# What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group.

A couple of good resources for this problem that gives some historical overview are:

• Passman, Donald S. The algebraic structure of group rings. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.

• Passman, Donald S. Group rings, crossed products and Galois theory. CBMS Regional Conference Series in Mathematics, 64. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1986.

The conjecture has been proven affirmative, when $G$ belongs to special classes of groups. I tried to write down some of the history:

• Ordered groups (A.I. Malcev 1948 and B.H. Neumann 1949)
• Supersolvable groups (E. Formanek 1973)
• Polycyclic-by-finite groups (K.A. Brown 1976, D.R. Farkas & R.L. Snider 1976)
• Unique product groups (J.M. Cohen, 1974)

Here are my questions:

1. Was Irving Kaplansky the first one to state this conjecture? Can someone provide me with a reference to a paper or book that claims this?
2. Since the publications of Passman's expository note (above) in 1986, has there been any major developments on the problem? Are there any new classes of groups that will yield a positive answer to the conjecture? Can someone help me to extend my list above?

The zero-divisor conjecture (let's denote it by "(Z)") is related to the following two conjectures:

(I): If $G$ is torsion-free, then $K[G]$ has no non-trivial idempotents.

(U): If $G$ is torsion-free, then $K[G]$ has no non-trivial units.

Now, if $G$ is torsion-free, then one can show that:

(U) $\Rightarrow$ (Z) $\Rightarrow$ (I).

Has there been any developments, since 1986, to any partial answers on conjecture (U)? Passman claims that "this is not even known for supersolvable groups". Is this still the case?

I want to point out that this post is related to another old MO-post.

• @Martin: in the case K= complex numbers, there are variants of the zero-divisor question with such a flavour (Linnell has worked on some of these). Also, again over the complex field, the related conjecture (I) that Johan mentions is in many cases deduced by knowing that the reduced group C*-algebra of the group in question does not have non-trivial idempotents (Kadison-Kaplansky). Lastly, answers for other fields have sometimes used embedding techniques which may have been inspired by operator-algebraic approaches to the complex version. – Yemon Choi Oct 31 '11 at 1:15
• I am pretty sure (U) is still wide open, even for super-solvable groups. The following is a presentation for maybe the "nicest" group for which this is open: $G=\langle x,y \mid (x^2)^y=x^{-2}, (y^2)^x = y^{-2} \rangle$. This groups is supersolvable (in particular polycyclic). It contains the subgroup $H=\langle x^2,y^2,[x,y]\rangle \cong \mathbb{Z}^3$, extended by the Klein Four group $V_4=C_2\times C_2\cong\langle xH, yH \rangle$. It is also a crystallographic group. – Max Horn Oct 31 '11 at 10:49
• @Martin and Johan: The Kaplansky-Kadison conjecture has been proved in 1997 by N. Higson and G. Kasparov, for the (huge) class of a-(T)-menable groups, or groups with the Haagerup property: these are groups admitting a metrically proper, affine isometric action on Hilbert space (examples: amenable, free, Coxeter, discrete subgroups of $SO(n,1)$ or $SU(m,1)$, countable subgroups of $GL_2(field)$...) – Alain Valette Oct 31 '11 at 14:22
• One more question: what do we know about this conjecture if $K$ is a finite field? – Joerg Sixt Nov 2 '11 at 9:30
• Recently posted to the arXiv is a purported counterexample to the unit conjecture for group rings: arxiv.org/pdf/2102.11818.pdf – Paul Gilmartin Feb 27 at 10:48

I suggest you have a look at Chapter 10 of W. Lueck's book, "$L^2$-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002. There he discusses the Atiyah conjecture (conjecture 10.3): if $K$ is a subfield of $\mathbb{C}$, a group $G$ satisfies the Atiyah conjecture with coefficients in $K$ if for any $m\times n$ matrix $A$ with coefficients in $K[G]$, the von Neumann dimension of the kernel of the operator $r_A:\ell^2(G)^m\rightarrow \ell^2(G)^n:x\mapsto Ax$, is an integer.

Lueck then proves (lemma 10.15) that if $G$ is torsion-free and satisfies the Atiyah conjecture with coefficients in $K$, then it satisfies Kaplansky's conjecture (Z). For $K=\mathbb{C}$ and $G$ amenable, the converse is true (lemma 10.16).

Lueck goes on to prove (Theorem 10.19) a remarkable result by P. A. Linnell (Division rings and group von Neumann algebras. Forum Math., 5(6):561-576,1993): Let $\cal{C}$ be the smallest class of groups containing free groups and closed under directed unions and extensions with elementary amenable quotients; if $G$ is in $\cal{C}$ and has finite subgroups of bounded order, then the Atiyah conjecture with coefficients in $\mathbb{C}$ holds.

For further reading and more recent results, try typing "Atiyah conjecture" on Google or in the ArXiV.

• @Alain: Thank you very much Alain! I was not aware of the book by Lück, but I will look into it right away! The Higson-Kasparov result I knew from before through your lovely note on the Baum-Connes conjecture. However, at the moment I am mainly interested in conjecture (Z). – Johan Öinert Oct 31 '11 at 16:22

Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $$K=\mathbb{F}_2$$ and virtually abelian $$G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \rangle$$ (as mentioned in David Craven's answer) in arXiv:2102.11818 .

• Very nice example! Did you find the example originally using some computer algebra system, too? – Carl-Fredrik Nyberg Brodda Feb 24 at 10:05
• Yes, I did use a computer. As with most of my computational projects, I used a computer algebra system (namely GAP) for specific things that it is very convenient for but mostly worked in a general-purpose language (namely python). – Giles Gardam Feb 25 at 9:42

It is still the case that (U) is not known for supersoluble groups. In fact, it is not known for the 'easiest' supersoluble group that isn't abelian-by-cyclic (for which (U) is known). The group is an extension of $$\mathbb{Z}^3$$ by the Klein four group.

Peter Pappas and I, in a paper in the early 2010s, gave some general statements about (U) for supersoluble groups, and a general strategy for how to prove (U) for this particular group, but it hasn't been done, even for this one group, in the intervening decade or so.