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The starting point of this question is the (presumably) well-known theorem (the proof I know is from Abelian $\ell$-adic representations and elliptic curves from J-P.Serre in which it is a lemma for $n=2$ and an exercise for $n>2$; which suggests that the result was already classical in the 60s).

Theorem I: If $G$ is a closed subgroup of $\operatorname{SL}_{n}(\mathbb Z_{p})$ which surjects on $\operatorname{SL}_{n}(\mathbb F_{p})$ and if $p≥5$, then $G=\operatorname{SL}_{n}(\mathbb Z_{p})$.

The theorem is optimal with respect to all hypotheses in the sense that there exists proper subgroups of $\operatorname{SL}_{2}(\mathbb Z_{p})$ mapping onto $\operatorname{SL}_{2}(\mathbb F_{p})$ when $p=2,3$ (a fact that played a role in the original proof of the modularity of semi-stable elliptic curves by A.Wiles, if I am not mistaken) and in the sense that for all $n≥2$ and all prime $p$, there exist discrete valuation rings $A$ of mixed characteristic $(0,p)$ such that $\operatorname{SL}_{n}(A)$ contains proper subgroups mapping onto $\operatorname{SL}_{n}(A/\mathfrak m)$ (just take a completely ramified $A$ over $\mathbb Z_{p}$ and consider $\operatorname{SL}_{2}(\mathbb Z_{p})$ inside $\operatorname{SL}_{2}(A)$).

A slightly less well-known fact is that the theorem admits the following generalization, due to N.Boston.

Theorem II: Let $A$ be a complete local noetherian ring with finite residual characteristic $p\neq 2$. If $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m^2)$, then $G=\operatorname{SL}_{n}(A)$.

This is in the appendix of On p-adic analytic families of Galois representations. Compositio Math. 59 (1986), no. 2, 231–264 by B.Mazur and A.Wiles and again, this theorem is optimal in the sense that there exists a proper subgroup of $\operatorname{SL}_{2}(\mathbb Z_{2})$ surjecting on $\operatorname{SL}_{2}(\mathbb Z/4\mathbb Z)$.

Now my actual question. Let $A$ be a complete local noetherian ring (UPDATE: domain, actually) of mixed characteristic $(0,p)$ with $p≠2$ UPDATE: which one can assume to be a discrete valuation ring if necessary. Suppose $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m)$.

Among the pre-images in $G$ of non-identity unipotents elements in $\operatorname{SL}_{n}(A/\mathfrak m)$, is it true that there exists a unipotent element (that is to say an element $\sigma$ such that $\sigma-1$ is nilpotent)?

Granted theorem II, this is obviously true if $G$ maps onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$. Without this hypothesis, it looks dubious to me but nevertheless, the obvious counterexamples to theorem I coming from ramified rings do not provide counterexamples to this claim and I confess that I don't quite know how to construct other counter-examples. Taking this into account, another more general point of view on the question would be the following.

What are the subgroups of $\operatorname{SL}_{n}(A)$ which do not map onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$ but which do map onto $\operatorname{SL}_{n}(A/\mathfrak m)$?

Perhaps group cohomology of $\operatorname{SL}_{n}$ would help then. Any positive result, even in the case $n=2$ and $p>3$ would already be of interest to me.

UPDATE: Jim Humphreys asks in comments what I mean by a unipotent element in a matrix group. I meant an element $\sigma$ such that $\sigma-1$ is nilpotent, but I realize now that this might not be so great a definition when the ring of coefficients is not a domain, or at least reduced. So perhaps it is better to take $A$ a domain in the above, or even a discrete valuation ring. Also, even though Tim Dokchister answered the first question in the negative, I would be interested to know if someone has something to say for infinite closed subgroup $G$.

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could you provide a link or reference to Boston's result? – YCor Nov 21 '13 at 21:50
@YvesCornulier Cher Yves, bien sûr. J'ai aussi pris la liberté de t'envoyer l'article par mail. – Olivier Nov 22 '13 at 8:06
It seems there are counterexamples to the first question, at least for $p=5$. The group $\text{SL}_2({\mathbb F}_5)$ has a 2-dimensional symplectic representation with character in ${\mathbb Q}(\sqrt 5)$ and Schur index 2, so it looks like it is realizable over $A={\mathbb Z}_5(\zeta_5)$. So $\text{SL}_2({\mathbb F}_5)$ injects in $\text{SL}_2(A)$ and reduces, I suppose, onto $\text{SL}_2(A/{\mathfrak m})$. But it is just a finite group, so it has no unipotent elements. – Tim Dokchitser Nov 22 '13 at 9:09
@TimDokchitser Dear Tim, thanks a lot for this great counterexample. Any idea about non-finite subgroups? Also, will you consider posting this as an answer? After all, it does completely answer the most general version of question I. – Olivier Nov 22 '13 at 9:21
A simple example for the second question is $A=\mathbf{Z}_p[[t]]$ (with $\mathbf{Z}_p$ the ring of $p$-adic integers), $G=\mathrm{SL}_n(\mathbf{Z}_p)$. In general, I don't think you expect better than a result assuming surjectivity of the reduction modulo $pA+\mathfrak{m}^2$. – YCor Nov 22 '13 at 9:23
up vote 4 down vote accepted

For the first question, at least for $p=5, n=2$ there is a counterexample.:

The group $\text{SL}_2({\mathbb F}_5)$ has a 2-dimensional symplectic representation with character in ${\mathbb Q}(\sqrt 5)$ and Schur index 2, so it is realizable over $A={\mathbb Z}_5(\zeta_5)$. So $\text{SL}_2({\mathbb F}_5)$ injects in $\text{SL}_2(A)$ and reduces onto $\text{SL}_2(A/{\mathfrak m})$. But it is just a finite group, so it has no non-identity unipotent elements.

I don't know whether it generalizes to higher $p$ and $n$ though, and whether there are examples with `interesting' infinite closed subgroups of $\text{SL}_2(A)$.

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Can you say more explicitly what you mean by "unipotent" element in a matrix group? – Jim Humphreys Nov 23 '13 at 17:21
Just to indicate why this is not in contradiction with my answer: in Tim's example, $A=\mathbf{Z}_5[\zeta_5]$, and $\mathfrak{m}$ is generated by $\{5,\zeta_5-1\}$. In his examples, lifts of elements of order 5 are conjugate (at least over an extension) to the matrix $\begin{pmatrix}\zeta_5 & 1\\0 & \zeta_5^{-1}\end{pmatrix}$, which is not unipotent. – YCor Nov 25 '13 at 14:26
Thanks again for providing this counterexample. – Olivier Nov 27 '13 at 13:17

In view of Boston's result it is enough to deal with an Artinian commutative local ring $(B,\mathfrak{m})$ such that $\mathfrak{m}^2=0$ and $p\in\mathfrak{m}$. In this setting, I think the best you can expect is to deal with an assumption of surjectivity of the map to $SL_n(B/(\mathfrak{m}^2+pB)$. Indeed it is hopeless to expect that surjectivity of the map to $SL_n(A/\mathfrak{m})$ is enough, in view of the case of $G=SL_n(\mathbf{Z}_p)$ for $A=\mathbf{Z}_p[[t]]$.

Here is a proof that if $p\ge 5$, if $n\ge 2$ and $G\subset SL_n(B)$, and $G$ has a surjective projection to $SL_n(B/pB)$, then $G$ has a surjective projection to $SL_n(B)$. Let $N$ be an elementary matrix (with 0 on the diagonal), and let us show that $1+pN\in G$. By hypothesis, there exists a matrix $X$ such that $1+N+pX\in G$. Hence $(1+N+pX)^p\in G$. Now since $p\ge 3$, modulo $p^2$, we have the formal equality $(1+x)^p=1+px+px^{p-1}+1$ (we have to be careful still that the similar formula $(a+b)^p=\dots$ is valid only if $a$ and $b$ commute).

$$(1+N+pX)^p=(1+(N+pX))^p=1+p(N+pX)+p(N+pX)^{p-1}+(N+pX)^p$$ $$=1+pN+pN^{p-1}+N^p+p(N^{p-1}X+N^{p-2}XN+\dots+ NXN^{p-2}+XN^{p-1})$$ now since $N^2=0$ and $p\ge 5$, this gives $(1+N+pX)^p=1+pN$.

On the other hand, the mapping $X\mapsto 1+pX$ induces a surjective homomorphism of the additive group $\mathfrak{sl}_n(B/pB)$ (the matrices with trace 0) to the kernel of the reduction $SL_n(B)\to SL_n(B/pB)$. Now as a normal subgroup, this kernel is generated as a normal subgroup by the image of elementary matrices, just noticing that $e_{ij}(1)e_{ji}(x)e_{ij}(-1)e_{ji}(-x)e_{ij}(-x)=d_{ij}(x)$, where $d_{ij}(x)-1$ is the matrix with 0 every where, except $x$ at $ii$ and $-x$ at $jj$.

We can do probably do something by hand for $p=2,3$ if we exclude a few small values of $n$.

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Thanks a lot. But in some sense this answers a dual question to mine: it provides a criteria for the group $G$ to be as large as it can be whereas I wonder if it has non-trivial unipotent elements even when it is not as large as it can be (while still surjecting on $\operatorname{SL}_2(k)$). – Olivier Nov 26 '13 at 10:52
@Olivier: I don't guess what you want exactly (I don't understand your sentence in the comment). This answer provides an example answering your second question, and the sequel also shows it's generic in a sense. Tim provided another example, which unlike mine is a finite group (and the domain being 1-dimensional) and hence does not even have unipotent elements. – YCor Nov 26 '13 at 12:57
Cher Yves, what I want is not a criteria for a group to be the whole of $\textrm{SL}_n$, what I want is a criteria to know if there are non-trivial unipotent elements in a group even if it is not $\textrm{SL}_n$. Tim did answer this question, but his example is a finite group and though I'm glad to know it, I care about infinite subgroups much more. – Olivier Nov 26 '13 at 19:53
@Olivier: OK, I sort of see your new question (which is not the one you initially asked), but it's still vague: you want a criterion for a subgroup of $SL_n(A)$ ($A$ complete local noetherian domain) to contain a nontrivial unipotent element? a criterion in terms of the reduction modulo $\mathfrak{m}$?... – YCor Nov 26 '13 at 20:42
It is also probably possible to artificially convert Tim's example into an infinite one: replace his ring $A$ by $A[[t]]$ (or a bigger one if helpful). Then consider Tim's finite subgroup $F$ whose reduction modulo the maximal ideal is onto, and replace it by the subgroup generated by $F$ and a ``generic enough" matrix whose reduction modulo $t$ is the identity. Then I don't expect such a group to have nontrivial unipotent elements (a careful check should take some effort)... – YCor Nov 26 '13 at 20:52

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