Knots, i.e. embeddings of spheres

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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$${{Stub}} == Introduction == <wikitex>; We work in a smooth category. In particular, terms embedding and smooth embedding or map and smooth map are used interchangeably. For a [[Embeddings_in_Euclidean_space:_an_introduction_to_their_classification#Introduction|general introduction to embeddings]] as well as the [[Embeddings_in_Euclidean_space:_an_introduction_to_their_classification#Notation and conventions|notation and conventions]] used on this page, we refer to \cite[$\S\S$1, $\S$$\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

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

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

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

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$:

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

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:

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

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$$4k$-submanifold of $D^{6k+1}$$D^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$, signature of $V$$V$ is zero, and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then there is an embedding $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G(\partial D^{4k})=\partial V$$G(\partial D^{4k})=\partial V$.

Proof of Theorem 6.1 using Lemma 6.2. By the first three paragraphs of the proof of Theorem 3.1 in [Ha62], for any embedding $g:S^{4k-1}\to \partial D^{6k+1}$$g:S^{4k-1}\to \partial D^{6k+1}$ such that $\varkappa(g)=0$$\varkappa(g)=0$ there is a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold $V$$V$ of $D^{6k+1}$$D^{6k+1}$ with zero signature such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$ and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then from Lemma 6.2 it follows that there is an extension $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G|_{S^{4k-1}} = g$$G|_{S^{4k-1}} = g$. From Smale Theorem it follows that $g$$g$ is isotopic to the standard embedding. $\Box$$\Box$

To prove Lemma 6.2 we need Lemma 6.3, Lemma 6.4 and Lemma 6.5.

All the manifolds below can have non-empty boundaries.!!!

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

1. There is a homotopy $k_t$$k_t$ such that $k_0 = k$$k_0 = k$ and $k_1(P)$$k_1(P)$ is an embedding.
2. Suppose in addition that there is a map $y: Q \rightarrow W$$y: Q \rightarrow W$ from a connected $q$$q$-manifold $Q$$Q$ such that the algebraic intersection number of $k(P)$$k(P)$ and $y(Q)$$y(Q)$ is zero. Then there is a homotopy $y_t$$y_t$ relative to the boundary such that $y_0 = y$$y_0 = y$ and $y_1(Q)$$y_1(Q)$ does not intersect $k(P)$$k(P)$. If $y$$y$ is an embedding, the homotopy $y_t$$y_t$ can be chosen so that $y_1$$y_1$ is an embedding.

Lemma 6.4. Let $V$$V$ be a $(2k-1)$$(2k-1)$-connected $4k$$4k$-manifold, and $x_1, \ldots, x_m \in H_{2k}(V)$$x_1, \ldots, x_m \in H_{2k}(V)$ are 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_m: \, S^{2k} \rightarrow V$$g_1, \ldots, g_m: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_m$$x_1, \ldots, x_m$ [such that $g_i(S^{2k})$$g_i(S^{2k})$ represents!!! $x_i$$x_i$ for every $i$$i$].

Proof. As $V$$V$ is $(2k-1)$$(2k-1)$-connected, the Hurewicz map $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 $v_i: S^{2k} \rightarrow V$$v_i: S^{2k} \rightarrow V$ represent the homotopy class $h^{-1}(x_i)$$h^{-1}(x_i)$. Now we perform the following inductive procedure. At the $i$$i$-th step of the procedure assume that the maps $g_j$$g_j$ from the assertion are already constructed for any $j < i$$j < i$, and we construct $g_i$$g_i$. First, we apply item 1 of Lemma 6.3 to $v_i$$v_i$. Observe that item 1 of Lemma 6.3 is applicable as $V$$V$ is simply-connected. Therefore, we may suppose that $v_i$$v_i$ itself is an embedding. Further, we apply item 2 of Lemma 6.3, setting $g_j$$g_j$ as $k$$k$, $v_i$$v_i$ as $y$$y$, and $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ as $W$$W$, for any $j < i$$j < i$. Observe that Lemma 6.3 is applicable as $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected provided $2k \geq 3$$2k \geq 3$ and $V$$V$ is simply connected. Also, as $x_i \cap_V x_j = 0$$x_i \cap_V x_j = 0$ for any $i$$i$, $j$$j$, the algebraic intersection number for all the pairs of maps above is zero. As the result, $v_i$$v_i$ is replaced with a homotopic embedding $g_i$$g_i$, and the images of $g_1, \ldots, g_i$$g_1, \ldots, g_i$ are disjoint. After the step $m$$m$ we obtain the desired set of embeddings.

$\square$$\square$

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let $V$$V$ be an orientable $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$, and $g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$$g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$ be embeddings with pairwise disjoint images such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$. Assume that either $g_i(S^{2k})$$g_i(S^{2k})$ has zero algebraic self-intersection in $V$$V$ for every $i$$i$, or $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$ for every $i$$i$, or $V$$V$ has a framing in $B^{6k+1}$$B^{6k+1}$. Then there are embeddings $G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$$G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$ extending $g_1,\ldots,g_s$$g_1,\ldots,g_s$ such that $G_i(S^{2k}\times D^{2k})\subset V$$G_i(S^{2k}\times D^{2k})\subset V$ for every $i$$i$.

Moreover, one can smoothen $V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$$V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ to obtain a smooth submanifold of $B^{6k+1}$$B^{6k+1}$.

Proof. By general position, it suffices to prove the existence of $G_1,\ldots,G_s$$G_1,\ldots,G_s$ for $s=1$$s=1$. So assume that $s=1$$s=1$. Since $g_1(S^{2k})$$g_1(S^{2k})$ has zero algebraic self-intersection in $V$$V$, the Euler class of the normal bundle of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ is zero. Hence $g_1(S^{2k})$$g_1(S^{2k})$ has a framing in $V$$V$.

In this paragraph we define the obstruction $\zeta\in\pi_{2k}(V_{4k,2k})$$\zeta\in\pi_{2k}(V_{4k,2k})$ to extending a normal framing of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ to a normal $2k$$2k$-framing of $G_1$$G_1$ in $B^{6k+1}$$B^{6k+1}$. Identify all the normal spaces!!! of $G_1(D^{2k+1})$$G_1(D^{2k+1})$ with the normal space at $G(0)$$G(0)$. The framing of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is orthogonal to $G(D^{2k+1})$$G(D^{2k+1})$. The image of this framing under the identification defines the required obstruction $\zeta$$\zeta$.

old Consider the exact sequence of the bundle $SO_{2k}\to SO_{4k}\to V_{4k,2k}$$SO_{2k}\to SO_{4k}\to V_{4k,2k}$. Since $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$, we have $\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$$\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$, cf. [Haefliger1962, 3.4, Lemma]. For $k$$k$ odd from the exact sequence of the bundle we obtain $\zeta=0$$\zeta=0$.

Consider a part of the long exact sequence associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$$S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k})\ar[r]^{i} \ar[d]^{\equiv} & \pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[r]^{\partial}\ar[d] & \pi_{2k-1}(SO_{2k})\ar[d]^{\equiv} \\ \pi_{2k}(SO_{2k})\ar[r]^{i'} & \pi_{2k}(SO_{6k+1})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ From Milnor [??, proof of 5.11] it follows that $\partial$$\partial$ is injective, therefore $i$$i$ is surjective. As $s$$s$ is surjective, from commutativity of the diagram $i'$$i'$ is also surjective. Therefore, $\partial '$$\partial '$ is injective. By Lemma 3.4 from [Ha62], $\partial' \zeta$$\partial' \zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$$g(S^{2k})$, which is trivial. Therefore, $\partial ' \zeta =0$$\partial ' \zeta =0$ and $\zeta = 0$$\zeta = 0$ as $\partial '$$\partial '$ is injective. Thus, the obstruction is always zero.

$\square$$\square$

Proof. Let us smoothen $V'$$V'$.

Fix some $i$$i$. From $\zeta_i=0$$\zeta_i=0$ it follows that there are orthonormal vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to $v_j$$v_j$ along $\phi_i^{-1}(g_i(S^{2k}))$$\phi_i^{-1}(g_i(S^{2k}))$ for any $j\leq 2k$$j\leq 2k$. Since every fiber of $D{'}^{2k}$$D{'}^{2k}$ is trivial, there are orthonormal vector fields $v_{2k+1}, \ldots, v_{4k}$$v_{2k+1}, \ldots, v_{4k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $v_j\perp v_m$$v_j\perp v_m$ for any $j\leq 2k$$j\leq 2k$ and $2k<m\leq 4k$$2k<m\leq 4k$. Let us choose an extensions of $v_1,\ldots, v_{4k}$$v_1,\ldots, v_{4k}$ to smole neighborhood of $D'$$D'$ such that

1. $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$;
2. the map $\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$$\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$, defined by formula $\psi(x, y)=\gamma_y(||y||)$$\psi(x, y)=\gamma_y(||y||)$, where $x\in D{'}^{2k}$$x\in D{'}^{2k}$, $y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$$y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$ and where $\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$$\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$ is a solution of Cauchy problem

$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$ is the diffeomorphism between $D{'}^{2k}\times \mathbb{R}^{4k}$$D{'}^{2k}\times \mathbb{R}^{4k}$ and image of $\psi$$\psi$.

1. $\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$$\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$.

We take as the extension of $g_i$$g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$$G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$$\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$$-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$.

$\square$$\square$

Below the symbol $[ \cdot ]$$[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.

Proof of Lemma 6.2 using Lemmas 6.4, 6.5.. By the fourth paragraph of the proof of Theorem 3.1 in [Ha62], there is a basis $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ in $H_{2k}(V)$$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$$f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images such that $f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$$f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$.

Denote by $f_{i}':S^{2k}\to D^{6k+1}\backslash V$$f_{i}':S^{2k}\to D^{6k+1}\backslash V$ for $i\leq s$$i\leq s$ the result of shifting of $f^i$$f^i$ by the first vector of the framing of $V$$V$. Since $\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$$\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$, we have $f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$. Since $\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$$\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, we have $\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$$\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$ for $m<2k$$m<2k$. Since $f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$, we have $[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$$[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$. Therefore there are extensions of $f_1, \ldots, f_s$$f_1, \ldots, f_s$ to maps!!! $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$$g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$.

Denote by $V'$$V'$ the manifold as in Lemma 6.5 for $V$$V$ and $g_1, \ldots, g_s$$g_1, \ldots, g_s$ as described above UNAPPLICABLE. Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$$H_{2k}(V)$, it follows that $H_{2k}(V')\simeq 0$$H_{2k}(V')\simeq 0$. Hence $\pi_1(V')\simeq 0$$\pi_1(V')\simeq 0$ and $H_m(V')\simeq 0$$H_m(V')\simeq 0$ for $0< m\leq 2k$$0< m\leq 2k$. From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$$V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $V'\cong D^{4k}$$V'\cong D^{4k}$. Then we set by $G$$G$ a diffeomorphism between $D^{4k}$$D^{4k}$ and $V'\subset D^{6k+1}$$V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$$G|_{S^{4k-1}}=g$. $\Box$$\Box$

References

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, $\S]{Skopenkov2016c}. == 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 [[Embeddings_in_Euclidean_space:_an_introduction_to_their_classification#Unknotting_theorems|Unknotting Spheres Theorem 2.3]] of \cite{Skopenkov2016c} and \cite[Remark 1.1]{Skopenkov2016f}). {{beginthm|Example}}\label{e:gen} (a) Analogously to [[3-manifolds_in_6-space#Examples|the Haefliger trefoil knot]] for any $l>1$ one constructs a smooth embedding $t:S^{2l-1}\to\Rr^{3l}$, see \cite[$\S]{Skopenkov2016h}. For $l$ even $t$ is not smoothly isotopic to the standard embedding; $t$ represents a generator of $E_D^{3l}(S^{2l-1})\cong\Zz$ \cite{Haefliger1962}. It would be interesting to know if for $l>1$ odd this embedding is a generator of $E_D^{3l}(S^{2l-1})\cong\Zz_2$. The last phrase of \cite{Haefliger1962t} suggests that this is true for $l=3$. (b) For any $k=1,3,7$ let $\eta\in\pi_{4k-1}(S^{2k})$ be the homotopy class of the Hopf map. Denote by $\zeta:\pi_{4k-1}(S^{2k})\to E_D^{6k}(S^{4k-1}\sqcup S^{4k-1})$ [[High_codimension_links#Examples|the Zeeman map]], see \cite[Definition 2.2]{Skopenkov2016h}. The embedded connected sum $\#\zeta\eta$ of the components of (a representative of) $\zeta\eta$ is not smoothly isotopic to the standard embedding; $\#\zeta\eta$ is a generator of $E_D^{6k}(S^{4k-1})\cong\Z$ \cite[Corollary 2.13]{Skopenkov2015a}. {{endthm}} == 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 m\ge3n+4$, and by analyzing what obstructs carrying this proof for m=3n+3$. By \cite[2.1, 2.2]{Haefliger1962} any embedding $f:S^{4k-1}\to S^{6k}$ has a framing extendable to a framed embedding $\overline f:V\to D^{6k+1}$ of a k$-manifold $V$ whose boundary is $S^{4k-1}$, and whose signature is zero. For an integer k$-cycle $c$ in $V$ let $\lambda^*(c)\in\Z$ be the linking number of $f(V)$ with a slight shift of $\overline f(c)$ along the first vector of the framing. This defines a map $\lambda^*:H_{2k}(V;\Z)\to\Z$. This map is a homomorphism (as opposed to the Arf map defined in a similar way \cite{Pontryagin1959}). Then by Lefschetz duality there is a unique $\lambda\in H_{2k}(V,\partial;\Z)$ such that $\lambda^*[c]=\lambda\cap_V[c]$ for any $[c]\in H_{2k}(V;\Z)$. Since $V$ has a normal framing, its intersection form $\cap_V$ is even. (Indeed, represent a class in $H_{2k}(V;\Z)$ by a closed oriented k$-submanifold $c$. Then $\rho_2[c]\cap_V[c]=\overline{w_{2k}}(c\subset V)=\rho_2[c]\cap_VPDw_{2k}(V)=0$ because $V$ has a normal framing.) Hence $\lambda\cap_V\lambda$ is an even integer. Define $$\varkappa(f):=\lambda\cap_V\lambda/2.$$ 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 $$\varkappa(f) = \sum\limits_{j=1}^s \lambda^*(\beta_j)\lambda^*(\alpha_j).$$ For an alternative definition via Seifert surfaces in k$-space, discovered in \cite{Guillou&Marin1986}, \cite{Takase2004}, see \cite[the Kreck Invariant Lemma 4.5]{Skopenkov2016t}. For a definition by Kreck, and for a generalization to 3-manifolds see \cite[$\S]{Skopenkov2016t}. ''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 $f$.'' \cite[Theorem 2.6]{Haefliger1962} Analogously one defines $\lambda(V)$ and $\varkappa(V):=\lambda(V)\cap_V\lambda(V)/2$ for a framed k$-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 \cite[2.9]{Haefliger1962} (cf. \cite[2.2 and 2.3]{Haefliger1962}) $\varkappa(f)$ is also independent of the framing of $f$ extendable to a framing of some k$-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 \cite{Haefliger1966}, \cite[$\S]{Skopenkov2005}. == Classification == ; {{beginthm|Theorem|\cite[Corollary in p. 44]{Levine1965}, \cite{Haefliger1966}}}\label{t:leha} For $m-n\ge3$ the group $E^m_D(S^n)$ is finite unless $n=4k-1$ and $m\le6k$, when $E^m_D(S^n)$ is the sum of $\Z$ and a finite group. {{endthm}} {{beginthm|Theorem|(Haefliger-Milgram)}}\label{t:hami} 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 $k\ge2$: $$\begin{array}{c|c|c|c|c|c|c|c} (m,n) &2m\ge3n+4 &(6k,4k-1) &(6k+3,4k+1) &(7,4) &(6k+4,4k+2) &(12k+7,8k+4) & (12k+1,8k)\ \hline E^m_D(S^n)&0 &\Z &\Z_2 &\Z_{12} &0 &\Z_4 &\Z_2\oplus\Z_2 \end{array}$$ {{endthm}} Proof for the first four columns, and for the fifth column when $k$ is odd, are presented in \cite[8.15]{Haefliger1966} (see also $\S; some proofs are deduced from that paper using simple calculations, cf. \cite[$\S]{Skopenkov2005}; there is a typo in \cite[8.15]{Haefliger1966}: $C^{3k}_{4k−2}=0$ should be $C^{4k}_{8k−2}=0$). The remaining results follow from \cite[8.15]{Haefliger1966} and \cite[Theorem F]{Milgram1972}. Alternative proofs for the cases $(m,n)=(7,4),(6,3)$ are given in \cite{Skopenkov2005}, \cite{Crowley&Skopenkov2008}, \cite{Skopenkov2008}. {{beginthm|Theorem|\cite[Corollary G]{Milgram1972}}}\label{t:mi} We have $E^m_D(S^n)=0$ if and only if either m\ge3n+4$, or $(m,n)=(6k+4,4k+2)$, or $(m,n)=(3k,2k)$ and $k\equiv3,11\mod12$, or $(m,n)=(3k+2,2k+2)$ and $k\equiv14,22\mod24$. {{endthm}} For a description of 2-components of $E^m_D(S^n)$ see \cite[Theorem F]{Milgram1972}. Observe that no reliable reference (containing complete proofs) of results announced in \cite{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. \cite{Skopenkov2005}, \cite{Crowley&Skopenkov2008}, \cite{Skopenkov2008}. For $m\ge n+3$ the group $E^m_D(S^n)$ has been described as follows, in terms of exact sequences \cite{Haefliger1966}, cf. \cite{Levine1965}, \cite{Haefliger1966a}, \cite{Milgram1972}, \cite{Habegger1986}. {{beginthm|Theorem|\cite{Haefliger1966}}}\label{t:knots} For $q\ge3$ there is the following exact sequence of abelian groups: $$ \ldots \to \pi_{n+1}(SG,SO) \xrightarrow{~u~} E^{n+q}_D(S^n) \xrightarrow{~a~} \pi_n(SG_q,SO_q) \xrightarrow{~s~} \pi_n(SG,SO) \xrightarrow{~u~} E^{n+q-1}_D(S^{n-1})\to \ldots~.$$ Here $SG_q$ is the space of maps $S^{q-1} \to S^{q-1}$ of degree \S1, $\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

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

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

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

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$:

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

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:

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

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$$4k$-submanifold of $D^{6k+1}$$D^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$, signature of $V$$V$ is zero, and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then there is an embedding $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G(\partial D^{4k})=\partial V$$G(\partial D^{4k})=\partial V$.

Proof of Theorem 6.1 using Lemma 6.2. By the first three paragraphs of the proof of Theorem 3.1 in [Ha62], for any embedding $g:S^{4k-1}\to \partial D^{6k+1}$$g:S^{4k-1}\to \partial D^{6k+1}$ such that $\varkappa(g)=0$$\varkappa(g)=0$ there is a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold $V$$V$ of $D^{6k+1}$$D^{6k+1}$ with zero signature such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$ and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then from Lemma 6.2 it follows that there is an extension $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G|_{S^{4k-1}} = g$$G|_{S^{4k-1}} = g$. From Smale Theorem it follows that $g$$g$ is isotopic to the standard embedding. $\Box$$\Box$

To prove Lemma 6.2 we need Lemma 6.3, Lemma 6.4 and Lemma 6.5.

All the manifolds below can have non-empty boundaries.!!!

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

1. There is a homotopy $k_t$$k_t$ such that $k_0 = k$$k_0 = k$ and $k_1(P)$$k_1(P)$ is an embedding.
2. Suppose in addition that there is a map $y: Q \rightarrow W$$y: Q \rightarrow W$ from a connected $q$$q$-manifold $Q$$Q$ such that the algebraic intersection number of $k(P)$$k(P)$ and $y(Q)$$y(Q)$ is zero. Then there is a homotopy $y_t$$y_t$ relative to the boundary such that $y_0 = y$$y_0 = y$ and $y_1(Q)$$y_1(Q)$ does not intersect $k(P)$$k(P)$. If $y$$y$ is an embedding, the homotopy $y_t$$y_t$ can be chosen so that $y_1$$y_1$ is an embedding.

Lemma 6.4. Let $V$$V$ be a $(2k-1)$$(2k-1)$-connected $4k$$4k$-manifold, and $x_1, \ldots, x_m \in H_{2k}(V)$$x_1, \ldots, x_m \in H_{2k}(V)$ are 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_m: \, S^{2k} \rightarrow V$$g_1, \ldots, g_m: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_m$$x_1, \ldots, x_m$ [such that $g_i(S^{2k})$$g_i(S^{2k})$ represents!!! $x_i$$x_i$ for every $i$$i$].

Proof. As $V$$V$ is $(2k-1)$$(2k-1)$-connected, the Hurewicz map $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 $v_i: S^{2k} \rightarrow V$$v_i: S^{2k} \rightarrow V$ represent the homotopy class $h^{-1}(x_i)$$h^{-1}(x_i)$. Now we perform the following inductive procedure. At the $i$$i$-th step of the procedure assume that the maps $g_j$$g_j$ from the assertion are already constructed for any $j < i$$j < i$, and we construct $g_i$$g_i$. First, we apply item 1 of Lemma 6.3 to $v_i$$v_i$. Observe that item 1 of Lemma 6.3 is applicable as $V$$V$ is simply-connected. Therefore, we may suppose that $v_i$$v_i$ itself is an embedding. Further, we apply item 2 of Lemma 6.3, setting $g_j$$g_j$ as $k$$k$, $v_i$$v_i$ as $y$$y$, and $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ as $W$$W$, for any $j < i$$j < i$. Observe that Lemma 6.3 is applicable as $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected provided $2k \geq 3$$2k \geq 3$ and $V$$V$ is simply connected. Also, as $x_i \cap_V x_j = 0$$x_i \cap_V x_j = 0$ for any $i$$i$, $j$$j$, the algebraic intersection number for all the pairs of maps above is zero. As the result, $v_i$$v_i$ is replaced with a homotopic embedding $g_i$$g_i$, and the images of $g_1, \ldots, g_i$$g_1, \ldots, g_i$ are disjoint. After the step $m$$m$ we obtain the desired set of embeddings.

$\square$$\square$

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let $V$$V$ be an orientable $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$, and $g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$$g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$ be embeddings with pairwise disjoint images such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$. Assume that either $g_i(S^{2k})$$g_i(S^{2k})$ has zero algebraic self-intersection in $V$$V$ for every $i$$i$, or $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$ for every $i$$i$, or $V$$V$ has a framing in $B^{6k+1}$$B^{6k+1}$. Then there are embeddings $G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$$G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$ extending $g_1,\ldots,g_s$$g_1,\ldots,g_s$ such that $G_i(S^{2k}\times D^{2k})\subset V$$G_i(S^{2k}\times D^{2k})\subset V$ for every $i$$i$.

Moreover, one can smoothen $V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$$V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ to obtain a smooth submanifold of $B^{6k+1}$$B^{6k+1}$.

Proof. By general position, it suffices to prove the existence of $G_1,\ldots,G_s$$G_1,\ldots,G_s$ for $s=1$$s=1$. So assume that $s=1$$s=1$. Since $g_1(S^{2k})$$g_1(S^{2k})$ has zero algebraic self-intersection in $V$$V$, the Euler class of the normal bundle of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ is zero. Hence $g_1(S^{2k})$$g_1(S^{2k})$ has a framing in $V$$V$.

In this paragraph we define the obstruction $\zeta\in\pi_{2k}(V_{4k,2k})$$\zeta\in\pi_{2k}(V_{4k,2k})$ to extending a normal framing of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ to a normal $2k$$2k$-framing of $G_1$$G_1$ in $B^{6k+1}$$B^{6k+1}$. Identify all the normal spaces!!! of $G_1(D^{2k+1})$$G_1(D^{2k+1})$ with the normal space at $G(0)$$G(0)$. The framing of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is orthogonal to $G(D^{2k+1})$$G(D^{2k+1})$. The image of this framing under the identification defines the required obstruction $\zeta$$\zeta$.

old Consider the exact sequence of the bundle $SO_{2k}\to SO_{4k}\to V_{4k,2k}$$SO_{2k}\to SO_{4k}\to V_{4k,2k}$. Since $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$, we have $\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$$\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$, cf. [Haefliger1962, 3.4, Lemma]. For $k$$k$ odd from the exact sequence of the bundle we obtain $\zeta=0$$\zeta=0$.

Consider a part of the long exact sequence associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$$S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k})\ar[r]^{i} \ar[d]^{\equiv} & \pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[r]^{\partial}\ar[d] & \pi_{2k-1}(SO_{2k})\ar[d]^{\equiv} \\ \pi_{2k}(SO_{2k})\ar[r]^{i'} & \pi_{2k}(SO_{6k+1})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ From Milnor [??, proof of 5.11] it follows that $\partial$$\partial$ is injective, therefore $i$$i$ is surjective. As $s$$s$ is surjective, from commutativity of the diagram $i'$$i'$ is also surjective. Therefore, $\partial '$$\partial '$ is injective. By Lemma 3.4 from [Ha62], $\partial' \zeta$$\partial' \zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$$g(S^{2k})$, which is trivial. Therefore, $\partial ' \zeta =0$$\partial ' \zeta =0$ and $\zeta = 0$$\zeta = 0$ as $\partial '$$\partial '$ is injective. Thus, the obstruction is always zero.

$\square$$\square$

Proof. Let us smoothen $V'$$V'$.

Fix some $i$$i$. From $\zeta_i=0$$\zeta_i=0$ it follows that there are orthonormal vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to $v_j$$v_j$ along $\phi_i^{-1}(g_i(S^{2k}))$$\phi_i^{-1}(g_i(S^{2k}))$ for any $j\leq 2k$$j\leq 2k$. Since every fiber of $D{'}^{2k}$$D{'}^{2k}$ is trivial, there are orthonormal vector fields $v_{2k+1}, \ldots, v_{4k}$$v_{2k+1}, \ldots, v_{4k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $v_j\perp v_m$$v_j\perp v_m$ for any $j\leq 2k$$j\leq 2k$ and $2k<m\leq 4k$$2k<m\leq 4k$. Let us choose an extensions of $v_1,\ldots, v_{4k}$$v_1,\ldots, v_{4k}$ to smole neighborhood of $D'$$D'$ such that

1. $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$;
2. the map $\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$$\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$, defined by formula $\psi(x, y)=\gamma_y(||y||)$$\psi(x, y)=\gamma_y(||y||)$, where $x\in D{'}^{2k}$$x\in D{'}^{2k}$, $y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$$y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$ and where $\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$$\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$ is a solution of Cauchy problem

$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$ is the diffeomorphism between $D{'}^{2k}\times \mathbb{R}^{4k}$$D{'}^{2k}\times \mathbb{R}^{4k}$ and image of $\psi$$\psi$.

1. $\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$$\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$.

We take as the extension of $g_i$$g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$$G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$$\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$$-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$.

$\square$$\square$

Below the symbol $[ \cdot ]$$[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.

Proof of Lemma 6.2 using Lemmas 6.4, 6.5.. By the fourth paragraph of the proof of Theorem 3.1 in [Ha62], there is a basis $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ in $H_{2k}(V)$$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$$f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images such that $f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$$f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$.

Denote by $f_{i}':S^{2k}\to D^{6k+1}\backslash V$$f_{i}':S^{2k}\to D^{6k+1}\backslash V$ for $i\leq s$$i\leq s$ the result of shifting of $f^i$$f^i$ by the first vector of the framing of $V$$V$. Since $\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$$\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$, we have $f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$. Since $\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$$\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, we have $\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$$\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$ for $m<2k$$m<2k$. Since $f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$, we have $[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$$[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$. Therefore there are extensions of $f_1, \ldots, f_s$$f_1, \ldots, f_s$ to maps!!! $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$$g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$.

Denote by $V'$$V'$ the manifold as in Lemma 6.5 for $V$$V$ and $g_1, \ldots, g_s$$g_1, \ldots, g_s$ as described above UNAPPLICABLE. Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$$H_{2k}(V)$, it follows that $H_{2k}(V')\simeq 0$$H_{2k}(V')\simeq 0$. Hence $\pi_1(V')\simeq 0$$\pi_1(V')\simeq 0$ and $H_m(V')\simeq 0$$H_m(V')\simeq 0$ for $0< m\leq 2k$$0< m\leq 2k$. From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$$V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $V'\cong D^{4k}$$V'\cong D^{4k}$. Then we set by $G$$G$ a diffeomorphism between $D^{4k}$$D^{4k}$ and $V'\subset D^{6k+1}$$V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$$G|_{S^{4k-1}}=g$. $\Box$$\Box$

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$. 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 \cite{Haefliger1966}, see also \cite[$\S]{Skopenkov2005}. {{endthm}} == 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 [[Embeddings_in_Euclidean_space:_an_introduction_to_their_classification#Introduction|Remark 1.2]] of \cite{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 \cite{Farber1981}, \cite{Farber1983}, \cite{Kearton1983}, \cite{Farber1984}. On the other hand, if one studies embeddings up to the weaker relation of [[Isotopy|''concordance'']], then much is known. See e.g. \cite{Levine1969a} and \cite{Ranicki1998}. == Proof of classification of (4k-1)-knots in 6k-space == ; \begin{theorem}\label{haef} The Haefliger invariant $\varkappa:E_D^{6k}(S^{4k-1})\to\Zz$ is injective for $k>1$. \end{theorem} The proof is a certain simplification of \cite{Haefliger1962}. We present an exposition structured to make it more accessible to non-specialists. \begin{lemma}\label{l:V_to_disk} Let $V$ be a framed $(2k-1)$-connected k$-submanifold of $D^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$, signature of $V$ is zero, and $\varkappa(V) = 0$. Then there is an embedding $G:D^{4k}\to D^{6k+1}$ such that $G(\partial D^{4k})=\partial V$. \end{lemma} ''Proof of Theorem \ref{haef} using Lemma \ref{l:V_to_disk}.'' By the first three paragraphs of the proof of Theorem 3.1 in \cite{Ha62}, for any embedding $g:S^{4k-1}\to \partial D^{6k+1}$ such that $\varkappa(g)=0$ there is a framed $(2k-1)$-connected k$-submanifold $V$ of $D^{6k+1}$ with zero signature such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$ and $\varkappa(V) = 0$. Then from Lemma \ref{l:V_to_disk} it follows that there is an extension $G:D^{4k}\to D^{6k+1}$ such that $ G|_{S^{4k-1}} = g$. From Smale Theorem it follows that $g$ is isotopic to the standard embedding. $\Box$ To prove Lemma \ref{l:V_to_disk} we need Lemma \ref{whitney}, Lemma \ref{embeddings} and Lemma \ref{l:multi_spherical_modification}. We formulate below a suitable version of the Whitney Lemma. Note that all the manifolds below can have non-empty boundaries. {{beginthm|Lemma|[Whitney lemma]}}\label{whitney} Let $k: P \rightarrow W$ be a map from a connected $p$-manifold $P$ to a simply connected $(p+q)$-manifold $W$. If $p, q \geq 3$, then # There is a homotopy $k_t$ such that $k_0 = k$ and $k_1(P)$ is an embedding. # Suppose in addition that there is a map $y: Q \rightarrow W$ from a connected $q$-manifold $Q$ such that the algebraic intersection number of $k(P)$ and $y(Q)$ is zero. Then there is a homotopy $y_t$ relative to the boundary such that $y_0 = y$ and $y_1(Q)$ does not intersect $k(P)$. If $y$ is an embedding, the homotopy $y_t$ can be chosen so that $y_1$ is an embedding. {{endthm}} \begin{lemma} \label{embeddings} Let $V$ be a $(2k-1)$-connected k$-manifold, and $x_1, \ldots, x_m \in H_{2k}(V)$ are such that $x_i \cap_V x_j = 0$ for every $i,j$. !!!Then there are embeddings $g_1, \ldots, g_m: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_m$ [such that $g_i(S^{2k})$ represents!!! $x_i$ for every $i$]. \end{lemma} {{beginproof}} As $V$ is $(2k-1)$-connected, the Hurewicz map $h: \pi_{2k}(V) \rightarrow H_{2k}(V)$ is an isomorphism. For an element $x_i \in H_{2k}(V)$, let $v_i: S^{2k} \rightarrow V$ represent the homotopy class $h^{-1}(x_i)$. Now we perform the following inductive procedure. At the $i$-th step of the procedure assume that the maps $g_j$ from the assertion are already constructed for any $j < i$, and we construct $g_i$. First, we apply item 1 of Lemma \ref{whitney} to $v_i$. Observe that item 1 of Lemma \ref{whitney} is applicable as $V$ is simply-connected. Therefore, we may suppose that $v_i$ itself is an embedding. Further, we apply item 2 of Lemma \ref{whitney}, setting $g_j$ as $k$, $v_i$ as $y$, and $V \setminus \bigcup\limits_{l In this paragraph we define the obstruction $\zeta\in\pi_{2k}(V_{4k,2k})$ to extending a normal framing of $g_1(S^{2k})$ in $V$ to a normal k$-framing of $G_1$ in $B^{6k+1}$. Identify all the normal spaces!!! of $G_1(D^{2k+1})$ with the normal space at $G(0)$. The framing of $g(S^{2k})$ in $V$ is orthogonal to $G(D^{2k+1})$. The image of this framing under the identification defines the required obstruction $\zeta$. old Consider the exact sequence of the bundle $SO_{2k}\to SO_{4k}\to V_{4k,2k}$. Since $g_i(S^{2k})$ has a framing in $V$, we have $\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$, cf. \cite[3.4, Lemma]{Haefliger1962}. For $k$ odd from the exact sequence of the bundle we obtain $\zeta=0$. Consider a part of the long exact sequence associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k})\ar[r]^{i} \ar[d]^{\equiv} & \pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[r]^{\partial}\ar[d] & \pi_{2k-1}(SO_{2k})\ar[d]^{\equiv} \ \pi_{2k}(SO_{2k})\ar[r]^{i'} & \pi_{2k}(SO_{6k+1})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ From Milnor [??, proof of 5.11] it follows that $\partial$ is injective, therefore $i$ is surjective. As $s$ is surjective, from commutativity of the diagram $i'$ is also surjective. Therefore, $\partial '$ is injective. By Lemma 3.4 from \cite{Ha62}, $\partial' \zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$, which is trivial. Therefore, $\partial ' \zeta =0$ and $\zeta = 0$ as $\partial '$ is injective. Thus, the obstruction is always zero. {{endproof}} {{beginproof}} Let us smoothen $V'$. Fix some $i$. From $\zeta_i=0$ it follows that there are orthonormal vector fields $v_1, \ldots, v_{2k}$ on $D{'}^{2k}$ such that $\phi_i^{-1}(V)$ is tangent to $v_j$ along $\phi_i^{-1}(g_i(S^{2k}))$ for any $j\leq 2k$. Since every fiber of $D{'}^{2k}$ is trivial, there are orthonormal vector fields $v_{2k+1}, \ldots, v_{4k}$ on $D{'}^{2k}$ such that $v_j\perp v_m$ for any $j\leq 2k$ and k\S1, $\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

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

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

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

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$:

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

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:

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

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$$4k$-submanifold of $D^{6k+1}$$D^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$, signature of $V$$V$ is zero, and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then there is an embedding $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G(\partial D^{4k})=\partial V$$G(\partial D^{4k})=\partial V$.

Proof of Theorem 6.1 using Lemma 6.2. By the first three paragraphs of the proof of Theorem 3.1 in [Ha62], for any embedding $g:S^{4k-1}\to \partial D^{6k+1}$$g:S^{4k-1}\to \partial D^{6k+1}$ such that $\varkappa(g)=0$$\varkappa(g)=0$ there is a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold $V$$V$ of $D^{6k+1}$$D^{6k+1}$ with zero signature such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$ and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then from Lemma 6.2 it follows that there is an extension $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G|_{S^{4k-1}} = g$$G|_{S^{4k-1}} = g$. From Smale Theorem it follows that $g$$g$ is isotopic to the standard embedding. $\Box$$\Box$

To prove Lemma 6.2 we need Lemma 6.3, Lemma 6.4 and Lemma 6.5.

All the manifolds below can have non-empty boundaries.!!!

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

1. There is a homotopy $k_t$$k_t$ such that $k_0 = k$$k_0 = k$ and $k_1(P)$$k_1(P)$ is an embedding.
2. Suppose in addition that there is a map $y: Q \rightarrow W$$y: Q \rightarrow W$ from a connected $q$$q$-manifold $Q$$Q$ such that the algebraic intersection number of $k(P)$$k(P)$ and $y(Q)$$y(Q)$ is zero. Then there is a homotopy $y_t$$y_t$ relative to the boundary such that $y_0 = y$$y_0 = y$ and $y_1(Q)$$y_1(Q)$ does not intersect $k(P)$$k(P)$. If $y$$y$ is an embedding, the homotopy $y_t$$y_t$ can be chosen so that $y_1$$y_1$ is an embedding.

Lemma 6.4. Let $V$$V$ be a $(2k-1)$$(2k-1)$-connected $4k$$4k$-manifold, and $x_1, \ldots, x_m \in H_{2k}(V)$$x_1, \ldots, x_m \in H_{2k}(V)$ are 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_m: \, S^{2k} \rightarrow V$$g_1, \ldots, g_m: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_m$$x_1, \ldots, x_m$ [such that $g_i(S^{2k})$$g_i(S^{2k})$ represents!!! $x_i$$x_i$ for every $i$$i$].

Proof. As $V$$V$ is $(2k-1)$$(2k-1)$-connected, the Hurewicz map $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 $v_i: S^{2k} \rightarrow V$$v_i: S^{2k} \rightarrow V$ represent the homotopy class $h^{-1}(x_i)$$h^{-1}(x_i)$. Now we perform the following inductive procedure. At the $i$$i$-th step of the procedure assume that the maps $g_j$$g_j$ from the assertion are already constructed for any $j < i$$j < i$, and we construct $g_i$$g_i$. First, we apply item 1 of Lemma 6.3 to $v_i$$v_i$. Observe that item 1 of Lemma 6.3 is applicable as $V$$V$ is simply-connected. Therefore, we may suppose that $v_i$$v_i$ itself is an embedding. Further, we apply item 2 of Lemma 6.3, setting $g_j$$g_j$ as $k$$k$, $v_i$$v_i$ as $y$$y$, and $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ as $W$$W$, for any $j < i$$j < i$. Observe that Lemma 6.3 is applicable as $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected provided $2k \geq 3$$2k \geq 3$ and $V$$V$ is simply connected. Also, as $x_i \cap_V x_j = 0$$x_i \cap_V x_j = 0$ for any $i$$i$, $j$$j$, the algebraic intersection number for all the pairs of maps above is zero. As the result, $v_i$$v_i$ is replaced with a homotopic embedding $g_i$$g_i$, and the images of $g_1, \ldots, g_i$$g_1, \ldots, g_i$ are disjoint. After the step $m$$m$ we obtain the desired set of embeddings.

$\square$$\square$

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let $V$$V$ be an orientable $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$, and $g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$$g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$ be embeddings with pairwise disjoint images such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$. Assume that either $g_i(S^{2k})$$g_i(S^{2k})$ has zero algebraic self-intersection in $V$$V$ for every $i$$i$, or $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$ for every $i$$i$, or $V$$V$ has a framing in $B^{6k+1}$$B^{6k+1}$. Then there are embeddings $G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$$G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$ extending $g_1,\ldots,g_s$$g_1,\ldots,g_s$ such that $G_i(S^{2k}\times D^{2k})\subset V$$G_i(S^{2k}\times D^{2k})\subset V$ for every $i$$i$.

Moreover, one can smoothen $V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$$V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ to obtain a smooth submanifold of $B^{6k+1}$$B^{6k+1}$.

Proof. By general position, it suffices to prove the existence of $G_1,\ldots,G_s$$G_1,\ldots,G_s$ for $s=1$$s=1$. So assume that $s=1$$s=1$. Since $g_1(S^{2k})$$g_1(S^{2k})$ has zero algebraic self-intersection in $V$$V$, the Euler class of the normal bundle of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ is zero. Hence $g_1(S^{2k})$$g_1(S^{2k})$ has a framing in $V$$V$.

In this paragraph we define the obstruction $\zeta\in\pi_{2k}(V_{4k,2k})$$\zeta\in\pi_{2k}(V_{4k,2k})$ to extending a normal framing of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ to a normal $2k$$2k$-framing of $G_1$$G_1$ in $B^{6k+1}$$B^{6k+1}$. Identify all the normal spaces!!! of $G_1(D^{2k+1})$$G_1(D^{2k+1})$ with the normal space at $G(0)$$G(0)$. The framing of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is orthogonal to $G(D^{2k+1})$$G(D^{2k+1})$. The image of this framing under the identification defines the required obstruction $\zeta$$\zeta$.

old Consider the exact sequence of the bundle $SO_{2k}\to SO_{4k}\to V_{4k,2k}$$SO_{2k}\to SO_{4k}\to V_{4k,2k}$. Since $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$, we have $\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$$\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$, cf. [Haefliger1962, 3.4, Lemma]. For $k$$k$ odd from the exact sequence of the bundle we obtain $\zeta=0$$\zeta=0$.

Consider a part of the long exact sequence associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$$S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k})\ar[r]^{i} \ar[d]^{\equiv} & \pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[r]^{\partial}\ar[d] & \pi_{2k-1}(SO_{2k})\ar[d]^{\equiv} \\ \pi_{2k}(SO_{2k})\ar[r]^{i'} & \pi_{2k}(SO_{6k+1})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ From Milnor [??, proof of 5.11] it follows that $\partial$$\partial$ is injective, therefore $i$$i$ is surjective. As $s$$s$ is surjective, from commutativity of the diagram $i'$$i'$ is also surjective. Therefore, $\partial '$$\partial '$ is injective. By Lemma 3.4 from [Ha62], $\partial' \zeta$$\partial' \zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$$g(S^{2k})$, which is trivial. Therefore, $\partial ' \zeta =0$$\partial ' \zeta =0$ and $\zeta = 0$$\zeta = 0$ as $\partial '$$\partial '$ is injective. Thus, the obstruction is always zero.

$\square$$\square$

Proof. Let us smoothen $V'$$V'$.

Fix some $i$$i$. From $\zeta_i=0$$\zeta_i=0$ it follows that there are orthonormal vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to $v_j$$v_j$ along $\phi_i^{-1}(g_i(S^{2k}))$$\phi_i^{-1}(g_i(S^{2k}))$ for any $j\leq 2k$$j\leq 2k$. Since every fiber of $D{'}^{2k}$$D{'}^{2k}$ is trivial, there are orthonormal vector fields $v_{2k+1}, \ldots, v_{4k}$$v_{2k+1}, \ldots, v_{4k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $v_j\perp v_m$$v_j\perp v_m$ for any $j\leq 2k$$j\leq 2k$ and $2k<m\leq 4k$$2k<m\leq 4k$. Let us choose an extensions of $v_1,\ldots, v_{4k}$$v_1,\ldots, v_{4k}$ to smole neighborhood of $D'$$D'$ such that

1. $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$;
2. the map $\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$$\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$, defined by formula $\psi(x, y)=\gamma_y(||y||)$$\psi(x, y)=\gamma_y(||y||)$, where $x\in D{'}^{2k}$$x\in D{'}^{2k}$, $y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$$y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$ and where $\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$$\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$ is a solution of Cauchy problem

$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$ is the diffeomorphism between $D{'}^{2k}\times \mathbb{R}^{4k}$$D{'}^{2k}\times \mathbb{R}^{4k}$ and image of $\psi$$\psi$.

1. $\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$$\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$.

We take as the extension of $g_i$$g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$$G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$$\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$$-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$.

$\square$$\square$

Below the symbol $[ \cdot ]$$[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.

Proof of Lemma 6.2 using Lemmas 6.4, 6.5.. By the fourth paragraph of the proof of Theorem 3.1 in [Ha62], there is a basis $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ in $H_{2k}(V)$$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$$f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images such that $f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$$f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$.

Denote by $f_{i}':S^{2k}\to D^{6k+1}\backslash V$$f_{i}':S^{2k}\to D^{6k+1}\backslash V$ for $i\leq s$$i\leq s$ the result of shifting of $f^i$$f^i$ by the first vector of the framing of $V$$V$. Since $\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$$\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$, we have $f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$. Since $\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$$\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, we have $\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$$\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$ for $m<2k$$m<2k$. Since $f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$, we have $[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$$[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$. Therefore there are extensions of $f_1, \ldots, f_s$$f_1, \ldots, f_s$ to maps!!! $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$$g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$.

Denote by $V'$$V'$ the manifold as in Lemma 6.5 for $V$$V$ and $g_1, \ldots, g_s$$g_1, \ldots, g_s$ as described above UNAPPLICABLE. Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$$H_{2k}(V)$, it follows that $H_{2k}(V')\simeq 0$$H_{2k}(V')\simeq 0$. Hence $\pi_1(V')\simeq 0$$\pi_1(V')\simeq 0$ and $H_m(V')\simeq 0$$H_m(V')\simeq 0$ for $0< m\leq 2k$$0< m\leq 2k$. From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$$V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $V'\cong D^{4k}$$V'\cong D^{4k}$. Then we set by $G$$G$ a diffeomorphism between $D^{4k}$$D^{4k}$ and $V'\subset D^{6k+1}$$V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$$G|_{S^{4k-1}}=g$. $\Box$$\Box$

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< m\leq 2k$. From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $V'\cong D^{4k}$. Then we set by $G$ a diffeomorphism between $D^{4k}$ and $V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$. $\Box$ == References == {{#RefList:}} [[Category:Manifolds]] [[Category:Embeddings of manifolds]]\S1, $\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

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

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

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

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$:

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

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:

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

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$$4k$-submanifold of $D^{6k+1}$$D^{6k+1}$ such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$, signature of $V$$V$ is zero, and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then there is an embedding $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G(\partial D^{4k})=\partial V$$G(\partial D^{4k})=\partial V$.

Proof of Theorem 6.1 using Lemma 6.2. By the first three paragraphs of the proof of Theorem 3.1 in [Ha62], for any embedding $g:S^{4k-1}\to \partial D^{6k+1}$$g:S^{4k-1}\to \partial D^{6k+1}$ such that $\varkappa(g)=0$$\varkappa(g)=0$ there is a framed $(2k-1)$$(2k-1)$-connected $4k$$4k$-submanifold $V$$V$ of $D^{6k+1}$$D^{6k+1}$ with zero signature such that $S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$$S^{4k-1}\cong \partial V \subset \partial D^{6k+1}$ and $\varkappa(V) = 0$$\varkappa(V) = 0$. Then from Lemma 6.2 it follows that there is an extension $G:D^{4k}\to D^{6k+1}$$G:D^{4k}\to D^{6k+1}$ such that $G|_{S^{4k-1}} = g$$G|_{S^{4k-1}} = g$. From Smale Theorem it follows that $g$$g$ is isotopic to the standard embedding. $\Box$$\Box$

To prove Lemma 6.2 we need Lemma 6.3, Lemma 6.4 and Lemma 6.5.

All the manifolds below can have non-empty boundaries.!!!

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

1. There is a homotopy $k_t$$k_t$ such that $k_0 = k$$k_0 = k$ and $k_1(P)$$k_1(P)$ is an embedding.
2. Suppose in addition that there is a map $y: Q \rightarrow W$$y: Q \rightarrow W$ from a connected $q$$q$-manifold $Q$$Q$ such that the algebraic intersection number of $k(P)$$k(P)$ and $y(Q)$$y(Q)$ is zero. Then there is a homotopy $y_t$$y_t$ relative to the boundary such that $y_0 = y$$y_0 = y$ and $y_1(Q)$$y_1(Q)$ does not intersect $k(P)$$k(P)$. If $y$$y$ is an embedding, the homotopy $y_t$$y_t$ can be chosen so that $y_1$$y_1$ is an embedding.

Lemma 6.4. Let $V$$V$ be a $(2k-1)$$(2k-1)$-connected $4k$$4k$-manifold, and $x_1, \ldots, x_m \in H_{2k}(V)$$x_1, \ldots, x_m \in H_{2k}(V)$ are 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_m: \, S^{2k} \rightarrow V$$g_1, \ldots, g_m: \, S^{2k} \rightarrow V$ with pairwise disjoint images representing $x_1, \ldots, x_m$$x_1, \ldots, x_m$ [such that $g_i(S^{2k})$$g_i(S^{2k})$ represents!!! $x_i$$x_i$ for every $i$$i$].

Proof. As $V$$V$ is $(2k-1)$$(2k-1)$-connected, the Hurewicz map $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 $v_i: S^{2k} \rightarrow V$$v_i: S^{2k} \rightarrow V$ represent the homotopy class $h^{-1}(x_i)$$h^{-1}(x_i)$. Now we perform the following inductive procedure. At the $i$$i$-th step of the procedure assume that the maps $g_j$$g_j$ from the assertion are already constructed for any $j < i$$j < i$, and we construct $g_i$$g_i$. First, we apply item 1 of Lemma 6.3 to $v_i$$v_i$. Observe that item 1 of Lemma 6.3 is applicable as $V$$V$ is simply-connected. Therefore, we may suppose that $v_i$$v_i$ itself is an embedding. Further, we apply item 2 of Lemma 6.3, setting $g_j$$g_j$ as $k$$k$, $v_i$$v_i$ as $y$$y$, and $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ as $W$$W$, for any $j < i$$j < i$. Observe that Lemma 6.3 is applicable as $V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$$V \setminus \bigcup\limits_{l<j}g_l(S^{2k})$ is simply connected provided $2k \geq 3$$2k \geq 3$ and $V$$V$ is simply connected. Also, as $x_i \cap_V x_j = 0$$x_i \cap_V x_j = 0$ for any $i$$i$, $j$$j$, the algebraic intersection number for all the pairs of maps above is zero. As the result, $v_i$$v_i$ is replaced with a homotopic embedding $g_i$$g_i$, and the images of $g_1, \ldots, g_i$$g_1, \ldots, g_i$ are disjoint. After the step $m$$m$ we obtain the desired set of embeddings.

$\square$$\square$

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let $V$$V$ be an orientable $4k$$4k$-submanifold of $B^{6k+1}$$B^{6k+1}$, and $g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$$g_1,\ldots,g_s:D^{2k+1}\to B^{6k+1}$ be embeddings with pairwise disjoint images such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$. Assume that either $g_i(S^{2k})$$g_i(S^{2k})$ has zero algebraic self-intersection in $V$$V$ for every $i$$i$, or $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$ for every $i$$i$, or $V$$V$ has a framing in $B^{6k+1}$$B^{6k+1}$. Then there are embeddings $G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$$G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1}$ extending $g_1,\ldots,g_s$$g_1,\ldots,g_s$ such that $G_i(S^{2k}\times D^{2k})\subset V$$G_i(S^{2k}\times D^{2k})\subset V$ for every $i$$i$.

Moreover, one can smoothen $V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$$V':=V\backslash ???(\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ to obtain a smooth submanifold of $B^{6k+1}$$B^{6k+1}$.

Proof. By general position, it suffices to prove the existence of $G_1,\ldots,G_s$$G_1,\ldots,G_s$ for $s=1$$s=1$. So assume that $s=1$$s=1$. Since $g_1(S^{2k})$$g_1(S^{2k})$ has zero algebraic self-intersection in $V$$V$, the Euler class of the normal bundle of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ is zero. Hence $g_1(S^{2k})$$g_1(S^{2k})$ has a framing in $V$$V$.

In this paragraph we define the obstruction $\zeta\in\pi_{2k}(V_{4k,2k})$$\zeta\in\pi_{2k}(V_{4k,2k})$ to extending a normal framing of $g_1(S^{2k})$$g_1(S^{2k})$ in $V$$V$ to a normal $2k$$2k$-framing of $G_1$$G_1$ in $B^{6k+1}$$B^{6k+1}$. Identify all the normal spaces!!! of $G_1(D^{2k+1})$$G_1(D^{2k+1})$ with the normal space at $G(0)$$G(0)$. The framing of $g(S^{2k})$$g(S^{2k})$ in $V$$V$ is orthogonal to $G(D^{2k+1})$$G(D^{2k+1})$. The image of this framing under the identification defines the required obstruction $\zeta$$\zeta$.

old Consider the exact sequence of the bundle $SO_{2k}\to SO_{4k}\to V_{4k,2k}$$SO_{2k}\to SO_{4k}\to V_{4k,2k}$. Since $g_i(S^{2k})$$g_i(S^{2k})$ has a framing in $V$$V$, we have $\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$$\partial\zeta_i=0\in\pi_{2k-1}(SO_{2k})$, cf. [Haefliger1962, 3.4, Lemma]. For $k$$k$ odd from the exact sequence of the bundle we obtain $\zeta=0$$\zeta=0$.

Consider a part of the long exact sequence associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$$S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k})\ar[r]^{i} \ar[d]^{\equiv} & \pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[r]^{\partial}\ar[d] & \pi_{2k-1}(SO_{2k})\ar[d]^{\equiv} \\ \pi_{2k}(SO_{2k})\ar[r]^{i'} & \pi_{2k}(SO_{6k+1})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ From Milnor [??, proof of 5.11] it follows that $\partial$$\partial$ is injective, therefore $i$$i$ is surjective. As $s$$s$ is surjective, from commutativity of the diagram $i'$$i'$ is also surjective. Therefore, $\partial '$$\partial '$ is injective. By Lemma 3.4 from [Ha62], $\partial' \zeta$$\partial' \zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$$g(S^{2k})$, which is trivial. Therefore, $\partial ' \zeta =0$$\partial ' \zeta =0$ and $\zeta = 0$$\zeta = 0$ as $\partial '$$\partial '$ is injective. Thus, the obstruction is always zero.

$\square$$\square$

Proof. Let us smoothen $V'$$V'$.

Fix some $i$$i$. From $\zeta_i=0$$\zeta_i=0$ it follows that there are orthonormal vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to $v_j$$v_j$ along $\phi_i^{-1}(g_i(S^{2k}))$$\phi_i^{-1}(g_i(S^{2k}))$ for any $j\leq 2k$$j\leq 2k$. Since every fiber of $D{'}^{2k}$$D{'}^{2k}$ is trivial, there are orthonormal vector fields $v_{2k+1}, \ldots, v_{4k}$$v_{2k+1}, \ldots, v_{4k}$ on $D{'}^{2k}$$D{'}^{2k}$ such that $v_j\perp v_m$$v_j\perp v_m$ for any $j\leq 2k$$j\leq 2k$ and $2k<m\leq 4k$$2k<m\leq 4k$. Let us choose an extensions of $v_1,\ldots, v_{4k}$$v_1,\ldots, v_{4k}$ to smole neighborhood of $D'$$D'$ such that

1. $\phi_i^{-1}(V)$$\phi_i^{-1}(V)$ is tangent to vector fields $v_1, \ldots, v_{2k}$$v_1, \ldots, v_{2k}$;
2. the map $\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$$\psi: D{'}^{2k}\times \mathbb{R}^{4k}\to\mathbb{R}^{6k+1}$, defined by formula $\psi(x, y)=\gamma_y(||y||)$$\psi(x, y)=\gamma_y(||y||)$, where $x\in D{'}^{2k}$$x\in D{'}^{2k}$, $y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$$y=(y_1, \ldots, y_{4k})\in \mathbb{R}^{4k}$ and where $\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$$\gamma_y:\mathbb{R}\to\mathbb{R}^{6k+1}$ is a solution of Cauchy problem

$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$$\begin{cases} \gamma_y'(t)=\frac{y_1}{||y||}v_1(t)+\ldots +\frac{y_{4k}}{||y||}v_{4k}(t) \\ \gamma_y(0)=x, \end{cases}$ is the diffeomorphism between $D{'}^{2k}\times \mathbb{R}^{4k}$$D{'}^{2k}\times \mathbb{R}^{4k}$ and image of $\psi$$\psi$.

1. $\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$$\psi(D{'}^{2k}\times \mathbb{R}^{4k})\cap \phi^{-1}(V)\cong S^{2k}\times \mbox{Int} D^{2k}$.

We take as the extension of $g_i$$g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$$G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$$\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$$-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$.

$\square$$\square$

Below the symbol $[ \cdot ]$$[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.

Proof of Lemma 6.2 using Lemmas 6.4, 6.5.. By the fourth paragraph of the proof of Theorem 3.1 in [Ha62], there is a basis $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ in $H_{2k}(V)$$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$$f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images such that $f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$$f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$.

Denote by $f_{i}':S^{2k}\to D^{6k+1}\backslash V$$f_{i}':S^{2k}\to D^{6k+1}\backslash V$ for $i\leq s$$i\leq s$ the result of shifting of $f^i$$f^i$ by the first vector of the framing of $V$$V$. Since $\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$$\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$, we have $f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$. Since $\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$$\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, we have $\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$$\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$ for $m<2k$$m<2k$. Since $f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$$f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$, we have $[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$$[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$. Therefore there are extensions of $f_1, \ldots, f_s$$f_1, \ldots, f_s$ to maps!!! $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$$g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$$g_i(D^{2k+1})\cap V = g_i(S^{2k})$.

Denote by $V'$$V'$ the manifold as in Lemma 6.5 for $V$$V$ and $g_1, \ldots, g_s$$g_1, \ldots, g_s$ as described above UNAPPLICABLE. Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$$\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$$H_{2k}(V)$, it follows that $H_{2k}(V')\simeq 0$$H_{2k}(V')\simeq 0$. Hence $\pi_1(V')\simeq 0$$\pi_1(V')\simeq 0$ and $H_m(V')\simeq 0$$H_m(V')\simeq 0$ for $0< m\leq 2k$$0< m\leq 2k$. From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$$V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$. Hence $V'\cong D^{4k}$$V'\cong D^{4k}$. Then we set by $G$$G$ a diffeomorphism between $D^{4k}$$D^{4k}$ and $V'\subset D^{6k+1}$$V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$$G|_{S^{4k-1}}=g$. $\Box$$\Box$

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