Knots, i.e. embeddings of spheres

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# There is a homotopy $k_t$ such that $k_0 = k$ and $k_1(P)$ is an embedding.???
# 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.
+
# 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}}
{{endthm}}
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{{beginthm|Lemma}} \label{l:smoothen}
{{beginthm|Lemma}} \label{l:smoothen}
Denote by $G_1,\ldots, G_s$ the embeddings from Lemma \ref{l:multi_spherical_modification}.
Denote by $G_1,\ldots, G_s$ the embeddings from Lemma \ref{l:multi_spherical_modification}.
−
There is smooth manifold $V'\subset D^{6k+1}$ such that $V'$ is homeomorphic to the $V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ and $\partial V'=\partial V$.
+
There is smooth manifold $V'\subset D^{6k+1}$ such that $V'$ is homeomorphic to the $V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ and $\partial V'=\partial V$.
{{endthm}}
{{endthm}}
{{beginproof}}
{{beginproof}}
−
For any $G_i$ choose smole neighborhood $U_i$ of $G_i(D^{2k+1})$ and smooth diffeomorphism $\phi_i: \mathbb{R}^{6k+1}\to U_i$ such that $\phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k})$ is defined by $x^2\leq 1, y^2\leq 1, z=0$ in some coordinates $(x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k})$ of $\mathbb{R}^{6k+1}$.
+
For any $G_i$ choose smole??? neighborhood $U_i$ of $G_i(D^{2k+1})$ and smooth diffeomorphism $\phi_i: \mathbb{R}^{6k+1}\to U_i$ such that $\phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k})$ is defined by $x^2\leq 1, y^2\leq 1, z=0$ in some coordinates $(x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k})$ of $\mathbb{R}^{6k+1}$.
Denote by $a:\mathbb{R}\to \mathbb{R}$ a smooth monotonous map such that $a(x)=0$ for $x\leq \frac{1}{2}$ and $a(x)=1$ for $x\geq 1$.
Denote by $a:\mathbb{R}\to \mathbb{R}$ a smooth monotonous map such that $a(x)=0$ for $x\leq \frac{1}{2}$ and $a(x)=1$ for $x\geq 1$.
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Therefore, map $f:V\to V'$ defined by equation $f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases}$ is the homeomorphism.
Therefore, map $f:V\to V'$ defined by equation $f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases}$ is the homeomorphism.
Since $V'$ coincide with $V$ outside the $\bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2})$ and $V'$ is smooth in $\bigcup U_i$, we have that $V'$ is smooth.
Since $V'$ coincide with $V$ outside the $\bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2})$ and $V'$ is smooth in $\bigcup U_i$, we have that $V'$ is smooth.
−
+
{{endproof}}
<!--Let us smoothen $V'$.
<!--Let us smoothen $V'$.
Hence for $i\leq s$ there are disjoint neighborhoods $U_i$ of $g_i(D^{2k+1})$ and homeomorphisms $\phi_i:\mathbb{R}^{6k+1}\to U_i$ such that
Hence for $i\leq s$ there are disjoint neighborhoods $U_i$ of $g_i(D^{2k+1})$ and homeomorphisms $\phi_i:\mathbb{R}^{6k+1}\to U_i$ such that
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We take as the extension of $g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$. -->
We take as the extension of $g_i$ an embedding $G_i:D^{2k+1}\times D^{2k-1}\to D^{6k+1}$ such that $\phi^{-1}(G_i(D^{2k+1}\times D^{2k-1}))$ is defined by $-a(x^2+y^2)\leq -x^2+y^2\leq a(x^2+y^2)$. -->
−
{{endproof}}
Below the symbol $[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.
Below the symbol $[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context.
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Therefore there are extensions of $f_1, \ldots, f_s$ to maps $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$.
Therefore there are extensions of $f_1, \ldots, f_s$ to maps $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$.
−
Set $\varepsilon>0$ such that $g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V$.
+
???Set $\varepsilon>0$ such that $g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V$.
As $\lambda^*(g_{i*}[S^{2k}])=0$ for any $i$, the algebraic intersection number of $g_i(\varepsilon D^{2k+1})$ and $V$ is zero.
As $\lambda^*(g_{i*}[S^{2k}])=0$ for any $i$, the algebraic intersection number of $g_i(\varepsilon D^{2k+1})$ and $V$ is zero.
Applying item 2 of Lemma\ref{whitney} to $g_i|_{\varepsilon D^{2k+1}}$ as $y$, embedding of $V$ into $D^{6k+1}$ as $k$ and $D^{6k+1}$ as $W$ we may suppose that $g_i(\varepsilon D^{2k+1})$ does not intersect $V$.
Applying item 2 of Lemma\ref{whitney} to $g_i|_{\varepsilon D^{2k+1}}$ as $y$, embedding of $V$ into $D^{6k+1}$ as $k$ and $D^{6k+1}$ as $W$ we may suppose that $g_i(\varepsilon D^{2k+1})$ does not intersect $V$.
Hance for any $i$ renamed by $g_i:D^{2k+1}\to D^{6k+1}\backslash V $ a map which coincides with $g_i$ on $D^{2k+1}\backslash \varepsilon D^{2k+1}$ and $g_i(\varepsilon D^{2k+1})\cap V=\emptyset$.
Hance for any $i$ renamed by $g_i:D^{2k+1}\to D^{6k+1}\backslash V $ a map which coincides with $g_i$ on $D^{2k+1}\backslash \varepsilon D^{2k+1}$ and $g_i(\varepsilon D^{2k+1})\cap V=\emptyset$.
−
Denote by $G_1, \ldots, G_s$ the maps as in Lemma \ref{l:multi_spherical_modification} for $V$ and $g_1, \ldots, g_s$ as described above.
+
???Denote by $G_1, \ldots, G_s$ the maps as in Lemma \ref{l:multi_spherical_modification} for $V$ and $g_1, \ldots, g_s$ as described above.
Denote by $V'$ the manifold as in Lemma \ref{l:smoothen}.
Denote by $V'$ the manifold as in Lemma \ref{l:smoothen}.
−
Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$, it follows that $H_{2k}(V')= 0$.
+
Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$, ???it follows that $H_{2k}(V')= 0$.
Since $\pi_1(V)= 0$ we have $\pi_1(V')= 0$.
Since $\pi_1(V)= 0$ we have $\pi_1(V')= 0$.
−
Hence $H_m(V')= 0$ for $0< m\leq 2k$.
+
Hence??? $H_m(V')= 0$ for $0< m\leq 2k$.
From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$.
From Generalized Poincare conjecture proved by Smale it follows that $V'\cup_{\partial V'=\partial D^{4k}} D^{4k}\cong S^{4k}$.
Hence $V'\cong D^{4k}$.
Hence $V'\cong D^{4k}$.
−
Then we set by $G$ a diffeomorphism between $D^{4k}$ and $V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$.
+
Then we set by??? $G$ a ???diffeomorphism between $D^{4k}$ and $V'\subset D^{6k+1}$ such that $G|_{S^{4k-1}}=g$.
$\Box$
$\Box$

Revision as of 18:44, 16 February 2024

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Contents

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, \S1, \S3].

2 Examples

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

Example 2.1. (a) Analogously to the Haefliger trefoil knot for any l>1 one constructs a smooth embedding t:S^{2l-1}\to\Rr^{3l}, see [Skopenkov2016h, \S5]. 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 [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 [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}) the Zeeman map, see [Skopenkov2016h, Definition 2.2]. 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 [Skopenkov2015a, Corollary 2.13].

3 Invariants

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

By [Haefliger1962, 2.1, 2.2] 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 4k-manifold V whose boundary is S^{4k-1}, and whose signature is zero. For an integer 2k-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 [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 2k-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
\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.

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

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

For an alternative definition via Seifert surfaces in 6k-space, discovered in [Guillou&Marin1986], [Takase2004], see [Skopenkov2016t, the Kreck Invariant Lemma 4.5]. For a definition by Kreck, and for a generalization to 3-manifolds see [Skopenkov2016t, \S4].

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

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

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

For definition of the attaching invariant E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q) see [Haefliger1966], [Skopenkov2005, \S3].

4 Classification

Theorem 4.1 [Levine1965, Corollary in p. 44], [Haefliger1966]. 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.

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

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

Proof for the first four columns, and for the fifth column when k is odd, are presented in [Haefliger1966, 8.15] (see also \S6; some proofs are deduced from that paper using simple calculations, cf. [Skopenkov2005, \S3]; there is a typo in [Haefliger1966, 8.15]: C^{3k}_{4k−2}=0 should be C^{4k}_{8k−2}=0). The remaining results follow from [Haefliger1966, 8.15] and [Milgram1972, Theorem F]. Alternative proofs for the cases (m,n)=(7,4),(6,3) are given in [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

Theorem 4.3 [Milgram1972, Corollary G]. We have E^m_D(S^n)=0 if and only if either 2m\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.

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

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

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

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

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

Here SG_q is the space of maps S^{q-1} \to S^{q-1} of degree 1. Restricting a map from SO_q to S^{q-1} \subset \Rr^q identifies SO_q as a subspace of SG_q. Define SG:=SG_1\cup\ldots\cup SG_q\cup\ldots. Analogously define SO. Let s be the stabilization homomorphism. The attaching invariant a and the map u are defined in [Haefliger1966], see also [Skopenkov2005, \S3].

5 Some remarks on codimension 2 knots

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

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

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

Theorem 6.1. The Haefliger invariant \varkappa:E_D^{6k}(S^{4k-1})\to\Zz is injective for 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 be a framed (2k-1)-connected 4k-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.

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} such that \varkappa(g)=0 there is a framed (2k-1)-connected 4k-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 6.2 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 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 be a map from a connected p-manifold P to a simply connected (p+q)-manifold W. If p, q \geq 3, then

  1. There is a homotopy k_t such that k_0 = k and k_1(P) is an embedding.???
  2. 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.

Lemma 6.4. Let V be a (2k-1)-connected 4k-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, respectively.

Proof. 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, applying item 1 of Lemma 6.3 to v_i, we may suppose that v_i itself is an embedding. Further, we apply item 2 of Lemma 6.3, ???setting g_j as k, v_i as y, and V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) as W, for any j < i. ???Observe that Lemma 6.3 is applicable as V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) is simply connected provided 2k \geq 3 and V is simply connected. Also, as x_i \cap_V x_j = 0 for any i, j, the algebraic intersection number for all the pairs of maps above??? is zero. As the result, v_i is replaced with a homotopic embedding g_i, and the images of g_1, \ldots, g_i are disjoint. After the step m we obtain the ???desired set of embeddings.

\square

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let V be an orientable 4k-submanifold of B^{6k+1}, and ???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}) are orthogonal to V, and over every g_i(S^{2k}) the manifold V has a framing whose first vector is tangent to g_i(D^{2k+1}). Assume that g_i(S^{2k}) has zero algebraic self-intersection in V for every i. Then there are embeddings G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1} extending g_1,\ldots,g_s such that G_i(S^{2k}\times D^{2k})\subset V for every i.

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

Identify all the normal spaces!!! of G_1(D^{2k+1}) with the normal space at G_1(0). The normal framing a of g_1(S^{2k}) in V is orthogonal to G_1(D^{2k+1}). So a defines a map S^{2k}\to V_{4k,2k}. Let \zeta=\in\pi_{2k}(V_{4k,2k}) be the homotopy class of this map. This is the obstruction to extending a to a normal 2k-framing of G_1 in B^{6k+1} (so apriori \zeta=\zeta(a)). It suffices to prove that \zeta=0.

Consider the exact sequence of the bundle SO_{a+b}/SO_b = V_{a+b,a}: \pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b). By the following well-known assertion, \partial\zeta=0: if \alpha:S^q\to V_{a,b} is a map, then \partial[\alpha] is the obstruction to trivialization of the orthogonal complement to the field of b-frames in S^q\times\R^a corresponding to \alpha. By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{4k}) is a finite group. Since \pi_{2k}(V_{4k,2k})\cong\Z, we obtain that j=0 for a=b=2k. This and \partial\zeta=0 imply that \zeta=0.

(In [Haefliger1962, the last but one paragraph of 3.5] perhaps one has to replace `\partial\zeta is the obstruction to trivializing the normal bundle of g(S^{2k}) in V' by `\partial\zeta is the obstruction to trivializing te orthogonal complement to the f_1-direction of the normal bundle of V restricted to g(S^{2k})'?)

(No, as we use different obstruction; our is orthogonal to Haefliger's)

Alternatively: Consider a map of the long exact sequences associated to the inclusion S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}:

\xymatrix{\pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[d] \\ \pi_{2k}(SO_{4k})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}

By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{2k+1}) is a finite group. Since \pi_{2k}(S^{2k})\cong\Z, we obtain that p=0. Since s is surjective, we obtain that p'=0.

Hence \partial' is injective. The class \partial'\zeta is??? the obstruction to trivializing the orthogonal complement of g(S^{2k}), cf. [Haefliger1962, 3.4, Lemma].


Suggestion: interpreting \partial'\zeta as an obstruction, we derive that \partial'\zeta is zero, as V is framed and the first vector of the framing is tangent to G_1.

Therefore, \partial ' \zeta =0. Since \partial ' is injective, we obtain \zeta = 0.

\square

Lemma 6.6. Denote by G_1,\ldots, G_s the embeddings from Lemma 6.5. There is smooth manifold V'\subset D^{6k+1} such that V' is homeomorphic to the V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2}) and \partial V'=\partial V.

Proof. For any G_i choose smole??? neighborhood U_i of G_i(D^{2k+1}) and smooth diffeomorphism \phi_i: \mathbb{R}^{6k+1}\to U_i such that \phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k}) is defined by x^2\leq 1, y^2\leq 1, z=0 in some coordinates (x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k}) of \mathbb{R}^{6k+1}.

Denote by a:\mathbb{R}\to \mathbb{R} a smooth monotonous map such that a(x)=0 for x\leq \frac{1}{2} and a(x)=1 for x\geq 1. Denote by g:D^{2k+1}\times S^{2k-1}\to \mathbb{R}^{6k+1} the smooth map, defined by the formula g(x, y):=(x,y\cdot a(x^2), 0). Take by V' the manifold V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}). V' is homeomorphic to V. Therefore, map f:V\to V' defined by equation f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases} is the homeomorphism. Since V' coincide with V outside the \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}) and V' is smooth in \bigcup U_i, we have that V' is smooth.

\square

Below the symbol [ \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\} in H_{2k}(V) such that \alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0, \alpha_i\cap \beta_j=\delta_{i, j} and \lambda^*(\alpha_i)=0 for any i, j. From Lemma 6.4 it follows that there are embeddings f_1, \ldots, f_s:S^{2k}\to V with pairwise disjoint images such that 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 for i\leq s the result of shifting of f_i by the first vector of the framing of V. Since \lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0, we have 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, we have \pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0 for m<2k. Since 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). Therefore there are extensions of f_1, \ldots, f_s to maps 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}).

???Set \varepsilon>0 such that g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V. As \lambda^*(g_{i*}[S^{2k}])=0 for any i, the algebraic intersection number of g_i(\varepsilon D^{2k+1}) and V is zero. Applying item 2 of Lemma6.3 to g_i|_{\varepsilon D^{2k+1}} as y, embedding of V into D^{6k+1} as k and D^{6k+1} as W we may suppose that g_i(\varepsilon D^{2k+1}) does not intersect V. Hance for any i renamed by g_i:D^{2k+1}\to D^{6k+1}\backslash V a map which coincides with g_i on D^{2k+1}\backslash \varepsilon D^{2k+1} and g_i(\varepsilon D^{2k+1})\cap V=\emptyset.

???Denote by G_1, \ldots, G_s the maps as in Lemma 6.5 for V and g_1, \ldots, g_s as described above. Denote by V' the manifold as in Lemma 6.6. Since \left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\} is a basis in H_{2k}(V), ???it follows that H_{2k}(V')= 0. Since \pi_1(V)= 0 we have \pi_1(V')= 0. Hence??? H_m(V')= 0 for 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}. 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

  • [Crowley&Skopenkov2008] D. Crowley and A. Skopenkov, A classification of smooth embeddings of 4-manifolds in 7-space, II, Intern. J. Math., 22:6 (2011) 731-757. Available at the arXiv:0808.1795.
  • [Farber1981] M. Sh. Farber, Classification of stable fibered knots, Mat. Sb. (N.S.), 115(157):2(6) (1981) 223–262.
  • [Farber1983] M. Sh. Farber, The classification of simple knots, Russian Math. Surveys, 38:5 (1983).

  • [Farber1984] M. Sh. Farber, An algebraic classification of some even-dimensional spherical knots I, II, Trans. Amer. Math. Soc. 281 (1984), 507-528; 529-570.
  • [Fomenko&Fuchs2016] A. T. Fomenko and D. B. Fuks, Homotopical Topology. Translated from the Russian. Graduate Texts in Mathematics, 273. Springer-Verlag, Berlin, 2016. DOI 10.1007/978-3-319-23488-5.

, $\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, \S3].

2 Examples

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

Example 2.1. (a) Analogously to the Haefliger trefoil knot for any l>1 one constructs a smooth embedding t:S^{2l-1}\to\Rr^{3l}, see [Skopenkov2016h, \S5]. 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 [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 [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}) the Zeeman map, see [Skopenkov2016h, Definition 2.2]. 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 [Skopenkov2015a, Corollary 2.13].

3 Invariants

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

By [Haefliger1962, 2.1, 2.2] 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 4k-manifold V whose boundary is S^{4k-1}, and whose signature is zero. For an integer 2k-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 [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 2k-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
\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.

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

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

For an alternative definition via Seifert surfaces in 6k-space, discovered in [Guillou&Marin1986], [Takase2004], see [Skopenkov2016t, the Kreck Invariant Lemma 4.5]. For a definition by Kreck, and for a generalization to 3-manifolds see [Skopenkov2016t, \S4].

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

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

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

For definition of the attaching invariant E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q) see [Haefliger1966], [Skopenkov2005, \S3].

4 Classification

Theorem 4.1 [Levine1965, Corollary in p. 44], [Haefliger1966]. 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.

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

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

Proof for the first four columns, and for the fifth column when k is odd, are presented in [Haefliger1966, 8.15] (see also \S6; some proofs are deduced from that paper using simple calculations, cf. [Skopenkov2005, \S3]; there is a typo in [Haefliger1966, 8.15]: C^{3k}_{4k−2}=0 should be C^{4k}_{8k−2}=0). The remaining results follow from [Haefliger1966, 8.15] and [Milgram1972, Theorem F]. Alternative proofs for the cases (m,n)=(7,4),(6,3) are given in [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

Theorem 4.3 [Milgram1972, Corollary G]. We have E^m_D(S^n)=0 if and only if either 2m\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.

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

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

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

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

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

Here SG_q is the space of maps S^{q-1} \to S^{q-1} of degree 1. Restricting a map from SO_q to S^{q-1} \subset \Rr^q identifies SO_q as a subspace of SG_q. Define SG:=SG_1\cup\ldots\cup SG_q\cup\ldots. Analogously define SO. Let s be the stabilization homomorphism. The attaching invariant a and the map u are defined in [Haefliger1966], see also [Skopenkov2005, \S3].

5 Some remarks on codimension 2 knots

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

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

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

Theorem 6.1. The Haefliger invariant \varkappa:E_D^{6k}(S^{4k-1})\to\Zz is injective for 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 be a framed (2k-1)-connected 4k-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.

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} such that \varkappa(g)=0 there is a framed (2k-1)-connected 4k-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 6.2 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 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 be a map from a connected p-manifold P to a simply connected (p+q)-manifold W. If p, q \geq 3, then

  1. There is a homotopy k_t such that k_0 = k and k_1(P) is an embedding.???
  2. 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.

Lemma 6.4. Let V be a (2k-1)-connected 4k-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, respectively.

Proof. 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, applying item 1 of Lemma 6.3 to v_i, we may suppose that v_i itself is an embedding. Further, we apply item 2 of Lemma 6.3, ???setting g_j as k, v_i as y, and V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) as W, for any j < i. ???Observe that Lemma 6.3 is applicable as V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) is simply connected provided 2k \geq 3 and V is simply connected. Also, as x_i \cap_V x_j = 0 for any i, j, the algebraic intersection number for all the pairs of maps above??? is zero. As the result, v_i is replaced with a homotopic embedding g_i, and the images of g_1, \ldots, g_i are disjoint. After the step m we obtain the ???desired set of embeddings.

\square

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let V be an orientable 4k-submanifold of B^{6k+1}, and ???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}) are orthogonal to V, and over every g_i(S^{2k}) the manifold V has a framing whose first vector is tangent to g_i(D^{2k+1}). Assume that g_i(S^{2k}) has zero algebraic self-intersection in V for every i. Then there are embeddings G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1} extending g_1,\ldots,g_s such that G_i(S^{2k}\times D^{2k})\subset V for every i.

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

Identify all the normal spaces!!! of G_1(D^{2k+1}) with the normal space at G_1(0). The normal framing a of g_1(S^{2k}) in V is orthogonal to G_1(D^{2k+1}). So a defines a map S^{2k}\to V_{4k,2k}. Let \zeta=\in\pi_{2k}(V_{4k,2k}) be the homotopy class of this map. This is the obstruction to extending a to a normal 2k-framing of G_1 in B^{6k+1} (so apriori \zeta=\zeta(a)). It suffices to prove that \zeta=0.

Consider the exact sequence of the bundle SO_{a+b}/SO_b = V_{a+b,a}: \pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b). By the following well-known assertion, \partial\zeta=0: if \alpha:S^q\to V_{a,b} is a map, then \partial[\alpha] is the obstruction to trivialization of the orthogonal complement to the field of b-frames in S^q\times\R^a corresponding to \alpha. By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{4k}) is a finite group. Since \pi_{2k}(V_{4k,2k})\cong\Z, we obtain that j=0 for a=b=2k. This and \partial\zeta=0 imply that \zeta=0.

(In [Haefliger1962, the last but one paragraph of 3.5] perhaps one has to replace `\partial\zeta is the obstruction to trivializing the normal bundle of g(S^{2k}) in V' by `\partial\zeta is the obstruction to trivializing te orthogonal complement to the f_1-direction of the normal bundle of V restricted to g(S^{2k})'?)

(No, as we use different obstruction; our is orthogonal to Haefliger's)

Alternatively: Consider a map of the long exact sequences associated to the inclusion S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}:

\xymatrix{\pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[d] \\ \pi_{2k}(SO_{4k})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}

By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{2k+1}) is a finite group. Since \pi_{2k}(S^{2k})\cong\Z, we obtain that p=0. Since s is surjective, we obtain that p'=0.

Hence \partial' is injective. The class \partial'\zeta is??? the obstruction to trivializing the orthogonal complement of g(S^{2k}), cf. [Haefliger1962, 3.4, Lemma].


Suggestion: interpreting \partial'\zeta as an obstruction, we derive that \partial'\zeta is zero, as V is framed and the first vector of the framing is tangent to G_1.

Therefore, \partial ' \zeta =0. Since \partial ' is injective, we obtain \zeta = 0.

\square

Lemma 6.6. Denote by G_1,\ldots, G_s the embeddings from Lemma 6.5. There is smooth manifold V'\subset D^{6k+1} such that V' is homeomorphic to the V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2}) and \partial V'=\partial V.

Proof. For any G_i choose smole??? neighborhood U_i of G_i(D^{2k+1}) and smooth diffeomorphism \phi_i: \mathbb{R}^{6k+1}\to U_i such that \phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k}) is defined by x^2\leq 1, y^2\leq 1, z=0 in some coordinates (x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k}) of \mathbb{R}^{6k+1}.

Denote by a:\mathbb{R}\to \mathbb{R} a smooth monotonous map such that a(x)=0 for x\leq \frac{1}{2} and a(x)=1 for x\geq 1. Denote by g:D^{2k+1}\times S^{2k-1}\to \mathbb{R}^{6k+1} the smooth map, defined by the formula g(x, y):=(x,y\cdot a(x^2), 0). Take by V' the manifold V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}). V' is homeomorphic to V. Therefore, map f:V\to V' defined by equation f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases} is the homeomorphism. Since V' coincide with V outside the \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}) and V' is smooth in \bigcup U_i, we have that V' is smooth.

\square

Below the symbol [ \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\} in H_{2k}(V) such that \alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0, \alpha_i\cap \beta_j=\delta_{i, j} and \lambda^*(\alpha_i)=0 for any i, j. From Lemma 6.4 it follows that there are embeddings f_1, \ldots, f_s:S^{2k}\to V with pairwise disjoint images such that 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 for i\leq s the result of shifting of f_i by the first vector of the framing of V. Since \lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0, we have 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, we have \pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0 for m<2k. Since 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). Therefore there are extensions of f_1, \ldots, f_s to maps 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}).

???Set \varepsilon>0 such that g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V. As \lambda^*(g_{i*}[S^{2k}])=0 for any i, the algebraic intersection number of g_i(\varepsilon D^{2k+1}) and V is zero. Applying item 2 of Lemma6.3 to g_i|_{\varepsilon D^{2k+1}} as y, embedding of V into D^{6k+1} as k and D^{6k+1} as W we may suppose that g_i(\varepsilon D^{2k+1}) does not intersect V. Hance for any i renamed by g_i:D^{2k+1}\to D^{6k+1}\backslash V a map which coincides with g_i on D^{2k+1}\backslash \varepsilon D^{2k+1} and g_i(\varepsilon D^{2k+1})\cap V=\emptyset.

???Denote by G_1, \ldots, G_s the maps as in Lemma 6.5 for V and g_1, \ldots, g_s as described above. Denote by V' the manifold as in Lemma 6.6. Since \left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\} is a basis in H_{2k}(V), ???it follows that H_{2k}(V')= 0. Since \pi_1(V)= 0 we have \pi_1(V')= 0. Hence??? H_m(V')= 0 for 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}. 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

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  • [Farber1981] M. Sh. Farber, Classification of stable fibered knots, Mat. Sb. (N.S.), 115(157):2(6) (1981) 223–262.
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  • [Farber1984] M. Sh. Farber, An algebraic classification of some even-dimensional spherical knots I, II, Trans. Amer. Math. Soc. 281 (1984), 507-528; 529-570.
  • [Fomenko&Fuchs2016] A. T. Fomenko and D. B. Fuks, Homotopical Topology. Translated from the Russian. Graduate Texts in Mathematics, 273. Springer-Verlag, Berlin, 2016. DOI 10.1007/978-3-319-23488-5.

$. 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}. 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 $x_1, \ldots, x_m$, respectively. \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, applying item 1 of Lemma \ref{whitney} to $v_i$, 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 Identify all the normal spaces!!! of $G_1(D^{2k+1})$ with the normal space at $G_1(0)$. The normal framing $a$ of $g_1(S^{2k})$ in $V$ is orthogonal to $G_1(D^{2k+1})$. So $a$ defines a map $S^{2k}\to V_{4k,2k}$. Let $\zeta=\in\pi_{2k}(V_{4k,2k})$ be the homotopy class of this map. This is the obstruction to extending $a$ to a normal k$-framing of $G_1$ in $B^{6k+1}$ (so apriori $\zeta=\zeta(a)$). It suffices to prove that $\zeta=0$. Consider the exact sequence of the bundle $SO_{a+b}/SO_b = V_{a+b,a}$: $\pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b)$. By the following well-known assertion, $\partial\zeta=0$: ''if $\alpha:S^q\to V_{a,b}$ is a map, then $\partial[\alpha]$ is the obstruction to trivialization of the orthogonal complement to the field of $b$-frames in $S^q\times\R^a$ corresponding to $\alpha$.'' By \cite[Corollary in $\S.4]{Fomenko&Fuchs2016} $\pi_{2k}(SO_{4k})$ is a finite group. Since $\pi_{2k}(V_{4k,2k})\cong\Z$, we obtain that $j=0$ for $a=b=2k$. This and $\partial\zeta=0$ imply that $\zeta=0$. (In \cite[the last but one paragraph of 3.5]{Haefliger1962} perhaps one has to replace `$\partial\zeta$ is the obstruction to trivializing the normal bundle of $g(S^{2k})$ in $V$' by `$\partial\zeta$ is the obstruction to trivializing te orthogonal complement to the $f_1$-direction of the normal bundle of $V$ restricted to $g(S^{2k})$'?) (No, as we use different obstruction; our is orthogonal to Haefliger's) Alternatively: Consider a map of the long exact sequences associated to the inclusion $S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}$: $\xymatrix{\pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[d] \ \pi_{2k}(SO_{4k})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}$ By \cite[Corollary in $\S.4]{Fomenko&Fuchs2016} $\pi_{2k}(SO_{2k+1})$ is a finite group. Since $\pi_{2k}(S^{2k})\cong\Z$, we obtain that $p=0$. Since $s$ is surjective, we obtain that $p'=0$. Hence $\partial'$ is injective. The class $\partial'\zeta$ is??? the obstruction to trivializing the orthogonal complement of $g(S^{2k})$, cf. \cite[3.4, Lemma]{Haefliger1962}. Suggestion: interpreting $\partial'\zeta$ as an obstruction, we derive that $\partial'\zeta$ is zero, as $V$ is framed and the first vector of the framing is tangent to $G_1$. Therefore, $\partial ' \zeta =0$. Since $\partial '$ is injective, we obtain $\zeta = 0$. {{endproof}} {{beginthm|Lemma}} \label{l:smoothen} Denote by $G_1,\ldots, G_s$ the embeddings from Lemma \ref{l:multi_spherical_modification}. There is smooth manifold $V'\subset D^{6k+1}$ such that $V'$ is homeomorphic to the $V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2})$ and $\partial V'=\partial V$. {{endthm}} {{beginproof}} For any $G_i$ choose smole??? neighborhood $U_i$ of $G_i(D^{2k+1})$ and smooth diffeomorphism $\phi_i: \mathbb{R}^{6k+1}\to U_i$ such that $\phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k})$ is defined by $x^2\leq 1, y^2\leq 1, z=0$ in some coordinates $(x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k})$ of $\mathbb{R}^{6k+1}$. Denote by $a:\mathbb{R}\to \mathbb{R}$ a smooth monotonous map such that $a(x)=0$ for $x\leq \frac{1}{2}$ and $a(x)=1$ for $x\geq 1$. Denote by $g:D^{2k+1}\times S^{2k-1}\to \mathbb{R}^{6k+1}$ the smooth map, defined by the formula $g(x, y):=(x,y\cdot a(x^2), 0)$. Take by $V'$ the manifold $V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2})$. $V'$ is homeomorphic to $V$. Therefore, map $f:V\to V'$ defined by equation $f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases}$ is the homeomorphism. Since $V'$ coincide with $V$ outside the $\bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2})$ and $V'$ is smooth in $\bigcup U_i$, we have that $V'$ is smooth. {{endproof}} Below the symbol $[ \cdot ]$ denotes the integral fundamental class of a manifold or the homotopy class of a map, depending on the context. ''Proof of Lemma \ref{l:V_to_disk} using Lemmas \ref{embeddings}, \ref{l:multi_spherical_modification}.''. By the fourth paragraph of the proof of Theorem 3.1 in \cite{Ha62}, there is a basis $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ in $H_{2k}(V)$ such that $\alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0$, $\alpha_i\cap \beta_j=\delta_{i, j}$ and $\lambda^*(\alpha_i)=0$ for any $i, j$. From Lemma \ref{embeddings} it follows that there are embeddings $f_1, \ldots, f_s:S^{2k}\to V$ with pairwise disjoint images such that $f_{1*}[S^{2k}]=\alpha_1, \ldots, f_{s*}[S^{2k}]=\alpha_s$. Denote by $f_{i}':S^{2k}\to D^{6k+1}\backslash V$ for $i\leq s$ the result of shifting of $f_i$ by the first vector of the framing of $V$. Since $\lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0$, we have $f_{i*}'([S^{2k}])=0 \in H_{2k}(D^{6k+1}\backslash V)$. Since $\mbox{dim} D^{6k+1}-\mbox{dim} V=6k+1-4k=2k+1$, we have $\pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0$ for $m<2k$. Since $f_{i*}'([S^{2k}])=0\in H_{2k}(D^{6k+1}\backslash V)$, we have $[f_i']=0\in \pi_{2k}(D^{6k+1}\backslash V)$. Therefore there are extensions of $f_1, \ldots, f_s$ to maps $g_1, \ldots, g_s:D^{2k+1}\to B^{6k+1}$ such that $g_i(D^{2k+1})\cap V = g_i(S^{2k})$. ???Set $\varepsilon>0$ such that $g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V$. As $\lambda^*(g_{i*}[S^{2k}])=0$ for any $i$, the algebraic intersection number of $g_i(\varepsilon D^{2k+1})$ and $V$ is zero. Applying item 2 of Lemma\ref{whitney} to $g_i|_{\varepsilon D^{2k+1}}$ as $y$, embedding of $V$ into $D^{6k+1}$ as $k$ and $D^{6k+1}$ as $W$ we may suppose that $g_i(\varepsilon D^{2k+1})$ does not intersect $V$. Hance for any $i$ renamed by $g_i:D^{2k+1}\to D^{6k+1}\backslash V $ a map which coincides with $g_i$ on $D^{2k+1}\backslash \varepsilon D^{2k+1}$ and $g_i(\varepsilon D^{2k+1})\cap V=\emptyset$. ???Denote by $G_1, \ldots, G_s$ the maps as in Lemma \ref{l:multi_spherical_modification} for $V$ and $g_1, \ldots, g_s$ as described above. Denote by $V'$ the manifold as in Lemma \ref{l:smoothen}. Since $\left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\}$ is a basis in $H_{2k}(V)$, ???it follows that $H_{2k}(V')= 0$. Since $\pi_1(V)= 0$ we have $\pi_1(V')= 0$. Hence??? $H_m(V')= 0$ for 1, \S3].

2 Examples

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

Example 2.1. (a) Analogously to the Haefliger trefoil knot for any l>1 one constructs a smooth embedding t:S^{2l-1}\to\Rr^{3l}, see [Skopenkov2016h, \S5]. 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 [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 [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}) the Zeeman map, see [Skopenkov2016h, Definition 2.2]. 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 [Skopenkov2015a, Corollary 2.13].

3 Invariants

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

By [Haefliger1962, 2.1, 2.2] 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 4k-manifold V whose boundary is S^{4k-1}, and whose signature is zero. For an integer 2k-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 [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 2k-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
\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.

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

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

For an alternative definition via Seifert surfaces in 6k-space, discovered in [Guillou&Marin1986], [Takase2004], see [Skopenkov2016t, the Kreck Invariant Lemma 4.5]. For a definition by Kreck, and for a generalization to 3-manifolds see [Skopenkov2016t, \S4].

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

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

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

For definition of the attaching invariant E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q) see [Haefliger1966], [Skopenkov2005, \S3].

4 Classification

Theorem 4.1 [Levine1965, Corollary in p. 44], [Haefliger1966]. 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.

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

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

Proof for the first four columns, and for the fifth column when k is odd, are presented in [Haefliger1966, 8.15] (see also \S6; some proofs are deduced from that paper using simple calculations, cf. [Skopenkov2005, \S3]; there is a typo in [Haefliger1966, 8.15]: C^{3k}_{4k−2}=0 should be C^{4k}_{8k−2}=0). The remaining results follow from [Haefliger1966, 8.15] and [Milgram1972, Theorem F]. Alternative proofs for the cases (m,n)=(7,4),(6,3) are given in [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

Theorem 4.3 [Milgram1972, Corollary G]. We have E^m_D(S^n)=0 if and only if either 2m\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.

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

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

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

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

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

Here SG_q is the space of maps S^{q-1} \to S^{q-1} of degree 1. Restricting a map from SO_q to S^{q-1} \subset \Rr^q identifies SO_q as a subspace of SG_q. Define SG:=SG_1\cup\ldots\cup SG_q\cup\ldots. Analogously define SO. Let s be the stabilization homomorphism. The attaching invariant a and the map u are defined in [Haefliger1966], see also [Skopenkov2005, \S3].

5 Some remarks on codimension 2 knots

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

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

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

Theorem 6.1. The Haefliger invariant \varkappa:E_D^{6k}(S^{4k-1})\to\Zz is injective for 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 be a framed (2k-1)-connected 4k-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.

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} such that \varkappa(g)=0 there is a framed (2k-1)-connected 4k-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 6.2 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 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 be a map from a connected p-manifold P to a simply connected (p+q)-manifold W. If p, q \geq 3, then

  1. There is a homotopy k_t such that k_0 = k and k_1(P) is an embedding.???
  2. 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.

Lemma 6.4. Let V be a (2k-1)-connected 4k-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, respectively.

Proof. 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, applying item 1 of Lemma 6.3 to v_i, we may suppose that v_i itself is an embedding. Further, we apply item 2 of Lemma 6.3, ???setting g_j as k, v_i as y, and V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) as W, for any j < i. ???Observe that Lemma 6.3 is applicable as V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) is simply connected provided 2k \geq 3 and V is simply connected. Also, as x_i \cap_V x_j = 0 for any i, j, the algebraic intersection number for all the pairs of maps above??? is zero. As the result, v_i is replaced with a homotopic embedding g_i, and the images of g_1, \ldots, g_i are disjoint. After the step m we obtain the ???desired set of embeddings.

\square

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let V be an orientable 4k-submanifold of B^{6k+1}, and ???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}) are orthogonal to V, and over every g_i(S^{2k}) the manifold V has a framing whose first vector is tangent to g_i(D^{2k+1}). Assume that g_i(S^{2k}) has zero algebraic self-intersection in V for every i. Then there are embeddings G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1} extending g_1,\ldots,g_s such that G_i(S^{2k}\times D^{2k})\subset V for every i.

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

Identify all the normal spaces!!! of G_1(D^{2k+1}) with the normal space at G_1(0). The normal framing a of g_1(S^{2k}) in V is orthogonal to G_1(D^{2k+1}). So a defines a map S^{2k}\to V_{4k,2k}. Let \zeta=\in\pi_{2k}(V_{4k,2k}) be the homotopy class of this map. This is the obstruction to extending a to a normal 2k-framing of G_1 in B^{6k+1} (so apriori \zeta=\zeta(a)). It suffices to prove that \zeta=0.

Consider the exact sequence of the bundle SO_{a+b}/SO_b = V_{a+b,a}: \pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b). By the following well-known assertion, \partial\zeta=0: if \alpha:S^q\to V_{a,b} is a map, then \partial[\alpha] is the obstruction to trivialization of the orthogonal complement to the field of b-frames in S^q\times\R^a corresponding to \alpha. By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{4k}) is a finite group. Since \pi_{2k}(V_{4k,2k})\cong\Z, we obtain that j=0 for a=b=2k. This and \partial\zeta=0 imply that \zeta=0.

(In [Haefliger1962, the last but one paragraph of 3.5] perhaps one has to replace `\partial\zeta is the obstruction to trivializing the normal bundle of g(S^{2k}) in V' by `\partial\zeta is the obstruction to trivializing te orthogonal complement to the f_1-direction of the normal bundle of V restricted to g(S^{2k})'?)

(No, as we use different obstruction; our is orthogonal to Haefliger's)

Alternatively: Consider a map of the long exact sequences associated to the inclusion S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}:

\xymatrix{\pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[d] \\ \pi_{2k}(SO_{4k})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}

By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{2k+1}) is a finite group. Since \pi_{2k}(S^{2k})\cong\Z, we obtain that p=0. Since s is surjective, we obtain that p'=0.

Hence \partial' is injective. The class \partial'\zeta is??? the obstruction to trivializing the orthogonal complement of g(S^{2k}), cf. [Haefliger1962, 3.4, Lemma].


Suggestion: interpreting \partial'\zeta as an obstruction, we derive that \partial'\zeta is zero, as V is framed and the first vector of the framing is tangent to G_1.

Therefore, \partial ' \zeta =0. Since \partial ' is injective, we obtain \zeta = 0.

\square

Lemma 6.6. Denote by G_1,\ldots, G_s the embeddings from Lemma 6.5. There is smooth manifold V'\subset D^{6k+1} such that V' is homeomorphic to the V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2}) and \partial V'=\partial V.

Proof. For any G_i choose smole??? neighborhood U_i of G_i(D^{2k+1}) and smooth diffeomorphism \phi_i: \mathbb{R}^{6k+1}\to U_i such that \phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k}) is defined by x^2\leq 1, y^2\leq 1, z=0 in some coordinates (x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k}) of \mathbb{R}^{6k+1}.

Denote by a:\mathbb{R}\to \mathbb{R} a smooth monotonous map such that a(x)=0 for x\leq \frac{1}{2} and a(x)=1 for x\geq 1. Denote by g:D^{2k+1}\times S^{2k-1}\to \mathbb{R}^{6k+1} the smooth map, defined by the formula g(x, y):=(x,y\cdot a(x^2), 0). Take by V' the manifold V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}). V' is homeomorphic to V. Therefore, map f:V\to V' defined by equation f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases} is the homeomorphism. Since V' coincide with V outside the \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}) and V' is smooth in \bigcup U_i, we have that V' is smooth.

\square

Below the symbol [ \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\} in H_{2k}(V) such that \alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0, \alpha_i\cap \beta_j=\delta_{i, j} and \lambda^*(\alpha_i)=0 for any i, j. From Lemma 6.4 it follows that there are embeddings f_1, \ldots, f_s:S^{2k}\to V with pairwise disjoint images such that 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 for i\leq s the result of shifting of f_i by the first vector of the framing of V. Since \lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0, we have 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, we have \pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0 for m<2k. Since 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). Therefore there are extensions of f_1, \ldots, f_s to maps 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}).

???Set \varepsilon>0 such that g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V. As \lambda^*(g_{i*}[S^{2k}])=0 for any i, the algebraic intersection number of g_i(\varepsilon D^{2k+1}) and V is zero. Applying item 2 of Lemma6.3 to g_i|_{\varepsilon D^{2k+1}} as y, embedding of V into D^{6k+1} as k and D^{6k+1} as W we may suppose that g_i(\varepsilon D^{2k+1}) does not intersect V. Hance for any i renamed by g_i:D^{2k+1}\to D^{6k+1}\backslash V a map which coincides with g_i on D^{2k+1}\backslash \varepsilon D^{2k+1} and g_i(\varepsilon D^{2k+1})\cap V=\emptyset.

???Denote by G_1, \ldots, G_s the maps as in Lemma 6.5 for V and g_1, \ldots, g_s as described above. Denote by V' the manifold as in Lemma 6.6. Since \left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\} is a basis in H_{2k}(V), ???it follows that H_{2k}(V')= 0. Since \pi_1(V)= 0 we have \pi_1(V')= 0. Hence??? H_m(V')= 0 for 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}. 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

  • [Crowley&Skopenkov2008] D. Crowley and A. Skopenkov, A classification of smooth embeddings of 4-manifolds in 7-space, II, Intern. J. Math., 22:6 (2011) 731-757. Available at the arXiv:0808.1795.
  • [Farber1981] M. Sh. Farber, Classification of stable fibered knots, Mat. Sb. (N.S.), 115(157):2(6) (1981) 223–262.
  • [Farber1983] M. Sh. Farber, The classification of simple knots, Russian Math. Surveys, 38:5 (1983).

  • [Farber1984] M. Sh. Farber, An algebraic classification of some even-dimensional spherical knots I, II, Trans. Amer. Math. Soc. 281 (1984), 507-528; 529-570.
  • [Fomenko&Fuchs2016] A. T. Fomenko and D. B. Fuks, Homotopical Topology. Translated from the Russian. Graduate Texts in Mathematics, 273. Springer-Verlag, Berlin, 2016. DOI 10.1007/978-3-319-23488-5.

< 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, \S3].

2 Examples

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

Example 2.1. (a) Analogously to the Haefliger trefoil knot for any l>1 one constructs a smooth embedding t:S^{2l-1}\to\Rr^{3l}, see [Skopenkov2016h, \S5]. 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 [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 [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}) the Zeeman map, see [Skopenkov2016h, Definition 2.2]. 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 [Skopenkov2015a, Corollary 2.13].

3 Invariants

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

By [Haefliger1962, 2.1, 2.2] 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 4k-manifold V whose boundary is S^{4k-1}, and whose signature is zero. For an integer 2k-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 [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 2k-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
\displaystyle \varkappa(f):=\lambda\cap_V\lambda/2.

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

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

For an alternative definition via Seifert surfaces in 6k-space, discovered in [Guillou&Marin1986], [Takase2004], see [Skopenkov2016t, the Kreck Invariant Lemma 4.5]. For a definition by Kreck, and for a generalization to 3-manifolds see [Skopenkov2016t, \S4].

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

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

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

For definition of the attaching invariant E^{n+q}_D(S^n)\to\pi_n(G_q,SO_q) see [Haefliger1966], [Skopenkov2005, \S3].

4 Classification

Theorem 4.1 [Levine1965, Corollary in p. 44], [Haefliger1966]. 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.

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

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

Proof for the first four columns, and for the fifth column when k is odd, are presented in [Haefliger1966, 8.15] (see also \S6; some proofs are deduced from that paper using simple calculations, cf. [Skopenkov2005, \S3]; there is a typo in [Haefliger1966, 8.15]: C^{3k}_{4k−2}=0 should be C^{4k}_{8k−2}=0). The remaining results follow from [Haefliger1966, 8.15] and [Milgram1972, Theorem F]. Alternative proofs for the cases (m,n)=(7,4),(6,3) are given in [Skopenkov2005], [Crowley&Skopenkov2008], [Skopenkov2008].

Theorem 4.3 [Milgram1972, Corollary G]. We have E^m_D(S^n)=0 if and only if either 2m\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.

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

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

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

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

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

Here SG_q is the space of maps S^{q-1} \to S^{q-1} of degree 1. Restricting a map from SO_q to S^{q-1} \subset \Rr^q identifies SO_q as a subspace of SG_q. Define SG:=SG_1\cup\ldots\cup SG_q\cup\ldots. Analogously define SO. Let s be the stabilization homomorphism. The attaching invariant a and the map u are defined in [Haefliger1966], see also [Skopenkov2005, \S3].

5 Some remarks on codimension 2 knots

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

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

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

Theorem 6.1. The Haefliger invariant \varkappa:E_D^{6k}(S^{4k-1})\to\Zz is injective for 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 be a framed (2k-1)-connected 4k-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.

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} such that \varkappa(g)=0 there is a framed (2k-1)-connected 4k-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 6.2 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 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 be a map from a connected p-manifold P to a simply connected (p+q)-manifold W. If p, q \geq 3, then

  1. There is a homotopy k_t such that k_0 = k and k_1(P) is an embedding.???
  2. 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.

Lemma 6.4. Let V be a (2k-1)-connected 4k-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, respectively.

Proof. 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, applying item 1 of Lemma 6.3 to v_i, we may suppose that v_i itself is an embedding. Further, we apply item 2 of Lemma 6.3, ???setting g_j as k, v_i as y, and V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) as W, for any j < i. ???Observe that Lemma 6.3 is applicable as V \setminus \bigcup\limits_{l<j}g_l(S^{2k}) is simply connected provided 2k \geq 3 and V is simply connected. Also, as x_i \cap_V x_j = 0 for any i, j, the algebraic intersection number for all the pairs of maps above??? is zero. As the result, v_i is replaced with a homotopic embedding g_i, and the images of g_1, \ldots, g_i are disjoint. After the step m we obtain the ???desired set of embeddings.

\square

Lemma 6.5.[cf. Proposition 3.3 in [Ha62]] Let V be an orientable 4k-submanifold of B^{6k+1}, and ???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}) are orthogonal to V, and over every g_i(S^{2k}) the manifold V has a framing whose first vector is tangent to g_i(D^{2k+1}). Assume that g_i(S^{2k}) has zero algebraic self-intersection in V for every i. Then there are embeddings G_1,\ldots,G_s:D^{2k+1}\times D^{2k}\to B^{6k+1} extending g_1,\ldots,g_s such that G_i(S^{2k}\times D^{2k})\subset V for every i.

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

Identify all the normal spaces!!! of G_1(D^{2k+1}) with the normal space at G_1(0). The normal framing a of g_1(S^{2k}) in V is orthogonal to G_1(D^{2k+1}). So a defines a map S^{2k}\to V_{4k,2k}. Let \zeta=\in\pi_{2k}(V_{4k,2k}) be the homotopy class of this map. This is the obstruction to extending a to a normal 2k-framing of G_1 in B^{6k+1} (so apriori \zeta=\zeta(a)). It suffices to prove that \zeta=0.

Consider the exact sequence of the bundle SO_{a+b}/SO_b = V_{a+b,a}: \pi_q(SO_{a+b}) \overset{j}\to \pi_q(V_{a+b,a}) \overset{\partial}\to \pi_{q-1}(SO_b). By the following well-known assertion, \partial\zeta=0: if \alpha:S^q\to V_{a,b} is a map, then \partial[\alpha] is the obstruction to trivialization of the orthogonal complement to the field of b-frames in S^q\times\R^a corresponding to \alpha. By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{4k}) is a finite group. Since \pi_{2k}(V_{4k,2k})\cong\Z, we obtain that j=0 for a=b=2k. This and \partial\zeta=0 imply that \zeta=0.

(In [Haefliger1962, the last but one paragraph of 3.5] perhaps one has to replace `\partial\zeta is the obstruction to trivializing the normal bundle of g(S^{2k}) in V' by `\partial\zeta is the obstruction to trivializing te orthogonal complement to the f_1-direction of the normal bundle of V restricted to g(S^{2k})'?)

(No, as we use different obstruction; our is orthogonal to Haefliger's)

Alternatively: Consider a map of the long exact sequences associated to the inclusion S^{2k} = SO_{2k+1}/SO_{2k} \rightarrow SO_{4k}/SO_{2k} = V_{4k, 2k}:

\xymatrix{\pi_{2k}(SO_{2k+1}) \ar[r]^{p = 0} \ar[d]^{s} & \pi_{2k}(S^{2k}) \ar[d] \\ \pi_{2k}(SO_{4k})\ar[r]^{p'=0} & \pi_{2k}(V_{4k, 2k})\ar[r]^{\partial '} & \pi_{2k-1}(SO_{2k})}

By [Fomenko&Fuchs2016, Corollary in \S25.4] \pi_{2k}(SO_{2k+1}) is a finite group. Since \pi_{2k}(S^{2k})\cong\Z, we obtain that p=0. Since s is surjective, we obtain that p'=0.

Hence \partial' is injective. The class \partial'\zeta is??? the obstruction to trivializing the orthogonal complement of g(S^{2k}), cf. [Haefliger1962, 3.4, Lemma].


Suggestion: interpreting \partial'\zeta as an obstruction, we derive that \partial'\zeta is zero, as V is framed and the first vector of the framing is tangent to G_1.

Therefore, \partial ' \zeta =0. Since \partial ' is injective, we obtain \zeta = 0.

\square

Lemma 6.6. Denote by G_1,\ldots, G_s the embeddings from Lemma 6.5. There is smooth manifold V'\subset D^{6k+1} such that V' is homeomorphic to the V\backslash (\bigcup G_i(???S^{2k}\times D^{2k-1})) \bigcup G_i(D^{2k+1}\times S^{2k-2}) and \partial V'=\partial V.

Proof. For any G_i choose smole??? neighborhood U_i of G_i(D^{2k+1}) and smooth diffeomorphism \phi_i: \mathbb{R}^{6k+1}\to U_i such that \phi_i^{-1}\circ G_i(D^{2k+1}\times D^{2k}) is defined by x^2\leq 1, y^2\leq 1, z=0 in some coordinates (x, y, z)=(x_1,\ldots, x_{2k+1}, y_1, \ldots, y_{2k}, z_1, \ldots, z_{2k}) of \mathbb{R}^{6k+1}.

Denote by a:\mathbb{R}\to \mathbb{R} a smooth monotonous map such that a(x)=0 for x\leq \frac{1}{2} and a(x)=1 for x\geq 1. Denote by g:D^{2k+1}\times S^{2k-1}\to \mathbb{R}^{6k+1} the smooth map, defined by the formula g(x, y):=(x,y\cdot a(x^2), 0). Take by V' the manifold V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}). V' is homeomorphic to V. Therefore, map f:V\to V' defined by equation f(x)=\begin{cases} x\text{, if }x\in V\backslash (\bigcup G_i(S^{2k}\times D^{2k-1})) \\g\circ G_i^{-1}\circ \phi_i^{-1}(x)\text{, if }x\in G_i(S^{2k}\times D^{2k-1}) \end{cases} is the homeomorphism. Since V' coincide with V outside the \bigcup \phi_i\circ g(D^{2k+1}\times S^{2k-2}) and V' is smooth in \bigcup U_i, we have that V' is smooth.

\square

Below the symbol [ \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\} in H_{2k}(V) such that \alpha_i\cap \alpha_j=\beta_i\cap \beta_j = 0, \alpha_i\cap \beta_j=\delta_{i, j} and \lambda^*(\alpha_i)=0 for any i, j. From Lemma 6.4 it follows that there are embeddings f_1, \ldots, f_s:S^{2k}\to V with pairwise disjoint images such that 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 for i\leq s the result of shifting of f_i by the first vector of the framing of V. Since \lambda(f_{i*}[S^{2k}])=\lambda(\alpha_i)=0, we have 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, we have \pi_m(D^{6k+1}\backslash \mbox{Int}{V}) = 0 for m<2k. Since 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). Therefore there are extensions of f_1, \ldots, f_s to maps 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}).

???Set \varepsilon>0 such that g_i(\varepsilon D^{2k+1})\cap V= g_i(\mbox{Int} D^{2k+1})\cap V. As \lambda^*(g_{i*}[S^{2k}])=0 for any i, the algebraic intersection number of g_i(\varepsilon D^{2k+1}) and V is zero. Applying item 2 of Lemma6.3 to g_i|_{\varepsilon D^{2k+1}} as y, embedding of V into D^{6k+1} as k and D^{6k+1} as W we may suppose that g_i(\varepsilon D^{2k+1}) does not intersect V. Hance for any i renamed by g_i:D^{2k+1}\to D^{6k+1}\backslash V a map which coincides with g_i on D^{2k+1}\backslash \varepsilon D^{2k+1} and g_i(\varepsilon D^{2k+1})\cap V=\emptyset.

???Denote by G_1, \ldots, G_s the maps as in Lemma 6.5 for V and g_1, \ldots, g_s as described above. Denote by V' the manifold as in Lemma 6.6. Since \left\{\alpha_1, \ldots, \alpha_s, \beta_1, \ldots, \beta_s\right\} is a basis in H_{2k}(V), ???it follows that H_{2k}(V')= 0. Since \pi_1(V)= 0 we have \pi_1(V')= 0. Hence??? H_m(V')= 0 for 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}. 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

  • [Crowley&Skopenkov2008] D. Crowley and A. Skopenkov, A classification of smooth embeddings of 4-manifolds in 7-space, II, Intern. J. Math., 22:6 (2011) 731-757. Available at the arXiv:0808.1795.
  • [Farber1981] M. Sh. Farber, Classification of stable fibered knots, Mat. Sb. (N.S.), 115(157):2(6) (1981) 223–262.
  • [Farber1983] M. Sh. Farber, The classification of simple knots, Russian Math. Surveys, 38:5 (1983).

  • [Farber1984] M. Sh. Farber, An algebraic classification of some even-dimensional spherical knots I, II, Trans. Amer. Math. Soc. 281 (1984), 507-528; 529-570.
  • [Fomenko&Fuchs2016] A. T. Fomenko and D. B. Fuks, Homotopical Topology. Translated from the Russian. Graduate Texts in Mathematics, 273. Springer-Verlag, Berlin, 2016. DOI 10.1007/978-3-319-23488-5.

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