The too naive and vague version of my question is the following: given a collection of integer symplectic matrices all of the same size (say 2n by 2n), how can I tell if they generate the full symplectic group Sp(2n,Z)?

Here is the fleshed-out version. Here Sp(2n,Z) means the group of matrices preserving the form

$ J = \left( \begin{array}{cc} 0&I \\ -I&0& \end{array} \right)$

and I stick to the case of Sp(4,Z). It is a theorem of Stanek that this group is generated by the following three matrices:

$ T = \left( \begin{array}{cccc} 1&0&1&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right), \quad R = \left( \begin{array}{cccc} 1&0&0&0 \\ 1&1&0&0 \\ 0&0&1&-1 \\ 0&0&0&1 \end{array} \right), \quad D=\left( \begin{array}{cccc} 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{array} \right). $

Now it so happens (in some geometric situation I care about) that I have two other elements A and B of Sp(4,Z), given as follows:

$ A = \left( \begin{array}{cccc} 0&0&-1&0 \\ 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \end{array} \right), \quad B = \left( \begin{array}{cccc} 0&0&0&-1 \\ 0&0&-1&5 \\ 5&1&2&5 \\ 1&0&0&2 \end{array} \right). $

Then my more concrete question is the following:

Is {T,A,B} also a generating set for Sp(4,Z)?

To make that question seem a little more reasonable, here is some

Evidence: The images of T, A, and B in Sp(4,F_p) do in fact generate Sp(4,F_p) for all primes p up to 47. (My computer stopped cooperating at that point.)

It seems plausible to me that if the subgroup $\left< T, A, B \right>$ was actually proper, then this would be detected by its image in one of the groups Sp(4,F_p), and moreover that the first prime p at which this happened wouldn't be too big (for example, perhaps not bigger than the largest absolute value of coefficients of A or B). But maybe that is completely off the mark. (If so, please tell me so!)

Of course I don't expect a definitive answer to my specific question here, but anything that can be said about approaches to this kind of question would be much appreciated.

Finally let me mention that, for the geometric question I'm interested in, it would be sufficient to know that $\left< T, A, B \right>$ is merely a finite-index subgroup of Sp(4,Z). (Hence the arithmetic groups tag.) But the stronger statement seems to be true, as explained above, so I phrased the question that way.

Edit (May 26): As pointed out by Derek Holt in his answer, it is not true that the reduction of my group mod 2 gives all of Sp(4,F_2). I must have made a mistake in that computation. In any case, he also shows that my group has finite index (indeed, index 6) in Sp(4,Z), settling the question satisfactorily. But any more remarks about generating sets for the symplectic group are still most welcome!

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    $\begingroup$ Have you already tried to use the relations in Stanek's (freely available) paper ams.org/journals/proc/1963-014-05/S0002-9939-1963-0153748-8/… ? $\endgroup$ May 25, 2011 at 10:35
  • $\begingroup$ Dear Tom, thanks for your suggestion. Could I ask you to elaborate a little? Those relations are all between matrices of specific types (rotations, translations, semi-involutions); I don't know, for instance, how B decomposes as a product of such. $\endgroup$
    – user5117
    May 25, 2011 at 12:29
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    $\begingroup$ The matrix $B$ can be written as $B = \left(\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \cdot J^{-1} \cdot \left(\begin{array}{rrrr} 1 & 0 & -5 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \cdot J \cdot \left(\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$. $\endgroup$ May 25, 2011 at 13:54
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    $\begingroup$ The matrix $B$ is of the form $B = \begin{pmatrix} 0 & X \\ -X^{-1} & B \end{pmatrix}$, and the essential observation is that $\begin{pmatrix} I & X \\ 0 & I \end{pmatrix} \begin{pmatrix} I & 0 \\ -X^{-1} & I \end{pmatrix} = \begin{pmatrix} 0 & X \\ -X^{-1} & I \end{pmatrix}$. The fact that $\begin{pmatrix} 0 & X \\ -X^{-1} & I \end{pmatrix} \begin{pmatrix} I & Y \\ 0 & I \end{pmatrix} = \begin{pmatrix} 0 & X \\ -X^{-1} & I - X^{-1}Y \end{pmatrix}$ takes care of the last factor. $\endgroup$ May 25, 2011 at 14:42
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    $\begingroup$ Here's a related question: mathoverflow.net/questions/9628/… $\endgroup$
    – Ian Agol
    May 25, 2011 at 15:33

3 Answers 3


In fact, if you reduce mod 2, then you find that the index of the subgroup generated by $T,A,B$ in ${\rm Sp}(4,2)$ is 6, not 1.

I have now carried out Igor's suggested approach and a coset enumeration (which I did in MAGMA) shows that the subgroup has index 6 in ${\rm Sp}(4,\mathbb{Z})$.

Bender's presentation of ${\rm Sp}(4,\mathbb{Z})$ is on the generators

$K = \left(\begin{array}{rrrr}1&0&0&0\\\\ 1&-1&0&0\\\\ 0&0&1&1\\\\ 0&0&0&-1 \end{array}\right),\ \ \ L=\left(\begin{array}{rrrr}0&0&-1&0\\\\ 0&0&0&-1\\\\ 1&0&1&0\\\\ 0&1&0&0 \end{array}\right).$

and the relations are as follows. (I won't typeset this in case anyone wants to cut and paste!)



K*L^7*K*L^5*K*L = L*K*L^5*K*L^7*K,

L^2*K*L^4*K*L^5*K*L^7*K = K*L^5*K*L^7*K*L^2*K*L^4,

L^3*K*L^3*K*L^5*K*L^7*K = K*L^5*K*L^7*K*L^3*K*L^3,

(L^2*K*L^5*K*L^7*K)^2 = (K*L^5*K*L^7*K*L^2)^2,

L*(L^6*K*L^5*K*L^7*K)^2 = (L^6*K*L^5*K*L^7*K)^2*L,

(K*L^5)^5 = (L^6*K*L^5*K*L^7*K)^2.

I used a combination of brute force search and useful intermediate matrices (including those used by Tom De Medts in his comment to the original post), to get words for $T,A,B$ in terms of $K,L$. They are easily checked once you have them.

T = (L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2*L,

A = L^-3*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2,

B = (L*K*L^-2*K*L^5*K*L^5*K*L^2*L*K*L^-2*K*L^5*K*L^5*K*L^2*L^-1)^5*L^2*K*L^4*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^2*T^-5*L^2*K*L^4*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2*L^-4*K*L^-2.

Then a coset enumeration shows almost instantly that the index of the subgroup generated by $T,A,B$ in the group defined by Bender's presentation is 6.

The fact that the index is 6 in the original problem does not actually depend on the fact that this is a complete presentation of ${\rm Sp}(4,\mathbb{Z})$ - only that the relations of the presentation are satisfied in ${\rm Sp}(4,\mathbb{Z})$.

  • $\begingroup$ Dear Derek, thanks for this most helpful answer! By the way, I am not sure why my calculation for the reduction mod 2 gave the wrong answer. In any case, as stated in the question, I really only care about my group having finite index in Sp(4,Z), so your answer is great news. $\endgroup$
    – user5117
    May 26, 2011 at 13:32
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    $\begingroup$ There's a cute subgroup of $Sp(4, \mathbb{Z}_2)$ of index 6, which is the stablizer of an odd theta-characteristic (seen as a quadratic form on the 2-torsion point). I wonder if it is this one. $\endgroup$
    – IMeasy
    Jun 9, 2014 at 9:27

Not exactly an answer, but an omnibus comment:

  1. Re @Charles' answer: from the work of T. Weigel, it is known that if the surjection is onto for SOME prime greater than three, than the group is Zariski-dense. So, the OP's experiments show that the group IS Zariski-dense. Weigel's paper is a bit hard to find (also a bit hard to read), but it is here.

  2. There are groups (first examples were constructed by Steve Humphries in the 1980s) which are infinite index in $Sp(4, \mathbb{Z}),$ and such that the all the congruence projections are onto.

2a. It is an (unpublished) observation of mine that this property is in fact true with positive probability (that is, if you pick, in some reasonably well defined sense, a pair of matrices in $Sp(2n, \mathbb{Z}),$ the group they generate is almost certainly a free zariski dense, group which is profinitely dense with positive probability.

  1. There is a presentation of $SP(4, \mathbb{Z})$ with two generators and not so many relations:

BENDER, P. “Presentation of Symplectic Group Sp(4,Z) with 2 Generatrices and 8 Definitive Relations.” Journal of Algebra 65, no. 2 (1980): 328-331.

If you can express your matrices in terms of these generators, you can try GAP's "finitely presented groups" package, which can sometimes tell you the answer.

  1. Your general question is most likely undecidable, see: (un)decidability in matrix groups

  2. If the second matrix is a transvection or even just unipotent (trace is 4, so it might be), then you can use to Chris Hall's theorem Big symplectic or orthogonal monodromy modulo $\ell. $ C Hall - Duke Mathematical Journal, 2008 to check whether it surjects onto any given congruence quotient.

  • $\begingroup$ Dear Igor, thanks very much for your helpful comment! It's very interesting to know how little my "evidence" above might really be worth! $\endgroup$
    – user5117
    May 26, 2011 at 6:17
  • $\begingroup$ Dear Igor, could you please give a reference for Weigel´s result? thanks, khosro $\endgroup$
    – user19119
    Nov 8, 2011 at 17:58
  • $\begingroup$ Dear Igor, Concerning (2), is it possible to have a finitely generated subgroup of $\mathrm{Sp}(2g,\mathbb{Z})$ of infinite index surject onto $\mathrm{Sp}(2g,\mathbb{Z}/p^k)$ for all primes $p$ and all $k\geq 1$? $\endgroup$ Oct 2, 2020 at 15:42

By looking mod p for some primes you are actually testing a condition more relevant to determining whether the subgroup generated is Zariski-dense in the symplectic group. This is a kind of heuristic on "strong approximation", and I shall explain where it comes from. One comment is that if the subgroup is not Zariski-dense you should be able to see that in the adjoint representation. But that is a weak sort of condition, only useful for ruling out some cases.

So this is the train of thought. Given that the symplectic group is simply connected any Zariski-dense subgroup of rational matrices has full reduction mod p when p is large; and they tell me that experimentally p probably doesn't have to be so large. (My paper with Vaserstein and Weisfeiler, which probably is "effective", though the numerical gap between saying that and what I have just said is presumably somewhat ridiculous.)

My guess is you need to think in terms of some more group-theoretic method.

  • $\begingroup$ Dear Charles, thanks for this helpful response. $\endgroup$
    – user5117
    May 26, 2011 at 6:17

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