# Unit conjecture disproved!

There are three conjectures about group rings of torsion-free groups that are attributed to Kaplansky. To state them, let $$K$$ be a field, $$G$$ be a torsion-free group and denote by $$K[G]$$ the corresponding group ring.

1. The unit conjecture states that every unit in $$K[G]$$ is of the form $$kg$$ for $$k \in K\setminus\{0\}$$ and $$g \in G$$.
2. The zero divisor conjecture states that the only zero divisor in $$K[G]$$ is $$0$$.
3. The idempotent conjecture states that $$0$$ and $$1$$ are the only idempotents in $$K[G]$$.

Torsion-freeness of $$G$$ is a necessary condition in the conjecture since otherwise one gets obvious counter-examples:

1. Let $$g$$ be an element of order $$2$$ in $$G$$. Then for any $$k \in K$$ we have $((1-k)+kg)\cdot((1-k)-kg) = 1-2k\,.$ Hence, if $$1-2k$$ is a unit in $$K$$, then the element $$(1-k)+kg$$ is a unit in $$K[G]$$ which is non-trivial for $$k \notin \{0,1\}$$.
2. If $$g$$ has order $$n$$ in $$G$$, then $(1-g)(1+g+g^2+ \cdots + g^{n-1}) = 1-g^n = 0$ in $$K[G]$$ and consequently $$1-g$$ is a non-trivial zero divisor in $$K[G]$$.
3. If the order $$n$$ of $$g$$ is invertible in $$K$$, then $$n^{-1}\cdot\sum_{k=0}^{n-1} g^k$$ is a non-trivial idempotent in $$K[G]$$.

It is known that the unit conjecture implies the zero divisor conjecture, and the zero divisor conjecture implies the idempotent conjecture.

1. The strongest of these conjectures, the unit conjecture, is currently only proven in those cases where one can first show a stronger, purely group-theoretic property: having unique products. The group $$G$$ is said to have unique products if for every choice of finite, non-empty subsets $$A,B \subset G$$ there are elements $$a \in A$$ and $$b \in B$$ such that the product $$ab$$ can not be expressed as any other product of elements from $$A$$ and $$B$$. Standard examples of groups with unique products are orderable groups.
2. The zero divisor conjecture is proven for all elementary amenable groups and holds over $$\mathbb{C}$$ for groups satisfying the Atiyah conjecture.
3. The idempotent conjecture is known to hold for many groups since it follows from the isomorphism conjectures: for $$K[G]$$ it follows from the corresponding Farrell-Jones conjecture and for $$\mathbb{C}[G]$$ from both the Baum-Connes and the Burghelea conjecture.

Today I want to report on recent progress on the unit conjecture. Namely, if was disproven! You can find the corresponding preprint of Giles Gardam on the arXiv:2102.11818. In his counter-example the group is virtually abelian (i.e., from the general point of view of geometric group theory an extremely ‘simple’ group) and the field is $$\mathbb{F}_2$$. Note that since the group is virtually abelian, the zero divisor conjecture holds for it.

I am very interested in the follow-up question whether one can also construct counter-examples to the unit conjecture over the field $$\mathbb{C}$$ (or any other field of characteristic $$0$$)? The motivation for asking this stems from the fact that the other two conjectures (the zero divisor and the idempotent conjecture) are known in the case of $$\mathbb{C}$$ for many more groups than in the general case (i.e., for arbitrary fields $$K$$) since in this case they follow from other, more analytic conjectures (i.e., the Atiyah and the Baum-Connes conjecture). But there is currently no analytic conjecture known which implies the unit conjecture.

edit (April 14th, 2021): There is a connection of this counterexample to Hantzsche-Wendt manifolds about which I blogged shortly after this post (https://blog.spp2026.de/hantzsche-wendt-manifolds/). In fact, if I understand it correctly, the group in Giles counterexample is exactly the Bieberbach group of the (unique) three-dimensional Hantzsche-Wendt manifold.

edit (June 7th, 2021): As expected, new counter-examples quickly arose; especially we have now also counter-examples in every prime characteristic of the field arXiv:2106.02147.