Knots, i.e. embeddings of spheres

2 Examples

There are smooth embeddings $S^{2l-1}\to\Rr^{3l}$ 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]).

3 Invariants

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

$$\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.$$

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

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

For an alternative definition via Seifert surfaces in $6k$-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$4].

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

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

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

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

4 Classification

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

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

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

For a description of 2-components of $E^m_D(S^n)$ 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 $E^m_D(S^n)$ are $E^8_D(S^5)$ and $E^{11}_D(S^7)$. Hopefully application of Kreck surgery could be useful to find these groups, cf. [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

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

Theorem 4.4 [Haefliger1966]. For $q\ge3$ 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 $SG_q$ is the space of maps $S^{q-1} \to S^{q-1}$ of degree $1$. Restricting a map from $SO_q$ to $S^{q-1} \subset \Rr^q$ identifies $SO_q$ as a subspace of $SG_q$. Define $SG:=SG_1\cup\ldots\cup SG_q\cup\ldots$. Analogously define $SO$. Let $s$ be the stabilization homomorphism. The attaching invariant $a$ and the map $u$ are defined in [Haefliger1966], see also [Skopenkov2005, $\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 $\Rr^3$), 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

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

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

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

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

Proof. As $V$ is $(2k-1)$-connected, $h: \pi_{2k}(V) \rightarrow H_{2k}(V)$ is an isomorphism. For an element $x_i \in H_{2k}(V)$, let $\widetilde{x_i}: S^{2k} \rightarrow V$ be a representative of the homotopy class $h^{-1}(x_i)$. Make the following inductive procedure. At the $i$-th step, $i=1,\ldots,s$, assume that the maps $g_1, \ldots, g_{i-1}$ are already constructed, and we construct $g_i$.

Applying item 1 of Lemma 6.3 to $\widetilde{x_i}$, we may assume that $\widetilde{x_i}$ is an embedding.

Since $2k \geq 3$ and $V$ is simply connected, $W:=V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected for any $j$. The algebraic intersection number of $g_j(S^{2k})$ and $\widetilde{x_i}(S^{2k})$ is zero for any $j$. Hence we can apply item 2 of Lemma 6.3 to $u=g_j$ and $v=\widetilde{x_i}$ and $W$ as above for any $j < i$. As the result, $\widetilde{x_i}$ is replaced by a homotopic embedding $g_i$, and the images of $g_1, \ldots, g_i$ are pairwise disjoint.

After the step $s$ we obtain a required set of embeddings.

$\square$

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

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

Consider the exact sequence of the bundle $SO_{a+b}/SO_b = V_{a+b,a}$: $\pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b)$. By the following well-known assertion, $\partial\zeta=0$:

if $\alpha\in\pi_q(V_{a,b})$, then $\partial\alpha$ is the obstruction to trivialization of the orthogonal complement to the field of

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$b$-frames in $S^q\times\R^a$ corresponding to a representative of $\alpha$.

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

$\square$

Lemma 6.6 is essentialy proved in [Haefliger1962, $\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 $\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$ in $H_{2k}(V)$ such that $\alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0$, $\alpha_i\cap \beta_j=\delta_{i, j}$ and $\lambda^*(\alpha_i)=0$ for any $i, j$. From Lemma 6.4 it follows that there are embeddings $f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images representing $x_1, \ldots, x_s$, respectively. [!!!such that $f_{i*}[S^{2k}]=\alpha_i$ for every $i=1,\ldots,s$]

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

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

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

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2 Examples

There are smooth embeddings $S^{2l-1}\to\Rr^{3l}$ 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]).

3 Invariants

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

$$\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.$$

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

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

For an alternative definition via Seifert surfaces in $6k$-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$4].

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

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

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

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

4 Classification

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

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

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

For a description of 2-components of $E^m_D(S^n)$ 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 $E^m_D(S^n)$ are $E^8_D(S^5)$ and $E^{11}_D(S^7)$. Hopefully application of Kreck surgery could be useful to find these groups, cf. [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

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

Theorem 4.4 [Haefliger1966]. For $q\ge3$ 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 $SG_q$ is the space of maps $S^{q-1} \to S^{q-1}$ of degree $1$. Restricting a map from $SO_q$ to $S^{q-1} \subset \Rr^q$ identifies $SO_q$ as a subspace of $SG_q$. Define $SG:=SG_1\cup\ldots\cup SG_q\cup\ldots$. Analogously define $SO$. Let $s$ be the stabilization homomorphism. The attaching invariant $a$ and the map $u$ are defined in [Haefliger1966], see also [Skopenkov2005, $\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 $\Rr^3$), 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

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

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

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

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

Proof. As $V$ is $(2k-1)$-connected, $h: \pi_{2k}(V) \rightarrow H_{2k}(V)$ is an isomorphism. For an element $x_i \in H_{2k}(V)$, let $\widetilde{x_i}: S^{2k} \rightarrow V$ be a representative of the homotopy class $h^{-1}(x_i)$. Make the following inductive procedure. At the $i$-th step, $i=1,\ldots,s$, assume that the maps $g_1, \ldots, g_{i-1}$ are already constructed, and we construct $g_i$.

Applying item 1 of Lemma 6.3 to $\widetilde{x_i}$, we may assume that $\widetilde{x_i}$ is an embedding.

Since $2k \geq 3$ and $V$ is simply connected, $W:=V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected for any $j$. The algebraic intersection number of $g_j(S^{2k})$ and $\widetilde{x_i}(S^{2k})$ is zero for any $j$. Hence we can apply item 2 of Lemma 6.3 to $u=g_j$ and $v=\widetilde{x_i}$ and $W$ as above for any $j < i$. As the result, $\widetilde{x_i}$ is replaced by a homotopic embedding $g_i$, and the images of $g_1, \ldots, g_i$ are pairwise disjoint.

After the step $s$ we obtain a required set of embeddings.

$\square$

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

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

Consider the exact sequence of the bundle $SO_{a+b}/SO_b = V_{a+b,a}$: $\pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b)$. By the following well-known assertion, $\partial\zeta=0$:

if $\alpha\in\pi_q(V_{a,b})$, then $\partial\alpha$ is the obstruction to trivialization of the orthogonal complement to the field of

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$b$-frames in $S^q\times\R^a$ corresponding to a representative of $\alpha$.

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

$\square$

Lemma 6.6 is essentialy proved in [Haefliger1962, $\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 $\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s$ in $H_{2k}(V)$ such that $\alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0$, $\alpha_i\cap \beta_j=\delta_{i, j}$ and $\lambda^*(\alpha_i)=0$ for any $i, j$. From Lemma 6.4 it follows that there are embeddings $f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images representing $x_1, \ldots, x_s$, respectively. [!!!such that $f_{i*}[S^{2k}]=\alpha_i$ for every $i=1,\ldots,s$]

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

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

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

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