A geometric solution of the Kervaire Invariant One problem

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A geometric solution of the Kervaire Invariant One problem Petr M. Akhmet ev 19 May 2009

Let f : M n 1 R n, n = 4k + 2, n 2 be a smooth generic immersion of a closed manifold of codimension 1. Let g : N n 2 R n be the immersion of the double points intersection of g. The Kervaire invariant of f is defined by Θ sf (f) = η n 2 2 N ; [N n 2 ], where η N = w 2 (N n 2 ) is the second normal Stiefel-Whitney class of N n 2.

In particular, if n = 2, Θ sf (f) is the parity of the number of self-intersection points of the curve f on the plane R 2. Let us denote by Imm sf (n 1, 1) the cobordism group of immersions in the codimension 1 of (non-oriented) closed (n 1) manifold ( sf stands for skew-framed).

Theorem The Kervaire invariant is a well-defined homomorphism: Θ sf : Imm sf (4k + 1, 1) Z/2. 1. This homomorphism is trivial if 4k + 2 2 l 2, l 2. 2. For k = 0, 1, 3, 7, 15 the homomorphism Θ sf is an epimorphism.

Part 1 is the Theorem by W. Browder, The Kervaire invariant of framed manifolds and its generalization, Ann. of Math., (2) 90 (1969) 157-186. A special case of this theorem was discovered by M. Kervaire, A manifold which does not admit any differentiable structure, Comment. Math. Helv., 34 (1960) 257-270. Part 2 (in the case k = 7) was proved by M.E. Mahowald and M.C. Tangora, Some differentials in the Adams spectral sequence, Topology 6 (1967), 349-369. The case k = 15 was proved by M. G. Barratt, J. D. S. Jones and M. E. Mahowald, Relations amongst Toda brackets and the Kervaire invariant in dimension 62, J.London Math.Soc. (2) 30 (1984), no. 3, 533-550.

Main Theorem There exists an integer l 0, such that for an arbitrary l l 0, the Kervaire invariant Θ sf : Imm sf (2 l 3, 1) Z/2 is the trivial homomorphism. The proof of this theorem is based on the approach proposed by P.J.Eccles Codimension One Immersions and the Kervaire Invariant One Problem, Math. Proc. Cambridge Phil. Soc., vol.90 (1981) 483-493.

A commutative diagram that we use to define dihedral structure of self-intersection manifold of skew-framed immersions: Imm sf (n 1, 1) J k sf Imm sf (n k, k) δ Z/2 [2] δ k Z/2 [2] Imm Z/2[2] (n 2, 2) J k Z/2 [2] Imm Z/2[2] (n 2k, 2k).

A commutative diagram that we use to define the Kervaire invariant in the codimension k: Imm sf (n k, k) δ k Z/2 [2] Imm Z/2[2] (n 2k, 2k) Θ k sf Z/2 skew framed immersions Θ k Z/2 [2] Z/2 dihedral immersions.

Structure groups of immersions Let us consider the following collection of (d 1) sets Υ d, Υ d 1,..., Υ 2, where each set consists of proper coordinate subspaces of R 2d 1. The set of the subspaces Υ i, 2 i d, (we will use only the case d = 6) consists of 2 i 1 coordinate subspaces generated by the basis vectors: ((e 1,... e 2 d i),..., (e 2 d 1 2 d i +1,..., e 2 d 1)).

Let us denote by Z/2 [d] the subgroup Z/2 Σ 2 d 1 O(2 d 1 ) under the following condition: the transformation Z/2 [d] R 2d 1 R 2d 1 admits the invariant collection of sets Υ d, Υ d 1,..., Υ 2. In particular, in the case d = 2 we get that Υ 2 contains only one collection of subspaces and this collection is ((e 1 ), (e 2 )). Therefore Z/2 [2] is a dihedral group.

Imm Z/2[2] (n 2, 2) Imm Z/2[2] (n 2k, 2k) δ Z/2 [3] δ k Z/2 [3] Imm Z/2[3] (n 4, 4) J k Z/2 [3] Imm Z/2[3] (n 4k, 4k) δ Z/2 [4] δ k Z/2 [4] Imm Z/2[4] (n 8, 8) J k Z/2 [4] Imm Z/2[4] (n 8k, 8k) δ Z/2 [5] δ k Z/2 [5] Imm Z/2[5] (n 16, 16) J k Z/2 [5] Imm Z/2[5] (n 16k, 16k) δ Z/2 [6] δ k Z/2 [6] Imm Z/2[6] (n 32, 32) J k Z/2 [6] Imm Z/2[6] (n 32k, 32k).

Imm Z/2[2] (n 2k, 2k) Z/2 δ k Z/2 [3] Imm Z/2[3] (n 4k, 4k) δ k Z/2 [4] Imm Z/2[4] (n 8k, 8k) δ k Z/2 [5] Imm Z/2[5] (n 16k, 16k) δ k Z/2 [6] Θ k Z/2 [3] Z/2 Θ k Z/2 [4] Z/2 Θ k Z/2 [5] Z/2 Imm Z/2[6] (n 32k, 32k) Θ k Z/2 [6] Z/2.

I d İd [2] Abelian Z/2 structure i [3] I a İd Z/2 [3] cyclic Abelian structure i [4] I a İa Z/2 [4] bicyclic structure i [5] Q İa Z/2 [5] quaternionic cyclic structure i [6] Q Q Z/2 [6] biquaternionic structure.

Dihedral group The dihedral group (of the order 8) Z/2 [2] O(2): {a, b a 4 = b 2 = e, [a, b] = a 2 }. Let {f 1, f 2 } be the standard base of the plane R 2. The element a is represented by the rotation through the angle π: 2 f 1 f 2 ; f 2 f 1. The element b is represented by the permutation of the base vectors f 1 f 2 ; f 2 f 1.

Elementary 2-group The elementary subgroup I d İd Z/2 [2] of the rank 2: {a 2, b a 4 = b 2 = e, [a 2, b] = e}. This group preserves the vectors f 1 + f 2, f 1 f 2. Let τ [2] H 2 (Z/2 [2] ; Z/2) be the universal class, i d d (τ [2]) H 2 (I d İd; Z/2) is the pull-back of τ [2] under the inclusion i d d : I d İd Z/2 [2]. i d d (τ [2]) = κ d κd, κ d H 1 (I d İd; Z/2). p d : I d Id I d, κ d = p d (t d), e t d I d Z/2, and κd H1 (I d İd; Z/2) is defined analogously.

f : M n k R n, x = [(f, Ξ, κ M )] Imm sf (n k, k), κ M is a line bundle over M n k, Ξ is a skew-framing of the normal bundle of f, i.e. an isomorphism Ξ : ν f = kκ M. y = δ k Z/2 [2] (x) = (g, Ψ, η N ) Imm Z[2] (n 2k, 2k), g : N n 2k R n, η N is a Z/2 [2] bundle over N n 2k, Ψ is a dihedral framing of the normal bundle of f, i.e. an isomorphism Ψ : ν g = kη N.

Definition of Abelian structure A skew-framed immersion (f, Ξ, κ M ) admits an Abelian structure if there exists a map η d d,n : N n 2k K(I d İd, 1) (Eilenberg-Mac Lain space), satisfying the following condition: η n 2k 2 N ; [N n 2k ] = Θ k Z/2 [2] (y) = η 15k N η n 32k 2 d d ; [N]. η d d = η d d,n (i d d (τ [2])) H 2 (N n 2k ; Z/2), [N] is the fundamental class of N n 2k, η N H 2 (N n 2k ; Z/2) is the characteristic class of Z/2 [2] framing.

Definition of a Desuspension A skew-framed cobordism class x = [(f, Ξ, κ M )] Imm sf (n k, k) admits a desuspension of order q, if this class is represented by a triple, such that κ M = I κ (q), κ (q) : M n k RP n k q 1, I : RP n k q 1 RP.

Desuspension theorem For an arbitrary q there exists an integer l 0 = l 0 (q), such that an arbitrary element x Imm sf (2 l 3, 1), l l 0, admits a desuspension of order q.

Abelian structure immersion theorem Let q be an arbitrary integer divisible by 16, and let n = 2 l 2 with l is sufficiently large. Put k = k(q) = n + 2 32 q 16. Let us assume that x Imm sf (n k, k) admits a desuspension of order q. Then the class x is represented by a triple (f, Ξ, κ M ), such that this skew-framed immersion admits an Abelian structure.

Cyclic group The cyclic index 2 subgroup of the order 4: I a = {a a 4 = e} Z/2 [2]. Bicyclic group The bicyclic index 2 11 subgroup of the order 16: I a İa Z/2 [2] Z/2 [2] Z/2 [4].

The universal cohomology class of the bicyclic group There exists τ [4] H 8 (Z/2 [4] ; Z/2) (the universal class), i a ȧ(τ [4] ) H 8 (I a İa; Z/2) is the pull-back of τ [4] under the inclusion i a ȧ : I a İa Z/2 [4]. i a ȧ(τ [4] ) = ηaη 2 ȧ, 2 η a, ηȧ H 2 (I a İa; Z/2). Here η a, ηȧ are defined similar to κ a, κȧ.

x Imm sf (n k, k), y = δ Z/2 [2](x) Imm Z/2[2] (n 2k, 2k), z = δ Z/2 [3] δ Z/2 [2] Imm Z/2[3] (n 4k, 4k), u = δ Z [4] δ Z [3] δ Z [2](x) u = [(h, Λ, ζ L )] Imm Z/2[4] (n 8k, 8k), h : L n 8k R n, ζ L is a Z/2 [4] bundle over L n 8k, Λ is a an 8 dimensional Z/4 [4] -framing of the normal bundle of h, i.e. an isomorphism Λ : ν h kζ L.

Definition of bicyclic structure A Z/2 [3] immersion [(g, Ψ, η N )] = δ Z/2 [3] δ Z/2 [2](x) Imm Z/2[3] (n 4k, 4k) admits a bicyclic structure if there exists a map ζ a ȧ,l : L n 8k K(I a İa, 1) (Eilenberg-Mac Lain space), satisfying the following equation: Θ k Z/2 (z) = π [3] d d,a ȧ,l (ζ L) 3k n 32k 2 ζ d d,l ; [ L d d ]. Here the cohomology class ζ d d,l ζ a ȧ,l. is defined by means of

In the previous formula: L n 8k is the double-point Z/2 [4] manifold of g [ L d d ] is the fundamental class of the corresponding 4-sheeted cover π d d,a ȧ n 8k : L d d L n 8k, induced from the 4-sheeted cover of Eilenberg-Maclain spaces K(I d İd, 1) K(I a İa, 1) by the map ζ a ȧ,l : L n 8k K(I a İa, 1) ζd d,l H2 ( L n 8k ; Z/2) is the universal cohomological d d I d İd-class, constructed by means of the map ζ Q Q,L

ζ L H 8 (L n 8k ; Z/2) is the top characteristic class of the Z/2 [4] framing and it is the pull-back of the universal class τ [4] H 8 (Z/2 [4] ; Z/2) under the classifying map L n 8k K(Z/2 [4], 1) of the corresponded Z/2 [4] bundle previously denoted by ζ L. π d d,a ȧ,l (ζ L) H 8 ( L n 8k d d ; Z/2) is the pull-back of the class ζ L under the 4-sheeted cover π d d,a ȧ n 8k : L d d L n 8k.

Bicyclic structure immersion Theorem Let us assume that x Imm sf (n k, k), k = n 2s +2 32, s 6, admits a desuspension of the order q = 2s 2 2. Then the class z = δ k Z [3] δ k [2](x) Imm Z/2[3] (n 4k, 4k) is represented by a triple (h, Λ, ζ L ), such that this skew-framed immersion admits an Abelian structure.

Quaternionic group The quaternionic group of the order 8: Q = {i, j, k ij = k = ji, jk = i = kj, ki = j = ik, i 2 = j 2 = k 2 = 1}. This is an index 16 subgroup Q Z/2 [3]. The standard representation χ + : Q Z/2 [3] transforms the quaternions i, j, k into the following matrices:

i = 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0,

j = 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0,

k = 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0.

Biquaternionic group The biquaternionic index 2 57 subgroup of the order 64: Q Q Z/2 [3] Z/2 [3] Z/2 [6].

The universal cohomology class of the biquaternionic group There exists τ [6] H 32 (Z/2 [6] ; Z/2) (the universal class), i Q Q (τ [6]) H 32 (Q Q; Z/2) is the pull-back of τ [6] under the inclusion i Q Q : Q Q Z/2 [6]. i Q Q (τ [6]) = ζqζ 4 4 Q, ζ Q, ζ Q H4 (Q Q; Z/2). Here ζ Q, ζ Q to η a, ηȧ. are defined similar

x Imm sf (n k, k), y = δ Z/2 [2](x) Imm Z/2[2] (n 2k, 2k), z = δ Z/2 [3] δ Z/2 [2] Imm Z/2[3] (n 4k, 4k), u = δ Z [4] δ Z [3] δ Z [2](x) Imm Z/2[3] (n 4k, 4k), v = δ Z [5] δ Z [4] δ Z [3] δ Z [2](x) Imm Z/2[4] (n 8k, 8k), w = δ Z [6] δ Z [5] δ Z [4] δ Z [3] δ Z [2](x) Imm Z/2[5] (n 16k, 16k).

Definition of biquaternionic structure A Z/2 [5] immersion [(h, Λ, ζ L )] = δ Z/2 [5] δ Z/2 [4] δ Z/2 [3] δ Z/2 [2](x) Imm Z/2[5] (n 16k, 16k) admits a biquaternionic structure if there exists a map ω Q Q,K : Kn 32k K(Q Q, 1) (Eilenberg-Mac Lain space satisfying the following equation: Θ k Z/2 (w) = ω n 32k [5] 2 d d,k ; [ K d d ]. Here the cohomology class ω d d,k ω Q Q,K. is defined by means of

In the previous formula: K n 32k is the double-point Z/2 [6] manifold of h [ K d d ] is the fundamental class of the 16-sheeted cover π d d,q n 32k Q : K d d K n 32k, induced from the 16-sheeted cover of Eilenberg-Mac Lain spaces K(I d İd, 1) K(Q Q, 1) by the map ω Q Q,K : Kn 32k K(Q Q, 1) ω d d,k H2 n 32k ( K d d ; Z/2) is the universal cohomology I d İd-class, constructed by means of the map ω Q Q,K.

Biquaternionic structure immersion theorem Let k = n 2s +2 32, s 6, (q = 2s 2 2 ), q be an integer divisible by 16, and let n = 2 l 2 with l sufficiently large. Put k = k(q) = n + 2 32 q 16. Let us assume that x Imm sf (n k, k) admits a desuspension of the order q = 2s 2 2. Then the class w = δ k Z [5] δ k Z [ 4] δk Z [3] δ k Z [ 2] (x) ImmZ/2[5] (n 16k, 16k) is represented by a triple (h, Λ, ζ L ) such that this triple admits a biquaternionic structure.

Biquaternionic Kervaire Invariant Theorem Assume that w Imm Z/2[5] (n 16k, 16k), n = 2 l 2, k = 0 (mod 64), k > 0, n 32k > 0 admits a biquaternionic structure. Then Θ Z/2 [5](w) = 0. As a corollary we get Main Theorem There exists an integer l 0, such that for an arbitrary l l 0, the Kervaire invariant Θ sf : Imm sf (2 l 3, 1) Z/2 is the trivial homomorphism.

Proof of Biquaternionic Theorem Let w Imm Z/2[5] (n 16k, 16k), w = [(e, Ω, ω K )], S n 32k be the double point manifold of the immersion e, ω Q ω Q : Sn 32k K(Q, 1) K( Q, 1) be the biquaternionic map. Recall that n 32k = dim(s) 14. Let i T : T 14 S n 32k be a closed submanifold dual to the cohomology class (ω S;Q ω S; Q ) n 32k 14 8 H n 32k 14 (S n 32k ; Z/2), where ω S;Q = ω Q (ζ Q); ω S; Q = ω Q (ζ Q ) H4 (S n 32k ; Z/2).

The following (non-standard) representation χ : Q Z/2 [3] transforms the quaternions i, j, k into the following matrices: 0 1 0 0 1 0 0 0 i =, 0 0 0 1 0 0 1 0

j = 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0,

k = 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0.

Let us define the vector bundles ζ +, ζ be over S 15 /Q a. The bundle ζ + is defined by means of the representation χ +. The bundle ζ is defined by means of the representation χ. The bundle ζ + admits a complex structure. Note that c 1 (ζ + ) = 0, because the restriction of the bundle ζ + over S 3 /Q S 15 /Q is the trivial complex bundle and H 2 (S 15 /Q; Z) H 2 (S 3 /Q; Z) is an isomorphism.

Therefore, p 1 (2ζ + ) = c 2 1(2ζ + ) 2c 2 (2ζ + ) = 4c 2 1(ζ + ) 4c 2 (ζ + ) = 4ζ Q H 4 (K(Q, 1); Z). By the analogical computation: p 1 (2 ζ + ) = 4ζ Q H4 (K( Q, 1); Z).

The bundle ζ admits a complex structure. Note that c 1 (ζ ) = 0 by analogical calculations. Therefore, p 1 (ζ + ζ ) = c 2 1(ζ + ζ ) 2c 2 (ζ + ζ ) = c 2 1(ζ + ) + c 2 1(ζ ) + 2c 1 (ζ + )c 1 (ζ ) 2c 2 (ζ + ) 2c 2 (ζ ) = 0, because the Euler classes e(ζ + ) H 4 (S 15 /Q; Z) e(ζ ) H 4 (S 15 /Q; Z) are opposite: e(ζ + ) = e(ζ ).

The normal bundle ν T is stably isomorphic to the bundle lζ T,+ l ζ T,+, where l is an integer, l = 2 (mod 4). The bundle ζ T,+ is the 4-dimensional vector bundle over T defined as ζ T,+ = ωt,q(ζ + ), ω T,Q = ω Q T : T 14 K(Q, 1). The bundle ζ T,+ is the 4-dimensional vector bundle over T defined as ζ T,+ = ω T, Q (ζ +), ω T, Q = ω Q T : T 14 K( Q, 1).

Put T 14 to be T 14 with the opposite orientation. The normal bundle ν T is stably isomorphic to the bundle (l 1)ζ T,+ ζ T, l ζ T,+ (we will put after l = 2 for the shortness). The bundle ζ T,+ is the 4-dimensional vector bundle defined as ζ T,+ = ω T,Q(ζ + ), ω T,Q = ω Q T : T 14 K(Q, 1). The bundle ζ T, is the 4-dimensional vector bundle defined as ζ T, = ω T,Q(ζ ).

The bundle ζ T,+ is the 4-dimensional vector bundle defined as ζ T,+ = ω T, Q (ζ +), ω T, Q = ω T, Q T : T 14 K( Q, 1).

Let us assume that Θ Z/5 (w) = 1.. Then the decomposition of the cycle ω Q Q, ([T ]) in the standard base of H 14 (Q Q; Z) involves the element u 7 v 7, where u 7 H 7 (K(Q, 1); Z) = Z/8, v 7 H 7 (K( Q, 1); Z) = Z/8 are the generators, H 7 (K(Q, 1); Z) H 7 (K( Q, 1); Z) H 14 (K(Q Q, 1); Z).

Let F = id id : T 14 T 14 T 14 be the standard degree 0 map. Let us consider the following homology class: ℵ = (ω Q Q F ) ([p 1 (ν T )] op + [p 1 (ν T )] op ]) H 10 (K(Q Q, 1); Z), where the upper index op stands for Poincaré dual.

Let us prove that ℵ involves the element 4u 3 v 7 H 3 (K(Q, 1); Z) H 7 (K( Q, 1); Z) H 10 (K(Q Q, 1); Z). Without loss of the generality we may assume that ω Q Q, ([T ]) = u 7 v 7 + xu 3 v 11 +..., where x is an arbitrary integer. (For all last terms in this formula the characteristic class ℵ does not involve the element u 3 v 7 by the dimension reason). Under this assumption by the computation above we get: F ([p 1 (ν T )] op ) = 4u 3 v 7 + 4xu 3 v 7 + H 3 (K(Q, 1); Z) H 7 (K( Q, 1); Z) H 10 (K(Q Q, 1); Z), F ([p 1 (ν T )] op ) = 4xu 3 v 7 +....

Therefore the first (normal) Pontrjagin class satisfy the equation: 0 4u 3 v 7 + = (ω Q Q F ) ([p 1 (ν T )] op + [p 1 (ν T )] op ]). In particular, F is not cobordant to zero. But the mapping F is cobordant to zero by definition. Contradiction. Therefore Θ Z/5 (w) = 0.