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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. This is weird. By the epimophism part of FST, it is enough to understand the composite 2 Shc ◦ S hc. 2 2 2 We can explain the order: 2[Shc ◦ S hc] = [2Shc ◦ S hc] = [0 ◦ S hc]=0. But, the main problem of this stem is to show the nontriviality of this com- posite. 2 Proposition 3.1: [Shc ◦ S hc] 6=0. 3 Proof: HLES of the complex Hopf fibration shows that π∗ : π4(S ) → 2 3 π4(S )= Z2 is isomorphism and since π4(S ) is generated by [Shc] , we have that 2 π4(S ) is generated by [hc ◦ Shc] . On the other hand, we have a 2-local fibration S2 → ΩS3 → ΩS5 which gives the so-called EHP-sequence, [9], and thus an 2 3 isomorphism E : π4(S ) → π5(S ) where E is S, the suspension homomorphism. 2 The result follows since E([hc ◦ Shc]) = Shc ◦ S hc . The proof above is quite unsatisfactory for me. Because, I want to try to 3 6 find an explanation by means of the mapping cone space M = S ∪f B and 2 5 (6) related maps with it, where f = Shc ◦ S hc. Is M the 6-th skeleton ΩS 2-locally? The attaching map f and the space M are mysterious. S 2 Corollary 3.2: π2 = Z2 and it is generated by η . An alternative proof of Proposition 3.1 can be obtained from Theorem 7.2 of [1]. Adams claims that his dR invariant detects the nontriviality of η2. Another explanation for the non-triviality of η2 is the Pontrjagin approach, S [9]: πk is isomorphic to the cobordism group of framed k-dimensional manifolds ∞ fr S ∼ fr in R , denoted by Ωk . Under the isomorphism πk = Ωk , η corresponds to the class of the circle S1 and η2 corresponds to the class of the torus S1 × S1 . The non-triviality of η2 is detected by an invariant called the Arf-Kervaire invariant which is the Arf invariant of the quadratic form on the mod 2 first homology of the torus as far as I understand. You may see the famous Turkish mathematician Cahit Arf’s paper for the origin of this invariant, [2].
4 The Third Stem
We have the following ingredients: With the help of homotopy tables, FST S 5 7 implies π3 = π8(S )= Z24. There is a quaternionic Hopf Fibration hq : S → 4 3 5 S with fiber S and its suspension is the generator of π8(S ), so its infinite
4 S suspension ν is the generator of π3 . Furthermore, it is the image of x = 1 ∈ KO(S4)= Z under the J-homomorphism. Notice that x has infinite order while νghas finite order. Why is d [Shq] = 0 (or dν = 0) if and only if 24 divides d? We can mimic the proof of Proposition 2.2 to deduce that [Shq] 6= 0 in the following way: We have a quaternionic Hopf line bundle ηq over the quaternionic projective space 1 4 2 HP = S . The mapping cone of hq is the Thom complex T (ηq) = HP , the quaternionic projective plane. Then, by looking at K(S4HP 2) and the effect 5 9 of Adams operations on it, we may deduce that S ∪eShq B is not homotopy equivalent to S5 ∨ S9. But this solves only the case d = 1 of the question we just stated. Instead, we will use the Adams conjecture to find the upper bound. It is such a deep theorem that it is still called as a conjecture. Proposition 4.1: The order of ν divides 24. Proof: You may see Diba˘g’s article [4] for a short elementary and self- contained proof of the Adams conjecture. For CW -complexes, it implies for sufficiently big natural number N,
N k k (ψR(x) − x)=0 in the J-group J(X) for all x ∈ KO(X) and natural numbers k. If wee take Xe= S4, x =1 ∈ KOg(S4)= Z, we get kN (k2 − 1)x = 0 in J(S4) . The common multiple of the numbersg kN (k − 1)(k + 1) is 24 while N ise big. So, 24x = 0 in J(S4) and under the J-homomorphism we get 24ν = 0. 5 9 The Propositione above implies that the space S ∪24Shq B is split, hence 4 8 5 9 4 8 we get S(S ∪24hq B )= S ∨ S .On the other hand, S ∪24hq B is the Thom 24 4 1 24 space T (ηq ). In KO(S ) = KO(HP ), we have ηq +92 = 24ηq. The Thom 25 23 space T (24ηq) is the stunted quaternionic projective space HP HP , [3]. It follows from these facts that for some natural number N we have
SN (HP 25 HP 23)= S96+N ∨ S100+N as an homotopy equivalence equality. For the lower bound, we must prove the following Proposition. We will refer to [6] mentioned in Proposition 2.2. Proposition 4.2: 12ν 6= 0. Proof: By a similar argument made just after the Preposition 4.1, we deduce that if 12ν were zero then HP 13 HP 11 would be S48 ∨S52. We must show that this is not possible. This can easily follow as a very special case of Theorem 1.2 of [6]: The spaces HP n+k HP k−1 and HP n+l HP l−1 have the same stable homotopy type if k − l = O(Bn) where Bn is the J-order of the Hopf bundle of n HP . If we take n = 1, B1 = 24. Now, let k = 12 and l = 0. Then it follows 13 11 1 0 4 that HP HP can not be stably homotopic to HP+ = S ∨ S which is stably homotopic to S48 ∨ S52. S Corollary 4.3: π2 = Z24 and it is generated by ν.
5 4k Adams, [1], generalizes the last result as J(S ) = Zm(2k) where m(2k) is B2k e the denominator of and B is the 2k-th Bernoulli number and this group 4k 2k S is a direct summand of π4k−1. Pontrjagin approach is mind blowing at this stem, [9]. Under the isomor- S ∼ fr 3 phism π3 = Ω3 , ν corresponds to the class of the three dimensional sphere S and it is told that deletion of 24 three dimensional balls around the singularities of a K3-surface corresponds to the null cobordism 24 S3 =0. As a final note, I recall some relations. Two of them are η4 = ην =0. If we cheat a little without giving a constructive proof, these relations follow from the S 3 fact that π4 =0. Another relation is η = 12ν. The explanation of this relation without spectral sequences must be hard.
5 Appendix: The Group π1(SO) Let SO(n) be the group of n × n special orthogonal matrices. For each n ∈ Z+, we have natural group inclusions in : SO(n) → SO(n + 1) which give fibrations SO(n) → SO(n + 1) → Sn by group quotient. Hence, due to HLES of these fibrations, the maps in are (n−1)-connected maps so that πk(SO(n)) = πk(SO) for all n ≥ k +2 where SO is the infinite dimensional special orthogonal group, obtained by taking the direct limit of the group SO(n) when n → ∞. Similar arguments goes for the orthogonal group O(n) = Z2 × SO(n) and the infinite dimensional orthogonal group O = Z2 × SO. Note that πk(O)= πk(SO) for all k ≥ 1 but π0(SO) = 0 whereas π0(O)= Z2. For k = 1, we must take at least n = 3 to calculate π1(SO) = Z2. Really, π1(SO(1)) = 0 and π1(SO(2)) = Z, therefore this bound is strict. Proposition A.1: SO(3) = RP 3. Proof: S3 ⊂ H can be considered as the space of unit quaternions and each q ∈ S3 defines a rotation x → qxq about the origin of the real vector space of pure imaginary quaternions Ri ⊕ Rj ⊕ Rk =∼ R3. This map turns out to be a surjective group homomorphism S3 → SO(3) with kernel {±1} .Hence we get 3 an isomorphism S /Z2 → SO(3). Corollary A. 2: π1(SO)= Z2. 3 Proof: HLES of the fibration Z2 → S → SO(3) gives π1(SO(3)) = π0(Z2)= Z2.The result follows by stability. Another but a more difficult proof of Proposition A.1 can be given in the following way. Each rotation of R3 about the origin is determined by a vector which is called the axis of the rotation and a rotation angle θ ∈ [0, π]. When the rotation angle is π, antipodal directions give the same rotation. Consider the closed ball in R3 with center the origin and radius π. Identify the antipodal points of the boundary and get a quotient space X. X is homeomorphic to RP 3. Define φ : X → SO(3) by sending [x,y,z] ∈ X to the rotation with direction determined by the vector (x,y,z) and rotation angle x2 + y2 + z2. This map p is a homeomorphism.
6 The generator of the group π1(SO(3)) is described by using the space X and the map φ above in the following way. Consider the loop γ : I = [0, 1] → X defined by γ(t) = [0, 0, π cos πt]. Proposition A.3: φ∗([γ]) is the generator of π1(SO(3)). Proof of Proposition A.3 will be given later in this section. The loop γ corre- sponds to the set of rotations about z-axis, in other words, image of γ in SO(3) is exactly SO(2). The following proposition is equivalent to the Proposition A.3 and it is an alternative proof.
Proposition A.4: π1(SO(3)) =
−10 0 −1 0 0 0 − cos2πt − sin 2πt and 0 − cos2πt sin 2πt 0 − sin 2πt cos2πt 0 sin 2πt cos2πt in SO(3), respectively. They are inverse paths to each other as expected. The cos πt sin πt 0 path γ of course corresponds to − sin πt cos πt 0 . One can also write 0 01 the homotopy matrices from γ to α or β in SO(3). Proof of Proposition A.3 : It is enough to show that [α] is not trivial. Suppose that it is trivial with an homotopy to the constant loop at the pole. By com- pressing this homotopy to the boundary we obtain a homotopy in the boundary. But the boundary is RP 2 and thus [α] is trivial in RP 2. Now, a similar argument can be repeated for RP 2concluding that [α] is trivial in RP 1 = S1. But, [α] is the generator of the fundamental group of this last boundary by construction. A contradiction. Vector bundle interpretation can be stated as π1(SO)= π1(O)= π2(BO)= 2 2 2 KO(S ) = Z2. There is a realification homomorphism r : K(S ) → KO(S ). Sinceg S2 =∼ CP 1, we have K(S2)= K(CP 1)= Z generated bye η − 1, reductiong of the Hopf line bundle η : Ee (η) → CeP 1. The homomorphism r is surjective and
7 hence KO(S2) is generated by r(η) − 2, the reduction of the realification of the Hopf lineg bundle and its order in this group is 2.
References
[1] J. Frank Adams. On the groups J(X) - IV. Topology, 5:21–71, 1966. [2] C. Arf. Untersuchungen uber quadratische Formen in Korpern der Charak- teristik 2. I.J. Reine Angew. Math., 183:148–167, 1941. [3] M. F. Atiyah.Thom complexes, Proc. Lond. Math. Soc. 11 (1961). 291-310. [4] I.˙ Diba˘g. The Adams Conjecture, Proc. A.M.S. 87(2) (1983), 367-374. [5] I.˙ Diba˘g. Primary decomposition of the J -groups of complex projective and lens spaces, Topology and its Applications, 153 (2006), 2484–2498. [6] S. Feder, S. Gitler. The Classification of Stunted Projective Spaces by Stable Homotopy Type, Transactions of AMS, Volume 225 (1977), 59-81. [7] A. Hatcher. Algebraic Topology. Cambridge University Press, 2002. [8] M. Kırdar. KO-Rings and J-Groups of Lens Spaces. Ph.D Thesis, Ankara, 1998. [9] Guozhen Wang, Zhouli Xu. A survey of computations of homotopy groups of Spheres and Cobordisms, 2010.
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