aa r X i v : . [ m a t h . G R ] F e b A COUNTEREXAMPLE TO THE UNIT CONJECTURE FOR GROUP RINGS
GILES GARDAMA
BSTRACT . The unit conjecture, commonly attributed to Kaplansky, predicts that if K is a field and G is a torsion-free group then the only units of the group ring K [ G ] arethe trivial units, that is, the non-zero scalar multiples of group elements. We give aconcrete counterexample to this conjecture; the group is virtually abelian and the fieldis order two.
1. I
NTRODUCTION
Three long-standing open problems on the group rings of torsion-free groups are com-monly attributed to Kaplansky: the unit conjecture, the zero divisor conjecture and theidempotent conjecture. Let K be a field and G be a torsion-free group and consider thegroup ring K [ G ] . The unit conjecture states that every unit in K [ G ] is of the form kg for k ∈ K \ { } and g ∈ G , the zero divisor conjecture states that K [ G ] has no non-trivialzero divisors, and the idempotent conjecture states that K [ G ] has no idempotents otherthan 0 and 1. The unit conjecture implies the zero divisor conjecture [Pas85, Lemma13.1.2], which in turn implies the idempotent conjecture; these implications hold foreach individual group ring K [ G ] .In this paper we disprove the strongest of these three conjectures, namely the unitconjecture. Theorem A.
Let P be the torsion-free group h a , b | ( a ) b = a − , ( b ) a = b − i and setx = a , y = b , z = ( ab ) . Setp = ( + x )( + y )( + z − ) q = x − y − + x + y − z + zr = + x + y − z + xyzs = + ( x + x − + y + y − ) z − . Then p + qa + rb + sab is a non-trivial unit in the group ring F P. The unit conjecture and the zero divisor conjecture were formulated by Higman in hisunpublished 1940 thesis [Hig40a, p. 77] (see also [San81, p. 112]), in which he proved
Mathematics Subject Classification.
Key words and phrases.
Group rings, unit conjecture. he unit conjecture for locally indicable groups (the corresponding paper [Hig40b] waswritten earlier and omits the conjectures). The zero divisor conjecture appeared inprint in the report of a 1956 talk of Kaplansky [Kap57, Problem 6]; for integral grouprings it appears in the original 1965 Kourovka Notebook as a “well-known problem”[KM18, 1.3]. The unit conjecture was later posed alongside the zero divisor conjectureby Kaplansky [Kap70] having also been asked for integral group rings by Smirnovand Bovdi. Further context is provided in Section 2, where our focus on the group P isexplained. Theorem A is proved in Section 3. Remark.
After the fact, Theorem A is of course readily verified using computer alge-bra. Since it admits a short human-readable proof, we present such a proof.2. B
ACKGROUND
The zero divisor conjecture and the idempotent conjecture have turned out to be sus-ceptible to analytic and K -theoretic methods, despite having being posed with verylittle evidence; for example, the zero divisor conjecture holds for elementary amenablegroups [KLM88, Theorem 1.4] and holds over C for groups satisfying the Atiyah con-jecture (see [Lin93], [Lüc02, Lemma 10.39]), and the idempotent conjecture over C fol-lows from either the Baum–Connes or Farrell–Jones conjecture (see [Val02, §6.3] and[BLR08, Theorem 1.12] respectively). In contrast, the unit conjecture has only been es-tablished as a consequence of the stronger combinatorial and purely group-theoreticproperty of having unique products. A group G is said to have unique products if forevery choice of finite subsets A , B ⊂ G the set A · B = { ab : a ∈ A , B ∈ B } containssome element uniquely expressible as ab for a ∈ A , b ∈ B . This implies the a priori stronger ‘two unique products’ property [Str80] and thus the unit conjecture.The first example of a torsion-free group without unique products was constructed byRips and Segev using small cancellation techniques [RS87]. Shortly thereafter, Promis-low provided an elementary example [Pro88] in the (torsion-free) virtually abeliangroup P variously known as the Hantzsche–Wendt group, Passman group, Promislowgroup, or Fibonacci group F (
2, 6 ) . The group P is the unique 3-dimensional crystallo-graphic group with finite abelianization; specifically, it is a non-split extension1 → Z → P → Z /2 × Z /2 → A = B and | A | = P have non-trivialunits, filtering potential units according to a complexity measure called length L thatcomes from the word length in the infinite dihedral quotient of P . They established theunit conjecture for P amongst elements of length L ≤ ave to entertain the possibility that the unit conjecture is equivalent to unique prod-ucts; if it is, and failure of unique products begets non-trivial units, then the aforemen-tioned theorem shows that this cannot happen “locally” in the corresponding sense.Our counterexample has length L = a ).The zero divisor conjecture is known for P so this counterexample does not directlysuggest a line of attack. If the zero divisor conjecture is true, then we have establishedthat it is not true for the perhaps more combinatorial considerations of the unit conjec-ture. On the other hand, if the zero divisor conjecture is false, then this paper removesa serious psychological impediment to finding a counterexample.3. T HE COUNTEREXAMPLE
Setup.
In this paper we will work with the structure of P as an extension of Z by a Z /2 × Z /2 quotient. In order to facilitate calculations, we will describe thisextension very explicitly, including its defining cocycle (factor set) and action. Onecould perform this computation in various ways; our approach is flavoured with Bass–Serre theory.We adopt the convention of conjugation acting on the right: s t : = t − st . We introducenew variables x = a and y = b into the presentation h x , b , y , a | x b = x − , y a = y − , x = a , b = y i and observe that this expresses P as an amalgam of two Klein bottle groups, namely h x , b | x b = x − i and h y , a | y a = y − i , along their isomorphic index-2 Z subgroups h x , b i = h a , y i . Being normal in each factor, this subgroup is normal in the amalgam,with corresponding quotient D ∞ = Z /2 ∗ Z /2. We pick z = abab as a lift to P of agenerator of the infinite cyclic group [ D ∞ , D ∞ ] . As x z = ( a ) ( abab ) = ( a − ) ( ab ) = x andsimilarly y z = y , we see that in fact h x , y , z i ∼ = Z is the kernel of P → Z /2 × Z /2.Write Q for the quotient Z /2 × Z /2. The action of P by conjugation on h x , y , z i inducesan action of Q in which the non-trivial elements act as conjugation by a , b and ab . Theaction of a and b on h x , y i can be read off the presentation; for the action on z note that z ab = z and z a z = bab ( a b ) ab = bab ( ba − ) ab = b ( ab ) a − b = b ( b − a ) a − b = Q . x a = x y a = y − z a = z − x b = x − y b = y z b = z − x ab = x − y ab = y − z ab = z he set-theoretic section σ : Q → P with image { a , b , ab } defines the cocycle f : Q × Q → Z by f ( g , h ) = σ ( g ) σ ( h ) σ ( gh ) − . In order to compute it we just need to knowhow to push an a past a b . One of the defining relations tells us that b − a = a − b − and thus bab − a − = b ( b − a ) a − b − a − = b ( a − b − ) a − b − a − = yx − z − = x − yz − .With this identity in hand we determine f ( g , h ) a b ab a x xb x − yz − y x − z − ab y − z y − z where the table reads left-to-right (the rows give g and the columns give h ).3.2. Proof of Theorem A.
Proof.
Let α = p + qa + rb + sab . Since P is virtually abelian it certainly satisfies thezero divisors conjecture, so it suffices to check that α is left-invertible in order to showit is a unit (as α ′ α = α ′ ( αα ′ − ) = α ′ = p ′ + q ′ a + r ′ b + s ′ ab for p ′ , q ′ , r ′ , s ′ ∈ F [ x ± , y ± , z ± ] then referring to the table of values of the cocycle f letsus write down the general form of the product α ′ α as follows, in terms of the 4 uniqueLaurent polynomials in F [ x ± , y ± , z ± ] such that α ′ α = ( α ′ α ) + ( α ′ α ) a a + ( α ′ α ) b b +( α ′ α ) ab ab : ( α ′ α ) = p ′ p + xq ′ q a + yr ′ r b + zs ′ s ab ( α ′ α ) a = p ′ q + q ′ p a + x − z − r ′ s b + y − s ′ r ab ( α ′ α ) b = p ′ r + xq ′ s a + r ′ p b + y − zs ′ q ab ( α ′ α ) ab = p ′ s + q ′ r a + x − yz − r ′ q b + s ′ p ab .We shall prove that the following choices p ′ , q ′ , r ′ , s ′ make α ′ inverse to α . In this ta-ble we also record the action of ab and b on the polynomials p , q , r , s . Geometrically,the action of ab corresponds to rotation by π about the z -axis, so the middle columnexpresses rotational symmetry in Z ⊗ R about axes parallel to the z -axis; the thirdcolumn of equations is simply the second column conjugated by a . p ′ : = x − p a p ab = x − y − p p b = x − yp a q ′ : = x − q q ab = yq q b = y − q a r ′ : = y − r r ab = x − r r b = x − r a s ′ : = z − s a s ab = s s b = s a he symmetry equations satisfied by the polynomials already do half of the computa-tion of α ′ α , without needing to consider the actual polynomials: substituting in imme-diately gives (over F ) ( α ′ α ) a = x − p a q + x − qp a + x − z − y − rs a + y − z − s a x − r = ( α ′ α ) b = x − p a r + xx − qs a + y − rx − yp a + y − zz − s a yq = ( α ′ α ) ab . We see that ( α ′ α ) ab = x − p a s + x − qr a + x − yz − y − ry − q a + z − s a x − y − p = x − ( p a s + qr a + y − z − q a r + y − z − ps a )= x − ( ξ + y − z − ξ a ) where ξ : = p a s + qr a . Define ψ : F [ x ± , y ± , z ± ] → F [ x ± , y ± , z ± ] by ψ ( γ ) = y − z − γ a . We show that ψ ( ξ ) = ξ , so that ( α ′ α ) ab vanishes as claimed, by modifyingeach of the summands of ξ by ǫ : = ( + x )( + y − )( + z ) . Note that p a s + ǫ = ( + x )( + y − )( + z )( x + x − + y + y − ) z − is invariant under ψ as ( + y − )( + z − ) is invariant and the remaining factor ( + x )( x + x − + y + y − ) is centralized by a . For qr a we have no choice but to expand itout; it is convenient to bracket the terms according to the power of x : qr a = ( x − y − + ( y − z + z ) + x )( + yz − + x ( + y − z − )= x − ( y − + z − ) + ( y − + y − z − + y − z + + z + y )+ x ( y − z + y − + z + y − + + yz − ) + x ( + y − z − ) .We now see that qr a + ǫ = x − ( y − + z − ) + ( y − z − + y ) + x ( y − + yz − ) + x ( + y − z − ) is likewise invariant under ψ , completing the proof that ( α ′ α ) ab =
0, as ψ ( ξ ) = ψ ( p a + ǫ + qr a + ǫ ) = ξ .It remains to show that ( α ′ α ) =
1. Substituting in, we see that ( α ′ α ) = x − pp a + qq a + x − rr a + ss a .For convenience, write x = x + x − , y = y + y − and z = z + z − . We expand out qq a and x − rr a according to the power of z . Let us write them down in factorized formfirst: qq a = (cid:16) x − y − + x + ( y − + ) z (cid:17) (cid:16) ( y + ) z − + ( x − y + x ) (cid:17) , ( x − r ) r a = (cid:16) x − + + ( x − y − + y ) z (cid:17) (cid:16) ( y + xy − ) z − + + x (cid:17) . bserve that the coefficient of z in qq a is ( x − y − + x )( x − y + x ) + ( y − + )( y + ) = x + x − = x .Similarly, the coefficient of z in x − rr a is y . Thus, multiplying out and collectingterms: qq a + x − rr a = (( x − + x − y − + xy + x ) + ( x − y + y − + y + xy − )) z − + x + y + (( x − + xy − + x − y + x ) + ( x − y − + y − + y + xy )) z = ( xy + xy − + x − y + x − y − + x + x − + y + y − )( z − + z ) + x + y = xyz + xz + yz + x + y .Exploiting characteristic 2 to get ( + x )( + x − ) = x and similarly for y and z , wesee that x − pp a = x − ( + x )( + y )( + z − )( + x )( + y − )( + z ) = xyz .Furthermore ss a = ( + ( x + y ) z − )( + ( x + y ) z ) = + ( x + y )( z + z − ) + ( x + y ) = + xz + yz + x + y .Thus ( α ′ α ) = xyz + ( xyz + xz + yz + x + y ) + ( + xz + yz + x + y ) = (cid:3) Acknowledgements.
This work was supported by the Deutsche Forschungsgemein-schaft (DFG, German Research Foundation) – Project-ID 427320536 – SFB 1442, as wellas under Germany’s Excellence Strategy EXC 2044–390685587, Mathematics Münster:Dynamics–Geometry–Structure and within the Priority Programme SPP2026 ‘Geome-try at Infinity’.I thank Arthur Bartels for helpful conversations and Martin Bridson for comments on adraft. I am grateful to Kenneth A. Brown and Sven Raum for bringing to my attentionthat Higman formulated the conjecture.R
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