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On the First, the Second and the Third Stems of the Stable Homotopy

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On the First, the Second and the Third Stems of the Stable Homotopy On the First, the Second and the Third Stems of the Stable Homotopy Groups of Spheres Mehmet Kırdar Tekirda˘gNamık Kemal University, Mathematics Department, Tekirda˘g, Turkey July 14, 2021 Abstract We describe the first stem of the stable homotopy groups of spheres by using some Puppe sequences, Thom complexes, K-Theory and Adams Operations following the ideas of J. Frank Adams. We also touch upon the second and the third stems in this perspective. Neither spectral squences nor Steenrod operations are used. Key Words. Stable, Homotopy, J-Homomorphism, Hopf Mathematics Subject Classification. [2020] 55Q45, 55R25 1 Introduction S n The computation of the stable homotopy group of spheres π = lim πn+k(S ), k n→∞ k ≥ 0, has been one of the most interesting questions of mathematics. See [9] S for a survey. πk is called the k-th stem. Skipping the zeroth stem, the purpose of this study is to make a little survey for the first, the second and the third stems. It is well-known that the first stem S π1 is Z2 and it is generated by η which is the image of the generator η of π1(SO) arXiv:2107.06103v1 [math.AT] 11 Jul 2021 S S under the J-homomorphism J : π1(SO) → π1 . The second stem π2 is also Z2 and it is generated by η2 in the ring of the stable homotopy groups of spheres. The third stem is amazingly Z24 and it is generated by ν which also comes from a J-homomorphism like η. Firstly, we introduce the complex Hopf fibration and describe and compute S π1 together with π1(SO). We prove the non-triviality of the suspension of the Hopf fibration by using complex K-theory and Adams operations on the Puppe sequence of this fibration. Its order is also explained by using the Puppe S S sequence of its double. Then, we will touch upon the stems π2 and π3 . We will mostly follow the ideas in [1]. But, neither spectral sequences nor Steenrod operations will be involved. In the appendix, we studied π1(SO) in some details. 1 2 The First Stem One of the most interesting objects in mathematics is a Hopf fibration. There are various definitions/constructions for them. 3 2 For example, the complex Hopf fibration hc : S → S is defined by sending a unit quaternion q ∈ S3 to qiq ∈ S2 ⊂ Ri ⊕ Rj ⊕ Rk where q is the conjugate of q. Fibers of this fibration are S1 and any two fibers are linked. It can be seen as the double cover S3 → SO(3) → SO(3)/SO(2) = S2 of the fibration given in the Appendix. It also can be seen as the map S3 ⊂ C2 −{0} → CP 1 = S2 which sends (z1,z2) to [z1,z2]. Homotopy Long Exact Sequence (HLES) of this fibration gives the following Proposition 2.1: 2 i) π3(S )= Z generated by [hc] the Hopf fibration itself. 2 3 ii) πk(S )= πk(S ) for k ≥ 3. The famous and deep Freudenthal suspension theorem (FST) for spheres n n+1 2 gives isomorphisms πk(S ) =∼ πk+1(S ) for k ≤ 2n−2. Unfortunately, π3(S ) 3 3 is not in the stability region but π4(S ) is. Therefore π4(S ) is the group which is isomorphic to the first stem. On the other hand, FST for spheres n n+1 gives epimorphisms π2n−1(S ) → π2n(S ) so that we have an epimorphism 2 3 3 π3(S ) → π4(S ). Therefore, π4(S ) is either trivial or a nontrivial cyclic group generated by [Shc]. In fact, according to the homotopy table of [7] on page 339, 3 it is nontrivial and π4(S )= Z2. Therefore, the latter argument is true and we are done. But, we want to explain the non-triviality of [Shc] and the relation 2 [Shc]= 3 0 without using the information π4(S )= Z2. Proposition 2.2: [Shc] 6=0. 2 Proof: The suspension of the mapping cone of the map hc, that is S(S ∪hc 4 B ), is homotopy equivalent to the mapping cone of the map Shc, that is to 3 5 S ∪Shc B , due to the Puppe sequence of hc, [1]. But, it is well-known that the 2 4 mapping cone S ∪hc B is the Thom complex of the Hopf line bundle η over CP 1 and it is the complex projective space CP 2. See [3] for Thom complexes. 2 If Shc was trivial, SCP would be split, that is, it would be the wedge sum S3 ∨ S5 and thus S2CP 2 would be S4 ∨ S6, etc. We must show that SCP 2 and S3 ∨ S5 or S2CP 2 and S4 ∨ S6 are distinct, that is, not homotopy equivalent. The reader may think that these distinctions are obvious. In fact, for example, they are a very special case of Theorem 1.1. of [6], S. Feder and S. Gitler’s article about the classification of the stable homotopy types of the stunted projective spaces. But I will try to do an awkward proof for the latter distinction. Z(µν) ⊕ Z(µ2ν) By Bott periodicity, we have K(S2CP 2)= K(CP 2)⊗K(S2)= (ν2,µ3) e e e 2 where µ is the reduction of the complex Hopf line bundle ηc over CP and ν ∈ K(S2) is the Bott element .On the other hand, K(S4 ∨ S6) = K(S4) ⊕ K(S6e) = Z(x) ⊕ Z(y) where virtually dim x = 4, dimey = 6. These ringse are reallye isomorphic. 2 Suppose we an homotopy equivalence φ : S2CP 2 → S4 ∨ S6 so that we ∗ have an induced ring isomorphism φ : K(S4) ⊕ K(S6) → K(S2CP 2). We may ∗ assume that φ (x)=+µν. e e e ∗ ∗ Now, we must have φ (ψp(x)) = ψp(φ (x)) due the neutrality of the complex Adams operation ψp for some natural number p> 1. But then, by properties of ∗ Adams operations, [6], φ (p2x) = ψp(µν) and p2µν = ψp(µ)ψp(ν) = pνψp(µ). Hence we get ψp(µ) = pµ. But this is a contradiction. Because, ψp(µ) = p(p−1) 2 pµ + 2 µ . Next Proposition is also well-known to experts. Proposition 2.3: 2 [Shc]=0. 2 4 2 2 4 Proof: S ∪2hc B is the Thom space of η like S ∪hc B is the Thom space of η. But, in KO(S2) = KO(CP 1) we calculate r(η2)+2=2r(η) = 4 and it is trivial. Therefore, it follows that SN T (r(η2)+2) = SN T (4) for some natural number N where equality means the homotopy equivalence. N+2 2 4 N+4 N+6 Hence we get, S (S ∪2hc B )= S ∨ S . N+2 2 On the other hand, due to Puppe sequence of 2hc, we have S (S ∪2hc 4 N+4 N+6 N+2 B )= S ∪2S hc B . N+2 N+2 Thus 2 [Shc]=2 S hc = 2S hc = 0. We must remark here that what makes the splitting possible is actually the J-order of the bundle, not the KO-order, in the above proposition. But, since these orders are equal, we have no problem here. But, in the third stem, it will be different. We also note that we did not mention about the J-homomorphism so far! 3 Corollary 2.4: π4(S )= Z2 and it is generated by [Shc] . S Corollary 2.5: π1 = Z2 and it is generated by η, the infinite suspension of the Hopf fibration. J- Homomorphism proof of Corollary 2.5 is briefly explained in the following way: From a map Sr −→ O(n), r ≥ 0, one can construct a map Sr+n → Sn by the ”Hopf construction”. One can show that this construction results n in a homomorphism J : πr(O(n)) → πr+n(S ) and also by taking limit, a r+1 S homomorphism J : πr(O) = KO(S ) → πr . And most interestingly, this homomorphism factors throughg a group denoted by J(Sr+1), the reduced J- group of the sphere Sr+1. e My supervisor Ibrahim˙ Diba˘gdevoted great efforts for determining the de- composition of J-groups of complex projective spaces and lens spaces as a direct- sum of cyclic groups.e These groups involve very complicated arithmetic. I think his efforts culminated in [5]. S In our case r = 1, one has the homomorphism J : π1(O) = π1(SO) → π1 . And furthermore, one can prove, that J(η) = η where η is the generator of π1(SO) as explained in the Appendix. Hence, our isomorphism is achieved. Adams, [1], introduced the d and e invariants to detect the image of the J-homomorphism for all r. He proved that for r > 0 and r ≡ 0, 1 (mod 8), J-homomorphism is injective and gives a Z2 direct sum part of the r-th stem. 3 3 The Second Stem S 4 FST implies that π2 = π6(S ). But according to the homotopy table of [7] 4 S again, π6(S )= Z2. Therefore, easily π2 = Z2. On the other hand, this stem is a phenomenon of the 20th century. First of all, KO(S3) = 0 and the generator of the second stem is not image of the J-homomorphism.g But, this generator is the direct limit of the composite n n+1 2 S hc ◦ S hc when n −→ ∞. In other words, it is η in the ring of stable ∞ homotopy group of spheres πS. But, we know that η is a image of the M k k=0 J-homomorphism.
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