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ALGEBRAIC TORSION 1. Introduction 1.1. Definition of K 1(R)

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ALGEBRAIC TORSION 1. Introduction 1.1. Definition of K 1(R) ALGEBRAIC TORSION 1. Introduction 1.1. Definition of K1(R). By a ring we always mean an associative ring R with 1 6= 0. For such a ring R define GL(R) to be the group of all N × N-matrices, that are the infinite identity matrix except for finitely many entries. More rigorously, GL(R) = lim GL(R; d), where we have the following maps in the directed system: ! GL(R; d) ! GL(R; d + 1) A 0 A 7! : 0 1 Given a 2 R and i 6= j 2 f1; : : : ; dg we denote by Eij the d × d-matrix which is zero except for the (i; j)-entry which equals 1. We refer to all matrices of form id +aEij with a 2 R as elementary matrices. We denote by E(R) the subgroup of GL(R) generated by all elementary matrices. In the following, given a group G and g; h 2 G we denote by [g; h] = ghg−1h−1 the commutator. Furthermore, we denote by n n o Y [G; G] := [gi; hi] g1; h1; : : : ; gn; hn 2 G i=1 the commutator subgroup of G. The following proposition gives a convenient descrip- tion of the commutator subgroup of GL(R). Proposition 1.1. E(R) = [GL(R); GL(R)]: Proof. For i 6= j 6= k one can easily verify that id +aEik = (id +aEij)(id +Ejk)(id −aEij)(id −Ejk) −1 −1 = (id +aEij)(id +Ejk)(id +aEij) (id +Ejk) : This shows that each elementary matrix in GL(d; A) with d ≥ 3 is indeed a commu- tator. Conversely, using the following three identities one can easily show that each com- mutator XYX−1Y −1 2 GL(d; R) is a product of elementary matrices within the Date: July 2, 2015. 2010 Mathematics Subject Classification. Primary 57M27; Secondary 57Q10. 1 2 ALGEBRAIC TORSION group GL(2d; R): XYX−1Y −1 0 X 0 Y 0 (YX)−1 0 (1) = ; 0 id 0 X−1 0 Y −1 0 YX X 0 id X id 0 id − id id 0 (2) = ; 0 X−1 0 id id −X−1 id 0 id id −X id d 2d id X Q Q (3) = (id +xijEij): 0 id i=1 j=d+1 The Whitehead group K1(R) is defined as the abelian group K1(R) := GL(R)=[GL(R); GL(R)] = GL(R)=E(R): For every d there exists a canonical homomorphism GL(R; d) ! K1(R), in particular there exists a homomorphism from the units of R into K1(R). By abuse of notation we denote the image of A 2 GL(R; d) in K1(R) by A as well. Lemma 1.2. For any two matrices A 2 GL(R; d1) and B 2 GL(R; d2) the product AB 2 K1(R) is given by A 0 AB = 2 K (R): 0 B 1 Proof. Without loss of generality we can assume that d1 = d2. Then AB 0 A 0 B 0 AB = = 2 K (R): 0 id 0 B 0 B−1 1 In the proof of Proposition 1.1 we had seen that the second matrix on the right hand side is trivial in K1(R). The proof of the following lemma is left as an exercise. Lemma 1.3. Let a; b be units in R. Then 0 a −a 0 and b 0 0 b represent the same element in K1(R). Corollary 1.4. If F is a commutative field, then the map ∗ K1(F) ! F := F n f0g [A] 7! det(A) is an isomorphism. In the following, given a commutative field F, we will use the isomorphism from ∗ Corollary 1.4 to identify K1(F) with F . ALGEBRAIC TORSION 3 Proof. It is straightforward to see that the map is a well-defined epimorphism. It remains to show that the map is injective. So let A be any matrix in GL(d; F). Using elementary linear algebra and Lemmas 1.2 and 1.3 one can show that A and the 1 × 1-matrix (det A) represent the same element in K1(F). It follows that the map under consideration is a monomorphism. In the following we will mostly work with the reduced Whitehead group K1(R) = K1(R)=subgroup of order 2 generated by (−1): On several occasions we will make use of the fact that a ring homomorphism ': R ! S induces group homomorphisms '∗ : K1(R) ! K1(S) and '∗ : K1(R) ! K1(S). Also, the maps GL(d; M(n × n; R)) ! GL(dn; R) A viewed as an d × d-matrix A with entries 7! nd × nd-matrix that are n × n-matrices over R with entries in R are homomorphisms that induce an isomorphism K1(M(n × n; R)) ! K1(R) that is sometimes referred to as Morita equivalence. 1.2. Definition of algebraic torsion. Henceforth we will assume that all rings in these lecture notes have the following property: r s if r 6= s 2 N0, then R is not isomorphic to R as an R{module. It is clear that any field, Z, and more generally any domain have this property. The following lemma gives another important example. Lemma 1.5. If G is a group, then the group ring Z[G] has the above property. r ∼ s Proof. We view Z as a Z[G]-module via the trivial action of G on Z. If Z[G] = Z[G] , then we also have r ∼ r ∼ s ∼ s Z = Z ⊗Z[G] Z[G] = Z ⊗Z[G] Z[G] = Z ; hence r = s. In the following, given a ring R we mean by an R-module a finitely generated left R-module, unless we say explicitly otherwise. We view elements in Rn as row vectors, and an n × m-matrix over R gives rise to a homomorphism Rn ! Rm of left R-modules via right multiplication on row vectors. A based R-module is a free R-module together with an (unordered) basis. Given a free R-module V and two bases d = fd1; : : : ; dng and e = fe1; : : : ; eng we denote by [d=e] 2 K1(R) the element represented by the matrix of the basis change, P i.e. [d=e] = (aij) where di = j aijej. We say that two bases are equivalent if [d=e] is trivial in K1(R). 4 ALGEBRAIC TORSION p Given a short exact sequence 0 ! A −!i B −! C ! 0 of free R-modules and bases A = faig for A and C = fcjg for C we pick liftsc ~j to B and we denote by AC the equivalence classes of bases represented by the ai and thec ~j. It is straightforward to verify that AC does not depend on the choice of the lifts. If B is also equipped with a basis B, then we say that the bases A; B and C are compatible, if B and AC are equivalent. Let (C∗; C∗) be a finite chain complex of based R{modules. In the following we will associate to (C∗; C∗) its torsion τ(C∗; C∗) 2 K1(R) [ {∗}: If C∗ is not acyclic, then we define its torsion τ(C∗; C∗) to be ∗. Now suppose that C∗ is acyclic. In the following we make the technical assumption that each Bi := Im(Ci+1) ⊂ Ci is a free R-module. For each i we pick a basis Bi of Bi. Since C∗ is acyclic the sequences 0 ! Bi ! Ci ! Bi−1 ! 0 are exact. We define the torsion (sometimes also called Reidemeister-torsion) of the based acyclic complex (C∗; C∗) to be Y (−1)i+1 τ(C∗; C∗) = [BiBi−1=Ci] 2 K1(R): It is straightforward to see that τ(C∗; C∗) is independent of the choice of the Bi. If the R{modules Bi are not free, then one can show that they are stably free and a stable basis will then make the definition work again. We refer to [Mi66, p. 369] or [Tu01, p. 13] for the full details.1 For future reference we record the simplest example in the following lemma. @k Lemma 1.6. If 0 ! Ck −! Ck−1 ! 0 is an isomorphism between free based R- modules of rank n, then (−1)k−1 τ(C∗) = [Bk] 2 K1(R); where Bk denotes the n × n-matrix over R which represents @k with respect to the given bases. The statement of the lemma is an immediate consequence of the definitions. It says in particular that the torsion of a chain complex of based R-modules can be viewed as the generalization of the determinant of a homomorphism between two based R-modules. Example. Let A be an n × n-matrix over Z with det(A) 6= 0. Then n A n τ 0 ! Q −! Q ! 0 = det(A) 1The torsion introduced in [Tu01] is the multiplicative inverse of the torsion introduced in [Mi66]. We follow the convention of [Mi66]. ALGEBRAIC TORSION 5 which up to sign equals the number of elements in the finite abelian group Zn=AZn. This explains, perhaps, the name torsion for the invariant τ(C∗). The term `torsion' was introduced by Franz [Fra35]. In the following we extend the multiplication of K1(R) to τ(C∗; C∗) 2 K1(R) [ {∗} by defining the product of 0 with any other element to be 0. The following elementary lemma now shows how the torsion depends on the choice of the bases. 0 Lemma 1.7. Let C∗ be a chain complex of free R-modules and let C∗ and C∗ be two choices of bases.
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