Math. Ann. (2008) 340:159–184 DOI 10.1007/s00208-007-0142-y

Mathematische Annalen

Homology of GLn : injectivity conjecture for GL4 Behrooz Mirzaii

Received: 14 June 2006 / Revised: 30 April 2007 / Published online: 24 July 2007 © Springer-Verlag 2007

Abstract The homology of GLn (R) and SLn (R) is studied, where R is a commutative ‘ring with many units’. Our main theorem states that the natural map H4 (GL3 (R), k) → H4 (GL4 (R), k) is injective, where k is a field with char(k) = 2, 3. For an algebraically closed field F, we prove a better result, namely, H4 (GL3 (F), Z) → H4 (GL4 (F), Z) is injective. We will prove a similar result replacing GL by SL. This is used to investigate the indecomposable part of the K -group K 4 (R). Mathematics Subject Classification (2000) 19D55 · 18G60 · 20J05 1 Introduction In the beginning of the 1970s two types of K -groups in algebra appeared: Quillen’s K -groups and Milnor’s K -groups. For a ring R, Quillen defined the K -group K n (R) as the n-th homotopy group of the space BGL(R)+ and Milnor defined the K -group K nM (F) of a field F as the n-th degree part of T (F ∗ )/a ⊗ (1 − a) : a ∈ F ∗ − {1}, where T (F ∗ ) is the tensor algebra of F ∗ . There is a canonical ring homomorphism K ∗M (F) → K ∗ (F), therefore a canonical homomorphism K nM (F) → K n (F). A useful approach to investigate K -groups of a field F is by means of their relation with integral homology groups of GL(F) and Milnor K -groups. Suslin’s stability theorem states that for an infinite field F the natural map Hi (GLn (F), Z) → Hi (GL(F), Z) is bijective if n ≥ i [18]. Using this Suslin constructed a map from Hn (GLn (F), Z) to Milnor’s K n -group K nM (F), denoted by sn , such that the sequence Hn (inc)

sn

Hn (GLn−1 (F), Z) −→ Hn (GLn (F), Z) −→ K nM (F) −→ 0

B. Mirzaii (B) Department of Pure Mathematics, Queen’s University Belfast, Belfast BT7 1NN, UK e-mail: [email protected]

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is exact. Combining these two results he constructed a map from K n (F) to K nM (F) such that the composite homomorphism K nM (F) → K n (F) → K nM (F) coincides with the multiplication by (−1)n−1 (n − 1)! [18, Sect. 4]. Now one might ask about the kernel of Hn (inc) in the above exact sequence. In this direction, Suslin posed a problem, which is now referred to as ‘a conjecture by Suslin’ (see [16, Problem 4.13] and [4, Remark 7.7], [7, Conjecture 2]). The name ‘Injectivity Conjecture’ seems appropriate. Conjecture 1 (Injectivity Conjecture) For any infinite field F the natural homomorphism

  Hn GLn−1 (F), Q → Hn GLn (F), Q is injective. Many of Suslin’s results were generalized to a wider class of rings, called ‘rings with many units’, by Nesterenko-Suslin [15] and Guin [10]. Of course it is also reasonable to formulate the above conjecture for this class of rings. The above conjecture is easy if n = 1, 2. For n = 3 the conjecture was proved positively by Sah [16] and Elbaz-Vincent [9]. The conjecture is proven in full for number fields by Borel and Yang [4]. The proof of this conjecture for n = 4, when R is a commutative ring with many units, is the main goal of this paper (Theorem 3). Here we take a general step towards this conjecture. Let R be a commutative ring with many units. We show that if Hm (GLm−1 (R), Q) → Hm (GLm (R), Q) is injective for m = n − 2, n − 1 and if the complex α∗ −inc∗

Hn (R ∗ 2 × GLn−2 (R), Q) −→ Hn (R ∗ × GLn−1 (R), Q) inc∗

−→ Hn (GLn (R), Q) → 0, is exact, where α : (a, b, A) → (b, a, A), then Hn (GLn−1 (R), Q) → Hn (GLn (R), Q) is injective. The proof follows from a careful analysis of some spectral sequences, connecting the homology of the groups R ∗ p × GLn− p (R) for different values of p. One of the main ingredients in the proof of this proposition is a construction of an explicit map from K nM (R) to Hn (GLn (R), Z), denoted by νn (this was only known for n = 2). This construction fits in our previous theory, that is, the composite homomorphism νn

sn

K nM (R) −→ Hn (GLn (R), Z) −→ K nM (R) coincides with the multiplication by (−1)n−1 (n − 1)!.

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When n = 4, indeed we prove that the complex α∗ −inc∗

inc∗

H4 (R ∗ 2 × GL2 (R), Z) −→ H4 (R ∗ × GL3 (R), Z) → H4 (GL4 (R), Z) → 0 is exact. This enables us to prove that if k is a field with char(k) = 2, 3, then H4 (GL3 (R), k) → H4 (GL4 (R), k) is injective. If F is an algebraically closed field, the same technique implies that H4 (GL3 (F), Z) → H4 (GL4 (F), Z) is injective. Also we show that for n = 4 a similar injectivity result is true if we replace GL with SL. As an application we will study the indecomposable part of K 4 (R). Namely we will prove that for an algebraically closed field F, K 4 (F)ind := coker(K 4M (F)→K 4 (F)) embeds in H4 (SL3 (F), Z). Here we establish some notation. In this note, by Hi (G) we mean the i-th integral homology of the group G. We use the bar resolution to define the homology of a group [5, Chap. I, Sect. 5]. Define c(g1 , g2 , . . . , gn ) =



sign(σ )[gσ (1) |gσ (2) | . . . |gσ (n) ] ∈ Hn (G),

σ ∈n

where gi ∈ G pairwise commute and n is the symmetric group of degree n. By GLn and SLn we mean GLn (R) and SLn (R), where R is a commutative ring with many units. Note that GL0 is the trivial group and GL1 = R ∗ . By R ∗ m we mean R ∗ × · · · × R ∗ (m-times) or the subgroup of R ∗ , {a m : a ∈ R ∗ }, depending on the context. This will not cause any confusion. The i-th factor of R ∗ m = R ∗ × · · · × R ∗ , (m-times), is denoted by Ri∗ . If A → A is a homomorphism of abelian groups, by A /A we mean coker(A → A ). 2 Rings with many units Definition 1 We say that a commutative ring R is a ring with many units if for any n ≥ 1 and for any finite number of surjective linear forms f i : R n → R, there exists a v ∈ R n such that, for all i, f i (v) ∈ R ∗ . Remark 1 (i) The study of ‘rings with many units’ is originated by W. van der Kallen in [21], where he shows that K 2 of such commutative rings behaves very much like K 2 of fields (see Proposition 2 in the following). (ii) It is easy to see that a ring with many units has stable range one, sr(R) = 1, [10, Sect. 1]. (iii) If R is a commutative ring with many units, then for any n ≥ 1, there exist n elements in R such that the sum of each nonempty subfamily belongs to R ∗ [10, Proposition 1.3]. Rings with this property are considered by Nesterenko and Suslin in [15] (see Proposition 1).

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Example 1 (i) An infinite field is a ring with many units. (ii) A commutative semilocal ring is a ring with many units if and only if all its residue fields are infinite fields. This is an example which one should keep in mind in this paper. (iii) The product of rings with many units is a ring with many units. (iv) Let F be an infinite field. Then any commutative finite dimensional F-algebra is a semilocal ring. Therefore it is a ring with many units. Here we give two known results which are used in the construction of spectral sequences in Sect. 5. Lemma 1 Let R be a commutative ring with many units. Let n ≥ 2 and assume Ti , 1 ≤ i ≤ l, are finitely many finite subsets of R n such that each Ti is a basis of a free summand of R n with k elements, where k ≤ n − 1. Then there is a vector v ∈ R n , such that Ti ∪ {v}, 1 ≤ i ≤ l, is a basis for a free summand of R n . Proof This is well-known and easy to prove. We leave the proof to the reader.



The next result is due to Suslin. Proposition 1 Let R be a commutative ring with many units. Let G i be a subgroup of GLn i , i = 1, 2, and assume that at least one of them contains the subgroup of diagonal matrices. Let M be a submodule of Mn 1 ,n 2 (R) such that G 1 M = M = M G 2 . Then the inclusion 

G1 0 0 G2



 →

G1 M 0 G2



induces isomorphism on homologies with coefficients in Z. Proof It is easy to see that if R is such a ring, then for any n ≥ 1, there exist n elements of R such that the sum of each nonempty subfamily belongs to R ∗ [see Remark 1 (iii)]. For the rest see [18, Theorem 1.9] and [15, Sect. 1]. 

Let R be a commutative ring. The n-th Milnor K -group of R is defined as an abelian group K nM (R) generated by elements {a1 , . . . , an }, ai ∈ R ∗ , i = 1, . . . , n, subject to the following relations       a1 , . . . , ai ai , . . . , an = a1 , . . . , ai , . . . , an + a1 , . . . , ai , . . . , an ,   a1 , . . . , an = 0 if there exist i, j, i = j, such that ai + a j = 0 or 1. Usual method shows that there is a natural map K nM (R) → K n (R).

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Proposition 2 Let R be a commutative ring with many units. Then (i) SK1 (R) = 0. (ii) Van der Kallen [21] K 2 (R)  K 2M (R) = R ∗ ⊗Z R ∗ /a ⊗ (1 − a) : a, 1 − a ∈ R ∗ . Proof

(i) By homology stability theorem [10, Theorem 1] K 1 (R) = H1 (GL(R))  H1 (GL1 (R))  R ∗ .

But we also have K 1 (R)  R ∗ × SK 1 (R). Thus SK1 (R) = 0. (ii) (Nesterenko-Suslin) By a theorem of Dennis [6, Corollary 8] K 2 (R) H2 (GL2 )/ H2 (GL1 ) (see also [15, Lemma 4.2]). By [10, Theorem 2] we have K 2M (R)  H2 (GL2 )/H2 (GL1 ). Therefore K 2M (R)  K 2 (R). For the rest see [10, Proposition 3.2.3].



In this paper we always assume that R is a commutative ring with many units. 3 Homology of SLn   −1  a0 a 0 induces an action A 01 0 1 of R ∗ on Hi (SLn ). In this article by Hi (SLn ) R ∗ we simply mean H0 (R ∗ , Hi (SLn )). groups It is easy to see that the natural map SLn → SL induces a map of homology   det(M)−1 0 Hi (SLn ) R ∗ → Hi (SL). The map δ : GL → SL given by M → 0 M induces a homomorphism Hi (δ) : Hi (GL) → Hi (SL) such that the composition

The action of R ∗ on SLn defined by a.A :=

Hi (inc)



Hi (δ)

Hi (SL) −→ Hi (GL) −→ Hi (SL) is the identity map. Therefore Hi (SL) embeds in Hi (GL). Lemma 2

(i) Let F be an infinite field. Then Hi (SLn (F)) F ∗  Hi (SL(F)) f or n ≥ i.

In particular if F is algebraically closed, then Hi (SLn (F))  Hi (SL(F)) for n ≥ i. (ii) Let A := Z[ n1 ] and let q ≤ n. Then for p ≥ 0, H p (R ∗ , Hq (SLn , A)) → H p (R ∗ , Hq (SL, A)), induced by the map of pair (id, inc) : (R ∗ , SLn ) → (R ∗ , SL), is an isomorphism.

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Proof The part (i) is rather well known (see [16, 2.7]). (ii) The proof is by induction on q. If q = 0, then the claim is trivial. So let q ≥ 1 and assume that the claim is true for q ≤ n − 1. First we prove that H0 (R ∗ , Hn (SLn , A))  H0 (R ∗ , Hn (SL, A)). From the map of extensions 1 −→ SL ⏐ n −→ ⏐ n −→ GL ⏐ ⏐ 1 −→ SL −→ GL −→

R⏐∗ −→ 1 ⏐ R ∗ −→ 1,

we get the map of Lyndon–Hochschild–Serre spectral sequences ∗ , H (SL , A)) ⇒ H E 2p,q = H p (R⏐ q n p+q (GL ⏐ n , A) ⏐ ⏐

E 2p,q = H p (R ∗ , Hq (SL, A)) ⇒ H p+q (GL, A). By induction hypothesis for q ≤ n − 1, E 2p,q = E 2p,q . From an easy comparison of spectral sequences it follows that for p + q ≤ n, q ≤ n − 1, ∞

E∞ p,q = E p,q .

The map of spectral sequences gives us a map of filtration 0 = F−1 ⊆ F0 ⊆ · · · ⊆ Fn−1 ⊆ Fn = Hn (GLn , A) ↓ ↓ ↓

⊆ F ⊆ · · · ⊆ F

0 = F−1 0 n−1 ⊆ Fn = Hn (GL, A), ∞ ∞

. Using homology stability theo Fi /Fi−1 , E i,n−i  Fi /Fi−1 such that E i,n−i rem [10, Theorem 1] and the above facts, one sees that Fi  Fi for all i. Since Hn (SL, A) → Hn (GL, A) is injective, F0 = H0 (R ∗ , Hn (SL, A)). It is also easy to see that F0 = H0 (R ∗ , Hn (SLn , A))/T for some subgroup T . From the homomorphism R ∗ × SLn → GLn , (b, B) → bB, we obtain the isomorphism

Hn (R ∗ × SLn , A) R ∗  Hn (GLn , A) (see the proof of Corollary 1 in the following). This shows that the group H0 (R ∗ , Hn (SLn , A)) embeds in Hn (GLn , A) and thus T = 0. Therefore H0 (R ∗ , Hn (SLn , A)) = F0  F0 = H0 (R ∗ , Hn (SL, A)). To complete the induction and thus the proof, we must show that H p (R ∗ , Hn (SLn , A))  H p (R ∗ , Hn (SL, A))

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165 (.)n

for all p. The short exact sequence 1 → µn,R → R ∗ → R ∗ n → 1 gives us the Lyndon–Hochschild–Serre spectral sequence 2 = Hr (R ∗ n , Hs (µn,R , T )) ⇒ Hr +s (R ∗ , T ), Er,s

where T = Hn (SLn , A). Since the order of µn,R is invertible in A, we have 2 = 0 for Hs (µn,R , T ) = 0 for s ≥ 1 [5, Chap. III, Proposition 10.1]. Thus Er,s n 2 s ≥ 1. The action of R ∗ on µn,R and T is trivial, so E0,0 = H0 (µn,R , T ) = H0 (µn,R , Hn (SLn , A)). From this and the above one deduces that ∞  H0 (µn,R , Hn (SLn , A))  H0 (R ∗ , Hn (SLn , A))  Hn (SL, A) E0,0 ∞  E 2 = H (R ∗ n , H (SL, A)). An easy analysis shows that and therefore Er,0 r n r,0

Hr (R ∗ n , Hn (SL, A))  Hr (R ∗ , Hn (SLn , A)). Once more from the exact sequence 1 → R ∗ n → R ∗ →R ∗ /R ∗ n → 1 one gets the Lyndon-Hochschild-Serre spectral sequence E r,s = Hr (R ∗ /R ∗ n , Hs (R ∗ n , S)) ⇒ Hr +s (R ∗ , S), 2

2 where S = Hn (SL, A). It is easy to see that E r,s = 0 for r ≥ 1 and ∞

E 0,s  E 0,s = H0 (R ∗ /R ∗ n , Hs (R ∗ n , S)) = Hs (R ∗ n , Hn (SL, A)). 2

This implies that Hs (R ∗ n , Hn (SL, A))  Hs (R ∗ , Hn (SL, A)). Hence for r ≥ 0, Hr (R ∗ , Hn (SLn , A))  Hr (R ∗ , Hn (SL, A)). It is not difficult to see that this isomorphism is induced by the map of pair ((.)n , inc) : (R ∗ , SLn ) → (R ∗ , SL). In a similar way, one can prove that the map of pair ((.)n , inc) : (R ∗ , SL) → (R ∗ , SL) induces the isomorphism Hr (R ∗ , Hn (SL, A))  Hr (R ∗ , Hn (SL, A)), for all r ≥ 0. Applying the functor Hr to the commutative diagram (id,inc)

, SL) (R ∗ ,⏐SLn ) −−−−−−→ (R ∗⏐ ⏐(id,inc) ⏐ n ((.) ,inc) ((.)n ,inc)

(R ∗ , SLn ) −−−−−−→ (R ∗ , SL) gives us the isomorphism that we are looking for.



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1  , then Theorem 1 (i) If B := Z n−1 Hn (SLn−1 , B) R ∗ → Hn (SL, B) → K nM (R) ⊗ B → 0 is exact. (ii) If F is an algebraically closed field, then Hn (SLn−1 (F)) → Hn (SLn (F)) → K nM (F) → 0 is exact. Proof As in the proof of the above lemma we have a map of spectral sequences E 2p,q = H p (R ∗ , Hq⏐(SLn−1 , B)) ⇒ H p+q (GL ⏐ n−1 , B) ⏐ ⏐ E 2p,q =

H p (R ∗ , Hq (SL, B))

⇒ H p+q (GL, B).

By Lemma 2, for q ≤ n − 1, E 2p,q = H p (R ∗ , Hq (SLn−1 , B))  H p (R ∗ , Hq (SL, B)) = E p,q . 2

The map of spectral sequences gives us a map of filtration 0 = F−1 ⊆ F0 ⊆ · · · ⊆ Fn−1 ⊆ Fn = Hn (GLn−1 , B) ↓ ↓ ↓

⊆ F ⊆ · · · ⊆ F

0 = F−1 0 n−1 ⊆ Fn = Hn (GL, B). From these we obtain F0 /F0  Fn /Fn  K nM (R) ⊗ B. But F0 = Hn (SL, B) and F0 = Hn (SLn−1 , B) R ∗ /T , for some subgroup T . Since F0 → F0 is induced by the natural map SLn−1 → SL, T ⊆ ker(Hn (SLn−1 , B) R ∗ →  Hn (SL, B)). This completes the proof of (i). The proof of (ii) is analogue to (i). Remark 2 (i) In the above theorem the condition that n − 1 is invertible in the coefficients ring B, can not be removed in general. For example it is not difficult to see that H3 (SL(R)) → K 3M (R) is not surjective (see [16, Proposition 2.15] or [12, Corollary 6.2]). (ii) Part (ii) of the above theorem gives a positive answer to a question of Barge and Morel [2, p. 196] for algebraically closed fields. Let K nM (R) → K n (R) be the natural map from the Milnor K -group to the Quillen K -group. Define K n (R)ind := coker(K nM (R) → K n (R)). This group is called the indecomposable part of K n (R).

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Corollary 1 Let k be a field such that (n−1)! ∈ k ∗ . Assume Hn (inc) : Hn (GLn−1 , k) → Hn (GLn , k) is injective. Then Hn (inc)

(i) Hn (SLn−1 , k) R ∗ −→ Hn (SL, k) is injective. (ii) K n (R)ind ⊗ k embeds in Hn (SLn−1 , k) R ∗ and Hn (SLn−1 , k) R ∗ /K n (R)ind ⊗ k  Hn (SL, k)/K n (R) ⊗ k. (iii) In Hn (GLn , k), K n (R) ⊗ k ∩ Hn (GLn−1 , k) coincides with K n (R) ⊗ k ∩ Hn (SLn−1 , k) R ∗ = K n (R)ind ⊗ k. Proof (i) This follows from the assumption, the commutativity of the diagram Hn (SLn−1 ⏐ k) ⏐ , k) R ∗ −→ Hn (SL, ⏐ ⏐ Hn (GLn−1 , k) −→ Hn (GL, k) and the injectivity of inc∗ : Hn (SLn−1 , k) R ∗ → Hn (GLn−1 , k). To prove the injectivity of inc∗ consider the map γn−1 : R ∗ × SLn−1 → GLn−1 given by (b, B) → bB. By tracing the kernel and cokernel of this map we obtain the exact sequence γn−1

1 → µn−1,R → R ∗ × SLn−1 → GLn−1 → R ∗ /R ∗ n−1 → 1. From this we get two short exact sequences 1 → µn−1,R → R ∗ × SLn−1 → im(γn−1 ) → 1, 1 → im(γn−1 ) → GLn−1 → R ∗ /R ∗ n−1 → 1. Writing the Lyndon–Hochschild–Serre spectral sequence of the above short exact sequences (with coefficients in k) and carrying out an easy analysis, one gets Hn (R ∗ × SLn−1 , k) R ∗  Hn (GLn−1 , k). Now by the Künneth theorem Hn (SLn−1 , k) R ∗ embeds in Hn (GLn−1 , k), which is induced by the natural map inc : SLn−1 → GLn−1 . (ii) By (i) and Theorem 1, the complex 0 −→ Hn (SLn−1 , k) R ∗ −→ Hn (SL, k) −→ K nM (R) ⊗ k −→ 0

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is exact. By construction of the Suslin’s map K n (R) → K nM (R) [18, Sect. 4], [10, Theorem 4.1.1] we have a splitting exact sequence 0 −→ K n (R)ind ⊗ k−→K n (R) ⊗ k−→K nM (R) ⊗ k −→ 0. The claims follow from applying the Snake lemma to the commutative diagram M 0 −→ K n (R)⏐ind ⊗ k −→ K n (R) ⏐ ⊗ k −→ K n (R) ⏐ ⊗ k −→ 0 ⏐g ⏐h ⏐ n n M ∗ 0 −→ Hn (SLn−1 , k) R −→ Hn (SL, k) −→ K n (R) ⊗ k −→ 0,

where h n , n ≥ 2, is the Hurewicz map K n (R) ⊗ k → Hn (SL, k) which is injective 

[20, Proposition 3, p. 507] and gn = h n | K n (R)ind . 4 Milnor K -groups We start this section with an easy lemma. Lemma 3 Let G be a group and let g1 , g2 , h 1 , . . . , h n ∈ G such that each pair commutes. Let C G (h 1 , . . . , h n ) be the subgroup of G consisting of all elements of G that commute with all h i , i = 1, . . . , n. If c(g1 , g2 ) = 0 in H2 (C G (h 1 , . . . , h n )), then c(g1 , g2 , h 1 , . . . , h n ) = 0 in Hn+2 (G). Proof The homomorphism C G (h 1 , . . . , h n )×h 1 , . . . , h n  →G defined by (g, h) → gh induces the map H2 (C G (h 1 , . . . , h n )) ⊗ Hn (h 1 , . . . , h n ) → Hn+2 (G). The claim follows from the fact that c(g1 , g2 , h 1 , . . . , h n ) is the image of c(g1 , g2 ) ⊗ c(h 1 , . . . , h n ) under this map. 

−(i−1)

Definition 2 Let Ai,n := diag(ai , . . . , ai , ai

, In−i ) ∈ GLn . We define

[a1 , . . . , an ] := c(A1,n , . . . , An,n ) ∈ Hn (GLn ). Proposition 3

(i) The map νn : K nM (R) → Hn (GLn ) defined by {a1 , . . . , an } → [a1 , . . . , an ]

is a homomorphism. (ii) Let sn : Hn (GLn ) → K nM (R) be the map defined by Suslin. Then sn ◦ νn coincides with the multiplication by (−1)(n−1) (n − 1)!. Proof (i) It is well know that the Hurewicz map h 2 : K 2M (R)  K 2 (R) → H2 (GL)

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is defined by {a, b} → c(diag(a, 1, a −1 ), diag(b, b−1 , 1)). It is easy to see that in H2 (GL), [a, b] = c(diag(a, 1, a −1 ), diag(b, b−1 , 1)). The stability isomorphism H2 (GL2 )  H2 (GL) implies that the map K 2M (R) = K 2 (R) → H2 (GL2 ), {a, b} → [a, b] is well defined. Now by Lemma 3, for a1 , 1 − a1 ∈ R ∗ [a1 , 1 − a1 , a3 , . . . , an ] = 0 (see Proposition 2). To complete the proof of (i) it is sufficient to prove that [a1 , . . . , an−2 , an−1 , an ] = −[a1 , . . . , an−2 , an , an−1 ]. This can be done in the following way; [a1 , . . . , an−2 , an−1 , an ] = c(A1,n , . . . , An−2,n , An−1,n , An,n )

−(n−1) , An,n = c A1,n , . . . , An−2,n , diag an−1 In−2 , an−1 , an−1

−(n−1) (n−1) + c A1,n , . . . , An−2,n , diag In−2 , an−1 , an−1 , An,n 

−(n−1) , = c A1,n , . . . , An−2,n , diag an−1 In−2 , an−1 , an−1 
 diag an In−2 , an−(n−2) , 1 

−(n−1) , + c A1,n , . . . , An−2,n , diag an−1 In−2 , an−1 , an−1 

diag In−2 , an(n−1) , an−(n−1)     −(n−1) (n−1) + c A1,n , . . . , An−2,n , diag In−2 , an−1 , an−1 , An,n 

= − a1 , . . . , an−2 , an , an−1

−(n−1) , + c A1,n , . . . , An−2,n , diag In−2 , an−1 , an−1

diag In−2 , an(n−1) , an−(n−1)

+ c A1,n , . . . , An−2,n , diag(an−1 In−2 , 1, 1), diag In−2 , an(n−1) , an−(n−1)



−(n−1) (n−1) + c A1,n , . . . , An−2,n , diag In−2 , an−1 , an−1 , diag In−2 , an , an−(n−1)


 −(n−1) (n−1) + c A1,n , . . . , An−2,n , diag In−2 , an−1 , an−1 , diag an In−2 , 1, 1 

= − a1 , . . . , an , an−1 .

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(ii) Let τn be the composite map K nM (R) → K n (R) → Hn (GLn ). Then sn ◦ τn coincides with the multiplication by (−1)(n−1) (n − 1)! [18, Sect. 4]. It is well known that the composite map τn

K nM (R) → Hn (GLn ) → Hn (GLn )/Hn (GLn−1 ) is an isomorphism and it is defined by {a1 , . . . , an } → (a1 ∪ · · · ∪ an ) mod Hn (GLn−1 ), where a1 ∪ a2 · · · ∪ an = c(diag(a1 , In−1 ), diag(1, a2 , In−2 ), . . . , diag(In−1 , an )) (see [15, Remark 3.27], [10, Theorem 2]). Also we know that sn factors as Hn (GLn ) → Hn (GLn )/Hn (GLn−1 ) → K nM (R). Our claim follows from the fact that modulo Hn (GLn−1 ) [a1 , . . . , an ] = (−1)n−1 (n − 1)!(a1 ∪ · · · ∪ an ). 

Remark 3 As we mentioned in the introduction, there is a natural homomorphism from K nM (R) to Hn (GLn (R), Z) that factors through K n (R). We do not know whether this map and νn coincide. 5 Homology of GLn Let Cl (R n ) and Dl (R n ) be the free abelian groups with a basis consisting of (v0 , . . . , vl ) and (w0 , . . . , wl ) respectively, where every min{l + 1, n} of vi ∈ R n and every min{l + 1, 2} of wi ∈ R n are a basis of a direct summand of R n . Here by vi  we mean the submodule of R n generated by vi . Let ∂0 : C0 (R n ) → C−1 (R n ) := Z,   i n i (vi ) → i n i and ∂l =

l  (−1)i di : Cl (R n ) → Cl−1 (R n ), l ≥ 1, i=0

where  di ((v0 , . . . , vl )) = (v0 , . . . , v i , . . . , vl ).

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 Define the differential ∂˜l = li=0 (−1)i d˜i : Dl (R n ) → Dl−1 (R n ) similar to ∂l . It is easy to see that the complexes C∗ : 0 ← C−1 (R n ) ← C0 (R n ) ← · · · ← Cl (R n ) ← · · · D∗ : 0 ← D−1 (R n ) ← D0 (R n ) ← · · · ← Dl (R n ) ← · · · are exact. We consider Ci (R n ) and Di (R n ) as left GLn -module in a natural way. If it is necessary we convert this action to the right action by the definition m.g := g −1 m. Take a free left GLn -resolution P∗ → Z of Z with trivial GLn -action. From the double complexes C∗ ⊗GLn P∗ and D∗ ⊗GLn P∗ , using Proposition 1, we obtain two first quadrant spectral sequences converging to zero with  E 1p,q (n) = E˜ 1p,q (n) =



Hq (R ∗ p × GLn− p ) Hq (GLn , C p−1 (R n ))

if 0 ≤ p ≤ n if p ≥ n + 1,

Hq (R ∗ p × GLn− p ) Hq (GLn , D p−1 (R n ))

if 0 ≤ p ≤ 2 if p ≥ 3.

For 1 ≤ p ≤ n, and q ≥ 0, d 1p,q (n) =

p

i+1 H (α ), q i, p i=1 (−1)

where

αi, p : R ∗ p × GLn− p→ R ∗ p−1 × GLn− p+1  ,  a 0 . (a1 , . . . , a p , A) → a1 , . . . , ai , . . . , a p , i 0 A 

idZ if p is odd , so E 2p,0 (n) = 0 for 0 if p is even 2 (n) = E 2 p ≤ n − 1. It is also easy to see that E n,0 n+1,0 (n) = 0. See the proof of [11, Theorem 3.5] for more details. Let k be a field and Ci (R n ) := Ci (R n ) ⊗ k. Consider the commutative diagram of complexes In particular for 0 ≤ p ≤ n, d 1p,0 (n) =



n n n 0←⏐ k ← C0 (R ⏐ ) ← ··· ⏐ ) ← C2 (R ⏐ ) ← C1 (R ⏐ ⏐ ⏐ ⏐ 0 ← 0 ← C0 (R n ) ← C1 (R n ) ← C2 (R n ) ← · · · ,

where the first vertical map is zero and the other vertical maps are just identity maps. This gives a map of the first quadrant spectral sequences 1 E 1p,q (n) ⊗ k → E p,q (n), 1 (n) ⇒ H 1 where E p,q p+q−1 (GLn , k) with E -terms

 1 E p,q (n) =

E 1p,q (n) ⊗ k 0

if p ≥ 1 if p = 0

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d 1p,q (n) ⊗ idk if p ≥ 2 . It is not difficult to see that 0 if p = 1 ∞ ∞ E∞ p,q ⊗ k = E p,q if p  = 1, q ≤ n and p + q ≤ n + 1. Hence E p,q = 0 if p  = 1, q ≤ n and p + q ≤ n + 1. We look at the second spectral sequence in a different way. The complex and differentials d1p,q (n) =

0 ← C0 (R n ) ← C1 (R n ) ← · · · ← Cl (R n ) ← · · · 1 induces a spectral sequence E 1p,q (n) ⇒ H p+q (GLn , k), where E 1p,q (n) = E p+1,q (n)

and d 1p,q (n) = d1p+1,q (n). Thus E ∞ p,q (n) = 0 if p ≥ 1, q ≤ n − 1 and p + q ≤ n.

Proposition 4 Let n ≥ 3 and let k be a field such that (n − 1)! ∈ k ∗ . Let the complex (n)

(n)

β2

β1

Hn (R ∗ 2 × GLn−2 , k) −→ Hn (R ∗ × GLn−1 , k) −→ Hn (GLn , k) → 0 be exact, where β2(n) = Hn (α1,2 )− Hn (α2,2 ) and β1(n) = Hn (inc). If the map Hm (inc) : Hm (GLm−1 , k) → Hm (GLm , k) is injective for m = n − 1, n − 2, then Hn (inc) : Hn (GLn−1 , k) → Hn (GLn , k) is injective. Proof The exactness of the above complex shows that the differentials d r,n−r +1 (n) : E r,n−r +1 (n) → E 0,n (n) r

r

r

are zero for r ≥ 2. This proves that E 20,n (n)  E ∞ 0,n (n). To complete the proof it is sufficient to prove that the group Hn (GLn−1 , k) is a summand of E 20,n (n). To prove this it is sufficient to define a map ϕ : Hn (R ∗ × GLn−1 , k) → Hn (GLn−1 , k) n such that d 11,n (Hn (R ∗ 2 × GLn−2 , k)) ⊆ ker(ϕ). Let Hn (R ∗ × GLn , k) = i=0 Si , where Si = Hi (R ∗ , k) ⊗ Hn−i (GLn−1 , k). For 2 ≤ i ≤ n, the stability theorem gives the isomorphisms Hi (R ∗ , k)⊗Hn−i (GLn−2 , k)  Si . Define ϕ : S0 → Hn (GLn−1 , k) the identity map and ϕ : Si  Hi (R ∗ , k) ⊗ Hn−i (GLn−2 , k) → Hn (GLn−1 , k), for 2 ≤ i ≤ n, the shuffle product. To complete the definition of ϕ we must define it on S1 . By a theorem of Suslin [18, 3.4] and [10, Theorem 2] and the assumption, we have the decomposition M (R) ⊗ k. Hn−1 (GLn−1 , k)  Hn−1 (GLn−2 , k) ⊕ K n−1 M (R) ⊗ k. Now we define So S1  H1 (R ∗ , k) ⊗ Hn−1 (GLn−2 , k) ⊕ H1 (R ∗ , k) ⊗ K n−1 ∗ the map ϕ : H1 (R , k) ⊗ Hn−1 (GLn−2 , k) → Hn (GLn−1 , k) the shuffle product, M (R) ⊗ k → H (GL ϕ : H1 (R ∗ , k) ⊗ K n−1 n n−1 , k) the composite map f

M (R) ⊗ k → H1 (R ∗ , k) ⊗ Hn−1 (GLn−1 , k) H1 (R ∗ , k) ⊗ K n−1 g

h

→ Hn (R ∗ × GLn−1 , k) → Hn (GLn−1 , k),

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1 where f = n−1 (id ⊗ νn−1 ), g is the shuffle product and h is induced by the map ∗ R × GLn−1 → GLn−1 , diag(a, A) → a A. By the Künneth theorem we have the decomposition

T0 T1 T2 T3 T4

= Hn (GLn−2 , k), n = i=1 Hi (R1∗ , k) ⊗ Hn−i (GLn−2 , k), n = i=1 Hi (R2∗ , k) ⊗ Hn−i (GLn−2 , k), = H1 (R1∗ , k) ⊗ H1 (R2∗ , k) ⊗ Hn−2 (GLn−2 , k), = i+ j≥3 Hi (R1∗ , k) ⊗ H j (R2∗ , k) ⊗ Hn−i− j (GLn−2 , k). i, j=0

By Lemma 3, T3 = T3 ⊕ T3

, where T3 = H1 (R1∗ , k) ⊗ H1 (R2∗ , k) ⊗ Hn−2 (GLn−3 , k),

M T3

= H1 (R1∗ , k) ⊗ H1 (R2∗ , k) ⊗ K n−2 (R) ⊗ k.

It is not difficult to see that d 11,n (T0 ⊕ T1 ⊕ T2 ⊕ T3 ⊕ T4 ) ⊆ ker(ϕ). Here one should use the stability theorem. To prove d 11,n (T3

) ⊆ ker(ϕ) we apply Proposition 3; d 1,n (a ⊗ b ⊗ {c1 , . . . , cn−2 }) 1

(−1)n−3 (b ⊗ c(diag(a, In−2 ), diag(1, C1,n−2 ), . . . , diag(1, Cn−2,n−2 )) (n − 3)! + a ⊗ c(diag(b, In−2 ), diag(1, C1,n−2 ), . . . , diag(1, Cn−2,n−2 ))) 1 (b ⊗ [c1 , . . . , cn−2 , a] + a ⊗ [c1 , . . . , cn−2 , b], = (n − 2)! − b ⊗ c(diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(a In−2 , 1)) =−

− a ⊗ c(diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(bIn−2 , 1))). Therefore d 11,n (a ⊗ b ⊗ {c1 , . . . , cn−2 }) = (x1 , x2 ) ∈ T3 ⊕ T3

, where 1 (b ⊗ c(diag(C1,n−2 ), . . . , diag(Cn−2,n−2 ), diag(a In−2 )) (n − 2)! + a ⊗ c(diag(C1,n−2 ), . . . , diag(Cn−2,n−2 ), diag(bIn−2 ))), x2 = (−1)n−2 (b ⊗ {c1 , . . . , cn−2 , a} + a ⊗ {c1 , . . . , cn−2 , b}). x1 = −

1 y, where We have φ(x1 ) = − (n−2)!

y = + c(diag(b, In−2 ), diag(1, C1,n−2 ), . . . , diag(1, Cn−2,n−2 ), diag(1, a In−2 )) + c(diag(a, In−2 ), diag(1, C1,n−2 ), . . . , diag(1, Cn−2,n−2 ), diag(1, bIn−2 ))

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and φ(x2 ) =

(−1)n−2 (−1)n−2 n−1 (n−2)! z

=

1 (n−1)! z,

where

z = c(diag(bIn−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(a In−2 , a −(n−2) )) + c(diag(a In−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(bIn−2 , b−(n−2) )) = c(diag(bIn−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(a In−2 , a)) + c(diag(bIn−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , a −(n−1) )) + c(diag(a In−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(bIn−2 , b)) + c(diag(a In−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , b−(n−1) )). Hence φ(x2 ) =

−1

(n−2)! z ,

where

z = c(diag(bIn−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , a)) + c(diag(a In−1 ), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , b)) = c(diag(bIn−2 , 1), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , a)) + c(diag(In−2 , b), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , a)) + c(diag(a In−2 , 1), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , b)) + c(diag(In−2 , a), diag(C1,n−2 , 1), . . . , diag(Cn−2,n−2 , 1), diag(In−2 , b)) = −y. Therefore ϕ(x2 ) = fact that d 11,n (Hn (R

−1

(n−2)! z ∗2

−1 = − (n−2)! y = −ϕ(x1 ). This completes the proof of the

× GLn−2 , k)) ⊆ ker(ϕ).



Thus it is reasonable to conjecture Conjecture 2 Let (n − 1)! ∈ k ∗ and let n ≥ 3. Then (n)

(n)

β2

β1

Hn (R ∗ 2 × GLn−2 , k) −→ Hn (R ∗ × GLn−1 , k) −→ Hn (GLn , k) → 0 is exact. Corollary 2 Let (n − 1)! ∈ k ∗ . If Conjecture 2 is true for all n ≥ 3, then Hn (inc) : Hn (GLn−1 , k) → Hn (GLn , k) is injective for all n. In particular if k = Q, then Conjecture 2 implies Conjecture 1. Proof This follows immediately from Proposition 4. (n)



Remark 4 (i) The surjectivity of β1 is already proven by Suslin [18] for fields and Guin [10, Proposition 3.1.1] for commutative rings with many units.

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(ii) Conjecture 2 is proven for n = 3 in [12, Corollary 3.5] and we prove it in this paper for n = 4. (iii) A similar result is not true for n = 2, that is (2)

β2

(2)

β1

H2 (R ∗ 2 × GL0 ) −→ H2 (R ∗ × GL1 ) −→ H2 (GL2 ) → 0 is not exact in general. In fact if R = F is an infinite field then (2)

(2)

ker(β1 )/im(β2 )  x ∧ (x − 1) − x ⊗ (x − 1) : x ∈ F ∗  is a subset of H2 (F ∗ )⊕(F ∗ ⊗ F ∗ )σ , where (F ∗ ⊗ F ∗ )σ = (F ∗ ⊗ F ∗ )/a ⊗b +b ⊗a : a, b ∈ F ∗ . To prove this let Q(F) be the free abelian group with the basis {[x] : x ∈ F ∗ − {1}}. Denote by p(F) the factor group of Q(F) by the subgroup generated by the elements of the form [x]−[y]+[y/x]−[(1−x −1 )/(1− y −1 )]+[(1−x)/(1− y)]. The homomorphism ψ : Q(F) → F ∗ ⊗ F ∗ , [x] → x ⊗(x −1) induces a homomorphism 2 (2)  p(F). It is p(F) → (F ∗ ⊗ F ∗ )σ , [19, Lemma 1.1]. By [19, Lemma 2.2], E 4,0 not difficult to see that the E 2p,q (2)-terms have the following form ∗ ∗ 2 (2) ∗ 0 E 1,2 0 0 0 0 ∗ 0 0 0 0 p(F) ∗ . 2 (2) ⊆ H (F ∗ )⊕(F ∗ ⊗ F ∗ ) . By [19, Lemma 2.4] An easy calculation shows that E 1,2 2 σ 3 (2) : E 3 (2) → E 3 (2)  E 2 (2) is defined by d 3 (2)([x]) = x ∧ (x − 1) − d4,0 4,0 1,2 1,2 4,0 3 (2) is x ⊗ (x − 1). Because the spectral sequence converges to zero we see that d4,0 2 (2) is generated by elements of the form x ∧(x −1)−x ⊗(x −1) ∈ surjective and so E 1,2 ∗ ∗ H2 (F ) ⊕ (F ⊗ F ∗ )σ .

6 Homology of GL4 The following lemma is useful when we work with generators of Ci (R n ) and Di (R n ) for i ≥ n. Lemma 4 Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ R n . (i) Any subset of {e1 , . . . , en , x} with m-elements, m ≤ n, is a basis of a free summand of R n if and only if xi ∈ R ∗ for all i = 1, . . . , n. (ii) Any subset of {e1 , . . . , en , x, y} with m-elements, m ≤ n, is a basis of a free summand of R n if and only if for all i = 1, . . . , n, xi ∈ R ∗ , yi = xi z i , where z i ∈ R ∗ and for all i = j, z i − z j ∈ R ∗ . Proof See [10, Lemma 3.3.3].



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Lemma 5 E˜ 2p,0 (4) is trivial for 0 ≤ p ≤ 6. 2 is the most difficult one, which we prove here. The rest is Proof The triviality of E˜ 6,0 much easier and we leave it to the reader. 2 . The proof is in four steps; Triviality of E˜ 6,0 Step 1 The sequence 0 → C∗ (R 4 )GL4 → D∗ (R 4 )GL4 → Q ∗ (R 4 )GL4 → 0 is exact, where Q ∗ (R 4 ) := D∗ (R 4 )/C∗ (R 4 ). Step 2 The group H5 (Q ∗ (R 4 )GL4 ) is trivial. Step 3 The map induced in homology by C∗ (R 4 )GL4 → D∗ (R 4 )GL4 is zero in degree 5. 2 is trivial. Step 4 The group E˜ 6,0 Proof of step 1 For i ≥ −1, Di (R 4 )  Ci (R 4 ) ⊕ Q i (R 4 ). This decomposition is compatible with the action of GL4 , so we get an exact sequence of GL4 -modules

0 → Ci (R 4 ) → Di (R 4 ) → Q i (R 4 ) → 0 which splits as a sequence of GL4 -modules. One can easily deduce the desired exact sequence from this. Note that this exact sequence does not split as complexes. Proof of step 2 The complex Q ∗ (R 4 ) induces a spectral sequence Eˆ 1p,q =



0 Hq (GL4 , Q p−1 (R 4 ))

if 0 ≤ p ≤ 2 if p ≥ 3,

2 = 0 and which converges to zero. To prove the claim it is sufficient to prove that Eˆ 6,0 2 2 to prove this it is sufficient to prove that Eˆ 4,1 = Eˆ 3,2 = 0. Let wi ∈ D2 (R 4 ), u i ∈ D3 (R 4 ) and vi ∈ D4 (R 4 ), where

w1 = (e1 , e2 , e3 ), w2 = (e1 , e2 , e1 + e2 ), and u 1 = (e1 , e2 , e3 , e4 ), u 3 = (e1 , e2 , e3 , e1 + e2 ),

u 2 = (e1 , e2 , e3 , e1 + e2 + e3 ), u 4 = (e1 , e2 , e3 , e2 + e3 ),

u 5 = (e1 , e2 , e3 , e1 + e3 ), u 6 = (e1 , e2 , e1 + e2 , e3 ), u 7,a = (e1 , e2 , e1 + e2 , e1 + ae2 ), a, a − 1 ∈ R ∗ , and, fix d ∈ R such that d, d − 1 ∈ R ∗ , v1 i, j v2

= (e1 , e2 , e1 + e2 , e3 , e1 + de2 ), = (e1 , e2 , e3 , e4 , ei + e j ), 1 ≤ i < j ≤ 4,

v3,a = (e1 , e2 , e3 , e1 + e2 , e1 + ae2 ), a, a − 1 ∈ R ∗ v4 = (e1 , e2 , e3 , e1 + e2 + e3 , e4 ),

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Then one can see that 1 Eˆ 3,q = Hq (StabGL4 (w2 )) 1 = Hq (StabGL4 (u 2 )) ⊕ · · · ⊕ Hq (StabGL4 (u 7,a )), Eˆ 4,q 1 = Hq (StabGL4 (v1 )) ⊕ · · · ⊕ Hq (StabGL4 (v4 )) ⊕ T. Eˆ 5,q 1 | 2 It is easy to see that dˆ4,2 H2 (StabGL4 (u 6 )) = H2 (inc), so it is surjective. Thus Eˆ 3,2 = 0. We simplify these groups using Proposition 1 but use the same notations. Easy calculation shows that 1 1 dˆ5,1 : H1 (StabGL4 (v1 )) → Eˆ 4,1 , y → (0, y, 0, 0, y, −y),

 1,3 1 1 → Eˆ 4,1 , y → (0, ∗, 0, y, 0, y), dˆ5,1 : H1 StabGL4 v2

(1) (2)

1 1 : H1 (StabGL4 (v3,a )) → Eˆ 4,1 , y → (0, 0, 0, 0, 0, y), (3) dˆ5,1

 2,3 1 1











→ Eˆ 4,1 , (a , b , b , c ) → (0, (b , b , c , a ), ∗, 0, 0, 0), dˆ5,1 : H1 StabGL4 v2 1 1 dˆ5,1 : H1 (StabGL4 (v4 )) → Eˆ 4,1 , y → (y, 0, 0, 0, 0, 0),

 3,4 1 1 ˆ ˆ → E 4,1 , (0, a , 0, 0) → (0, (a , 0, 0, −a ), 0, 0, 0, 0). d5,1 : H1 StabGL4 v2

(4) (5) (6)

Here we assume that a, b, . . . ∈ R ∗ , a , b , . . . ∈ H1 (R ∗ ) and under the isomorphism R ∗ → H1 (R ∗ ) we have a → a . 1 ). By (1), (2), (3), (4) and (5) we may assume Let x = (x2 , x3 , . . . , x7,a ) ∈ ker(dˆ4,1 = 0, x2 = 0, respectively. If x4 = (a , b , b , c ), x6 = 0, x5 = 0,x7,a =0, x3 1 (x ) = b , b , a 0 = 0. Hence b = 0 and c = −a . Now by (6), then dˆ4,1 4 0c 1 ((0, a , 0, 0)) = (0, 0, x , 0, 0, 0) = x. Therefore Eˆ 2 = 0. dˆ5,1 4 4,1 Proof of step 3 Here all the calculation takes place in C∗ (R 4 )GL4 and D∗ (R 4 )GL4 . For simplicity the image of v ∈ C∗ (R 4 ) in C∗ (R 4 )GL4 again is denoted by v. Consider the following commutative diagram C6 (R⏐4 )GL4 → C5 (R⏐4 )GL4 → C4 (R⏐4 )GL4 ⏐ ⏐ ⏐ 4 4 D6 (R )GL4 → D5 (R )GL4 → D4 (R 4 )GL4 . The generators of C5 (R 4 )GL4 are of the form
 xa,b,c = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + ae2 + be3 + ce4  , where a, a − 1, b, b − 1, c, c − 1, a − b, a − c, b − c ∈ R ∗ . Since C4 (R 4 )GL4 = Z, xa,b,c ∈ ker(∂5 ) and the elements of this form generate ker(∂5 ). Hence to prove this step it is sufficient to prove that xa,b,c ∈ im(∂˜6 ). Let X a,b,c be
 e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + ae2 + be3 + ce4 , e1 + e2  ,

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where a, b, c are as above. Then ∂˜6 (X a,b,c ) = v 1−b ,1−b − v a−b ,a−b + u 1−b − u 1−c + u 1 + xa,b,c , 1−c

a−c

a−c

a−b

a

where
 vg,h = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e2 + ge3 + he4  ,
 u l = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + le2 
 − e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + e2  , g, g −1, h, h −1, g −h, l ∈ R ∗ . So it is sufficient to prove that the elements vg,h −v p,q and u l are in the image if ∂˜6 . Let
 Ul = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + e2 , e1 + le2  ,
 Vl = e1 , e2 , e3 , e4 , e1 + e2 , e1 + le2 , e3 + e4  , where l, l − 1 ∈ R ∗ . Then ∂˜6 (Vl − Ul ) = u l . Set
 Tg,h = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e2 + ge3 + he4 , e2 + e3  , where g, g − 1, h, h − 1, g − h ∈ R ∗ . Then ∂˜6 (Tg,h − T p,q ) = −s

1 1−h

+s

g g−h

+s

1 1−q

−s

p p−q

− z 1−h + z 1−q g−h

p−q

+ y 1 − y 1 + vg,h − v p,q , g

p

where
 sa = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + ae3 + e4  ,
 z a = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e2 + ae3 
 − e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e2 + e3  ,
 ya = e1 , e2 , e3 , e4 , e1 + e2 + e3 , e1 + ae2 
 + e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + ae3 
 − e1 , e2 , e3 , e4 , e1 + e2 + e3 , e1 + e2  
− e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + e3  . So to prove that vg,h − v p,q ∈ im(∂˜6 ) it is sufficient to prove that sa − sb , z a , ya ∈ im(∂˜6 ), a, a − 1, b, b − 1, a − b ∈ R ∗ . Set
 Ya = e1 , e2 , e3 , e4 , e1 + e2 + e3 , e1 + ae2 , e1 + e2  ,
 Ya = e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + ae3 , e2 + e3  .

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Then ya = ∂˜6 (Ya + Ya ) − 2∂˜6 (V 1 ). To prove that z a ∈ im(∂˜6 ), set a


 Z a = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e2 + e3 , e2 + ae3  . By an easy calculation z a = ∂˜6 (Va − Z a ). If
 Sa = e1 , e2 , e3 , e4 , e1 + e2 + e3 + e4 , e1 + ae3 + e4 , e1 + e4  , then ∂˜6 (Sa − Sb ) = R

1 1 1−a , 1−b

+ sa − sb , where a, a − 1, b, b − 1, a − b ∈ R ∗ and


 Ra,b = e1 , e2 , e3 , e4 , e1 + e2 + e4 , e1 + ae2 + e4 
 − e1 , e2 , e3 , e4 , e1 + e2 + e4 , e1 + be2 + e4 
 − e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + ae3 + e4 
 + e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + be3 + e4  . Thus to prove that sa − sb ∈ im(∂˜6 ) it is sufficient to prove that Ra,b ∈ im(∂˜6 ). Let
 Q a = e1 , e2 , e3 , e4 , e1 + e2 + e4 , e1 + ae2 + e4 , e1 + e4 
 − e1 , e2 , e3 , e4 , e2 + e3 + e4 , e2 + ae3 + e4 , e1 + e4  ,
 Na,b = e1 , e2 , e3 , e4 , e1 + e4 , e1 + ae4 
 − e1 , e2 , e3 , e4 , e1 + e4 , e1 + be4  ,
 Pa = e1 , e2 , e3 , e4 , e3 + e4 , ae3 + e4  , where a, a − 1, b, b − 1 ∈ R ∗ . Then ∂˜6 (Q a − Q b ) = N

1 1 1−a , 1−b

+P

1 1−a

+P

1 1−b

+ Ra,b .

If
 Oa,b = e1 , e2 , e3 , e4 , e1 + e4 , e1 + ae4 , e2 + e3 
 − e1 , e2 , e3 , e4 , e1 + e4 , e1 + be4 , e2 + e3  ,
 Ma = e1 , e2 , e3 , e4 , e3 + e4 , ae3 + e4 , e1 + e2  , where a, a − 1, b, b − 1, a − b ∈ R ∗ , then ∂˜6 (Oa,b ) = Na,b and ∂˜6 (Ma ) = Pa . This completes the proof of step 3. Proof of Step 4 Writing the homological long exact sequence of the short exact sequence obtained in the first step, we get the exact sequence H5 (C∗ (R 4 )GL4 ) → H5 (D∗ (R 4 )GL4 ) → H5 (Q ∗ (R 4 )GL4 ). 2 By steps 2 and 3, H5 (D∗ (R 4 )GL4 ) = 0. So E˜ 6,0 = H5 (D∗ (R 4 )GL4 ) = 0. This 2 . 

completes the proof of the triviality of E˜ 6,0

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Lemma 6 E˜ 2p,1 (4) is trivial for 0 ≤ p ≤ 5. Lemma 7 E˜ 2p,2 (4) is trivial for 0 ≤ p ≤ 4. Lemma 8 E˜ 2p,3 (4) is trivial for 0 ≤ p ≤ 3. The proofs of these lemmas are lengthy calculation, without using any difficult techniques. For a proof, when R is a field, we refer the reader to the preprint version of this paper [13, Sect. 6]. The same proofs work over commutative rings with many units. The only modification that we need is to replace conditions like a ∈ F ∗ − {1}, a, b ∈ F ∗ , a = b, with a, a − 1 ∈ R ∗ , a, b, a − b ∈ R ∗ , etc. Remark 5 (i) Lemma 5 gives a positive answer to a question asked by Dupont for n = 3 (see [16, 4.12]). (ii) When k = Z/ p, p a prime, and F an algebraically closed field, then the proofs of Lemmas 6, 7 and 8 are easy, some even trivial as for i ≥ 0, H2i+1 (F ∗ , Z/ p) = 0, for example Lemmas 6 and 8 immediately follow. Theorem 2 The complex (4)

(4)

β2

β1

H4 (R ∗ 2 × GL2 ) → H4 (R ∗ × GL3 ) → H4 (GL4 ) → 0 is exact. Proof This follows from Lemmas 5, 6, 7 and 8 and the fact that the spectral sequence converges to zero. 

H4 (inc)

Theorem 3 (i) If char(k) = 2, 3, then H4 (GL3 , k) −→ H4 (GL4 , k) is injective. (ii) If F is an algebraically closed field, then H4 (inc)

H4 (GL3 (F)) −→ H4 (GL4 (F)) is injective. Proof (i) This follows from Theorem 2, [12, Theorem 5.4] and Proposition 4. (ii) Since F is an algebraically closed field, H2 (F ∗ ) and K 2M (F) are uniquely diZ ∗ M ∗ ∗ visible, so Tor Z 1 (F , H2 (F )) = Tor 1 (F , K 2 (F)) = 0. Also from H2 (GL3 (F))  H2 (GL2 (F)) = H2 (GL1 (F)) ⊕ K 2M (F) ∗ one sees that Tor Z 1 (F , H2 (GL3 (F))) = 0. Therefore by the Künneth theorem

H4 (F ∗ × GL3 (F))  H4 (F ∗ 2 × GL2 (F)) 

123

4 i=0



Hi (F ∗ ) ⊗ H4−i (GL3 (F)),

0≤i+ j≤4

Hi (F1∗ ) ⊗ H j (F2∗ ) ⊗ H4−i− j (GL2 (F)).

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181

Now the proof of the claim is similar to the proof of Proposition 4, using the fact that H2 (inc)

H2 (GL1 (F)) −→ H2 (GL2 (F)),

H3 (inc)

H3 (GL2 (F)) −→ H3 (GL3 (F))

are injective [12, Theorem 5.4]. The only place where we need a modification is the definition of the map f . First we define

 ν 3 : K 3M (F) → H3 (GL3 (F)), {a, b, c} → a 1/3 , b, c . This map is well defined as K 3M (F) is uniquely divisible. Set f := id⊗ν 3 : H1 (F ∗ )⊗ K 3M (F) → H1 (F ∗ ) ⊗ H3 (GL3 (F)). The rest of the proof can be done similar to the proof of Proposition 4. We leave the detail to the reader. 

Example 2 H4 (inc) : H4 (GL3 (R), Z[ 21 ]) → H4 (GL4 (R), Z[ 21 ]) is injective. It is well-known that K 3M (R) = {−1, −1, −1} ⊕ V , where {−1, −1, −1} is a group of order 2 generated by {−1, −1, −1} and V is a uniquely divisible group. So the proof of our claim is similar to the proof of Theorem 3. (We invert 2 in the coefficients ring in order to eliminate the 2-torsion elements that appear in the decomposition of H4 (R∗ × GL3 (R)) and H4 (R∗ 2 × GL2 (R)). This might not be necessary.) Corollary 3

(i) If char(k) = 2, 3, then 0 → H4 (GL3 , k) → H4 (GL4 , k) → K 4M (R) ⊗ k → 0.

is split exact. (ii) If F is an algebraically closed field, then 0 → H4 (GL3 (F)) → H4 (GL4 (F)) → K 4M (F) → 0. is split exact. Proof The part (i) follows from Theorem 3 and Suslin’s exact sequence (see the introduction). For part (ii) we also need to know that K 4M (F) is uniquely divisible. It is easy to see that a splitting map can be defined by 1 {a, b, c, d} → − [a, b, c, d], {a, b, c, d} → [a −1/6 , b, c, d], 6 

respectively. Proposition 5

(i) If char(k) = 2, 3, then 0 → H4 (SL3 , k) R ∗ → H4 (SL, k) → K 4M (R) ⊗ k → 0.

is split exact.

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(ii) If F is an algebraically closed field, then 0 → H4 (SL3 (F)) → H4 (SL4 (F)) → K 4M (F) → 0. is split exact. Proof The first part follows from Corollary 3, Theorem 1 and Corollary 1. For the proof of (ii), as in the proof of Corollary 1, it is sufficient to prove that the canonical map H4 (SL3 (F)) → H4 (GL3 (F)) is injective. For this we look at the Lyndon– Hochschild–Serre spectral sequence E 2p,q = H p (GL3 (F), Hq (µ3,F )) ⇒ H p+q (F ∗ × SL3 (F)) γ3

obtained from the extension 1 → µ3,F → F ∗×SL3 → GL3 → 0, where γ3 (a, A) := a A. It is well-known that Hi (F ∗ ), K iM (F) and K i (F) are divisible [8, Proposition 4.7], [3, Proposition 1.2], [17]. The Hurewicz map K i (F) → Hi (SL(F)), i ≥ 2, is an isomorphism for i = 2, 3 (see [19, Corollary 5.2] or [16, Proposition 2.5]). Since BGL(F)+ ∼ B F ∗ × BSL(F)+ [19, Lemma 5.3], one can see that Hi (GL3 (F)) is divisible for 1 ≤ i ≤ 3. Therefore the E 2 -terms of the spectral sequence are of the following form Z/3Z 0 0 0 0 Z/3Z 0 Z/3Z ∗ 0 0 0 0 Z/3Z 0 Z/3Z 0 ∗ Z H1 (GL3 ) H2 (GL3 ) H3 (GL3 ) H4 (GL3 ). An easy analysis of this spectral sequence shows that H3 (γ3 ) : H3 (F ∗ × SL3 (F)) → H3 (GL3 (F)) is surjective with the kernel of order dividing 9. By the Künneth theorem H3 (F ∗ × SL3 (F))  H3 (SL3 (F)) ⊕ F ∗ ⊗ H2 (SL3 (F)) ⊕ H3 (F ∗ ). Let ξ ∈ F ∗ be the third root of unity, that is ξ 3 = 1 and ξ = 1. If we use the bar resolution C∗ (G) to define the homology of a group G (see [5, p. 36]), one can see that χ (ξ ) := [ξ |ξ |ξ ] + [ξ |ξ 2 |ξ ] ∈ H3 (F ∗ ) has order 3. In fact in C∗ (G)G (G = F ∗ ) ∂4 ([ξ |ξ |ξ |ξ ] + [ξ |ξ 2 |ξ |ξ ] + [ξ |ξ |ξ |ξ 2 ] + [ξ 2 |ξ |ξ 2 |ξ ]) = 3χ (ξ ). In a similar way we can define χ (ξ I3 ) ∈ H3 (SL3 (F)) and χ (ξ I3 ) ∈ H3 (GL3 (F)). Now it is easy to see that H3 (γ3 )(χ (ξ I3 ), 0, 2χ (ξ )) = 0. Therefore the kernel of 2 or d 4 is trivial. In either case this implies that H3 (γ3 ) is not trivial. Thus d4,0 4,0 H4 (F ∗ × SL3 (F)) → H4 (GL3 (F)). is injective. Therefore H4 (SL3 (F)) → H4 (GL3 (F)) is injective.

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To give a splitting map one should note that in H4 (GL4 (F)), [a, b, c, d] is equal to c(diag(a, 1, a −1 , 1), diag(b, b−1 , 1, 1), diag(c, 1, c−1 , 1), diag(d, d, d, d −3 )). We name this new version of [a, b, c, d] by [[a, b, c, d]]. Then [[a, b, c, d]] ∈ H4 (SL4 (F)) and a splitting map can be defined by 1 {a, b, c, d} → − [[a, b, c, d]], {a, b, c, d} → [[a −1/6 , b, c, d]], 6 

respectively.

Corollary 4 (i) If char(k) = 2, 3, then K 4 (R)ind ⊗ k embeds in the group H4 (SL3 , k) R ∗ . (ii) If F is an algebraically closed field, then K 4 (F)ind embeds in H4 (SL3 (F)). Proof (i) This follows from Theorem 3 and Corollary 1. Part (ii) can be proven in a similar way using the facts that the Hurewicz map K 4 (F) → H4 (SL4 (F)) is injective [1, Theorem 7.23] and 0 → K 4M (F) → K 4 (F)→K 4 (F)ind → 0 is split exact, for K 4M (F) [3, Proposition 1.2] and K 4 (F) [17] are uniquely divisible. 

Remark 6 For n ≥ 5 the terms of the spectral sequence E˜ 1p,q (n) become very large and a computation similar to the one done in this section seems to be hopeless. We think that working with a third spectral sequence E

1p,q (n) [14], with a very close connection to E 1p,q (n), might be more fruitful. In studying this new spectral sequence one can see that Conjectures 1 and 2, at least over a field F, should be connected to certain “higher pre-Bloch groups” pn (F). On the other hand these higher pre-Bloch groups are connected to certain homology groups H1 (GLn (F), Hn−1 (X n , Z)). We believe that the Conjectures 1, 2 should follow from a conjecture about these groups [14, Conjecture 5.3, Proposition 5.1]. For example when F is algebraically closed and k = Z/ p, p a prime, the injectivity of the map Hn (GLn−1 (F), Z/ p) → Hn (GLn (F), Z/ p) is almost equivalent to a certain p-divisibility property of pn (F) [14, Theorem 3.4]. We refer the reader to [14] for more information in this direction. We believe that the proof of the above conjectures, for all n, should be by induction on n and because of Remark 4(iii) one needs to start the induction from n = 3 and n = 4. These cases are treated in [12] for n = 3 and in this article for n = 4. We gave up the hope of proving these cases directly using E

1p,q (3), E

1p,q (4), since we do not know how to compute some of the higher differentials of these spectral sequences.

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Acknowledgments Part of this work is done during my stay at the Max-Planck Institute in Bonn. I would like to thank them for their support and hospitality. I would like to thank W. van der Kallen for his interest in this work and for his valuable comments.

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