Mathematical Journal of Okayama University

Transcription

1 Mathematical Journal of Okayama University Volume 44, Issue Article 1 JANUARY 2002 On the Group π(σa B,X) Seiya Sasao Tokyo Institute of Technology Copyright c 2002 by the authors. Mathematical Journal of Okayama University is produced by The Berkeley Electronic Press (bepress).

2 Sasao: On the Group Math. J. Okayama Univ. 44(2002), ON THE GROUP π(σa B, X) Seiya SASAO Introduction We work in the homotopy category of pointed CW -complexes, and denote the suspension functor with Σ. The group π(σa B, X) is not abelian in general under the usual multiplication but we use the notation + for it. In 1 we shall describe the multiplication of the group π(σa B, X), and, as a by-product, obtain from the associativity of the multiplication that the bi-additivity of the generalized Whitehead product (GWP) does not hold in general (see Proposition 3.4 of [1]). In fact, as an example, we offer the following: Let i 1 and i 2 be inclusion maps from CP 2 and S n into CP 2 S n respectively. Then we have that [2Σi 1, Σi 2 ] 2[Σi 1, Σi 2 ] and [Σi 1, 2Σi 2 ] = 2[Σi 1, Σi 2 ] in π(σcp 2 S n, ΣCP 2 S n ). In 2 we investigate the group π(σx, ΣA B) and show that any element of this group can be determined by 4-components under some assumptions. In 3 we apply 2 to the case of X = A B, i.e. the group of self-maps of the space ΣA B. Specially we are interested in describing the composition of two elements with their components and give the special case of A = B = S n as an example. 1. π(σa B, X) Let i 1 and i 2 be inclusion maps: A, B A B, and let P A and P B be projections: A B A, B respectively. Lemma 1.1. Let π : A B A B be the projection. Any element f π(σa B, X) can be uniquely represented by the form f = ασp A + βσp B + γσπ for α π(σa, X), β π(σb, X) and γ π(σa B, X). Proof. In fact a representation can be obtained from a part of Puppe exact sequence of the cofibering: A B A B. Then clearly we have α = fσi A, β = fσi B and moreover the uniqueness of γ follows from the injectivity of (Σπ). 123 Produced by The Berkeley Electronic Press,

3 Mathematical Journal of Okayama University, Vol. 44 [2002], Iss. 1, Art S. SASAO Here we give a brief account of GWP of [1]. For two maps α: ΣA X and β : ΣB X, GWP [α, β] π(σa B, X) is defined by ασp A + βσp B = βσp B + ασp A + [α, β]σπ. Proposition 1.2. GWP has following properties: (1) Let us be α, β the commutator of α and β ( π(σa, X)) then we have α, β = [α, β]σd A, where d A is the diagonal map: A A A. (2) If f π(x, Y ) then f[α, β] = [fα, fβ]. (3) If σ k π(σy k, Z) and f k π(x k, Y k ) for k = 1, 2 then it holds that [σ 1 Σf 1, σ 2 Σf 2 ] = [σ 1, σ 2 ]Σf 1 f 2. (4) For four maps Σf π(σy, ΣA), Σg π(σy, ΣB), σ π(σa, X) and τ π(σb, X) we have σσf + τσg = τσg + σσf + [σ, τ]σ(f g)σd Y. (5) If X is a suspension (i.e. X = ΣX ) then d X = 0. Lemma 1.3. For α π(σa, X), β π(σb, X) and γ π(σa B, X) we have the following: (1) ασp A + βσp B = βσp B + ασp A + [α, β]σπ. (2) ασp A + γσπ = γσπ + ασp A + [α, γ]σϕ A Σπ, where ϕ A is defined by ϕ A (a b) = a a b. (3) βσp B + γσπ = γσπ + βσp B + [β, γ]σψ B Σπ, where ψ B is defined by ψ B (a b) = b a b. Proof. (1) is just the definition of GWP. Next, by applying (1) and (4) of lemma 1.2 to the diagram: ΣA B ΣP A ΣA α X ΣA B Σπ ΣA B γ X, we have that [ασp A, γσπ]σd A 1 B = [α, γ]σ(p A π)σd A 1 B = [α, γ]σϕ A Σπ. The case (3) is analogous to the case (2). Thus the proof is completed. Now let us represent f = ασp A + βσp B + γσπ with the triple (α, β, γ). Theorem 1.4. (α 1, β 1, γ 1 ) + (α 2, β 2, γ 2 ) = (α 1 + α 2, β 1 + β 2, δ + γ 2 ) for δ = δ 1 [β 2, δ 1 ]Σψ B and δ 1 = [α 2, β 1 ] + γ 1 [α 2, γ 1 ]Σϕ A. 2

4 Sasao: On the Group ON THE GROUP π(σa B, X) 125 Proof. For abbreviation we use notations: ᾱ = ασp A, β = βσp B, γ = γσπ and so on. Now by using lemma 1.3 we have equalities f 1 + f 2 = (α 1, β 1, γ 1 ) + (α 2, β 2, γ 2 ) = ᾱ 1 + β 1 + γ 1 + ᾱ 2 + β 2 + γ 2 = ᾱ 1 + β 1 + ᾱ 2 + γ 1 [α 2, γ 1 ]Σϕ A Σπ + β 2 + γ 2 = ᾱ 1 + ᾱ 2 + β 1 [α 2, β 1 ] + γ 1 [α 2, γ 1 ]Σϕ A + β 2 + γ 2 = ᾱ 1 + ᾱ 2 + β 1 + δ 1 + β 2 + γ 2 = ᾱ 1 + ᾱ 2 + β 1 + β 2 + δ 1 [β 2, δ 1 ]Σψ B + γ 2 = ᾱ 1 + ᾱ 2 + β 1 + β 2 + δ + γ 2 = (α 1 + α 2, β 1 + β 2, δ + γ 2 ). Thus the proof is completed. Corollary 1.5. If A and B are both suspensions then it holds that (α 1, β 1, γ 1 ) + (α 2, β 2, γ 2 ) = (α 1 + α 2, β 1 + β 2, [α 2, β 1 ] + γ 1 + γ 2 ). Proof. Since ϕ A and ψ B are trivial by (5) of lemma 1.2 the proof follows from Theorem 1.4. Corollary 1.6. For α 1, α 2 π(σa, X) and β π(σb, X) it holds that Proof. On the other hand [α 1 + α 2, β] = [α 2, [α 1, β]]σϕ A + [α 1, β] + [α 2, β]. (0, β, 0) + {(α 1, 0, 0) + (α 2, 0, 0)} = (0, β, 0) + (α 1 + α 2, 0, 0) = (α 1 + α 2, β, [α 1 + α 2, β]). {(0, β, 0) + (α 1, 0, 0)} + (α 2, 0, 0) = (α 1, β, [α 1, β]) + (α 2, 0, 0) = (α 1 + α 2, β, [α 2, β] [α 1, β] [α 2, [α 1, β]]σϕ A ). Thus the proof follows from the associativity of the addition of the group π(σa B, X). By (5) of lemma 1.2 and the above corollary 1.6 it is easy to obtain the following: Corollary 1.7 (Proposition 3.4 of [1]). If A is a suspension then it holds that [α 1 + α 2, β] = [α 1, β] + [α 2, β] Produced by The Berkeley Electronic Press,

5 Mathematical Journal of Okayama University, Vol. 44 [2002], Iss. 1, Art S. SASAO for α 1, α 2 π(σa, X) and β π(σb, X). Now we consider a special case: A = CP 2, B = S n and X = ΣCP 2 S n. Let i 1 and i 2 be inclusions: CP 2, S n CP 2 S n respectively, and let us be α = Σi 1 and β = Σi 2. Since ϕ A : CP 2 S n CP 2 CP 2 S n is defined by ϕ A (a b) = a a b, ϕ A can be regarded as Σ n d for the diagonal map d: CP 2 CP 2 CP 2. Since d : H 4 (CP 2 CP 2 ) H 4 (CP 2 ) is clearly an isomorphism Σ n d is non-trivial. On the other hand, accordingly to Hilton-Milnor Theorem ([3]) [α, [α, β]] is injective. Hence [α, [α, β]]σϕ A is non-trivial. Then these show [2Σi 1, Σi 2 ] 2[Σi 1, Σi 2 ] and [Σi 1, 2Σi 2 ] = 2[Σi 1, Σi 2 ]. Remark. The second equality follows from Corollary On the group π(σx, ΣA B) First by using lemma 1.1 we define ξ AB π(σa B, ΣA B) as follows: Corollary 2.1. Σπξ AB = 1 ΣA B. 1 ΣA B = Σi A ΣP A + Σi B ΣP B + ξ AB Σπ. Proof. Apply (Σπ) to the above equality. Then we have Σπ = Σπ(Σi A ΣP A + Σi B ΣP B + ξ AB Σπ) = Σπξ A B Σπ. Since (Σπ) is injective the proof is completed. Here we note that the representation of f π(σa B, X) in lemma 1.1 is given by f = f ΣA ΣP A + f ΣB ΣP B + fξ AB Σπ, where f K denotes the restriction of f on K. For example if h is a map A B X then Σhξ AB is essentially the Hopf-construction of f, i.e. C(h) (see [2]) and we have a representation: Σh = Σh A + Σh B + C(h)Σπ, where h is a map of type (h A, h B ). Secondly we define two maps ϕ π(y, ΣA B) and φ π(σa B, Y ) for Y = ΣA ΣB ΣA B by ϕ = Σi A Σi B ξ AB, φ = i ΣA ΣP A + i ΣB ΣP B + i ΣA B Σπ. Lemma 2.2. ϕ is a homotopy equivalence with φ as its inverse. Proof. Easy. 4

6 Sasao: On the Group ON THE GROUP π(σa B, X) 127 In the following of this section we assume that (1) A is a-connected and B is b-connected, (2) a b, (3) dim X 2a + b + 2. Then by Hilton-Milnor theorem, f π(σx, ΣA B) can be represented as follows: f = Σi A f A + Σi B f B + [Σi A, Σi B ]f C1 + ξ AB f C2, where f π(σx, Σ ) and f C π(σx, ΣA B). More precisely we have Lemma 2.3. f A = ΣP A f, f B = ΣP B f, f C1 = Σ(P A P B )H(f) and f C2 = Σπf, where H(f) denotes Hopf-invariant of f. Proof. First we note that [γ, δσπ]h(f) = 0 because this element is decomposed as follows: ΣX H(f) Σ(A B) (A B) Σ(1 π) Σ(A B) (A B) [γ,δ] ΣA B. Then the proof is deduced from our assumptions. Secondly apply f from the right to the equality. We obtain that f = (Σi A P A + Σi B P B + ξ AB Σπ)f = (Σi A P A + Σi B P B )f + ξ AB Σπf = Σi A P A f + Σi B P B f + [Σi A, Σi B ](ΣP A ΣP B )H(f) + ξ AB Σπf. Thus the proof is completed. 3. On the group π(σa B, ΣA B) In this section we assume that A and B are both n-connected, dim A dim B and dim A + dim B 3n + 2. Lemma 3.1. Our assumptions contain (1) A, B and A B are all suspensions, so π(σk, X) is abelian for K = A, B and A B, (2) dim B 2n + 1. Hence π(σ, ΣA B) = 0 for = A or B, (3) Σ: π(x, Y ) π(σx, ΣY ) is onto for any pair (X, Y ) of {A, B}. Proof. Easy. In 1, f π(σa B, ΣA B) has a representation: f = f A ΣP A + f B ΣP B + f C Σπ for f π(, ΣA B) and C = A B. Produced by The Berkeley Electronic Press,

7 Mathematical Journal of Okayama University, Vol. 44 [2002], Iss. 1, Art S. SASAO And moreover in 2, f has a representation: f = Σi A f 1 + Σi B f 2 + ξ AB f 3 + [Σi A, Σi B ]f 4 for f 1 π(σ, ΣA), f 2 π(σ, ΣB) and f 3, f 4 π(σ, ΣA B). Thus f π(σa B, ΣA B) has a form of a (3 4)-matrix (note lemma 3.1): (f k ) = f A1 f A2 0 0 f B1 f B f C1 f C2 f C3 f C4 We want to compute the composition gf for f, g π(σa B, ΣA B). Since we have that gf = (gf A )ΣP A + (gf B )ΣP B + (gf C )Σπ it is sufficient for our purpose to compute gf π(σ, ΣA B). Lemma 3.2. g C = gξ AB. Proof. First we have g = g1 ΣA B = g(σi A ΣP A + gσi B ΣP B + gξ AB Σπ) = g ΣA ΣP A + g ΣB ΣP B + gξ AB Σπ. On the other hand g = g A ΣP A + g B ΣP B + g C Σπ. Hence we have g C = gξ AB. Now we proceed to gf : gf = g(σi A f 1 + Σi B f 2 + ξ AB f 3 + [Σi A, Σi B ]f 4 ) = g ΣA f 1 + g ΣB f 2 + gξ AB f 3 + [g ΣA, g ΣB ]f 4 = g A f 1 + g B f 2 + g C f 3 + [g A, g B ]f 4. Lemma 3.3. [g A, g B ] = Σi A [g A1, g B1 ] + Σi B [g A2, g B2 ] + [Σi A, Σi B ](Σg A1 Σg B2 τσg A2 g B1), where g k = Σg k for {A, B} and k = 1, 2. Proof. Apply lemma 1.2 and Propositions 3.3, 3.4 of [1] to the equality: [g A, g B ] = [Σi A g A1 + Σi B g A2, Σi A g B1 + Σi B g B2 ], then the proof is completed. Lemma 3.4. We have g A f C1 = Σi A (g A1 f C1 ) + Σi B (g A2 f C1 ) + [Σi A, Σi B ]Σg A1 g A2H(f C,1 ). 6

8 Sasao: On the Group ON THE GROUP π(σa B, X) 129 Proof. Apply the distributive law to the equality: then the proof is completed. g A f C1 = (Σi A g A1 + Σi B g A2 )f C1, From these lemmas we have Theorem 3.5. If f = (f k ) and g = (g k ) then gf = h = (h k ) is given by (1) the case of = A, or B, (2) the case of = C = A B, h 1 = g A1 f 1 + g B1 f 2, h 2 = g A2 f 1 + g B2 f 2, h C1 = g A1 f C1 + g B1 f C2 + [g A1, g B1 ]f C4 + g C1 f C3, h C2 = g A2 f C1 + g B2 f C2 + [g A2, g B2 ]f C4 + g C2 f C3, h C3 = g C3 f C3, h C4 = Σg A1 g A2H(f C1 ) + Σg B1 g B2H(f C2 ) + (Σg A1 g B2 ΣτΣg A2 g B1)f C4 + g C4 f C3. As an example we take A=B =S n. Let us be f π(σs n S n, ΣS n S n ) with its matrix: f 11 f f 21 f , f 31 f 32 f 33 f 34 where f ij ({i, j} = {1, 2}), f 33, f 34 Z and f 31, f 32 π 2n+1 (S n+1 ). If h = gf then h ij is given by h 11 = g 11 f 11 + g 21 f 12, h 12 = g 12 f 11 + g 22 f 12, h 21 = g 11 f 21 + g 21 f 22, h 22 = g 12 f 21 + g 22 f 22, h 31 = g 11 f 31 + g 21 f 32 + f 33 g 31 + f 34 g 12 g 22 [ι n+1, ι n+1 ], h 32 = g 12 f 31 + g 22 f 32 + f 33 g 32 + f 34 g 12 g 22 [ι n+1, ι n+1 ], h 33 = g 33 f 33, h 34 = (g 11 g 22 ( 1) n g 12 g 21 )f 34 + g 34 f 33 + g 11 g 12 H(f 31 ) + g 21 g 22 H(f 32 ), where g f denotes (g ι n+1 )f. Here we describe some results obtained from the above table of the composition. Produced by The Berkeley Electronic Press,

9 Mathematical Journal of Okayama University, Vol. 44 [2002], Iss. 1, Art S. SASAO 1. f is a homotopy equivalence if and only if f 33 = ±1 and f 11 f 12 f 21 f 22 = ±1. 2. We denote the group of self-homotopy equivalences of ΣS n S n with ε(g n ) and orientation-preserving self-homotopy equivalences with ε 0 (G n ), where orientation-preserving means that the above determinant is 1 and f 33 = 1, then ε 0 (G n ) is a normal subgroup and the quotient is isomorphic to Z 2 Z ε 0 (G n ) contains a subgroup: = Z a 4. ε 0 (G n ) contains a subgroup: = π 2n+1 (S n+1 ) π 2n+1 (S n+1 ). α β 1 0 References [1] M. Arkowitz, The generalized Whitehead product, Pacific J. Math. 12(1962), [2] K. Morisugi, Hopf construction, Samelson products and suspension maps, Contemporary Math. 239(1999), [3] G. W. Whitehead, Elements of homotopy theory, Graduate texts in Math. 61, Springer-Verlag, Seiya Sasao Tokyo Institute of Technology Ookayama, Meguro-ku Tokyo, Japan (Received April 1, 2002 ) 8