# Variety of nilpotent Lie algebras or $p$-groups

Here's a couple of analogous questions, one in terms of finite-dimensional complex Lie algebras and one in terms of finite $p$-groups; I'd be interested in an answer to either:

1) Let $\mathcal{L}$ be an isomorphism-closed class of finite-dimensional nilpotent complex Lie algebras. Assume $\mathcal{L}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free Lie algebra is residually-$\mathcal{L}$). Does $\mathcal{L}$ necessarily consist of all finite-dimensional nilpotent complex Lie algebras?

2) Let $p$ be prime. Let $\mathcal{C}$ be an isomorphism-closed class of finite $p$-groups. Assume $\mathcal{C}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free group is residually-$\mathcal{C}$, or still equivalently the free group on 2 generators is residually-$\mathcal{C}$). Does $\mathcal{C}$ necessarily consist of all finite $p$-groups?

(Note: I ask both questions in the positive but I don't particularly expect a positive answer!)

• Is the condition "does not satisfy any common identity" clearly enough formulated? (By the way, perhaps you mean to write e.g. rather than i.e.: e.g. means "for example", while i.e. means "that is".) – Jim Humphreys Sep 21 '14 at 19:23
• I really mean "that is". A class $\mathcal{C}$ of Lie algebras satisfies a (nontrivial) common identity if there exists a free Lie algebra $\mathfrak{f}$ and $m\in\mathfrak{f}-\{0\}$ such that $m$ is an identity for every $\mathfrak{g}\in\mathcal{C}$ (i.e., every Lie algebra homomorphism $\mathfrak{f}\to\mathfrak{g}$ maps $m$ to 0). – YCor Sep 21 '14 at 19:47
• Thanks. The language isn't familiar to me, but I'm aware of similarities between the two theories. It might be surprising if your two questions had different answers, but I have no idea how to approach either of them. (By the way, I guess the first question has the same answer over any algebraically closed field of characteristic 0.) – Jim Humphreys Sep 21 '14 at 21:03
• Yes, the answer to the first question does not even depend on the ground field of characteristic zero. – YCor Sep 21 '14 at 21:35
• For question 2), it appears that there is a literature on "pseudo-varieties of finite algebras". In particular, Baldwin and Berman proved that a class of finite algebras closed under subalgebras, homomorphisms and finite products is the set of finite algebras in an ascending union of varieties. yadda.icm.edu.pl/yadda/element/… – Ian Agol Sep 21 '14 at 22:34

## 1 Answer

Here is some idea, it is not very precise. Let $F$ be the free group on two generators and let $F_p$ be its pro-$p$ completion. Let $w$ be an infinite word in $F_p$, i.e., an element of $F_p$ which is not in $F$. Let $W$ be the closed verbal subgroup of $F_p$ generated by $w$. I think that it is possible to choose $w$ so that $W \cap F$ is trivial. For example, the first congruence subgroup of $SL_2(\mathbb{Z}_p)$ ($p>2)$ satisfies a pro-$p$ identity due to Zubkov, but I think it has a dense free group.

Let $D_n$ be the $n$-dimension subgroups of $F_p$. Then if I recall correctly there is a canonical way to write $w=w_nu_n$, where $u_n \in D_n$. Take your variety to be all the groups that satisfy $w_n$ for some $n$. I would guess they will satisfy your requirement.

I am not sure this idea will work, but if you like it and need more reference, then please contact me.

• Well, there is still some stuff to check and some stuff to prove before we can conclude this. – Yiftach Barnea Sep 21 '14 at 22:56
• Let $P$ be an open pro-$p$-subgroup in $\mathrm{GL}_2(\mathbf{Z}_p)$. Clearly it has a free subgroup (e.g., in $\mathrm{SL}_2(\mathbf{Z}))$. On the other hand according to Zubkov (at least for $p>2$), $P$ satisfies a nontrivial pro-$p$-identity $I$. Let $\mathcal{C}$ be the class of finite $p$-groups satisfying the identity $I$ (or alternatively the smallest class of groups containing finite quotients of $P$ and stable under taking subgroups, quotients, and direct products). Then $\mathcal{C}$ satisfies no common group identity, but does not contain all $p$-groups, answering (2). – YCor Sep 21 '14 at 23:07
• Yes, this formulation is indeed more convincing. – Yiftach Barnea Sep 22 '14 at 7:18
• I think one can try for the Lie algebra case to look at the first congruence subalgebra of $sl_2(\mathbb{C}[[X]])$ and prove similar results. I am not sure how difficult this might be. – Yiftach Barnea Sep 22 '14 at 11:56
• About case $p=2$: arxiv.org/abs/1910.05805 (On Pro-2 Identities of 2×2 Linear Groups, David El-Chai Ben-Ezra, Efim Zelmanov) – YCor Oct 15 '19 at 5:19