices, Theorem VII.21 (reference taken from Andy Putman's comment below), while gross overkill would be Platonov-Rapinchuk's book.

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  • $\begingroup$ If I remember correctly, Newman's book only does SL(n,Z), though I could be remembering incorrectly and don't have it in front of me. $\endgroup$ Commented Feb 8, 2018 at 21:28
  • $\begingroup$ Actually, I was wrong -- I remembered that I have an electronic copy of the book, and the result is Theorem VII.21. $\endgroup$ Commented Feb 8, 2018 at 21:33
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    $\begingroup$ @AndyPutman That book is a treasure, IMHO. $\endgroup$
    – Igor Rivin
    Commented Feb 8, 2018 at 22:06
  • $\begingroup$ It's full of good stuff, all done in a really concrete and down-to-earth way. $\endgroup$ Commented Feb 8, 2018 at 22:10

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