Representational Consistency of Group Rings

Received: March 21, 2011 Accepted: April 14, 2011 doi:10.5539/jmr.v3n3p89 Abstract It is shown that if a group ring satisfies the idempotent conjecture and some other restrictions it will be representational consistent. As an application we show the real numbers as a field are representational consistent. This leads to the problem of whether a circle can be given a field structure that can be shown to be consistent.


Introduction
The study of consistency has been viewed historically as very important to mathematics.For example Hilbert's second problem asks whether the axioms of arithmetic are consistent.There is more than one definition of consistency, the most straightforward being that a statement cannot be both true and false at the same time.Another definition is ω-consistency which is the opposite of ω-inconsistency.ω-inconsistency means a statement does not hold for some number, but is still provable for every number individually.Tarski has proved that the real numbers as a field are consistent see (Blum and Smale, 2000, p1477-1495).
Representational consistency is defined in a general sense using the notion of structure or interpretation from model theory (Crossley et al., 1972, p13).
Definition 1.Let S be a nonempty set, and U a collection of relations on S , S , U is called the structure of S .Let R 1 : S → V, R 2 : S → W be two monomorphisms of a structure S , U and (1) This defines an operator U on mappings whose domain is S , but the range of U is {{v, w} | v ∈ V, w ∈ W} .
Definition 2. The structure S , U is called pair wise consistent if for all pairs of monomorphisms R 1 , R 2 the map U(R 1 , R 2 ) is one to one.If for any pair of monomorphisms R 1 , R 2 the map U(R 1 , R 2 ) is not one to one, the structure is said to be pair wise inconsistent.
Let S = kG, where G is a group, and k is a field, and U are the relations given by the ring axioms on S .It will be shown that the structure S , U is pair wise consistent if G satisfies the idempotent conjecture and some other conditions.For example if G belongs to the class of groups S (Emmanouil, 2001, p151-160), or the class of groups C (Emmanouil, 1998, p307-330) then it satisfies the idempotent conjecture.
Irving Kaplansky made the idempotent conjecture in his classic book on fields and rings (Kaplansky, 1972, p122).
If G is a torsion free group, and k is a field then the group ring kG has no non-trivial zero-divisors.
The idempotent conjecture is currently open, although there are many cases in which it is known to hold.The idempotent conjecture is true for • Torsion free abelian groups • Torsion free polycyclic by finite groups • Torsion free Noetherian groups • Solvable groups with finite Hirsch number In Emmanouil's paper (Emmanouil, 2001, p151-160) he introduces a class S of groups satisfying the idempotent conjectures, with nice closure properties.Consider for any group G the complex group algebra CG and the augmentation homomorphism ε : CG → C and the augmentation ideal I G = ker ε.For any left CG-module M let M G = M/I G M. The class S consists of groups that satisfy the following condition: For any non-zero projective CG-module P, we have P G 0. The class S contains all torsion-free abelian groups, and is closed under subgroups, extensions, direct products and free products.
In addition to these examples, a larger class of groups with good closure properties can be found in (Emmanouil, 1998, p307-330).The class C consists of those groups that satisfy the following condition: For any element g ∈ G of infinite order the image α g Q of the class α g in the cohomology ring ⊕ i H i N g , Q is nilpotent.A group G is said to be residually contained in C if for any element g ∈ G with g 1 there exists a normal subgroup K G, such that g K and G/K ∈ C. Emmanouil proves the following theorem.
Theorem 3. Let G be a group.If G ∈ C is torsion-free, then G satisfies the idempotent conjecture.
The class C includes all torsion and all abelian groups, and is closed under subgroups, finite direct products and free products (Emmanouil, 1998, p307-330) Proposition 3.5.However there are examples of groups that do not satisfy the cohomological condition defining class C (Emmanouil, 2001, p159-171).
Further examples of pair wise consistent rings such as non-commutative ordered rings and rings other than group rings can be found in the book in progress by the author (Bhavnagri, 2010, p1-133).

Showing consistency
To show that structures with representational consistency exist we use the theorem from (Bhavnagri, 2009, p4).Theorem 4. A set S with structure has (pairwise) representational consistency if The identity automorphism is denoted by id.
Proof.If the representations of S are pair wise inconsistent then for some φ 1 , φ 2 the mapping U(φ 1 , φ 2 ) is not one to one.Hence there are s, t ∈ S with s t such that Since φ 1 , φ 2 are monomorphisms they are one to one so Since Aut(S ) is a group we have from equation 2 From equations 4 and 5 we have φ −1 1 φ 2 φ −1 1 φ 2 (s) = s.Since Aut(S ) acts freely on S it must be that φ −1 1 φ 2 φ −1 1 φ 2 = id where id ∈ Aut(S ) is the identity.From equation 3 the sets in equation 2 cannot have only one element, as U(φ 1 , φ 2 ) is not one to one, so φ 1 φ 2 .
The group Aut(S ) has an involution.
What is striking about the theorem above, is that we did not need to know anything about the structure in order to prove the theorem.It could perhaps apply to a large class of structures, but it leads to a formidable problem which has been unsolved since 1997.Identify structures which satisfy the following equations.
In this paper we now assume S is a ring, not necessarily commutative, because it turns out that ring theory provides some solutions to the problem.

Involutions and idempotents
An element e of a structure is called an idempotent if e 2 = e.In a ring involutions and idempotents are equivalent provided the characteristic of the base field is not 2.If e 2 = e then let g = 2e − 1 then g 2 = 1 and g is an involution.Conversely if g is an involution, g 2 = 1 and let e = 1 2 (g + 1) so that e 2 = e, and e is an idempotent.Caution is necessary when applying the equivalence of idempotent and involutory elements in a ring.This is because an involution φ with respect to composition in Aut (S ) is φ (φ (s)) = s which is not the same as an involutory element is not the same as the composition 2 (2e − 1) − 1.
Lemma 5. Let S be a ring (with unit) and define g b : S → S with g b (s) = bs and b = 2a − 1.
(i) If a ∈ S is an idempotent then g b is an involution.
(ii) If an idempotent is invertible it is trivial.
(iii) An involution is its own inverse.
Proof.(i) By definition However g b is not an automorphism as that would require g b (s) g b (t) = g b (st) for all s, t ∈ S That would require the equality of bsbt and bst.
On the other hand if we conjugate by two involutory elements b 2 = 1 and d 2 = 1 with φ (s) = bsd then φ is an involution with respect to composition in Aut (S ).It is necessary that db = 1 if φ ∈ Aut (S ), such as an inner automorphism which has d = b −1 .
From Lemma 5 (iii) (2a − 1) 2 = 1 and we have (iii) For all s, x xsx −1 x −1 = s since φ is an inner automorphism involution.Right multiply by Now what we are interested in is the extent to which the converse of Lemma 6 (ii) holds.That will give us conditions for the last equation in Theorem 4. That is under what conditions does knowing a ring has no non-trivial idempotents tell us it has no non-trivial automorphism involutions?The answer is provided by (iv) of the same Lemma 6.
There can be non-trivial inner automorphism involutions s → xsx −1 with x not an involution.For example in the ring of 2 × 2 real matrices if x 2 = c1 and x −2 = 1 c 1 where c is a scalar.An example of a non-trivial involution in this ring is x = 1 1 1 −1 .However x 2 is in the centre of the ring S .The standard notation for the centre of a ring is Z (S ).It is well where I n denotes the n × n identity matrix.So an inner automorphism involution is just a scaled involution in M n (C).
In particular n = 1 gives C itself, which has i → −i as automorphism.So if j is such that ji j −1 = −i then ji = −i j so j ∈ H, not in C which is commutative.Even in H the map x → ixi −1 is an involution automorphism, but H has no non-trivial idempotents, its automorphisms are all inner and it is central.
Also observe that for an inner automorphism the third equation in Theorem 4 is φ There are also examples of group rings CG whose centre is a direct sum of C • I n i , for example if G is a finite group in the book (Emmanouil, 2006, p195).
Automorphism groups can have outer automorphisms, that is automorphisms that are not inner.However, there are conditions under which all automorphisms are inner automorphisms.We will find an example of such rings after we deal with a simpler case first.

Ordered rings
An ordered ring is a ring with an order < such that for any a, b, c An ordered ring has no idempotents other than 0 and 1.There are three cases.If 0 < x < 1 then multiplying by x we have The real numbers are an ordered field, so they have no non-trivial idempotents.Likewise the rational numbers are another example of an ordered field.Both fields have no non-trivial automorphisms.
Any ordered field has positive squares x 2 > 0. The complex numbers are not an ordered field, otherwise √ −1 would have positive square.

Inner automorphisms
If k is a field then any automorphism of the ring of matrices over k (denoted M n (k)) can be extended to an inner automorphism.This is a corollary of the theorem of extension of isomorphisms we will now introduce.
A ring R is called simple if it has no non-trivial ideals and R 2 (0).The trivial ideals are R and (0).A division ring is a ring in which every non-zero element has a multiplicative inverse.Any division ring is simple.
Lemma 7. The ring M n (D) where D is a division ring is simple.
The Jacobson density theorem follows, which can be used to prove the Wedderburn-Artin theorem.
Theorem 8. Let M be a simple right R-module and let D = End R (M).Let X be a finite subset of M and I = ann R (X) the annihilator of X.If u ∈ M is such that uI = 0 then u ∈ XD.
A ring is said to satisfy the descending chain condition on right ideals, if every properly descending chain of right ideals is finite.The next theorem is the Wedderburn-Artin theorem.
Theorem 9. A simple ring with descending chain condition on right ideals is the full ring of linear transformations on a finite-dimensional vector space over a division ring.
A central simple k-algebra is a simple k-algebra whose center is the field k.For example C is not central simple and On the other hand the quaternion algebra H is central simple, the center being R.
Lemma 10.Let C be an algebra over F. Let A, B be subalgebras with A central simple and B simple.Suppose further that A and B commute elementwise.Then either AB = 0 or AB is isomorphic to A ⊗ B.
Lemma 11.Let D be a division algebra over F and A a finite dimensional algebra with unit element over F. Then A ⊗ D satisfies the descending chain condition on right ideals.
Theorem 12. Let E be a vector space over a division ring D with center Z.Let R be the ring of all linear transformations on E and let B and C be simple subalgebras of R, finite dimensional over Z and containing the unit element of R. Then any isomorphism between B and C can be extended to an inner automorphism of R.
As a corollary to this theorem we have that if k is a field then any automorphism of M n (k) can be extended to an inner automorphism.
For further details about rings see for example the book by Irving Kaplansky (Kaplansky, 1972, p79-198).A similar result about all automorphisms being inner is also a corollary of the Skolem-Noether theorem (Drupieski and Hamblet, 2005, p1).

Showing inconsistency
We have seen thus far how to prove consistency, now we show a technique for proving inconsistency.We will show that the cyclic group is representationaly inconsistent.
Lemma 13.The representations of Z n are inconsistent if n > 2.
Proof.Let φ : Z n → Z n with φ : s m → s m(n−1)modn .Suppose n − 1 and n have a common factor p > 1.Then n − 1 = pr and n = ps for some integers r, s.Since n = n − 1 + 1, ps = pr + 1, so p (r − s) = −1.This would imply p and r − s equal +1 or −1, contradicting p > 1.Thus n − 1 and n are relatively prime, so that the multiples of φ (s 1 ) = s n−1 generate Z n .Thus φ is a structure preserving mapping, hence a monomorphism of Z n .Since φ (s n−1 ) = s 1 , The above lemma can be generalized.If any automorphism φ is an involution the representations will be inconsistent.Let φ (s 1 ) = s 2 and φ (s and the representations are inconsistent.
The complex numbers C have complex conjugation as an automorphism.The composition of two conjugations is the identity, and hence it is an involution.Complex multiplication is commutative so there is no distinction between automorphism and anti-automorphism.Consequently the field of complex numbers C is representationaly inconsistent.

Consistency of circle
The field of real numbers R is consistent in both the classical sense proved by Tarski and in the sense of representational consistency.This is because it is both ordered and has trivial automorphism group.On the other hand the field of complex numbers C is inconsistent in the sense of representational consistency.Let us now consider the problem of giving the circle S 1 a ring structure, so that we can determine whether it is consistent or not.
A periodic function f : R → S 1 such as f (θ) = e iθ maps the real numbers to the circle.Consider ways in which R can be endowed with a product that is ordered.The lexicographic order (x 1 , y 1 ) < (x 2 , y 2 ) if x 1 < x 2 or if x 1 = x 2 and y 1 < y 2 .
More generally an open problem would be whether all ordered structures are consistent.Also it is hoped some of the machinery developed for the idempotent conjecture can be carried over.