# Knots, i.e. embeddings of spheres

## 1 Introduction

We work in a smooth category. In particular, terms embedding and smooth embedding or map and smooth map are used interchangeably. For a general introduction to embeddings as well as the notation and conventions used on this page, we refer to [Skopenkov2016c, $\S$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}\S$1, $\S$$\S$3].

## 2 Examples

There are smooth embeddings $S^{2l-1}\to\Rr^{3l}$$S^{2l-1}\to\Rr^{3l}$ which are not smoothly isotopic to the standard embedding. They are PS (piecewise smoothly) isotopic to the standard embedding (by the Zeeman Unknotting Spheres Theorem 2.3 of [Skopenkov2016c] and [Skopenkov2016f, Remark 1.1]).

Example 2.1. (a) Analogously to the Haefliger trefoil knot for any $l>1$$l>1$ one constructs a smooth embedding $t:S^{2l-1}\to\Rr^{3l}$$t:S^{2l-1}\to\Rr^{3l}$, see [Skopenkov2016h, $\S$$\S$5]. For $l$$l$ even $t$$t$ is not smoothly isotopic to the standard embedding; $t$$t$ represents a generator of $E_D^{3l}(S^{2l-1})\cong\Zz$$E_D^{3l}(S^{2l-1})\cong\Zz$ [Haefliger1962].

It would be interesting to know if for $l>1$$l>1$ odd this embedding is a generator of $E_D^{3l}(S^{2l-1})\cong\Zz_2$$E_D^{3l}(S^{2l-1})\cong\Zz_2$. The last phrase of [Haefliger1962t] suggests that this is true for $l=3$$l=3$.

(b) For any $k=1,3,7$$k=1,3,7$ let $\eta\in\pi_{4k-1}(S^{2k})$$\eta\in\pi_{4k-1}(S^{2k})$ be the homotopy class of the Hopf map. Denote by $\zeta:\pi_{4k-1}(S^{2k})\to E_D^{6k}(S^{4k-1}\sqcup S^{4k-1})$$\zeta:\pi_{4k-1}(S^{2k})\to E_D^{6k}(S^{4k-1}\sqcup S^{4k-1})$ the Zeeman map, see [Skopenkov2016h, Definition 2.2]. The embedded connected sum $\#\zeta\eta$$\#\zeta\eta$ of the components of (a representative of) $\zeta\eta$$\zeta\eta$ is not smoothly isotopic to the standard embedding; $\#\zeta\eta$$\#\zeta\eta$ is a generator of $E_D^{6k}(S^{4k-1})\cong\Z$$E_D^{6k}(S^{4k-1})\cong\Z$ [Skopenkov2015a, Corollary 2.13].

## 3 Invariants

Let us define the Haefliger invariant $\varkappa:E^{6k}_D(S^{4k-1})\to\Z$$\varkappa:E^{6k}_D(S^{4k-1})\to\Z$. The definition is motivated by Haefliger's proof that any embedding $S^n\to S^m$$S^n\to S^m$ is isotopic to the standard embedding for $2m\ge3n+4$$2m\ge3n+4$, and by analyzing what obstructs carrying this proof for $2m=3n+3$$2m=3n+3$.

By [Haefliger1962, 2.1, 2.2] any embedding $f:S^{4k-1}\to S^{6k}$$f:S^{4k-1}\to S^{6k}$ has a framing extendable to a framed embedding $\overline f:V\to D^{6k+1}$$\overline f:V\to D^{6k+1}$ of a $4k$$4k$-manifold $V$$V$ whose boundary is $S^{4k-1}$$S^{4k-1}$, and whose signature is zero. For an integer $2k$$2k$-cycle $c$$c$ in $V$$V$ let $\lambda^*(c)\in\Z$$\lambda^*(c)\in\Z$ be the linking number of $f(V)$$f(V)$ with a slight shift of $\overline f(c)$$\overline f(c)$ along the first vector of the framing. This defines a map $\lambda^*:H_{2k}(V;\Z)\to\Z$$\lambda^*:H_{2k}(V;\Z)\to\Z$. This map is a homomorphism (as opposed to the Arf map defined in a similar way [Pontryagin1959]). Then by Lefschetz duality there is a unique $\lambda\in H_{2k}(V,\partial;\Z)$$\lambda\in H_{2k}(V,\partial;\Z)$ such that $\lambda^*[c]=\lambda\cap_V[c]$$\lambda^*[c]=\lambda\cap_V[c]$ for any $[c]\in H_{2k}(V;\Z)$$[c]\in H_{2k}(V;\Z)$. Since $V$$V$ has a normal framing, its intersection form $\cap_V$$\cap_V$ is even. (Indeed, represent a class in $H_{2k}(V;\Z)$$H_{2k}(V;\Z)$ by a closed oriented $2k$$2k$-submanifold $c$$c$. Then $\rho_2[c]\cap_V[c]=\overline{w_{2k}}(c\subset V)=\rho_2[c]\cap_VPDw_{2k}(V)=0$$\rho_2[c]\cap_V[c]=\overline{w_{2k}}(c\subset V)=\rho_2[c]\cap_VPDw_{2k}(V)=0$ because $V$$V$ has a normal framing.) Hence $\lambda\cap_V\lambda$$\lambda\cap_V\lambda$ is an even integer. Define
$\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.$

Since the signature of $V$$V$ is zero, there is a symplectic basis $\alpha_1,\ldots,\alpha_s,\beta_1,\ldots,\beta_s$$\alpha_1,\ldots,\alpha_s,\beta_1,\ldots,\beta_s$ in $H_{2k}(V;\Z)$$H_{2k}(V;\Z)$. Then clearly

$\displaystyle \varkappa(f) = \sum\limits_{j=1}^s \lambda^*(\beta_j)\lambda^*(\alpha_j).$

For an alternative definition via Seifert surfaces in $6k$$6k$-space, discovered in [Guillou&Marin1986], [Takase2004], see [Skopenkov2016t, the Kreck Invariant Lemma 4.5]. For a definition by Kreck, and for a generalization to 3-manifolds see [Skopenkov2016t, $\S$$\S$4].

Sketch of a proof that $\varkappa(f)$$\varkappa(f)$ is well-defined (i.e. is independent of $V$$V$, $\overline f$$\overline f$, and the framings), and is invariant under isotopy of $f$$f$. [Haefliger1962, Theorem 2.6] Analogously one defines $\lambda(V)$$\lambda(V)$ and $\varkappa(V):=\lambda(V)\cap_V\lambda(V)/2$$\varkappa(V):=\lambda(V)\cap_V\lambda(V)/2$ for a framed $4k$$4k$-submanifold $V$$V$ of $S^{6k+1}$$S^{6k+1}$. Since $\varkappa(V)$$\varkappa(V)$ is a characteristic number, it is independent of framed cobordism. So $\varkappa(V)$$\varkappa(V)$ defines a homomorphism $\Omega_{fr}^{4k}(6k+1)=\pi_{6k+1}(S^{2k+1})\to\Z$$\Omega_{fr}^{4k}(6k+1)=\pi_{6k+1}(S^{2k+1})\to\Z$. The latter group is finite by the Serre theorem. Hence the homomorphism is trivial.

Since $\varkappa(f)$$\varkappa(f)$ is a characteristic number, it is independent of framed cobordism of a framed $f$$f$ (and hence of the isotopy of a framed $f$$f$).

Therefore $\varkappa(f)$$\varkappa(f)$ is a well-defined invariant of a framed cobordism class of a framed $f$$f$. By [Haefliger1962, 2.9] (cf. [Haefliger1962, 2.2 and 2.3]) $\varkappa(f)$$\varkappa(f)$ is also independent of the framing of $f$$f$ extendable to a framing of some $4k$$4k$-manifold $V$$V$ having trivial signature. QED

For definition of the attaching invariant $E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q)$$E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q)$ see [Haefliger1966], [Skopenkov2005, $\S$$\S$3].

## 4 Classification

Theorem 4.1 [Levine1965, Corollary in p. 44], [Haefliger1966]. For $m-n\ge3$$m-n\ge3$ the group $E^m_D(S^n)$$E^m_D(S^n)$ is finite unless $n=4k-1$$n=4k-1$ and $m\le6k$$m\le6k$, when $E^m_D(S^n)$$E^m_D(S^n)$ is the sum of $\Z$$\Z$ and a finite group.

Theorem 4.2 (Haefliger-Milgram). We have the following table for the group $E^m_D(S^n)$$E^m_D(S^n)$; in the whole table $k\ge1$$k\ge1$; in the fifth column $k\ne2$$k\ne2$; in the last two columns $k\ge2$$k\ge2$:

$\displaystyle \begin{array}{c|c|c|c|c|c|c|c} (m,n) &2m\ge3n+4 &(6k,4k-1) &(6k+3,4k+1) &(7,4) &(6k+4,4k+2) &(12k+7,8k+4) & (12k+1,8k)\\ \hline E^m_D(S^n)&0 &\Z &\Z_2 &\Z_{12} &0 &\Z_4 &\Z_2\oplus\Z_2 \end{array}$

Proof for the first four columns, and for the fifth column when $k$$k$ is odd, are presented in [Haefliger1966, 8.15] (see also $\S$$\S$6; some proofs are deduced from that paper using simple calculations, cf. [Skopenkov2005, $\S$$\S$3]; there is a typo in [Haefliger1966, 8.15]: $C^{3k}_{4k-2}=0$$C^{3k}_{4k-2}=0$ should be $C^{4k}_{8k-2}=0$$C^{4k}_{8k-2}=0$). The remaining results follow from [Haefliger1966, 8.15] and [Milgram1972, Theorem F]. Alternative proofs for the cases $(m,n)=(7,4),(6,3)$$(m,n)=(7,4),(6,3)$ are given in [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

Theorem 4.3 [Milgram1972, Corollary G]. We have $E^m_D(S^n)=0$$E^m_D(S^n)=0$ if and only if either $2m\ge3n+4$$2m\ge3n+4$, or $(m,n)=(6k+4,4k+2)$$(m,n)=(6k+4,4k+2)$, or $(m,n)=(3k,2k)$$(m,n)=(3k,2k)$ and $k\equiv3,11\mod12$$k\equiv3,11\mod12$, or $(m,n)=(3k+2,2k+2)$$(m,n)=(3k+2,2k+2)$ and $k\equiv14,22\mod24$$k\equiv14,22\mod24$.

For a description of 2-components of $E^m_D(S^n)$$E^m_D(S^n)$ see [Milgram1972, Theorem F]. Observe that no reliable reference (containing complete proofs) of results announced in [Milgram1972] appeared. Thus, strictly speaking, the corresponding results are conjectures.

The lowest-dimensional unknown groups $E^m_D(S^n)$$E^m_D(S^n)$ are $E^8_D(S^5)$$E^8_D(S^5)$ and $E^{11}_D(S^7)$$E^{11}_D(S^7)$. Hopefully application of Kreck surgery could be useful to find these groups, cf. [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

For $m\ge n+3$$m\ge n+3$ the group $E^m_D(S^n)$$E^m_D(S^n)$ has been described as follows, in terms of exact sequences [Haefliger1966], cf. [Levine1965], [Haefliger1966a], [Milgram1972], [Habegger1986].

Theorem 4.4 [Haefliger1966]. For $q\ge3$$q\ge3$ there is the following exact sequence of abelian groups:

$\displaystyle \ldots \to \pi_{n+1}(SG,SO) \xrightarrow{~u~} E^{n+q}_D(S^n) \xrightarrow{~a~} \pi_n(SG_q,SO_q) \xrightarrow{~s~} \pi_n(SG,SO) \xrightarrow{~u~} E^{n+q-1}_D(S^{n-1})\to \ldots~.$

Here $SG_q$$SG_q$ is the space of maps $S^{q-1} \to S^{q-1}$$S^{q-1} \to S^{q-1}$ of degree $1$$1$. Restricting a map from $SO_q$$SO_q$ to $S^{q-1} \subset \Rr^q$$S^{q-1} \subset \Rr^q$ identifies $SO_q$$SO_q$ as a subspace of $SG_q$$SG_q$. Define $SG:=SG_1\cup\ldots\cup SG_q\cup\ldots$$SG:=SG_1\cup\ldots\cup SG_q\cup\ldots$. Analogously define $SO$$SO$. Let $s$$s$ be the stabilization homomorphism. The attaching invariant $a$$a$ and the map $u$$u$ are defined in [Haefliger1966], see also [Skopenkov2005, $\S$$\S$3].

## 5 Some remarks on codimension 2 knots

For the best known specific case, i.e. for codimension 2 embeddings of spheres (in particular, for the classical theory of knots in $\Rr^3$$\Rr^3$), a complete readily calculable classification (in the sense of Remark 1.2 of [Skopenkov2016c]) is neither known nor expected at the time of writing. However, there is a vast literature on codimension 2 knots. See e.g. the interesting papers [Farber1981], [Farber1983], [Kearton1983], [Farber1984].

On the other hand, if one studies embeddings up to the weaker relation of concordance, then much is known. See e.g. [Levine1969a] and [Ranicki1998].

## 6 Proof of classification of (4k-1)-knots in 6k-space

Theorem 6.1. The Haefliger invariant $\varkappa:E_D^{6k}(S^{4k-1})\to\Zz$$\varkappa:E_D^{6k}(S^{4k-1})\to\Zz$ is injective for $k>1$$k>1$.

The proof is a certain simplification of [Haefliger1962]. We present an exposition structured to make it more accessible to non-specialists.

Lemma 6.2. Let $V$$V$ be a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial B^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial B^{6k+1}$, signature of $V$$V$ is zero, and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then there is a submanifold $V'\subset B^{6k+1}$$V'\subset B^{6k+1}$ such that $V'\cong D^{4k}$$V'\cong D^{4k}$ and $\partial V'=\partial V$$\partial V'=\partial V$.

Proof of Theorem 6.1 using Lemma 6.2. By the first three paragraphs of the proof of Theorem 3.1 in [Haefliger1962], for any embedding $g:S^{4k-1}\to \partial B^{6k+1}$$g:S^{4k-1}\to \partial B^{6k+1}$ such that $\varkappa(g)=0$$\varkappa(g)=0$ there is a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold $V$$V$ of $B^{6k+1}$$B^{6k+1}$ with zero signature such that $g(S^{4k-1}) = \partial V \subset \partial B^{6k+1}$$g(S^{4k-1}) = \partial V \subset \partial B^{6k+1}$ and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then by Lemma 6.2 there is a submanifold $V'\subset B^{6k+1}$$V'\subset B^{6k+1}$ such that $D^{4k} \cong V'$$D^{4k} \cong V'$ and $\partial V'=\partial V$$\partial V'=\partial V$. Recall that isotopy classes of embeddings $S^q\to S^n$$S^q\to S^n$ are in 1--1 correspondence with $h$$h$-cobordism classes of oriented submanifolds of $S^n$$S^n$ diffeomorphic to $S^q$$S^q$ for $n\ge5$$n\ge5$, $n\ge q+3$$n\ge q+3$, cf. [Haefliger1966, 1.8], [Kervaire1965]. Hence $g$$g$ is isotopic to standart embedding. $\Box$$\Box$

To prove Lemma 6.2 we need Lemmas 6.3, 6.4 and 6.5. Below manifolds can have non-empty boundaries.

Lemma 6.3 [Whitney lemma; [Prasolov2007], $\S$$\S$22]. Let $u: P \rightarrow W$$u: P \rightarrow W$ be a map from a connected oriented $p$$p$-manifold $P$$P$ to a simply connected oriented $(p+q)$$(p+q)$-manifold $W$$W$. If $p, q \geq 3$$p, q \geq 3$, then

1. If $q \geq p$$q \geq p$, there is a homotopy $u_t$$u_t$ such that $u_0 = u$$u_0 = u$ and $u_1(P)$$u_1(P)$ is an embedding.
2. Suppose in addition that $u(\partial P) \subseteq \partial W$$u(\partial P) \subseteq \partial W$ and there is a map $v: Q \rightarrow W$$v: Q \rightarrow W$ with $v(\partial Q) \subseteq \partial W$$v(\partial Q) \subseteq \partial W$ from a connected oriented $q$$q$-manifold $Q$$Q$ such that the algebraic intersection number of $u(P)$$u(P)$ and $v(Q)$$v(Q)$ is zero. Then there is a homotopy $v_t$$v_t$ relative to the boundary such that $v_0 = v$$v_0 = v$ and $v_1(Q)$$v_1(Q)$ does not intersect $u(P)$$u(P)$. If $v$$v$ is an embedding, the homotopy $v_t$$v_t$ can be chosen so that $v_1$$v_1$ is an embedding.

Below we denote by $h: \pi_{m}(\cdot) \rightarrow H_{m}(\cdot)$$h: \pi_{m}(\cdot) \rightarrow H_{m}(\cdot)$ the Hurewicz map.

Lemma 6.4. Let $V$$V$ be a $(2k-1)$$(2k-1)$-connected $4k$$4k$-manifold, and let $x_1, \ldots, x_{s} \in H_{2k}(V)$$x_1, \ldots, x_{s} \in H_{2k}(V)$ be homology classes such that $x_i \cap_V x_j = 0$$x_i \cap_V x_j = 0$ for every $i,j$$i,j$. Then there are embeddings $g_1, \ldots, g_s: \, S^{2k} \rightarrow V$$g_1, \ldots, g_s: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_s$$x_1, \ldots, x_s$, respectively.

Proof. As $V$$V$ is $(2k-1)$$(2k-1)$-connected, $h: \pi_{2k}(V) \rightarrow H_{2k}(V)$$h: \pi_{2k}(V) \rightarrow H_{2k}(V)$ is an isomorphism. For an element $x_i \in H_{2k}(V)$$x_i \in H_{2k}(V)$, let $\widetilde{x_i}: S^{2k} \rightarrow V$$\widetilde{x_i}: S^{2k} \rightarrow V$ be a representative of the homotopy class $h^{-1}(x_i)$$h^{-1}(x_i)$. Applying item 1 of Lemma 6.3 to $\widetilde{x_i}$$\widetilde{x_i}$, we may assume that $\widetilde{x_i}$$\widetilde{x_i}$ is an embedding.

Make the following inductive procedure. At the $i$$i$-th step, $i=1,\ldots,s$$i=1,\ldots,s$, assume that the embeddings $g_1, \ldots, g_{i-1}$$g_1, \ldots, g_{i-1}$ are already constructed, and we construct $g_i$$g_i$. Since $2k \geq 3$$2k \geq 3$ and $V$$V$ is simply connected, $W:=V \setminus \bigcup\limits_{l$W:=V \setminus \bigcup\limits_{l is simply connected for any $j < i$$j < i$. The algebraic intersection number of $g_j(S^{2k})$$g_j(S^{2k})$ and $\widetilde{x_i}(S^{2k})$$\widetilde{x_i}(S^{2k})$ is zero for any $j$$j$. Hence we can apply item 2 of Lemma 6.3 to $u=g_j$$u=g_j$ and $v=\widetilde{x_i}$$v=\widetilde{x_i}$ and $W$$W$ as above for any $j < i$$j < i$. So $\widetilde{x_i}$$\widetilde{x_i}$ is replaced by a homotopic embedding $g_i$$g_i$, and the images of $g_1, \ldots, g_i$$g_1, \ldots, g_i$ are pairwise disjoint. After $s$$s$-th step we obtain a required set of embeddings.

$\square$$\square$

Lemma 6.5. Let $V$$V$ be an orientable $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$, and $g:D^{2k+1}\to B^{6k+1}$$g:D^{2k+1}\to B^{6k+1}$ be an embedding such that $g(D^{2k+1})\cap V = g(S^{2k})$$g(D^{2k+1})\cap V = g(S^{2k})$ and over $g(S^{2k})$$g(S^{2k})$ the manifold $V$$V$ has a framing whose first vector is tangent to $g(D^{2k+1})$$g(D^{2k+1})$. Assume that $g(S^{2k})$$g(S^{2k})$ has zero algebraic self-intersection in $V$$V$. Then $g$$g$ extends to an embedding $G: D^{2k+1}\times D^{2k}\to B^{6k+1}$$G: D^{2k+1}\times D^{2k}\to B^{6k+1}$ such that $G(S^{2k}\times D^{2k})\subset V$$G(S^{2k}\times D^{2k})\subset V$.

Proof. (A slightly different proof is presented in the proof of Proposition 3.3 in [Haefliger1962].) Since $g(S^{2k})$$g(S^{2k})$ has zero algebraic self-intersection in $V$$V$, the Euler class of the normal bundle of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is zero. Since over $g(S^{2k})$$g(S^{2k})$ the manifold $V$$V$ has a framing, we obtain that $g(S^{2k})$$g(S^{2k})$ has a framing in $V$$V$.

Identify all the normal spaces of $G(D^{2k+1})$$G(D^{2k+1})$ with the normal space at $G(0)$$G(0)$. The normal framing $a$$a$ of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is orthogonal to $G(D^{2k+1})$$G(D^{2k+1})$. So $a$$a$ defines a map $S^{2k}\to V_{4k,2k}$$S^{2k}\to V_{4k,2k}$. Let $\zeta \in\pi_{2k}(V_{4k,2k})$$\zeta \in\pi_{2k}(V_{4k,2k})$ be the homotopy class of this map. This is the obstruction to extending $a$$a$ to a normal $2k$$2k$-framing of $G$$G$ in $B^{6k+1}$$B^{6k+1}$ (so apriori $\zeta=\zeta(a)$$\zeta=\zeta(a)$). It suffices to prove that $\zeta=0$$\zeta=0$.

Consider the exact sequence of the bundle $SO_{a+b}/SO_b = V_{a+b,a}$$SO_{a+b}/SO_b = V_{a+b,a}$: $\pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b)$$\pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b)$. By the following well-known assertion, $\partial\zeta=0$$\partial\zeta=0$: if $\alpha\in\pi_q(V_{a,b})$$\alpha\in\pi_q(V_{a,b})$, then $\partial\alpha$$\partial\alpha$ is the obstruction to trivialization of the orthogonal complement to the field of $b$$b$-frames in $S^q\times\R^a$$S^q\times\R^a$ corresponding to a representative of $\alpha$$\alpha$.

Consider a map of the exact sequences associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \to V_{4k,2k} = SO_{4k}/SO_{2k}$$S^{2k} = SO_{2k+1}/SO_{2k} \to V_{4k,2k} = SO_{4k}/SO_{2k}$. The composition $\pi_{2k}(S^{2k})\overset{i}\to\pi_{2k}(V_{4k,2k})\overset{\partial}\to\pi_{2k-1}(SO_{2k})$$\pi_{2k}(S^{2k})\overset{i}\to\pi_{2k}(V_{4k,2k})\overset{\partial}\to\pi_{2k-1}(SO_{2k})$ is the boundary map $\partial'$$\partial'$. The group $\pi_{2k-1}(SO_{2k})$$\pi_{2k-1}(SO_{2k})$ is in natural 1--1 correspondence with the group of $2k$$2k$-bundles over $S^{2k}$$S^{2k}$. The image $\partial'\iota_{2k}$$\partial'\iota_{2k}$ is the tangent bundle $\tau$$\tau$ of $S^{2k}$$S^{2k}$. Since the Euler class of $n\tau$$n\tau$ is $2n\ne0$$2n\ne0$, the map $\partial'$$\partial'$ is injective. Since $i$$i$ is an isomorphism, the map $\partial$$\partial$ is injective. This and $\partial\zeta=0$$\partial\zeta=0$ imply that $\zeta=0$$\zeta=0$.

Alternatively, by [Fomenko&Fuchs2016, Corollary in $\S$$\S$25.4] $\pi_{2k}(SO_{4k})$$\pi_{2k}(SO_{4k})$ is a finite group (in [Fomenko&Fuchs2016, Corollary in $\S$$\S$25.4] the formula for $\pi_q(SO_{2m})$$\pi_q(SO_{2m})$ is correct, although the formula for $\pi_q(SO_{2m+1})$$\pi_q(SO_{2m+1})$ is incorrect because $\pi_3(SO_3)\cong\Z\oplus\Z$$\pi_3(SO_3)\cong\Z\oplus\Z$). Since $\pi_{2k}(V_{4k,2k})\cong\Z$$\pi_{2k}(V_{4k,2k})\cong\Z$, we obtain that $j=0$$j=0$ for $a=b=2k$$a=b=2k$. This and $\partial\zeta=0$$\partial\zeta=0$ imply that $\zeta=0$$\zeta=0$.

$\square$$\square$

Lemma 6.6. Let $V$$V$ be a $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$ and let $G:D^{2k+1}\times D^{2k}\to B^{6k+1}$$G:D^{2k+1}\times D^{2k}\to B^{6k+1}$ be an embedding such that $G(S^{2k}\times D^{2k})\subset V$$G(S^{2k}\times D^{2k})\subset V$. Then there is a smooth submanifold $V'\subset B^{6k+1}$$V'\subset B^{6k+1}$ homeomorphic to $V\backslash ( G(S^{2k}\times D^{2k})) \bigcup G(D^{2k+1}\times S^{2k-1})$$V\backslash ( G(S^{2k}\times D^{2k})) \bigcup G(D^{2k+1}\times S^{2k-1})$ and such that $V\backslash G(D^{2k+1}\times D^{2k})=V'\backslash G(D^{2k+1}\times D^{2k})$$V\backslash G(D^{2k+1}\times D^{2k})=V'\backslash G(D^{2k+1}\times D^{2k})$.

Lemma 6.6 is essentialy proved in [Haefliger1962, $\S$$\S$3.3].

Proof of Lemma 6.2 using Lemmas 6.4, 6.5. By the fourth paragraph of the proof of Theorem 3.1 in [Haefliger1962], there is a basis $\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$$\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$ in $H_{2k}(V)$$H_{2k}(V)$ such that $\alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0$$\alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0$, $\alpha_i\cap \beta_j=\delta_{i, j}$$\alpha_i\cap \beta_j=\delta_{i, j}$ and $\lambda^*(\alpha_i)=0$$\lambda^*(\alpha_i)=0$ for any $i, j$$i, j$. From Lemma 6.4 it follows that there are embeddings $f_1, \ldots, f_s:S^{2k}\to V$$f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images representing $x_1, \ldots, x_s$$x_1, \ldots, x_s$, respectively. [!!!such that $f_{i*}[S^{2k}]=\alpha_i$$f_{i*}[S^{2k}]=\alpha_i$ for every $i=1,\ldots,s$$i=1,\ldots,s$]

For $i=1,\ldots,s$$i=1,\ldots,s$ denote by $\alpha_i'\in\pi_{2k}(B^{6k+1}\backslash V)$$\alpha_i'\in\pi_{2k}(B^{6k+1}\backslash V)$ the homotopy class of the shift of $f_i$$f_i$ by the first vector of the framing of $V$$V$ on $f_i(S^{2k})$$f_i(S^{2k})$. Since $\lambda^*(\alpha_i)=0$$\lambda^*(\alpha_i)=0$, we have $h\alpha_i' =0 \in H_{2k}(B^{6k+1}\backslash V)$$h\alpha_i' =0 \in H_{2k}(B^{6k+1}\backslash V)$. Since $\mbox{dim} B^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$$\mbox{dim} B^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, the complement $B^{6k+1}\backslash V$$B^{6k+1}\backslash V$ is $(2k-1)$$(2k-1)$-connected. Hence by Hurewicz Theorem $h\alpha_i'=0$$h\alpha_i'=0$ implies $\alpha_i'=0$$\alpha_i'=0$. Therefore there are extensions $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$$g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ of $f_1, \ldots, f_s$$f_1, \ldots, f_s$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$.

Take $\varepsilon>0$$\varepsilon>0$ such that $g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V$$g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V$ for any $i\leq s$$i\leq s$. Take a tubular neighborhoods $U_i$$U_i$ of $g_i(S^{2k})$$g_i(S^{2k})$ such that $g_i(D^{2k+1})\backslash U_i=g_i(\varepsilon D^{2k+1})$$g_i(D^{2k+1})\backslash U_i=g_i(\varepsilon D^{2k+1})$ for any $i\leq s$$i\leq s$. The algebraic intersection number of $g_i(\varepsilon D^{2k+1})$$g_i(\varepsilon D^{2k+1})$ and $V\backslash U_i$$V\backslash U_i$ equals $\lambda^*(\alpha_i)=\lambda^*(g_{i*}[S^{2k}])=0$$\lambda^*(\alpha_i)=\lambda^*(g_{i*}[S^{2k}])=0$. We have $\pi_1(B^{6k+1}\backslash U_i)=\pi_1(B^{6k+1}\backslash g_i(S^{2k}))=0$$\pi_1(B^{6k+1}\backslash U_i)=\pi_1(B^{6k+1}\backslash g_i(S^{2k}))=0$. So we can apply item 2 of Lemma 6.3 to $v=g_i|_{\varepsilon D^{2k+1}}$$v=g_i|_{\varepsilon D^{2k+1}}$, $k:V\backslash U_i\to B^{6k+1}\backslash U_i$$k:V\backslash U_i\to B^{6k+1}\backslash U_i$ the inclusion, and $W=B^{6k+1}\backslash U_i$$W=B^{6k+1}\backslash U_i$. So we may assume that $g_i(\varepsilon D^{2k+1})$$g_i(\varepsilon D^{2k+1})$ does not intersect $V \backslash U_i$$V \backslash U_i$. Hence we may assume that $g_i(D^{2k+1})\cap V=g_i(S^{2k})$$g_i(D^{2k+1})\cap V=g_i(S^{2k})$.

Apply Lemma 6.5 to $g=g_1, \ldots, g_s$$g=g_1, \ldots, g_s$ one by one, and to the manifold $V$$V$. Denote by $G_1, \ldots, G_s$$G_1, \ldots, G_s$ the resulting maps. Define manifolds $V^{i}$$V^{i}$ for $0\leq i\leq s$$0\leq i\leq s$ inductively. Let $V^0:=V$$V^0:=V$, and let $V^{i}$$V^{i}$ be a manifold $V'$$V'$ obtained applying Lemma 6.6 for $V=V^{i-1}$$V=V^{i-1}$ and $G=G_i$$G=G_i$. By Lemma 6.6, $\pi_1(V)=\pi_1(V^s)= 0$$\pi_1(V)=\pi_1(V^s)= 0$ and $H_j(V)=H_j(V^s)= 0$$H_j(V)=H_j(V^s)= 0$ for $j<2k$$j<2k$. Since $\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$$\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$ is a symplectic basis in $H_{2k}(V)$$H_{2k}(V)$, it follows that $H_{2k}(V^s)= 0$$H_{2k}(V^s)= 0$. Then from Generalized Poincare conjecture proved by Smale it follows that $V^s\cup_{\partial V^s=\partial D^{4k}} D^{4k}\cong S^{4k}$$V^s\cup_{\partial V^s=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $D^{4k} \cong V^s$$D^{4k} \cong V^s$. Then take $V':=V^s$$V':=V^s$. $\Box$$\Box$

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