Ellis's conjecture on free multiplicative Lie algebras

Let $P$ be a group, let $L(P)$ denote its free multiplicative Lie algebra, let $\Gamma_n(P)$ be the normal subgroup generated by Lie-bracket commutators of weight $n$ , and let $\gamma_1(P)=P$ and $\gamma_{n+1}(P)=[\gamma_n(P),P]$ be the lower central series. Ellis's conjecture asks whether $\Gamma_n(P)\cong\gamma_n(P)$ for every group $P$ and every integer $n\ge 1$ . In particular, the conjecture includes the assertion for every free group $P$ .

References

Primary source

arXiv

Additional references

The free multiplicative Lie algebra L(P) for a finitely generated parafree group P — arXiv — Dessislava H. Kochloukova

group-theory homological-algebra

Posted 2 weeks ago · edited 2 weeks ago

Progress summary

Refreshed 2 weeks ago

Claimed progress

A new paper claims to extend the free-group result to some parafree groups, but it does not settle the conjecture in full.

Ellis's conjecture says that all basic universal $n$ -commutator identities for $n \ge 3$ follow from five fundamental commutator identities. For free groups, the corresponding multiplicative Lie product is known to be either trivial or the ordinary commutator.

Known results

For free groups, every Lie product is either identically $1$ or $[x,y]$ ; attributed to Ellis.
More generally, the conclusion holds when the relevant centralizer condition on nontrivial normal subgroups is satisfied.
A homological formulation gives the result when $H_2(G)=0$ , including free and simple groups.

September 2026 parafree-group extension

Dessislava H. Kochloukova's paper claims the isomorphism for finitely generated parafree groups satisfying the stated $H_2$ and $H_3$ vanishing conditions, thereby extending the result beyond free groups. The full conjecture outside this class is explicitly not claimed resolved.

Current status (as of September 2026): the conjecture is established for free groups and claimed for the stated finitely generated parafree homological class, while the general case remains open.

Sources

No solutions have been posted yet.