ON SYMMETRIC COMMUTATOR SUBGROUPS, BRAIDS, LINKS AND HOMOTOPY GROUPS

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1 ON SYMMETRIC COMMUTATOR SUBGROUPS, BRAIDS, LINKS AND HOMOTOPY GROUPS J. Y. LI AND J. WU Abstract. In this paper, we investigate some applications of commutator subgroups to homotopy groups and geometric groups. In particular, we show that the intersection subgroups of some canonical subgroups in certain link groups modulo their symmetric commutator subgroups are isomorphic to the (higher) homotopy groups. This gives a connection between links and homotopy groups. Similar results hold for braid and surface groups. 1. Introduction The purpose of this article is to investigate some applications of commutator subgroups to homotopy groups and geometric groups. Recall [16, p ] that a bracket arrangement of weight n in a group G is a map β n : G n G which is defined inductively as follows: β 1 = id G, β 2 (a 1, a 2 ) = [a 1, a 2 ] for any a 1, a 2 G, where [a 1, a 2 ] = a 1 1 a 1 2 a 1a 2. Suppose that the bracket arrangements of weight k are defined for 1 k < n with n 3. A map β n : G n G is called a bracket arrangement of weight n if β n is the composite G n = G k G n k βk β n k G G β 2 G for some bracket arrangements β k and β n k of weight k and n k, respectively, with 1 k < n. For instance, if n = 3, there are two bracket arrangements given by [[a 1, a 2 ], a 3 ] and [a 1, [a 2, a 3 ]]. Let R j be a sequence of subgroups of G for 1 j n. The fat commutator subgroup [[R 1, R 2,..., R n ]] is defined to be the subgroup of G generated by all of the commutators (1.1) β t (g i1,..., g it ), where 1) 1 i s n; 2) {i 1,..., i t } = {1,..., n}, that is each integer in {1, 2,, n} appears as at least one of the integers i s ; 3) g j R j ; 4) β t runs over all of the bracket arrangements of weight t (with t n) Mathematics Subject Classification. Primary 55Q40,20F12; Secondary 20F36, 57M25. Key words and phrases. symmetric commutator subgroup, homotopy group, link group, Brunnian braid, free group, surface group. Partially supported by the National Natural Science Foundation of China Partially supported by the Academic Research Fund of the National University of Singapore R

2 2 J. Y. LI AND J. WU For convenience, let [[R 1 ]] = R 1. The fat commutator subgroups have important applications in homotopy theory [1, 3, 6, 19]. Let G = F n be the free group of rank n with a basis x 1,..., x n. Let R i = x i be the normal closure of x i in F n for 1 i n and let R n+1 = x 1 x n be the normal closure of the product element x 1 x 2 x n in F n. According to [19, Theorem 1.7], the homotopy group π n+1 (S 2 ) is isomorphic to the quotient group (R 1 R 2 R n+1 )/[[R 1, R 2,..., R n+1 ]] for each n. To understand the fat commutator subgroup, we consider the symmetric commutator subgroup 1 [[R 1, R 2 ],..., R n ] S defined by [[R 1, R 2 ],..., R n ] S = σ Σ n [[R σ(1), R σ(2) ],..., R σ(n) ], where [[R σ(1), R σ(2) ],..., R σ(n) ] is the subgroup generated by the left iterated commutators [[[g 1, g 2 ], g 3 ],..., g n ] with g i R σ(i). For convenience, let [R 1 ] S = R 1. From the definition, the symmetric commutator subgroup is a subgroup of the fat commutator subgroup. Our first result states that the fat commutator subgroup is in fact the same as the symmetric commutator subgroup. Theorem 1.1. Let R j be any normal subgroup of a group G with 1 j n. Then [[R 1, R 2,..., R n ]] = [[R 1, R 2 ],..., R n ] S. The symmetric commutator subgroup can be simplified as follows: Theorem 1.2. Let R j be any normal subgroup of a group G with 1 j n. Then [[R 1, R 2 ],..., R n ] S = [[R 1, R σ(2) ],..., R σ(n) ], σ Σ n 1 where Σ n 1 acts on {2, 3,..., n}. Our next step is to give a generalization of [19, Theorem 1.7]. This will give more connections between homotopy groups and symmetric commutator subgroups. Let (X, A) be a pair of spaces. An n-partition of X relative to A means a sequence of subspaces (A 1,..., A n ) of X such that (1). A = A i A j for each 1 i < j n and (2). X = n A i. An n-partition (A 1,..., A n ) of X relative to A is called cofibrant if the inclusion i I A i j J is a cofibration for any I J {1, 2,..., n}. Note that for a cofibrant partition, each union A I = i I A i is the homotopy colimit of the diagram given by the inclusions A I A I for I I I. A j 1 Symmetric commutator subgroup was named by Roman Mikhailov during the communications.

3 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 3 Theorem 1.3. Let (X, A) be a pair of spaces and let (A 1,..., A n ) be a cofibrant n-partition of X relative to A with n 2. Suppose that i) For any proper subset I = {i 1,..., i k } {1, 2,..., n}, the union i I A i is a path-connected K(π, 1)-space. ii) The inclusion A A i induces an epimorphism of the corresponding fundamental groups for each 1 i n. Let R i be the kernel of π 1 (A) π 1 (A i ) for 1 i n. Then (1). For any proper subset I = {i 1,..., i k } {1, 2,..., n}, R i1 R ik = [[R i1, R i2 ],..., R ik ] S. (2). For any 1 < k n and any subset I = {i 1,..., i k } {1, 2,..., n}, there is an isomorphism of groups π k (X) k = R is / [[R, R ],..., R ] i1 i2 ik S, j J R j where J = {1, 2,..., n} I. In particular, π n (X) = (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S. For the special case where n = 2, Theorem 1.3 is the classical Brown-Loday Theorem [3, Corollary 3.4] with a generalization recently given in [6]. But the connectivity hypothesis on the subgroups in [6] seems difficult to check. Theorem 1.3 emphasizes the point that, for any space X that admits a cofibrant K(π, 1)-partition, the higher homotopy groups of X measure the difference between the intersection subgroups and the symmetric commutator subgroups for certain subgroups in the fundamental groups of the partition spaces. Moreover Theorem 1.3 admits many applications. A direct consequence is to give an interesting connection between link groups and higher homotopy groups. Corollary 1.4. Let M be a path-connected 3-manifold and L be a proper m-link in M with m 2. Suppose that for any nonempty sub-link L of L, the link complement M L is a K(π, 1)-space. Let Λ 1, Λ 2,..., Λ n be any subsets of {1, 2,..., m}, with n 2, such that (i) Λ i for each 1 i n. (ii) Λ i Λ j = for i j. (iii) n Λ i = {1, 2,..., m}. Let α j be the j th meridian of the link L and let R i = α j j Λ i be the normal closure of α j with j Λ i in π 1 (M L ). Then (1). For any proper subset I = {i 1,..., i k } {1, 2,..., n}, R i1 R ik = [[R i1, R i2 ],..., R ik ] S. (2). There is an isomorphism of groups (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S = πn (M). j J R j

4 4 J. Y. LI AND J. WU There are examples of links whose complements are K(π, 1)-space s. For instance, let M = S 3 and let π : S 3 S 2 be the Hopf fibration. Let q 1,..., q m be m distinct points in S 2. Let L = π 1 ({q 1, q 2,..., q m }). Then L is an m-link in S 3 with the property that S 3 L is a K(π, 1)-space for any non-empty sub-link L of L. Consider the special case where n = m with Λ i = {i} in Corollary 1.4. Then R i is the normal closure generated by the i th meridian in π 1 (M L ). Any element in the intersection subgroup R 1 R 2 R n can be represented by a 1-link l. The union L {l} gives an (n + 1)-link in M related to Brunnian links. Thus Corollary 1.4 gives a connection between higher homotopy groups and Brunnian links. For more applications of Theorem 1.3, we consider certain subgroups of surface groups and braid groups whose intersection subgroups modulo symmetric commutator subgroups are given by the homotopy groups. Our results on the braid groups are described as follows. Let M be a manifold. Recall that the m th ordered configuration space F (M, m) is defined by F (M, m) = {(z 1,..., z m ) M m z i z j for i j} with subspace topology. Recall that the Artin braid group B n is generated by σ 1,..., σ n 1 with defining relations given by σ i σ j = σ j σ i for i j 2 and σ i σ i+1 σ i = σ i+1 σ i σ i+1 for each i. The pure braid group P n is defined to be the kernel of the canonical quotient homomorphism from B n to the symmetric group Σ n, with a set of generators given by for 1 i < j n. Let A i,j = σ j 1 σ j 2 σ i+1 σ 2 i σ 1 i+1 σ 1 j 2 σ 1 j 1 A 0,j = (A j,j+1 A j,j+2 A j,n ) 1 (A 1,j A j 1,j ) 1 = (σ j σ j+1 σ n 2 σ 2 n 1σ n 2 σ j ) 1 (σ j 1 σ 2 σ 2 1σ 1 σ j 1 ) 1. Then we have the following: Theorem 1.5. Let Λ 1, Λ 2,..., Λ n be any subsets of {1, 2,..., m}, with n 2, such that (i) Λ i for each 1 i n. (ii) Λ i Λ j = for i j. (iii) n Λ i = {1, 2,..., m}. Let R i = A 0,j j Λ i be the normal closure in the pure braid group P m. Then (1). For any proper subset I = {i 1,..., i k } {1, 2,..., n}, R i1 R ik = [[R i1, R i2 ],..., R ik ] S. (2). There is an isomorphism of groups (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S = πn (F (S 2, m)). For instance, for n = 2 with m 3, (R 1 R 2 )/[R 1, R 2 ] = π 2 (F (S 2, m)) = 0, where π 2 (F (S 2, m)) = 0 for m 3 is given in [8, p. 244]. For m n 3, from [7, Theorem 1], we have π n (F (S 2, m)) = π n (S 2 ) and so (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S = πn (F (S 2, m)) = π n (S 2 ).

5 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 5 The braided descriptions of the homotopy groups π (S 2 ) have been studied in [1, 2, 4, 5, 15, 20]. Theorem 1.5 is a new braided description of the homotopy groups. The article is organized as follows. The proofs of Theorems 1.1 and 1.2 are given in section 2. In section 3, we provide the proof of Theorem 1.3. In section 4, we discuss some applications to the free groups and surface groups. The applications to the braid groups and the proof of Theorem 1.5 are given in section The Proofs of Theorems 1.1 and Some Lemmas. In this subsection, we give some useful lemmas on commutator subgroups. The following lemma is elementary. Lemma 2.1. Let G be a group and let A, B, C be normal subgroups of G. Then [AB, C] = [A, C][B, C]. The following classical theorem can be found in [16, Theorem 5.2, p.290]. Theorem 2.2 (Hall s Theorem). Let G be a group and let A, B, C be normal subgroups of G. Then any one of the subgroups [A, [B, C]], [[A, B], C] and [[A, C], B] is a subgroup of the product of the other two. Now let R 1,..., R n be subgroups of G. Recall that the symmetric commutator subgroup [[R 1, R 2 ],..., R n ] S is defined by [[R 1, R 2 ],..., R n ] S = σ Σ n [[R σ(1), R σ(2) ],..., R σ(n) ]. If each R i is a normal subgroup of G, then [[R σ(1), R σ(2) ],..., R σ(n) ] is normal in G and so is [[R 1, R 2 ],..., R n ] S. Lemma 2.3. Let R 1,..., R n be normal subgroups of G. Let g j R j for 1 j n. Then β n (g σ(1),..., g σ(n) ) [[R 1, R 2 ],..., R n ] S for any σ Σ n and any bracket arrangement β n of weight n. Proof. The proof is given by double induction. The first induction is on n. Clearly the assertion holds for n = 1. Suppose that the assertion holds for m with m < n. Given an element β n (g σ(1),..., g σ(n) ) as in the lemma. Then β n (g σ(1),..., g σ(n) ) = [β p (g σ(1),..., g σ(p) ), β n p (g σ(p+1),..., g σ(n) )] for some bracket arrangements β p and β n p with 1 p n 1. induction is on q = n p. If q = 1, we have β n 1 (g σ(1),..., g σ(n 1) ) [[R σ(1), R σ(2) ],..., R σ(n 1) ] S by the first induction and so with β n (g σ(1),..., g σ(n) ) = [β n 1 (g σ(1),..., g σ(n 1) ), g σ(n) ] [[[R σ(1), R σ(2) ],..., R σ(n 1) ] S, R σ(n) ] = by Lemma 2.1 ========== [[[R [ σ(1), R σ(2) ],..., R σ(n 1) ] S, R σ(n) ] ] [[R τ Σn 1 σ(τ(1)), R σ(τ(2)) ],..., R σ(τ(n 1)) ], R σ(n) τ Σ n 1 [[[R σ(τ(1)), R σ(τ(2)) ],..., R σ(τ(n 1)) ], R σ(n) ] [[R 1, R 2 ],..., R n ] S. The second

6 6 J. Y. LI AND J. WU Now suppose that the assertion holds for q = n p < q. By the first induction, we have and Thus β p (g σ(1),..., g σ(p) ) [[R σ(1), R σ(2) ],..., R σ(p) ] S β n p (g σ(p+1),..., g σ(n) ) [[R σ(p+1), R σ(p+2) ],..., R σ(n) ] S. β n (g σ(1),..., g σ(n) ) [ [[R σ(1), R σ(2) ],..., R σ(p) ] S, [[R σ(p+1), R σ(p+2) ],..., R σ(n) ] S ]. By Lemma 2.1, β n (g σ(1),..., g σ(n) ) lies in the product subgroup [ T = [[Rσ(τ(1)),..., R σ(τ(p)) ], [[R σ(ρ(p+1)),..., R σ(ρ(n)) ] ], τ Σ p ρ Σ n p where Σ n p acts on {p + 1,..., n}. By applying Hall s Theorem, we have [ [[Rσ(τ(1)),..., R σ(τ(p)) ], [[R σ(ρ(p+1)),..., R σ(ρ(n)) ] ] = [ [[R σ(τ(1)),..., R σ(τ(p)) ], [ [[R σ(ρ(p+1)),..., R σ(ρ(n 1)) ], R σ(ρ(n)) ]] [[ [[R σ(τ(1)),..., R σ(τ(p)) ], [[R σ(ρ(p+1)),..., R σ(ρ(n 1)) ] ], R σ(ρ(n)) ] [[ [[R σ(τ(1)),..., R σ(τ(p)) ], R σ(ρ(n)) ], [[Rσ(ρ(p+1)),..., R σ(ρ(n 1)) ] ]. Note that A = [[ [[R σ(τ(1)),..., R σ(τ(p)) ], [[R σ(ρ(p+1)),..., R σ(ρ(n 1)) ] ] ], R σ(ρ(n)) is generated by the elements of the form ] ] [[[[g σ(τ(1)),..., g σ(τ(p)) ], [[g σ(ρ(p+1)),..., g σ(ρ(n 1)) ], g σ(ρ(n)) with g j R j. By the second induction on case that q = 1, the above elements lie in [[R 1, R 2 ],..., R n ] S and so A [[R 1, R 2 ],..., R n ] S. Similarly, by the second induction hypothesis, [[ [[Rσ(τ(1)),..., R σ(τ(p)) ], R σ(ρ(n)) ], [[Rσ(ρ(p+1)),..., R σ(ρ(n 1)) ] ] is a subgroup of [[R 1, R 2 ],..., R n ] S. It follows that T [[R 1, R 2 ],..., R n ] S and so β n (g σ(1),..., g σ(n) ) [[R 1, R 2 ],..., R n ] S. Both the first and second inductions are finished, hence the result. Lemma 2.4. Let G be a group and let R 1,..., R n be normal subgroups of G. Let (i 1, i 2,..., i p ) be a sequence of integers with 1 i s n. Suppose that {i 1, i 2,..., i p } = {1, 2,..., n}. Then [[R i1, R i2 ],..., R ip ] [[R 1, R 2 ],..., R n ] S.

7 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 7 Proof. The proof is given by double induction. The first induction is on n. The assertion clearly holds for n = 1. Suppose that the assertion holds for n 1 with n > 1. From the hypothesis that {i 1, i 2,..., i p } = {1, 2,..., n}, we have p n. When p = n, (i 1,..., i n ) is a permutation of (1,..., n) and so [[R i1, R i2 ],..., R in ] [[R 1, R 2 ],..., R n ] S. Suppose that [[R j1, R j2 ],..., R jq ] [[R 1, R 2 ],..., R n ] S for any sequence (j 1,..., j q ) with q < p and {j 1,..., j q } = {1,..., n}. Let (i 1,..., i p ) be a sequence with {i 1,..., i p } = {1,..., n}. If i p {i 1,..., i p 1 }, then {i 1,..., i p 1 } = {1,..., n} and so [[R i1, R i2 ],..., R ip 1 ] [[R 1, R 2 ],..., R n ] S by the second induction hypothesis. It follows that [[R i1, R i2 ],..., R ip ] [[R 1, R 2 ],..., R n ] S. If i p {i 1,..., i p 1 }, we may assume that i p = n. Then {i 1,..., i p 1 } = {1,..., n 1} and so [[R i1, R i2 ],..., R ip 1 ] [[R 1, R 2 ],..., R n 1 ] S by the first induction hypothesis. From Lemma 2.3, we have [[R i1, R i2 ],..., R ip ] [[R 1, R 2 ],..., R n ] S. The inductions are finished, hence the result holds. Lemma 2.5. Let G be a group and let R 1,..., R n be normal subgroups of G with n 2. Let (i 1,..., i p ) and (j 1,..., j q ) be sequences of integers such that {i 1,..., i p } {j 1,..., j q } = {1, 2,..., n}. Then [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] [[R 1, R 2 ],..., R n ] S. Proof. The proof is given by double induction on n and q with n 2 and q 1. First we prove the assertion holds for n = 2. If {i 1,..., i p } = {1, 2} or {j 1,..., j q } = {1, 2}, we have [[R i1, R i2 ],..., R ip ] [[R 1, R 2 ] S or [[R j1, R j2 ],..., R jq ] [[R 1, R 2 ] S by Lemma 2.4 and so [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] [[R 1, R 2 ] S. Otherwise, i 1 = = i p and j 1 = = j q, since {i 1,..., i p } {j 1,..., j q } = {1, 2}, we may assume that i 1 = = i p = 1, j 1 = = j q = 2, then and so [[R i1, R i2 ],..., R ip ] R 1 and [[R j1, R j2 ],..., R jq ] R 2 [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] [[R 1, R 2 ] S. Suppose the assertion holds for n 1, that is [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] [[R 1, R 2 ],..., R n 1 ] S. when {i 1,..., i p } {j 1,..., j q } = {1, 2,..., n 1}. We will use the second induction on q to prove the assertion holds for n.

8 8 J. Y. LI AND J. WU When q = 1, the assertion follows by Lemma 2.4. Suppose that the assertion holds for q 1. By the Hall Theorem, [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] is a subgroup of the product [[[[R i1,..., R ip ], [[R j1,..., R jq 1 ]], R jq ] [[[[R i1,..., R ip ], R jq ], [[R j1,..., R jq 1 ]]. By the second induction we have [[[[R i1,..., R ip ], R jq ], [[R j1,..., R jq 1 ]] [[R 1, R 2 ],..., R n ] S. If {i 1,..., i p } {j 1,..., j q 1 } = {1, 2,..., n}, by the second induction and hence [[[R i1,..., R ip ], [[R j1,..., R jq 1 ]] [[R 1, R 2 ],..., R n ] S [[[[R i1,..., R ip ], [[R j1,..., R jq 1 ]], R jq ] [[R 1, R 2 ],..., R n ] S. If {i 1,..., i p } {j 1,..., j q 1 } {1, 2,..., n}, we may assume that and j q = n. By the first induction, {i 1,..., i p } {j 1,..., j q 1 } = {1, 2,..., n 1} [[[R i1,..., R ip ], [[R j1,..., R jq 1 ]] [[R 1, R 2 ],..., R n 1 ] S. Then [[[[R i1,..., R ip ], [[R j1,..., R jq 1 ]], R jq ] [[R 1, R 2 ],..., R n ] S. It follows that [[[R i1, R i2 ],..., R ip ], [[R j1, R j2 ],..., R jq ]] [[R 1, R 2 ],..., R n ] S. The double induction is finished, hence the result Proof of Theorem 1.1. Clearly [[R 1, R 2 ],..., R n ] S [[R 1, R 2,..., R n ]]. We prove by induction on n that The assertion holds for n = 1. Hypothesis 2.1. Suppose that [[R 1, R 2,..., R n ]] [[R 1, R 2 ],..., R n ] S. [[R 1, R 2,..., R s ]] [[R 1, R 2 ],..., R s ] S for any normal subgroups R 1,..., R s of G with 1 s < n. Let R 1,..., R n be any normal subgroups of G. By definition, [[R 1, R 2,..., R n ]] is generated by all commutators β t (g i1,..., g it ) of weight t such that {i 1, i 2,..., i t } = {1, 2,..., n} with g j R j. To prove that each generator β t (g i1,..., g it ) [[R 1, R 2 ],..., R n ] S, we start the second induction on the weight t of β t with t n. If t = n, then (i 1,..., i n ) is a permutation of (1,..., n) and so the assertion holds by Lemma 2.3. Now assume that the following hypothesis holds: Hypothesis 2.2. Let n k < t and let β k (g i 1,..., g i k ) be any bracket arrangement of weight k such that where 1) 1 i s n; 2) {i 1,..., i k } = {1,..., n}; 3) g j R j;

9 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 9 Then β k (g i 1,..., g i k ) [[R 1, R 2 ],..., R n ] S. Let β t (g i1,..., g it ) be any bracket arrangement of weight t with {i 1,..., i t } = {1,..., n} and g j R j for 1 j n. From the definition of bracket arrangement, we have β t (g i1,..., g it ) = [β p (g i1,..., g ip ), β t p (g ip+1,..., g it )] for some bracket arrangements β p and β t p of weight p and t p, respectively, with 1 p n 1. Let A = {i 1,..., i p } and B = {i p+1,..., i t }. Then both A and B are the subsets of {1,..., n} with A B = {1,..., n}. Suppose that the cardinality A = n or B = n. We may assume that A = n. By Hypothesis 2.2, β p (g i1,..., g ip ) [[R 1, R 2 ],..., R n ] S. Since [[R 1, R 2 ],..., R n ] S is a normal subgroup of G, we have β t (g i1,..., g it ) = [β p (g i1,..., g ip ), β t p (g ip+1,..., g it )] [[R 1, R 2 ],..., R n ] S. This proves the result in this case. Suppose that A < n and B < n. Let A = {l 1,..., l a } with 1 l 1 < l 2 < < l a n and 1 a < n, and let B = {k 1,..., k b } with 1 k 1 < k 2 < < k b and 1 b < n. Observe that By Hypothesis 2.1, Thus Similarly β p (g i1,..., g ip ) [[R l1, R l2,..., R la ]]. [[R l1, R l2,..., R la ]] = [[R l1, R l2 ],..., R la ] S. β p (g i1,..., g ip ) [[R l1, R l2 ],..., R la ] S. β t p (g ip+1,..., g it ) [[R k1, R k2 ],..., R kb ] S. It follows that the element β t (g i1,..., g it ) lies in the commutator subgroup [[[R l1, R l2 ],..., R la ] S, [[R k1, R k2 ],..., R kb ] S ]. From Lemma 2.5, we have [ [[Rlσ(1), R l σ(2) ],..., R l σ(a) ], [[R k τ(1), R k τ(2) ],..., R k τ(b) ]] [[R 1, R 2 ],..., R n ] S for all σ Σ a and τ Σ b because {l 1,..., l a } {k 1,..., k b } = A B = {1, 2,..., n}. It follows from Lemma 2.1 that Thus [[[R l1, R l2 ],..., R la ] S, [[R k1, R k2 ],..., R kb ] S ] [[R 1, R 2 ],..., R n ] S. β t (g i1,..., g it ) [[R 1, R 2 ],..., R n ] S. The inductions are finished, hence Theorem 1.1.

10 10 J. Y. LI AND J. WU 2.3. Proof of Theorem 1.2. Clearly σ Σ n 1 [[R 1, R σ(2) ],..., R σ(n) ] [[R 1, R 2 ],..., R n ] S. We prove by induction on n that [[R 1, R 2 ],..., R n ] S The assertion holds for n = 1. Suppose that [[R 1, R 2 ],..., R s ] S σ Σ n 1 [[R 1, R σ(2) ],..., R σ(n) ]. σ Σ s 1 [[R 1, R σ(2) ],..., R σ(s) ] for any normal subgroups R 1,..., R s of G with 1 s < n. Let R 1,..., R n be any normal subgroups of G. By definition, [[R 1, R 2 ],..., R n ] S = τ Σ n [[R τ(1), R τ(2) ],..., R τ(n) ]. It suffices to prove that for any τ Σ n, [[R τ(1), R τ(2) ],..., R τ(n) ] σ Σ n 1 [[R 1, R σ(2) ],..., R σ(n) ]. The assertion holds for τ(1) = 1. Suppose that the assertion holds for τ(k 1) = 1 with 1 k 1 < n. When τ(k) = 1, consider the subgroup [[[[[R τ(1),..., R τ(k 2) ], R τ(k 1) ], R 1 ], R τ(k+1),..., R τ(n) ]. Following from the Hall Theorem, [[[[[R τ(1),..., R τ(k 2) ], R τ(k 1) ], R 1 ], R τ(k+1),..., R τ(n) ] [[[[[R τ(1),..., R τ(k 2) ], R 1 ], R τ(k 1) ], R τ(k+1),..., R τ(n) ]] [[[R τ(k 1), R 1 ], [[R τ(1),..., R τ(k 2) ]], R τ(k+1),..., R τ(n) ]. By the second induction [[[[[R τ(1),..., R τ(k 2) ], R 1 ], R τ(k 1) ], R τ(k+1),..., R τ(n) ]] [[R 1, R σ(2) ],..., R σ(n) ]. σ Σ n 1 From Lemma 2.5 and the first induction, we have [[R τ(k 1), R 1 ], [[R τ(1),..., R τ(k 2) ]] [[R 1, R l1 ], R l2,..., R lk 1 ] S [[R 1, R σ(l1)],..., R σ(lk 1 )], σ Σ k 1 where {1, τ(1),..., τ(k 1)} = {1, l 1, l 2,..., l k 1 } with 1 < l 1 < l 2 < < l k 1. By Lemma 2.1 [[[R τ(k 1), R 1 ], [[R τ(1),..., R τ(k 2) ]], R τ(k+1),..., R τ(n) ] σ Σ k 1 [[[R 1, R σ(l1)],..., R σ(lk 1 )], R τ(k+1),..., R τ(n) ] σ Σ n 1 [[R 1, R σ(2) ],..., R σ(n) ]

11 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 11 because for each σ Σ k 1, the sequence (σ(l 1 ),..., σ(l k 1 ), τ(k + 1),..., τ(n)) is a permutation of (τ(1),..., τ(k 1), τ(k + 1),..., τ(n)) which is a permutation of (2,..., n). It follows that [[[[[R τ(1),..., R τ(k 2) ], R τ(k 1) ], R 1 ], R τ(k+1),..., R τ(n) ] σ Σ n 1 [[R 1, R σ(2) ],..., R σ(n) ] and so [[R 1, R 2 ],..., R n ] S [[R 1, R σ(2) ],..., R σ(n) ]. σ Σ n 1 This finishes the proof. 3. Proof of Theorem Ellis-Mikhailov Theorem. In this subsection, we review some terminology and the main result in [6]. Let G be a group. An m-tuple of normal subgroups (R 1,..., R m ) of G is called connected if either (1). m 2 or (2). m 3 with the property that: for all subsets I, J {1,, m} with I 2, J 1 ( ) (3.1) R i R i. i I j J R j = i I Let G be a group with normal subgroups R 1,..., R n. Let X(G; R 1,..., R n ) be the homotopy colimit of the cubical diagram obtained from classifying spaces B(G/ i I R i) with the maps B(G/ i I R i ) B(G/ j J R j i I R i ) induced by the canonical quotient homomorphism G/ i I R i G/ i I R i for I I, where I ranges over all proper subsets I {1,..., n}. In the following theorem and the rest of the article, the notation ( â ) means to remove letter a. Theorem 3.1. [6, Theorem 1] Let G be a group with normal subgroups R 1,..., R n with n 2. Let X = X(G; R 1,..., R n ). Suppose that the (n 1)-tuple is connected for each 1 i n. Then (R 1,..., ˆR i,..., R n ) π n (X) R 1 R n = I J={1,...,n},I J= [ i IR i, j J R j ].

12 12 J. Y. LI AND J. WU 3.2. Proof of Theorem 1.3. Theorem 1.3 is part of the following statement. Theorem 3.2. Let (X, A) be a pair of spaces and let (A 1,..., A n ) be a cofibrant n-partition of X relative to A with n 2. Suppose that i) For any proper subset I = {i 1,..., i k } {1, 2,..., n}, the union i I A i is a path-connected K(π, 1)-space. ii) The inclusion A A i induces an epimorphism of the corresponding fundamental groups for each 1 i n. Let R i be the kernel of π 1 (A) π 1 (A i ) for 1 i n. Then (1). For any proper subset I = {i 1,..., i k } {1, 2,..., n}, R i1 R ik = [[R i1, R i2 ],..., R ik ] S. (2). The (n 1)-tuple (R 1,..., ˆR i,..., R n ) is connected for each 1 i n. (3). There is an isomorphism of groups π n (X) = (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S. (4). For 1 < k < n, I = {i 1,..., i k } {1, 2,..., n} and J = {1, 2,..., n} I, there is an isomorphism π k (X) k = (R is / R j ) ([[R, R ],..., R ] i1 i2 ik S). j J Proof. We prove assertions (1)-(3) by induction on n. If n = 2, assertions (1) and (2) hold obviously and assertion (3) follows from the classical Brown-Loday Theorem [3]. Suppose that assertions (1)-(3) hold for all cofibrant m-partitions (B 1,..., B m ) of any space Y relative to B satisfying conditions (i) and (ii) with m < n. Let (X, A) be a pair of spaces and let (A 1,..., A n ) be a cofibrant n-partition of X relative to A with n > 2. (1). Let I = {i 1,..., i k } {1, 2,..., n} be a proper subset. If k = 1, then R i1 = [R i1 ] S by definition. We may assume that 2 k < n. Let B s = A is for s = 1,..., k and let Y = B 1 B k. Then (B 1,..., B k ) is a cofibrant k-partition of Y relative to A satisfying conditions (i) and (ii). Notice that Ker(π 1 (A) π 1 (B s )) = Ker(π 1 (A) π 1 (A is )) = R is. By induction hypothesis, there is an isomorphism π k (Y ) = (R i1 R i2 R ik )/[[R i1, R i2 ],..., R ik ] S. From condition (i), Y = A i1 A ik is a K(π, 1)-space (as k < n) and so π k (Y ) = 0 (as k 2). Thus R i1 R i2 R ik = [[R i1, R i2 ],..., R ik ] S, which is assertion (1). (2). We show that the (n 1)-tuple (R 1,..., R n 1 ) is connected. Let m = n 1. If m = 2, then (R 1, R 2 ) is connected by definition. Assume that m 3. Let I, J {1,, m} with I 2 and J 1. We have to show the connectivity condition that ( ) R i R i. i I j J R j = i I j J R j j J R j

13 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 13 Clearly ( ) R i i I j J R j i I R i Now we show the other direction. Let I = {i 1,..., i k } with 2 k < n. Let B = j J A j and let B s = B A is for s = 1, 2,..., k. Let Y = k B s = i I,j J j J A i A j. Then (B 1,..., B k ) is a cofibrant k-partition of Y relative to B. Clearly condition (i) holds. Let G = π 1 (A). Then π 1 (B) = G/ j J R j and Thus condition (ii) holds. Let π 1 (B s ) = G/(R is R j ). j J R j. N s = Ker(π 1 (B) π 1 (B s )) = Ker(G/( j J R j) G/(R is j J R j)) = (R is j J R j)/ j J R j = R is /(R is j J R j). From induction hypothesis, we have π k (Y ) = (N 1 N 2 N k )/[[N 1, N 2 ],..., N k ] S. Since Y = i I,j J A i A j with I, J {1, 2,..., n 1}, the space Y is a K(π, 1)- space from condition (i). Thus π k (Y ) = 0 and so (3.2) N 1 N 2 N k = [[N 1, N 2 ],..., N k ] S with N s = R is /(R is j J R j) in G/ j J R j. Consider the quotient homomorphism φ: G G/ j J R j. Then N s = R is /(R is j J R j ) = φ(r is ) and so φ ([[R i1, R i2 ],..., R ik ] S ) = [[φ(r i1 ), φ(r i2 )],..., φ(r ik )] S = [[N 1, N 2 ],..., N k ] S = N 1 N 2 N k by equation (3.2). From the fact that φ R i i I j J R j k N s

14 14 J. Y. LI AND J. WU together with equation (3.2), we have i I (R i ) j J R j φ 1 (φ ([[R i1, R i2 ],..., R ik ] S )) = [[R i1, R i2 ],..., R ik ] S j J ( R j k ) R i s j J R j = ( i I R ) i j J R j. This proves that (R 1,..., R n 1 ) is connected. Similarly, each (R 1,..., ˆR i,..., R n ) is connected for 1 i < n and hence assertion (2). (3). From assertion (2), each (R 1,..., ˆR i,..., R n ) is connected for 1 i n. Since (A 1,..., A n ) is a cofibrant partition of X relative to A, X is the homotopy colimit of the diagram given by the inclusions A A j1 A j2 A jq A i1 A i2 A ip with {j 1,..., j q } {i 1,..., i p } {1, 2,..., n}. From Van Kampen s theorem π 1 (A i1 A ik ) = π 1 (A)/ k R is. Thus X = X(π 1 (A); R 1, R 2,..., R n ). From Theorem 3.1, we have π n (X) = It suffices to show that Recall that I J={1,...,n},I J= R 1 R n I J={1,...,n},I J= [ i IR i, j J R j ]. [[R 1, R 2 ],..., R n ] S = R i, R j = [[R 1, R 2 ],..., R n ] S. i I j J σ Σ n [[R σ(1), R σ(2) ],..., R σ(n) ]. For each σ Σ n, let I = {σ(1),..., σ(n 1)}, J = {σ(n)}, then and so [[R σ(1), R σ(2) ],..., R σ(n) ] = [[[R σ(1), R σ(2) ],..., R σ(n 1) ], R σ(n) ] [[R 1, R 2 ],..., R n ] S [ i I R i, j J R j] I J={1,...,n},I J= R i, R j. i I j J Conversely let I = {i 1,..., i p } and J = {j 1,..., j q } with 1 p, q n 1, I J = {1,..., n} and I J =. By assertion (1), R j = [[R j1, R j2 ],..., R jq ] S. i I R i = [[R i1, R i2 ],..., R ip ] S and j J Thus [ i I R i, j J R j ] = [ [[R i1, R i2 ],..., R ip ] S, [[R j1, R j2 ],..., R jq ] S ] [[R 1, R 2 ],..., R n ] S

15 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 15 by Theorem 1.1 because [ [[Ri1, R i2 ],..., R ip ] S, [[R j1, R j2 ],..., R jq ] S ] [[R1, R 2,..., R n ]]. This finishes the proof of assertion (3). Assertion (4) follows from assertion (3) by constructing a new partition as follows: let B = j J A j and let B s = B A is for s = 1, 2,..., k. Then X = k B s = i I,j J A i A j. and (B 1,..., B k ) is a cofibrant k-partition of X relative to B. Condition (i) and (ii) hold similar as in the proof of assertion (2). Let N s = Ker(π 1 (B) π 1 (B s )) = R is /(R is j J R j). From assertion (3), we have ( k ) π k (X) = N s /[[N 1, N 2 ],..., N k ] S. To finish the proof, it suffices to show that ( k ) (3.3) N s /[[N 1, N 2 ],..., N k ] S = ( k (R i s j J R j))/([[r i1, R i2 ],..., R ik ] S j J R j). Let φ: G G/ j J R j be the quotient homomorphism. Then it is straightforward to check that φ 1 (N s ) = R is j J R j, φ 1 ( k N s ) = k (R i s j J R j), φ 1 ([[N 1, N 2 ],..., N k ] S ) = [[R i1, R i2 ],..., R ik ] S j J R j. Thus equation (3.3) holds, hence assertion (4). The proof is finished. 4. Applications to the Free Groups and Surface Groups 4.1. Subgroups of the Surface Groups. Let X = S be a path-connected compact 2-manifold with or without boundary. Let Q i be a set of finite points in S S, 1 i n, such that (1). Q i for each 1 i n and (2). Q i Q j = for i j. Let A = S ( n Q i) be the punctured surface and let A i = A Q i. Then (A 1,..., A n ) is a cofibrant n-partition 2 of S relative to A. For any subset {i 1,..., i k } {1, 2,..., n}, 2 One needs to do some modifications such that the cofibrant hypothesis holds: By replacing each punctured point by a small open disk in S, the resulting punctured surfaces become compact 2-manifolds and so the cofibrant hypothesis for the partition (A 1,..., A n) holds.

16 16 J. Y. LI AND J. WU the space k is a K(π, 1)-space because it is a surface punctured by at least one point. Observe that each homomorphism π 1 (A) π 1 (A i ) is an epimorphism. Let R i be the kernel of the epimorphism π 1 (A) π 1 (A i ). By Theorem 1.3, we have (4.1) π n (S) = (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S. A is We give a group theoretic interpretation of this isomorphism. Case 1. X = S 2. Let Q = n Q i with Q = {q 1, q 2,..., q m } with a choice of order that q i < q j for i < j. For each q Q, let c q be a generator in π 1 (A) represented by a small circle around the point q with a choice of orientation such that π 1 (A) admits the presentation m π 1 (A) = c q q Q c qj = 1. Then R i = c q q Q i. From equation (4.1), we have the following result. Theorem 4.1. Let G = x 1, x 2,..., x m x 1 x 2 x m = 1 be the free group of rank m 1 with m 2. Let n 2 and let P i be any subset of {x 1, x 2,..., x m }, 1 i n, such that (i) P i for each 1 i n; (ii) P i P j = for i j and (iii) n P i = {x 1, x 2,..., x m }. Let R i = P i be the normal closure of P i in G. Then there is an isomorphism of groups (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S = πn (S 2 ). If n = m with P i = {x i }, then there is an isomorphism of groups ( x 1 x 2 x n )/[[ x 1, x 2 ],..., x n ] S = πn (S 2 ), which is [19, Theorem 1.7]. One interesting point of Theorem 4.1 is that the factor group (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S only depends on the length of the partition (P 1,..., P n ) of {x 1,..., x m }, which does not seem obvious from the group theoretic point of view. Case 2. X = RP 2. Let Q = n Q i with j=1 Q = {q 1, q 2,..., q m }.

17 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 17 Then π 1 (A) admits a presentation π 1 (A) = a 1, c q q Q, a 2 1 = m c qi with R i = c q q Q i. Note that π n (RP 2 ) = π n (S 2 ) for n 2. From equation (4.1), we have the following result. Theorem 4.2. Let G = a 1, x 1, x 2,..., x m a 2 1 = x 1 x 2 x m with m 2. Let n 2 and let P i be any subset of {x 1, x 2,..., x m }, 1 i n, such that (i) P i for each 1 i n; (ii) P i P j = for i j and (iii) n P i = {x 1, x 2,..., x m }. Let R i = P i be the normal closure of P i in G. Then there is an isomorphism of groups (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S = πn (S 2 ). Remark. In the above theorem, G = π 1 (RP 2 Q m ) is a free group of rank m with the presentation given in the form as in the statement. Thus the subgroups R i are different from those given in Theorem 4.1. Case 3. X S 2 or RP 2. Let Q = n Q i with Q = {q 1, q 2,..., q m }. If X is an oriented surface of genus g with t boundary components, then π 1 (A) admits a presentation g t m π 1 (A) = a 1, b 1,..., a g, b g, d 1,..., d t, c q q Q [a i, b i ] = d j c qi with R i = c q q Q i. If X is a non-oriented surface of genus h with t boundary components, then π 1 (A) admits a presentation h t m π 1 (A) = a 1,..., a h, d 1,..., d t, c q q Q a 2 i = d j c qi with R i = c q q Q i. Note that π n (X) = 0 for n 2. From equation (4.1), we have the following result. Theorem 4.3. Let G be one of the following surface groups: (a) a 1, b 1,..., a g, b g, y 1,..., y t, x 1,..., x m g [a i, b i ] = t j=1 y j m x i with g > 0 or t > 0. (b) a 1,..., a h, y 1,..., y t, x 1,..., x m h a2 i = t j=1 y j m x i with h > 1 or t > 0. Let P i be any subset of {x 1, x 2,..., x m }, 1 i n, such that (i) P i for each 1 i n; (ii) P i P j = for i j and (iii) n P i = {x 1, x 2,..., x m }. j=1 j=1

18 18 J. Y. LI AND J. WU Let R i = P i be the normal closure of P i in G. Then R 1 R 2 R n = [[R 1, R 2 ],..., R n ] S Homotopy Groups of Higher Dimensional Spheres and Free Products of Surface Groups. It is a natural question whether one can get similar group theoretical descriptions of the (general) homotopy groups of higherdimensional spheres. We give some remarks that the homotopy groups of certain 2-dimensional complexes contains the homotopy groups of all of higher-dimensional spheres as summands. From this, we can answer the above question in some sense. Let X and Y be path-connected spaces. According to [10], there is a homotopy decomposition (4.2) Ω(X Y ) ΩX ΩY ΩΣ(ΩX ΩY ). Now let X = S 2 and let Y be any surface. Equation (4.2) implies the following homotopy decomposition (4.3) Ω(S 2 Y ) ΩS 2 ΩY Ω(ΣΩS 2 ΩY ). From the classical James Theorem [11], there is homotopy decomposition ΣΩΣZ ΣZ k, k=1 where Z k is the k-fold self smash product of Z. Note that the k-fold self smash product of S 1 is S k. There is a homotopy decomposition ΣΩS 2 = ΣΩΣS 1 ΣS k = S k. By substituting this decomposition into formula (4.3), there is a homotopy decomposition Ω(S 2 Y ) ΩS 2 ΩY Ω ( k=2 Sk ΩY ) ΩS 2 ΩY ( k=2 Ω(Sk ΩY ) ) other factors. If Y is a surface with Y S 2, RP 2 and D 2, then ΩY is homotopy equivalent to a discrete space of a countably infinite set. Then S k Y is a wedge of countably infinite copies of S k. Thus given any k 2, π q (S k ) is a summand of π q (S 2 Y ) for each q with infinite multiplicity. If Y = S 2 or RP 2, then we can repeat the above decomposition formula and obtain the fact that, given any k 2, π q (S k ) is a summand of π q (S 2 Y ) for each q (with infinite multiplicity). Hence it suffices to consider π (S 2 Y ) for any surface Y D 2. By the same arguments in the previous subsection, we can get partitions of the space S 2 Y by removing points from S 2 and Y. This gives the following result. Theorem 4.4. Let G 1 = x 11, x 12,..., x 1m x 11 x 12 x 1m = 1 and let G 2 be one of the following surface groups: g (a) a 1, b 1,..., a g, b g, y 1,..., y t, x 21,..., x 2m [a i, b i ] = t y j m x 2i. k=1 (b) a 1,..., a h, y 1,..., y t, x 21,..., x 2m h a 2 i = k=2 t j=1 j=1 y j m x 2i.

19 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 19 Let P ji be any subset of {x j1, x j2,..., x jm }, j = 1, 2 and 1 i n, such that (i) P ji for each 1 i n; (ii) P ji P jk = for i k and (iii) n P ji = {x j1, x j2,..., x jm }. Let R i = P 1i, P 2i be the normal closure of P 1i P 2i in the free product G = G 1 G 2. Then there is an isomorphism of groups π n (S 2 Y ) = (R 1 R 2 R n )/[[R 1, R 2 ],..., R n ] S, where Y is a surface such that the fundamental group of Y punctured by m points is the group G Applications to Braid Groups and the Proof of Theorem Applications to the Braid Groups. Recall that a braid of n strands is called Brunnian if deleting any one of the strands produces a trivial braid of (n 1)- strands. Let Brun n denote the Brunnian subgroup of the n th pure Artin braid group P n. (Note. An n-strand Brunnian braid is always a pure braid for n 3 according to [1, Proposition 4.2.2].) The group Brun n has been characterized by Levinson [13, 14] for n 4. A classical question proposed by Makanin [17] in 1980 is to determine a set of generators for Brun n. This question has been answered by Johnson [12] and Stanford [18]. An answer using fat commutators was given in [1, Theorem 8.6.1]. As an application, we gave a generalization of Levinson s result [14, Theorem 2]. Recall that the braid group B n is generated by σ 1,..., σ n 1 with defining relations given by (1). σ i σ j = σ j σ i for i j 2; (2). σ i σ i+1 σ i = σ i+1 σ i σ i+1. Following Levinson s notation in [14], let t i = σ i σ i+1 σ n 2 σ 2 n 1σ 1 n 2 σ 1 i for 1 i n 1. Intuitively t i is the braid that links strand i and strand n in front of all other strands. Let R i = t i be the normal closure of t i in P n. (Note. Levinson uses the notation θ i.) Theorem 5.1. For each n 2, Brun n = [[R 1, R 2 ],..., R n 1 ] S. When n = 4, we have Brun 4 = [[R 1, R 2 ], R 3 ] S = [[R 1, R 2 ], R 3 ] [[R 1, R 3 ], R 2 ], which is exactly Levinson s theorem [14, Theorem 2]. Let M be any manifold. Recall that the coordinate projection F (M, n) F (M, k) (z 1,..., z n ) (z i1,..., z ik ) for given i 1 < < i k is a fibration by Fadell-Neuwirth s Theorem [9]. The coordinate projection d j : F (R 2, n) F (R 2, n 1), (z 1,..., z n ) (z 1,..., ẑ j,..., z n ) induces a group homomorphism d j : P n = π 1 (F (R 2, n)) P n 1 = π 1 (F (R 2, n 1)) which is given by removing the j th strand.

20 20 J. Y. LI AND J. WU Proof of Theorem 5.1. The basepoint (q 1, q 2,..., q n ) of F (R 2, n) is chosen in the way that the points q i R 1 R 2 with the order q 1 < q 2 < < q n. Let Q n = {q 1,..., q n } and let Q n,i = {q 1,..., ˆq i,..., q n }. For each 1 i n 1, there is a commutative diagram of fibrations R 2 Q n 1 f F (R 2, n) d n F (R 2, n 1) g i d i R 2 Q h n 1,i i F (R 2, n 1) F (R 2, n 2). d n 1 By taking fundamental groups and using the fact that F (R 2, m) is a K(π, 1)-space, there is a commutative diagram of short exact sequences of groups d i π 1 (R 2 Q n 1 ) f Pn = π 1 (F (R 2, n)) d n Pn 1 = π 1 (F (R 2, n 1)) g i d i π 1 (R 2 Q n 1,i ) h i Pn 1 = π 1 (F (R 2 (d n 1 ), n 1)) Pn 2 = π 1 (F (R 2, n 2)). Let Since h i is a monomorphism, R i = Ker(g i : π 1 (R 2 Q n 1 ) π 1 (R 2 Q n 1,i )). f : R i Ker(d n : P n P n 1 ) Ker(d i : P n P n 1 ) is an isomorphism. Observe that R i is the normal closure of the elements in π 1 (R 2 Q n 1 ) represented by the small circles around the point q i. By using the terminology of braids, the small circles around the point q i is represented by t ± i. Thus Ker(d n : P n P n 1 ) Ker(d i : P n P n 1 ) R i. On the other hand, since d n (t i ) = d i (t i ) = 1, we have and so R i Ker(d n : P n P n 1 ) Ker(d i : P n P n 1 ) (5.1) Ker(d n : P n P n 1 ) Ker(d i : P n P n 1 ) = R i with the property that is an isomorphism. If w n 1 f : R i R i d i R i, then w Im(f ) and f 1 (w) R i 1 i n 1. Thus the monomorphism f induces an isomorphism f : n 1 R i n 1 R i. By applying Theorem 4.3 to the punctured disks, we have R 1 R 2 R n 1 = [[R 1, R 2],..., R n 1] S. for each

21 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 21 It follows that From equality (5.1), we have n 1 R i = hence the result. n 1 R 1 R 2 R n 1 = [[R 1, R 2 ],..., R n 1 ] S. Ker(d n ) Ker(d i ) = n Ker(d j : P n P n 1 ) = Brun n, 5.2. Proof of Theorem 1.5. The key point to proving Theorem 1.5 is to construct a K(π, 1)-partition for the configuration space F (S 2, n). Before giving the proof, we need some lemmas. Let q 1, q 2,... be strictly increasing real numbers that lie in the open interval ( 1, 1) R 1 R 2 S 2. The base point of the configuration spaces F (R 2, m) F (S 2, m) is chosen to be (q 1,..., q m ) for each m 1. Lemma 5.2. Let 1 i m and let j=1 B i = {(z 1,..., z m ) F (S 2, m) z p for p i}. Let f : F (R 2, m) B i be the inclusion. Then the homomorphism f : P m = π 1 (F (R 2, m)) π 1 (B i ) is an epimorphism with Ker(f ) = A 0,i Pm, the normal closure of A 0,i in P m. Proof. The assertion holds trivially for m = 1. We may assume that m 2. Consider the Fadell-Neuwirth fibration d i : F (S 2, m) F (S 2, m 1) (z 1,..., z n ) (z 1,..., ẑ i,..., z m ). Then B i = d 1 i (F (R 2, m 1)) by the construction of B i. Consider the commutative diagram of the fibrations R 2 Q m,i F (R 2, m) F (R 2, m 1) f f S 2 Q m,i Bi F (R 2, m 1), where Q m,i = {q 1,..., q i 1, q i+1,..., q m }. Thus there is a commutative diagram of short exact sequence of groups π 1 (R 2 Q m,i ) π 1 (F (R 2, m)) π 1 (F (R 2, m 1)) f f π 1 (S 2 Q m,i ) π1 (B i ) π1 (F (R 2, m 1)). Since f : π 1 (R 2 Q m,i ) π 1 (S 2 Q m,i ) is an epimorphism, the middle map f : π 1 (F (R 2, m)) π 1 (B i ) is an epimorphism with Ker(f ) = Ker(f ). Note that Ker(f ) is the normal closure of the element [ω i ] in π 1 (R 2 Q m,i ), where the loop ω i is given by the following left picture:

22 22 J. Y. LI AND J. WU 1 i-1 i i+1 m q 1 q i-1 q i+1 q m ω i A 0,i In the above picture, the braid A 0,i is obtained from the loop ω i by drawing it as a braid. Thus [ω i ] = A 0,i and so Ker(f ) = Ker(f ) = A 0,i Pm π 1 (R 2 Q m,i ) A 0,i Pm. On the other hand, since Ker(f ) is normal in P m with A 0,i Ker(f ), we have A 0,i Pm Ker(f ). It follows that hence the result holds. Ker(f ) = A 0,i Pm, Lemma 5.3. Let I = {i 1, i 2,..., i k } be a subset of {1, 2,..., m}. Let B I = {(z 1,..., z m ) F (S 2, m) z p for p I} and let f : F (R 2, m) B I be the inclusion. Then the homomorphism f : π 1 (F (R 2, m)) π 1 (B I ) is an epimorphism whose kernel Ker(f ) = A 0,i i I Pm. Proof. Note that B I = k B is. By Lemma 5.2, P m = π 1 (F (R 2, m)) π 1 (B is ) is an epimorphism whose kernel is given by A 0,is Pm for each 1 s k. From Van Kampen s Theorem, is an epimorphism with hence the result. Ker(f ) = f : P m = π 1 (F (R 2, m)) π 1 (B I ) k A 0,is = A 0,is 1 s k Pm, Proof of Theorem 1.5. Let A = F (R 2, m) be the subspace of F (S 2, m). Let A i = {(z 1, z 2,..., z m ) F (S 2, m) z p for p Λ i }, where the Λ i are as defined in the statement of the Theorem. Since Λ i Λ j = for i j, we have A i A j = F (R 2, m) = A

23 ON SYMMETRIC COMMUTATOR SUBGROUPS AND HOMOTOPY GROUPS 23 for i j. From the assumption that n Λ i = {1, 2,..., m}, we have n A i = F (S 2, m). Thus (A 1, A 2,..., A n ) is a cofibrant n-partition 3 of the space F (S 2, m) relative to F (R 2, m). For a subset I = {i 1, i 2,..., i k } {1, 2,..., n}, let A I = with A = A. We now check that A I is a K(π, 1)-space for I {1, 2,..., n}. Let ( k ) J = {j 1, j 2,..., j t } = {1, 2,..., m} Λ is k with j 1 < j 2 <... < j t. Since I {1, 2,..., n}, J is not empty. Let (z 1,..., z m ) F (S 2, m). Observe that (z 1,..., z m ) A I if and only if z js for 1 s t. Consider the Fadell-Neuwirth fibration [9] There is a pull-back diagram A is d I : F (S 2, m) F (S 2, t). A I F (S 2, m) with a fibration d I AI F (R 2, t) pull d I F (S 2, t) F (S 2 Q t, m t) AI F (R 2, t), where Q t is a set of t distinct points in S 2. Since t 1, F (S 2 Q t, m t) = F (R 2 Q t 1, m t) is a K(π, 1)-space. Together with the fact that the base space F (R 2, t) is a K(π, 1)- space, the total space A I is a K(π, 1)-space for any I {1, 2,..., n}. By Lemma 5.3, the inclusion f : A = F (R 2, m) A i induces an epimorphism whose kernel The assertion follows from Theorem 1.3. f : P m = π 1 (A) π 1 (A i ) R i = Ker(f ) = A 0,j j Λ i Pm. Acknowledgements. The authors would like to thank the referee for pointing out some errors in the original manuscript. The authors also would like to thank Fred Cohen for his help on the writing of this article. 3 Similar to footnote 2, one needs to do some modifications such that the cofibrant hypothesis holds: Take a triangulation on F (S 2, m) and deform linearly A and each A i into certain subcomplexes with necessary subdivisions for having a cofibrant n-partition of F (S 2, m).

24 24 J. Y. LI AND J. WU References [1] J. A. Berrick, F. R. Cohen, Y. L. Wong and J. Wu, Braids, configurations and homotopy groups, J. Amer. Math. Soc. 19(2006), [2] V. G. Bardakov, R. Mikhailov, V. V. Vershinin and J. Wu, Brunnian braids on surfaces, preprint. [3] R. Brown and J.-L. Loday, Van Kampen theorems for diagrams of spaces, Topology 26 (1987), [4] F. R. Cohen and J. Wu, On braids, free groups and the loop space of the 2-sphere, Progress in Mathematic Techniques 215 (2004), Algebraic Topology: Categorical Decompositions, [5] F. R. Cohen and J. Wu, On braid groups and homotopy groups, Geometry & Topology Monographs 13 (2008), [6] G. Ellis and R. Mikhailov, A colimit of classifying spaces, arxiv: v1 [Math. GR] 22 April [7] E. Fadell, Homotopy groups of configuration spaces and the string problem of Dirac, Duke Math. J. 29 (1962), [8] E. Fadell and J. Van Buskirk, The braid groups of E 2 and S 2, Duke Math. J. 29 (1962), [9] E. Fadell and L. Neuwirth, Configuration spaces, Math. Scand 10 (1962), [10] B. Gray, A note on the Hilton-Milnor theorem, Topology 10 (1971), [11] I. M. James, Reduced product spaces, Ann. Math. 62 (1953), [12] D. L. Johnson, Towards a characterization of smooth braids, Math. Proc. Cambridge Philos. Soc. 92 (1982), [13] H. W. Levinson, Decomposable braids and linkages, Trans. AMS 178(1973), [14] H. W. Levinson, Decomposable braids as subgroups of braid groups, Trans. AMS 202(1975), [15] J. Y. Li and J. Wu, Artin braid groups and homotopy groups, Proc. London Math. Soc. 99 (2009), [16] W. Magnus, A. Karrass and D. Solitar, Combinatorial group theory, Pure and Applied Mathematics XIII(1966). [17] G. S. Makanin, Kourovka Notebook (Unsolved Questions in Group Theory) Seventh edition. Novosibirsk. 1980, Question p. 78. [18] T. Stanford, Brunnian braids and some of their generalizations, arxiv: math/ v1 [19] J. Wu, Combinatorial descriptions of the homotopy groups of certain spaces, Math. Proc. Camb. Philos. Soc. 130 (2001), [20] J. Wu, A braided simplicial group, Proc. London Math. Soc. 84 (2002), Institute of Mathematics and Physics, Shijiazhuang Railway Institute, Shijiazhuang , China address: yanjinglee@163.com Department of Mathematics, National University of Singapore, Block S17 (SOC1), 10, Lower Kent Ridge Road, Republic of Singapore address: matwuj@nus.edu.sg URL: