Let $G \subseteq \mathrm{Sp}_{2g}(\mathbb{Z}_2)$ be a closed subgroup of the symplectic group over the $2$-adic integers whose image under the mod-$2$ homomorphism $\pi : \mathrm{Sp}_{2g}(\mathbb{Z}_2) \twoheadrightarrow \mathrm{Sp}_{2g}(\mathbb{F}_2)$ is isomorphic to the symmetric group $S_{2g + 1}$ (or, as a variant to this question, we could assume that the mod-$2$ image of $G$ is instead $S_{2g + 2}$). Is it known whether for $g \geq 2$ there exists such a $G$ which is not of finite index in $\mathrm{Sp}_{2g}(\mathbb{Z}_2)$, and if so, has an explicit example ever been given? (Note that Brumer and Kramer show as Theorem 2.1.1 in their article "Large $2$-adic Galois image and non-existence of certain abelian surfaces over $\mathbb{Q}$" that such a $G$ could not contain a transvection.)

I would be surprised if such a $G$ couldn't exist, because in that case I would expect someone to have proven this already; however, it would be nice to have a definitive answer one way or another. In the $g = 1$ case I'm able to come up with an explicit example of such a $G$, and in any case, we know it should exist as the $2$-adic Galois image associated to a CM elliptic curve given by a polynomial with full Galois group. For $g \geq 2$, I can show that there exists a $G \subseteq \mathrm{Sp}_{2g}(\mathbb{Z}_2)$ surjecting onto $S_{2g + 1}$ which does not contain $\mathrm{ker}(\pi)$ modulo $4$, but I'm finding it somewhat tricky to lift that to an infinite-index subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z}_2)$.

  • $\begingroup$ Actually, it occurs to me that the definition of "transvection" in the paper I quoted above includes $2^{2n}$th powers of transvections which lie in the kernel of reduction mod $2^{2n}$. This seems to contradict the fact that I can choose $G$ to be the inverse image of the group $\bar{G} \subset \mathrm{Sp}_{2g}(\mathbb{Z} / 4\mathbb{Z})$ I found which does not contain $\mathrm{ker}(\pi)$, yet this $G$ contains a transvection and according to Theorem 2.1.1 does contain $\mathrm{ker}(\pi)$. The authors state that this case is well known and don't provide a source. $\endgroup$ Jan 30, 2018 at 16:58
  • $\begingroup$ I expect the issue lies somewhere in the convention for defining "transvection". $\endgroup$ Jan 30, 2018 at 16:58
  • $\begingroup$ As a side note, how do you embed the symmetric group $S_{2g+2}$ inside $\mathrm{Sp}_{2g}$? This can't be done for $g=1$ for index reasons. $\endgroup$
    – user94041
    Jan 30, 2018 at 17:01
  • 1
    $\begingroup$ Indeed it can't be done for $g = 1$ but can be for $g \geq 2$. Let $V$ be the $2g$-dimensional $\mathbb{F}_2$-vector whose elements are given by partitions of a set of $2g + 2$ elements into $2$ even-cardinality subsets. The addition law is induced by symmetric differences. The intersection pairing is defined using the parity of intersections between even-cardinality subsets. Then each permutation in $S_{2g + 2}$ induces an automorphism of $V$ which preserves the intersection pairing. The resulting homomorphism $S_{2g + 2} \to \mathrm{Sp}_{2g}(\mathbb{F}_2)$ is injective iff $g \geq 2$. $\endgroup$ Jan 30, 2018 at 17:12
  • $\begingroup$ Do you want $G$ to be a closed subgroup? Otherwise, I think one can easily find subgroups with this property using strong approximation (even free subgroups). $\endgroup$
    – Ian Agol
    Jan 30, 2018 at 18:42


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