A counting invariant for maps into spheres and for zero loci of sections of vector bundles

Abstract

The set of unrestricted homotopy classes $[M,S^n]$ where M is a closed and connected spin $(n+1)$-manifold is called the n-th cohomotopy group $\pi ^n(M)$ of M. Using homotopy theory it is known that $\pi ^n(M) = H^n(M;{\mathbb {Z}}) \oplus {\mathbb {Z}}_2$. We will provide a geometrical description of the ${\mathbb {Z}}_2$ part in $\pi ^n(M)$ analogous to Pontryagin’s computation of the stable homotopy group $\pi _{n+1}(S^n)$. This ${\mathbb {Z}}_2$ number can be computed by counting embedded circles in M with a certain framing of their normal bundle. This is a similar result to the mod 2 degree theorem for maps $M \rightarrow S^{n+1}$. Finally we will observe that the zero locus of a section in an oriented rank n vector bundle $E \rightarrow M$ defines an element in $\pi ^n(M)$ and it turns out that the ${\mathbb {Z}}_2$ part is an invariant of the isomorphism class of E. At the end we show that if the Euler class of E vanishes this ${\mathbb {Z}}_2$ invariant is the final obstruction to the existence of a nowhere vanishing section.

1 Introduction

Pontryagin computed in [16] the (stable) homotopy group $\pi _{n+1}(S^n)$ ($n\ge 3$) using differential topology. Let us describe briefly his construction, since this paper will generalize his idea.

He showed that $\pi _{n+1}(S^n)$ is isomorphic to the bordism group of closed 1-dimensional submanifolds of ${\mathbb {R}}^{n+1}$ furnished with a framing on its normal bundle (a framing is a homotopy class of trivializations, see Sect. 2). We denote this bordism group by $\Omega _1^{\text {fr}}({\mathbb {R}}^{n+1})$. Let $(C,\varphi )$ be a representative of an element of $\Omega _1^{\text {fr}}({\mathbb {R}}^{n+1})$, i.e. C is a union of embedded circles in ${\mathbb {R}}^{n+1}$ and there are maps $\varphi _1,\ldots ,\varphi _n :C \rightarrow {\mathbb {R}}^{n+1}$ such that $(\varphi _1(x),\ldots ,\varphi _n(x))$ is a basis of $\nu (C)_x$ for every $x \in C$. Let $\varphi _{n+1}$ be a trivialization of the tangent bundle of C such that $(\varphi _1(x),\ldots ,\varphi _{n+1}(x))$ is a positive oriented basis of ${\mathbb {R}}^{n+1}$ for every $x \in C$. Without loss of generality we may assume that $\varphi _1,\ldots ,\varphi _{n+1}$ is pointwise an orthonormal basis. If $(e_1,\ldots ,e_{n+1})$ denotes the standard basis of ${\mathbb {R}}^{n+1}$, then consider the map $A=(a_{ij}) :C \rightarrow \mathrm {SO}(n+1)$ such that

$$\begin{aligned} \varphi _i(x) = \sum _{j=1}^{n+1} a_{ij}(x)e_j \end{aligned}$$

for $x \in C$. Let $\pi _1(\mathrm {SO}(n+1))$ be identified with ${\mathbb {Z}}_2$, then Pontryagin defines [16, Theorem 20]

$$\begin{aligned} \delta (C,\varphi ) := [A] + (n(C) \mod 2) \end{aligned}$$

where [A] denotes the homotopy class of A in $\pi _1(\mathrm {SO}(n+1))$ and n(C) is the number of connected components of C. He showed that $\delta$ is well-defined on $\Omega _1^{\text {fr}}({\mathbb {R}}^{n+1})$ and is an isomorphism of groups.

From a different point of view, one may consider his computation not as a computation of a homotopy group of $S^n$ but rather of a cohomotopy group of $S^{n+1}$. If X is a CW space then the cohomotopy set of X is defined as the set of (unrestricted) homotopy classes $\pi ^n(X):=[X,S^n]$, cf. [3, 18]. The set $\pi ^n(X)$ for X a finite CW complex of dimension $n+1$ carries naturally a group structure, which is described in the beginning of Sect. 4. Steenrod showed [19, Theorem 28.1, p. 318] that $\pi ^n(X)$ fits into a short exact sequence

$$\begin{aligned} 0 \longrightarrow H^{n+1}(X;{\mathbb {Z}}_2)/ \text {Sq}^2 \mu (H^{n-1}(X;{\mathbb {Z}})) \longrightarrow \pi ^n(X) \longrightarrow H^n(X;{\mathbb {Z}}) \longrightarrow 0, \end{aligned}$$

where $\mu :H^{*}(X;{\mathbb {Z}}) \rightarrow H^*(X;{\mathbb {Z}}_2)$ is the mod 2 reduction homomorphism (see also [17] for a more geometric proof of this sequence). Here the surjective map is the Hurewicz homomorphism which assigns to every $f \in \pi ^n(X)$ the cohomology class $f^*(\sigma ) \in H^n(X;{\mathbb {Z}})$ where $\sigma \in H^n(S^n,{\mathbb {Z}})$ is a fixed generator.

Moreover using methods of Larmore and Thomas [11] Taylor showed in [20, Theorem 6.2, Example 6.3] that the short exact sequence splits, provided the images of $\text {Sq}^2 :H^{n-1}(X;{\mathbb {Z}}_2) \rightarrow H^{n+1}(X;{\mathbb {Z}}_2)$ and $\text {Sq}^2 \circ \mu :H^{n-1}(X;{\mathbb {Z}}) \rightarrow H^{n+1}(X;{\mathbb {Z}}_2)$ coincide.

If $X=M$ is an oriented manifold then the second Wu class [24] is equal to the second Stiefel-Whitney class $w_2(M)$, hence $\text {Sq}^2(x) = w_2(M)\smile x$ for $x \in H^{n-1}(M;{\mathbb {Z}}_2)$. Therefore if M is spin then $\pi ^n(M)$ fits into the exact sequence

$$\begin{aligned} 0 \longrightarrow {\mathbb {Z}}_2 \longrightarrow \pi ^n(M) \longrightarrow H^n(M;{\mathbb {Z}}) \longrightarrow 0. \end{aligned}$$

(ST)

and (ST) splits by [20, Example 6.3] thus

$$\begin{aligned} \pi ^n(M) \cong H^n(M;{\mathbb {Z}}) \oplus {\mathbb {Z}}_2 \end{aligned}$$

as abelian groups. However the splitting map is constructed in a purely homotopy theoretic way and an aim of this article is to provide a geometric description in case M is a spin manifold.

This splitting map $\kappa :\pi ^n(M) \rightarrow {\mathbb {Z}}_2$ (see Definition 3.8) for (ST) will be constructed similarly to Pontryagin’s invariant $\delta$ from above. An important ingredient in Pontryagin’s construction was the canonical background framing given by the standard basis of ${\mathbb {R}}^{n+1}$, which allowed him to define the map $A:C \rightarrow \mathrm {SO}(n+1)$. In general if we replace $S^{n+1}$ or ${\mathbb {R}}^{n+1}$ by M, this background framing is not available any more. But this can be circumvented by using the spin structure of M, since over a circle every vector bundle with a spin structure defines a certain framing, cf. Lemma 3.1. Section 4 is devoted to determine geometrically the kernel of the Hurewicz map $\pi ^n(M) \rightarrow H^n(M;{\mathbb {Z}})$. Finally we show that the splitting map possesses a naturality property, cf. Proposition 4.3 and that for a map $f:M\rightarrow S^n$ the number $\kappa (f)$ can be described by a counting formula, cf. Corollary 4.4. This is an analogous result to the mod 2 Hopf theorem, see [15, §4]. It should be mentioned that in [8] the authors discuss the case $n=3$ and in [10] a similar construction of a ${\mathbb {Z}}_2$ invariant was used to classify quaternionic line bundles over closed spin 5-manifolds.

In Sect. 5 we will apply the results of Sects. 3 and 4 to the theory of vector bundles. Suppose $E \rightarrow M$ is an oriented vector bundle of rank n over a closed spin $(n+1)$-manifold M. Then any section of E which is transverse to the zero section defines by of its zero locus an element of $\Omega _1^{\text {fr}}(M)$ and this element is independent of the transverse section. Thus using $\kappa$ one defines an invariant $\kappa (E) \in \Omega _1^\text {fr}$ of the isomorphism class of the bundle $E\rightarrow M$. In Theorem 5.5 it is shown that $\kappa (E)$ can be regarded as the secondary obstruction to the existence of a nowhere vanishing section. As an application we provide in Example 5.6 a simple proof of the well-known fact, that the maximal number of linear independent vector fields on $S^{4k+1}$ is equal to 1. Finally we show that $\pi ^n(M)$ can be mapped injectively into the set of isomorphism classes of oriented rank n vector bundles over spin $(n+1)$-manifolds for $n=4$ and $n=8$, cf. Proposition 5.8.

2 Preliminaries

If not otherwise stated we denote by M an $(n+1)$-dimensional oriented, closed and connected manifold, where $n\ge 3$. Let N be a arbitrary manifold and $E \rightarrow N$ a trivial vector bundle over N of rank r. A trivialization of $E\rightarrow N$ are r sections $s_1,\ldots ,s_r :N \rightarrow E$, such that $(s_1(q),\ldots ,s_r(q))$ is a basis of the fiber $E_q$ for all $q \in N$. A framing $\varphi$ of $E\rightarrow N$ is a homotopy class of trivializations.

We recall now the notion of bordism classes of normally framed submanifold in M of dimension k (cf. [15, §7]). Let C be a k-dimensional closed submanifolds of M. We say that C is normally framed if the normal bundle of C is trivial and possesses a framing $\varphi$. Two such normally framed submanifolds $(C_0,\varphi _0)$ and $(C_1,\varphi _1)$ are framed bordant if there is a $(k+1)$-dimensional submanifold $\Sigma \subset M \times [0,1]$ such that

• (a) $\partial \Sigma \cap (M \times i) = C_i$ for $i=0,1$,

• (b) $\partial \Sigma = C_0 \cup C_1$,

• (c) $\Sigma$ is normally framed in $M \times [0,1]$ such that the framing restricted to the $\partial \Sigma \cap (M \times i)$ coincides with $\varphi _i$.

To be framed bordant is an equivalence relation and the set of equivalence classes is called the bordism classes of normally framed k-dimensional submanifolds in M denoted by $\Omega _k^{\text {fr}}(M)$. If $(C,\varphi )$ is a normally framed submanifold then we denote by $[C,\varphi ]$ its bordism class in $\Omega _k^{\text {fr}}(M)$.

The Pontryagin-Thom map provides a bijection between $\pi ^{n+1-k}(M)$ and $\Omega _k^{\text {fr}}(M)$ as follows (cf. [15, §7]): Fix an orientation on $S^n$ and let $f :M \rightarrow S^{n+1-k}$ represent an element of $\pi ^{n+1-k}(M)$. Choose a regular value $x_0 \in S^{n+1-k}$ and set $C_{x_0}:= f^{-1}(x_0)$. Moreover choosing an oriented basis of the tangent space $T_{x_0}S^{n+1-k}$ endows the normal bundle with a framing $\varphi _{x_0}$ by means of the derivative of f. The bordism class $[C_{x_0},\varphi _{x_0}] \in \Omega _{k}^{\text {fr}}(M)$ is well defined and the map

$$\begin{aligned} \pi ^{n+1-k}(M)\longrightarrow \Omega _k^{\text {fr}}(M), \quad [f]\mapsto [C_{x_0},\varphi _{x_0}]. \end{aligned}$$

is a bijection, see [15, Theorem B and A].

A stable framing of a real vector bundle $E \rightarrow C$ of rank r is an equivalence class of trivializations of

$$\begin{aligned} E \oplus \varepsilon ^l \end{aligned}$$

for some $l \in {\mathbb {N}}$ where two trivializations

$$\begin{aligned} \tau _1 :E \oplus \varepsilon ^{l_1} \rightarrow \varepsilon ^{r+l_1} \quad \text {and}\quad \tau _2 :E \oplus \varepsilon ^{l_2} \rightarrow \varepsilon ^{r+l_2}, \end{aligned}$$

are considered to be equivalent if there exists some $L>l_1,l_2$ such that the isomorphisms

$$\begin{aligned} \tau _1 \oplus \text {id}:E \oplus \varepsilon ^{l_1} \oplus \varepsilon ^{L-l_1} \rightarrow \varepsilon ^{L+r} \end{aligned}$$

and

$$\begin{aligned} \tau _2 \oplus \text {id}:E \oplus \varepsilon ^{l_2} \oplus \varepsilon ^{L-l_2} \rightarrow \varepsilon ^{L+r} \end{aligned}$$

are homotopic, cf. [5, Section 8.3]. If E is the tangent bundle of C, then a stable framing of TC is called a stable tangential framing. If E is the normal bundle of an embedding of C into a sphere whose dimension is big enough, then we call a stable framing a stable normal framing.

We define $\Omega _k^{\text {fr}}$ to be the bordism classes of stably (tangential) framed k-dimensional manifolds. More precisely two stably framed manifolds $(C_0,\varphi _0)$ and $(C_1,\varphi _1)$ where $\varphi _i :TC_i \oplus \varepsilon ^l \rightarrow \varepsilon ^{k+l}$ is an isomorphism are equivalent if there is a bordism $\Sigma$ between $C_0$ and $C_1$ such that the tangent bundle of $\Sigma$ possesses a stable framing and the restriction on $C_0$ and $C_1$ coincides with the framing $\varphi _0$ and $\varphi _1$ respectively. Note that $\Omega _k^{\text {fr}}$ is isomorphic to $\pi _k^S$, the k-stable homotopy group of spheres (cf. [5, Theorem 8.17]) and by the Pontryagin-Thom construction we have $\Omega _k^{\text {fr}} = \varinjlim _l \Omega _k^{\text {fr}}(S^l)$ where we use the equatorial embeddings $S^{l_1} \hookrightarrow S^{l_2}$ if $l_1 < l_2$ to construct well-defined maps $\Omega _k^{\text {fr}}(S^{l_1}) \rightarrow \Omega _k^{\text {fr}}(S^{l_2})$.

For this article the case $k=1$ will be of importance. In this case we have $\Omega _1^{\text {fr}}\cong \pi _1^S \cong {\mathbb {Z}}_2$. Consider a connected and closed 1-dimensional manifold $S_0$ and stable tangential framing

$$\begin{aligned} \varphi _0 :TS_0 \oplus \varepsilon ^n {\mathop {\longrightarrow }\limits ^{\sim }} \varepsilon ^{n+1}. \end{aligned}$$

From the discussion above, $(S_0,\varphi _0)$ defines a class in $\Omega _1^{\text {fr}}$ and can be realized as follows: Let $S_0 =\{(x_1,\ldots ,x_{n+1}) \in {\mathbb {R}}^{n+1} : x_1^2+x_2^2=1,\, x_i=0,\, i=3,\ldots ,n+1\}$. Denote by $e_1,\ldots ,e_{n+1}$ the canonical basis of ${\mathbb {R}}^{n+1}$ and $E_i(x)=e_i$ for $x\in {\mathbb {R}}^{n+1}$. Moreover let $V(x)=x$ for $x \in {\mathbb {R}}^n$. The normal bundle $\nu (S_0)$ of $S_0$ is trivialized by $V,E_3,\ldots ,E_{n+1}$ restricted to $S_0$. Using this normal framing we obtain a stable framing

$$\begin{aligned} TS_0 \oplus \varepsilon ^{n} \cong TS_0 \oplus \nu (S_0) \cong (T{\mathbb {R}}^{n+1})|_{S_0} \cong \varepsilon ^{n+1} \end{aligned}$$

where the latter framing is induced by $E_1,\ldots ,E_{n+1}$. Hence this defines an element in $\Omega _1^{\text {fr}}(S^{n+1})$ which represents the framed null bordism, since the framing of $\nu (S_0)$ can be extended to a properly embedded stably framed disc in $S^{n+1} \times [0,1]$. Clearly the non-trivial element of $\Omega _1^{\text {fr}}(S^{n+1})$ can be represented by twisting the normal framing $V, E_3,\ldots ,E_{n+1}$ with a map $S_0 \rightarrow \text {SO}(n)$ such that its homotopy class in $\pi _1\left( \text {SO}(n) \right) \cong {\mathbb {Z}}_2$ is not zero. Every stable tangential framing of a closed and connected 1-dimensional manifold can be obtained in this way.

Let $E \rightarrow N$ be an oriented vector bundle over a manifold N. After choosing a Euclidean bundle metric on E one obtains a vector bundle with structure group $\mathbf{SO }(n)$. Since the space of such Euclidean bundle metrics is contractible, every construction which depends up to homotopy from a metric is independent thereof.

We say that E is spinnable if the second Stiefel–Whitney class $w_2(E)$ is zero. This means that E can carry a spin structure, that is a lift of the classifying map $N \rightarrow B \text {SO}(n)$ to a map $N \rightarrow B \text {Spin}(n)$ in the fibration $K({\mathbb {Z}}_2,1) \rightarrow B \text {Spin}(n) \rightarrow B \text {SO}(n)$. Consequently E is a spin bundle if it is spinnable and a spin structure is fixed. If a spin structure is fixed on $E \rightarrow N$ then any other spin structure is in 1 : 1 correspondence with elements in $H^1(N;{\mathbb {Z}}_2)$, cf. [12, Chapter II, Theorem 1.7]

We write F(N) for the orthonormal frame bundle of an oriented manifold N, where the frames are consistent with the given orientation on N. If $V\subset N$ is a submanifold such that its normal bundle is framed then we obtain an embedding $F(V) \subset F(N)$. Thus a spin structure on N induces a spin structure on V, cf. [14]. In particular if V is the boundary of a spin manifold N, then V inherits a spin structure from N. Finally if $E \rightarrow N$ is a vector bundle with a spin structure and $V\subset N$ a submanifold, then clearly $E|_V \rightarrow V$ also inherits a spin structure from $E \rightarrow N$.

Let $E \rightarrow S^1$ be a spinnable vector bundle of rank $r \ge 3$ over the unit circle $S^1$. Then E has exactly two non-isomorphic spin structures. Clearly $E\rightarrow S^1$ can be extended to $E \rightarrow D^2$, where $D^2$ denotes the closed unit disc in ${\mathbb {R}}^2$ . Since $D^2$ is contractible $E\rightarrow D^2$ admits a unique spin structure. Restricting this structure to the boundary of $D^2$ gives a spin structure on $E\rightarrow S^1$, which will be called the standard spin structure ($D^2$ is equipped with the standard orientation of ${\mathbb {R}}^3$ and the orientation on $S^1$ is induced by the outward pointing normal of $S^1$). The other should be called the non-standard spin structure. In other words, the standard spin structure on $E \rightarrow S^1$ can be extended to $D^2$ in contrary to the non-standard one.

3 The index of framed circles

In this section we define the key invariant of this article. For its construction the following basic lemma is the crucial observation.

Lemma 3.1

Let $E \rightarrow S^1$ be a spinnable vector bundle of rank $\ge 3$. Then E is isomorphic to the trivial bundle and a choice of a spin structure on E determines a framing on E.

Proof

E is isomorphic to the trivial bundles since it is an orientable vector bundle over a circle. Fix a spin structure on E, i.e. let $F'(E)$ be a $\mathbf {Spin}(n)$-principal bundle over $S^1$ which is a two-sheeted cover over the frame bundle F(E) of E. Let $\pi :F'(E)\rightarrow F(E)$ be the projection which is equivariant with respect to the two-sheeted covering $\mathbf {Spin}(n)\rightarrow \mathbf{SO }(n)$. Clearly $F'(E)$ is the trivial $\mathbf {Spin}(n)$-principal bundle over $S^1$ and denote by $\sigma :S^1 \rightarrow F'(E)$ a global section. Then $\pi \circ \sigma$ is a global section of F(E) hence a trivialization of $E \rightarrow S^1$. Any other such global section ${{\widetilde{\sigma }}} :S^1 \rightarrow F(E)$ differs from $\sigma$ by a map $\varphi :S^1 \rightarrow \mathbf {Spin}(n)$. Since $\pi _1(\mathbf {Spin}(n))=1$ the map $\varphi$ has to be null-homotopic which means that the two trivializations $\pi \circ \sigma$ and $\pi \circ {{\widetilde{\sigma }}}$ have to be homotopic, thus they define the same framing on E. $\square$

In the same way one proves

Corollary 3.2

Let $\Sigma$ be a 1-dimensional CW-complex (not necessarily connected) and $E\rightarrow \Sigma$ a vector bundle of rank $\ge 3$ endowed with a spin structure. Then E is isomorphic to the trivial bundle and the spin structure induces a framing on E.

Definition 3.3

Let $E \rightarrow S^1$ be a spinnable vector bundle. The framing induced by the standard spin structure on E is called the standard framing and the one induced by the non-standard spin structure the non-standard framing.

Example 3.4

The spheres $S^{n+1}$ admit a unique spin structure which can be constructed as mentioned in the preliminaries, i.e. $S^{n+1}$ is the boundary of the closed unit ball $D^{n+2}$ in ${\mathbb {R}}^{n+2}$ which admits a unique spin structure.

Let $S_0 \subset S^{n+1}$ be the intersection of a 2-dimensional linear subspace $W\subset {\mathbb {R}}^{n+2}$ with $S^{n+1}$ and denote by $D_0^2 = W \cap D^{n+2}$. Thus after Lemma 3.1$TS^{n+1}|_{S_0}$ inherits a framing from the spin structure. Denoting by $\varphi _1,\ldots ,\varphi _{n+1}$ a trivialization of this framing, the framing

$$\begin{aligned} {\overline{\varphi }}:S_0 \rightarrow \text {SO}(n+2),\quad x \mapsto (x,\varphi _1(x),\ldots ,\varphi _{n+1}(x)) \end{aligned}$$

must be null homotopic in $\text {SO}(n+2)$ by the definition of the spin structures of $S^{n+1}$ and $TS^{n+1}|_{S_0}$ (such that it lifts to $\text {Spin}(n+2)$). Thus ${\overline{\varphi }}$ must be homotopic the constant framing $x \mapsto (e_1,\ldots ,e_{n+2})$, where $e_1,\ldots ,e_{n+2}$ denotes the canonical basis of ${\mathbb {R}}^{n+2}$. In particular this means that $TS^{n+1}|_{S_0}$ inherits the standard framing from the spin structure of $S^{n+1}$.

$\Omega _1^{\text {fr}}(M)$ possesses a group structure which can be expressed as follows: Having two 1-dimensional closed submanifolds C and $C'$ of M which are normally framed, they are framed bordant in M to framed submanifolds ${\tilde{C}}$ and ${\tilde{C}}'$ with empty intersection. Taking the equivalence class of the disjoint union ${\tilde{C}} \cup {\tilde{C}}'$ with the respective framings yields an abelian group structure on $\Omega _1^{\text {fr}}(M)$, cf. [15, Problem 17 and p. 50].

Next, we construct a homomorphism $\kappa :\Omega _1^{\text {fr}}(M) \rightarrow \Omega _1^{\text {fr}}$ for M a spin manifold. Therefore let $(C,\varphi _C)$ be a closed submanifold of dimension 1, such that its normal bundle $\nu (C)$ is framed by $\varphi _C$ (thus representing an element in $\Omega _1^{\text {fr}}(M)$). From Lemma 3.1 the bundle $TM|_C$ inherits a framing $\varphi _\sigma$ from the spin structure of M. Using also the framing of $\varphi _C$ we obtain a stable tangential framing

$$\begin{aligned} \varepsilon ^{n+1} \cong TM|_C \cong TC \oplus \nu (C) \cong TC \oplus \varepsilon ^n \end{aligned}$$

which we denote by $\varphi _{\text {st}}$.

Proposition 3.5

The bordism class $[C,\varphi _{\text {st}}] \in \Omega _1^{\text {fr}}$ depends only on the bordism class $[C,\varphi _C] \in \Omega _1^{\text {fr}}(M)$.

Proof

Let $(C',\varphi _{C'})$ be another normally framed closed 1-dimensional submanifold framed bordant to $(C,\varphi _C)$. Thus there is a bordism $\Sigma \subset M \times I$ between C and $C'$ such that the normal bundle of $\Sigma$ in $M\times I$ possess a framing $\varphi _\Sigma$. By definition restricting $\varphi _\Sigma$ to C and $C'$ yields $\varphi _C$ and $\varphi _{C'}$ respectively. Since $\Sigma$ is homotopy equivalent to a 1-dimensional CW-complex and since $M\times I$ inherits a unique spin structure from M we obtain a framing $\varphi _{\Sigma ,\sigma }$ on $T(M\times I)|_{\Sigma }$. Of course the framings $\varphi _{\Sigma ,\sigma }$ restricted to C and $C'$ are just the framings $\varphi _\sigma$ and $\varphi _\sigma '$ respectively (i.e. induced by the spin structure of $TM|_C$ and $TM|_{C'}$). Since

$$\begin{aligned} T(M\times I)|_\Sigma \cong T\Sigma \oplus \nu (\Sigma ) \end{aligned}$$

the framings $\varphi _{\Sigma ,\sigma }$ and $\varphi _\Sigma$ determine a stable framing $\varphi _{\Sigma ,\text {st}}$ of $T\Sigma$. Then $(\Sigma ,\varphi _{\Sigma ,\text {st}})$ is a stably framed bordism between $(C,\varphi _{\text {st}})$ and $(C',\varphi _{\text {st}}')$. $\square$

Remark 3.6

As described above, the group structure of $\Omega _{1}^{\text {fr}}(M)$ is given by disjoint union of submanifolds and their respective normal framings. Let $(C,\varphi )$ be a framed 1-dimensional closed submanifold of M and denote by $C=S_1 \cup \ldots \cup S_k$ the connected components of C. We may assume that the union is always disjoint. Thus $S_i$ is an embedded circle and $\varphi _i := \varphi |_{S_{i}}$ a normal framing of $S_i$. Consequently we have

$$\begin{aligned}{}[C,\varphi ] = \sum _{i=1}^{k}\; [S_i,\varphi _i] \end{aligned}$$

in $\Omega _1^{\text {fr}}(M)$.

Definition 3.7

Let $S \subset M$ be an embedded circle and $\varphi$ a framing of $\nu (S)$. We call the bordism class $[S,\varphi ] \in \Omega _1^{\text {fr}}(M)$ a framed circle of M. The corresponding stable class $[S,\varphi _{\text {st}}]\in \Omega _1^{\text {fr}}$ will be called the index of $[S,\varphi ]$ (with respect to the spin structure of M) and will be denoted by ${\text {ind}}(S,\varphi )$.

Definition 3.8

Let M be an $(n+1)$-dimensional closed spin manifold. Then we define a map

$$\begin{aligned} \kappa :\Omega _1^{\text {fr}}(M) \rightarrow \Omega _1^{\text {fr}},\quad [C,\varphi ] \mapsto \kappa \left( [C,\varphi ] \right) := \sum _{\begin{array}{c} S \subset C,\\ S\text { connected} \end{array}}^{} {\text {ind}}(S,\varphi |_S) = [C,\varphi _{\text {st}}]. \end{aligned}$$

We call $\kappa$ the degree map of M with respect to the chosen spin structure.

Remark 3.9

It follows from the construction that $\kappa$ is a homomorphism.

Examples 3.10

• (a) Consider $S^{n+1}$ with the induced spin structure from the unit disc of ${\mathbb {R}}^{n+2}$, cf. Example 3.4 respectively [14]. Let $S_0$ be the intersection of $S^{n+1}$ with a 2-dimensional linear subspace W of ${\mathbb {R}}^{n+2}$. We argued in Example 3.4 that $TS^{n+1}|_{S_0}$ inherits the standard framing. Choose the standard framing $\varphi _0$ on $\nu (S_0)$, then

  $$\begin{aligned} \kappa ([S_0,\varphi _0])=0. \end{aligned}$$

  Consequently the non-standard framing $\varphi _1$ of $\nu (S_0)$ yields

  $$\begin{aligned} \kappa ([S_0,\varphi _1])\ne 0. \end{aligned}$$

• (b) Let N be a closed, simply connected, spin manifold of dimension n. Then $M:=S^1 \times N$ admits two different spin structures since $H^1(S^1 \times N;{\mathbb {Z}}_2) \cong H^1(S^1;{\mathbb {Z}}_2) \cong {\mathbb {Z}}_2$. M is the boundary of $D^2 \times N$ which has up to isomorphism a unique spin structure. The two different spin structures on M can be described as follows: One can be extended from M to $D^2 \times N$ and the other not. We call the former one the standard spin structure and the latter one the non-standard spin structure of $S^1 \times N$.

  For $q_0 \in N$ consider the circle $S_0:= S^1 \times q_0 \subset S^1 \times N$. Clearly we have a canonical isomorphism

  $$\begin{aligned} \nu (S_0) \cong S_0 \times T_{q_0}N. \end{aligned}$$

  Thus choosing a basis in $T_{q_0}N$ gives a framing $\varphi _0$ on $\nu (S_0)$ which extends to a framing of $(D^2 \times q_0) \times T_{q_0}N$. This implies

  $$\begin{aligned} \kappa _0([S_0,\varphi _0]) =0 \end{aligned}$$

  for the standard spin structure and

  $$\begin{aligned} \kappa _1([S_0,\varphi _0]) \ne 0 \end{aligned}$$

  for the non-standard spin structure.

  For $q_1 \in N$ with $q_0 \ne q_1$ we consider $C = S^1 \times q_0 \cup S^1 \times q_1$ with fixed normal framing on $S^1 \times q_i$ which gives a framing $\varphi$ on C. Then $\kappa ([C,\varphi ])$ is independent of the chosen spin structure of M. This shows that in general $\kappa$ will depend on the spin structure. The next proposition will show how it depends on it.

Suppose $C\subset M$ is a closed 1-dimensional submanifold. Then C defines a ${\mathbb {Z}}_2$ fundamental homology class $[C] \in H_1(M;{\mathbb {Z}}_2)$. We denote by $w(C) \in H^n(M;{\mathbb {Z}}_2)$ the cohomology class which is the Poincaré dual of [C].

Proposition 3.11

Fix a spin structure $\sigma$ on M and denote by $\kappa$ the degree map of M with respect to the chosen spin structure. Choose another spin structure of M, which is represented by $\alpha \in H^1(M;{\mathbb {Z}}_2)$ and denote by $\kappa ^\alpha$ the corresponding degree map. Then we have

$$\begin{aligned} \kappa ([C,\varphi ]) = \kappa ^\alpha ([C,\varphi ]) + \delta (\alpha \smile w(C)), \end{aligned}$$

where $\delta :H^{n+1}(M;{\mathbb {Z}}_2) \rightarrow \Omega _1^{\text {fr}}$ is the unique isomorphism. Thus if $w(C) \smile \alpha =0$ then $\kappa \left( [C,\varphi ] \right) =\kappa ^{\alpha }\left( [C,\varphi ] \right)$.

Proof

Assume first that $(S,\varphi )$ is a framed circle and $i :S \rightarrow M$ is the inclusion. The spin structure $\sigma$ induces a spin structure on $TM|_S = i^*(TM)$ and the spin structure induced by $\alpha$ is represented by $i^*(\alpha ) \in H^1(S;{\mathbb {Z}}_2)$. Of course $TM|_S$ can have at most two different spin structures. From the definition of the index we have

$$\begin{aligned} {\text {ind}}(S,\varphi ) = {\text {ind}}_\alpha (S,\varphi ) + \delta (i^*(\alpha )) \end{aligned}$$

where ${\text {ind}}$ is defined by $\sigma$, ${\text {ind}}_\alpha$ by $\alpha$ and $\delta :H^1(S;{\mathbb {Z}}_2) \rightarrow \Omega _1^{\text {fr}}$ the unique isomorphism.

Let $[S] \in H_1(S;{\mathbb {Z}}_2)$ be the ${\mathbb {Z}}_2$ fundamental class of S, then $i^*(\alpha )\frown [S] \in H_0(S;{\mathbb {Z}}_2)$, which is mapped under $i_*$ to $\alpha \frown i_*([S]) \in H_0(M;{\mathbb {Z}}_2)$. Let $[M] \in H_n(M;{\mathbb {Z}}_2)$ denote the ${\mathbb {Z}}_2$ fundamental class of M. Then

$$\begin{aligned} \alpha \frown i_*([S]) = \alpha \frown \left( w(S) \frown [M] \right) =\left( \alpha \smile w(S) \right) \frown [M], \end{aligned}$$

where we used that $i_*([S])$ is Poincaré dual to w(S). Since $\cdot \frown [S]$ and $\cdot \frown [M]$ are isomorphisms by Poincaré duality as well as $i_*:H_0(S;{\mathbb {Z}}_2) \rightarrow H_0(M;{\mathbb {Z}}_2)$ because S and M are connected we infer

$$\begin{aligned} {\text {ind}}(S,\varphi ) = {\text {ind}}_\alpha (S,\varphi ) + \delta (\alpha \smile w(S)) \end{aligned}$$

where now $\delta :H^{n+1}(M;{\mathbb {Z}}_2) \rightarrow \Omega _1^{\text {fr}}$ is again the unique isomorphism.

Consider now $(C,\varphi )$ with the disjoint union $C= S_1 \cup \ldots S_k$ and $\varphi _j := \varphi |_{S_j}$, such that $S_j$ is connected. With the previous computations we have

$$\begin{aligned} \kappa ([C,\varphi ]) =\sum _{j=1}^{k} \left( {\text {ind}}_\alpha (S_j,\varphi _j) +\delta (\alpha \smile w(S_j) \right) ) = \kappa ^\alpha ([C,\varphi ]) + \delta (\alpha \smile w(C)). \end{aligned}$$

and the proposition follows. $\square$

We continue with the description of the ,,dual” short exact sequence to (ST). There is a natural group homomorphism $\Omega _1^{\text {fr}}(M) \rightarrow \Omega _1^{\text {SO}}(M)$, which assigns to every framed 1-submanifold $[C,\varphi ]$ the oriented bordism class induced by the orientation of framing $\varphi$. This is well-defined since every normally framed bordism in M is also an oriented bordism (M is oriented). By the seminal work of Thom [21] we have an isomorphism

$$\begin{aligned} \Omega _1^{\text {SO}}(M) \rightarrow H_1(M;{\mathbb {Z}}) \end{aligned}$$

which assigns every oriented submanifold its fundamental class in $H_1(M;{\mathbb {Z}})$. Thus we obtain a group homomorphism

$$\begin{aligned} \Phi :\Omega _1^{\text {fr}}(M) \rightarrow H_1(M;{\mathbb {Z}}) \end{aligned}$$

(1)

which is clearly surjective. The kernel of $\Phi$ is at most isomorphic to ${\mathbb {Z}}_2$ and elements of the kernel are represented by framed circles $(S,\varphi )$ such that S is oriented null-bordant, i.e. there is an embedded oriented disc $D\subset M \times I$ with the properties $\partial D =S$ and the orientations of $\partial D$ and S agree. We may equip the normal bundle of S with two framings. If both framings can be extended over D then the kernel is trivial and otherwise ${\mathbb {Z}}_2$.

Lemma 3.12

The restricted degree map $\kappa |_{\ker \Phi } :\ker \Phi \rightarrow \Omega _1^{\text {fr}}$ is an isomorphism.

Proof

Since $\kappa$ is a homomorphism it will map the neutral element of $\ker \Phi$ to that of $\Omega _1^{\text {fr}}$. Thus it suffices to show the following: Let $(S,\varphi )$ be a framed circle such that S is oriented null-bordant in M but $\varphi$ cannot be extended over the nullbordism. We have to show $\kappa ([S,\varphi ])\ne 0$, where 0 denotes the neutral element of $\Omega _1^{\text {fr}}$. We may assume that S lies in a chart of M.¹ Thus we may embed S into ${\mathbb {R}}^{n+1}$ endowed with a normal framing, which cannot be extended over a nullbordism in ${\mathbb {R}}^{n+1}$. Hence the index of $(S,\varphi )$ defines a non-trivial element in $\Omega _1^{\text {fr}}$ (note that since $w(S)=0$ the element $\kappa [(S,\varphi )]$ does not depend on the spin structure of M, cf. Lemma 3.11). $\square$

Thus we may identify $\ker \Phi$ with $\Omega _1^{\text {fr}}$ via $(\kappa |_{\ker \Phi })^{-1}$ and we obtain a short exact sequence

$$\begin{aligned} 0 \longrightarrow \Omega _1^{\text {fr}} \longrightarrow \Omega _1^{\text {fr}}(M) \longrightarrow H_1(M;{\mathbb {Z}}) \longrightarrow 0 \end{aligned}$$

and from Lemma 3.12$\kappa$ is a splitting map. Therefore we showed

Theorem 3.13

Let M be an $(n+1)$-dimensional closed spin manifold. Choose a spin structure on M. Then

$$\begin{aligned} \Omega _1^{\text {fr}}(M) \longrightarrow H_1(M;{\mathbb {Z}}) \oplus \Omega _1^{\text {fr}}, \quad [C,\varphi ] \mapsto ([C],\kappa ([C,\varphi ])) \end{aligned}$$

is an isomorphism of abelian groups.

We finish this section by giving an alternative way to compute the index of a framed circle in the spirit of Pontryagin [16]. Suppose $[S,\varphi ]$ is a framed circle, thus there are trivializations of $\nu (S)$ and $TM|_{S}$ such that we obtain the stable framing

$$\begin{aligned} \varepsilon ^{n+1} \cong TS \oplus \varepsilon ^n \end{aligned}$$

(where we can assume that the isomorphism is orientation preserving). Denote by $v_1,\ldots ,v_{n+1}$ and by $w_2,\ldots ,w_{n+1}$ the trivializations of $TM|_S$ and $\nu (S)$ respectively. Let $w_1$ be a trivialization of TS. Let $\Phi :TS \oplus \varepsilon ^n \rightarrow \varepsilon ^{n+1}$ be the isomorphism of the stable framing, then there is a matrix $A=(A_{ij}) :S \rightarrow \mathrm {GL}^{+}(n+1)$ (where $\text {GL}^+(n+1)$ is the set of all invertible real matrices of size $(n+1) \times (n+1)$ with positive determinant) such that

$$\begin{aligned} \Phi (w_i) = \sum _{j=1}^{n+1} A_{ij} \cdot v_j. \end{aligned}$$

Since $\mathrm {SO}(n+1)$ is a strong deformation retract of $\mathrm {GL}^+(n+1)$ we have $\pi _1 \left( \mathrm {GL}^+(n+1)\right) \cong {\mathbb {Z}}_2$. The map $A :S \rightarrow \mathrm {GL}^+(n+1)$ defines an element $[A] \in \pi _1\left( \mathrm {GL}^+(n+1) \right)$. Changing the homotopy classes of trivializations of $TM|_S$ and $\nu (S)$ does not change [A]. Furthermore [A] is also independent of the choice of trivializations of TS.

According to the Preliminaries (Section 2) any stable framing ${\text {ind}}(S,\varphi )$ can be represented by a framed circle $S_0$ in ${\mathbb {R}}^{n+1}$ such that

$$\begin{aligned} TS_0 \oplus \varepsilon ^n \cong TS_0 \oplus \nu (S_0) \cong (T{\mathbb {R}}^{n+1})|_{S_0} \cong \varepsilon ^{n+1} \end{aligned}$$

recovers the stable framing of $(S,\varphi )$. It follows that

$$\begin{aligned} {\text {ind}}(S,\varphi ) = \delta (S_0,\varphi _0), \end{aligned}$$

where $\delta$ is the invariant constructed by Pontryagin, [16, Theorem 20]. We will use a different notation: Let us denote by [A] the homotopy class constructed above from the stable framing and by $\overline{[A]}$ the element $[A]+1 \in \Omega _1^{\text {fr}}(S^n)\cong {\mathbb {Z}}_2$ where 1 is the non-trivial element. Thus we proved

Lemma 3.14

We identify $\pi _1\left( \mathrm {GL}^+(n+1) \right)$ with $\Omega _1^{\text {fr}}$ by the unique isomorphism ${\mathbb {Z}}_2 \rightarrow {\mathbb {Z}}_2$. Then

$$\begin{aligned} \overline{[A]} = {\text {ind}}(S,\varphi ). \end{aligned}$$

4 Computation of $\pi ^n(M)$

We start this section to explain the group structure of $\pi ^n(M)$. Let $j :S^{n} \vee S^n \rightarrow S^n \times S^n$ be the inclusion of the ($2n-1$)-skeleton of $S^n\times S^n$ (endowed with the standard CW structure) then, since M is $n+1$-dimensional CW complex, the induced map $j_\# :[M,S^n\vee S^n] \rightarrow [M,S^n\times S^n]$ is an isomorphism. For $f,g \in \pi ^n(M)$ the group structure is defined by

$$\begin{aligned} f + g := (\text {id}_{S^n} \vee \text {id}_{S^n})_\# \circ (j_\#)^{-1}(f\times g). \end{aligned}$$

This makes $\pi ^n(M)$ to an abelian group.

Now, let $f :M \rightarrow S^n$ be a differentiable map and $x_0 \in S^n$ a regular value. We orient $S^n$ by the normal vector field pointing outwards and the standard orientation of ${\mathbb {R}}^{n+1}$.

Let $\Psi :\pi ^n(M) \rightarrow H^n(M;{\mathbb {Z}})$ be the map $\Psi ([f]):=f^{*}\sigma$ where $\sigma \in H^n(S^n;{\mathbb {Z}})$ is a fixed generator. We define the analogous degree map $\kappa :\pi ^n(M) \rightarrow \pi _1^S$, where $\pi _1^S$ is the first stable homotopy group of spheres, as follows: $\kappa$ is the composition of

$$\begin{aligned} \pi ^n(M) \overset{\sim }{\longrightarrow } \Omega _1^{\text {fr}}(M) \overset{\kappa }{\longrightarrow } \Omega _1^{\text {fr}} \overset{\sim }{\longrightarrow } \pi _1^S. \end{aligned}$$

where the first and the last isomorphism is again induced by the Pontryagin-Thom isomorphism.

Theorem 4.1

Let M be a closed $(n+1)$-dimensional spin manifold. Then

• (a) The generator of $\ker \Psi \cong {\mathbb {Z}}_2$ is given by the homotopy class of the map $\eta \circ \omega :M\rightarrow S^{n}$, where $\eta$ represents a generator of $\pi _{n+1}(S^n)$ and $\omega :M \rightarrow S^{n+1}$ is a map of odd degree. Thus $\ker \Psi \cong \pi _{n+1}(S^n)$.

• (b) Identifying $\pi _1^S$ with $\pi _{n+1}(S^n)$ the degree map $\kappa :\pi ^n(M) \rightarrow \pi _1^S$ splits the short exact sequence (ST). Thus we have

  $$\begin{aligned} \pi ^n(M) \longrightarrow H^n(M;{\mathbb {Z}}) \oplus \pi _{n+1}(S^n),\quad [f] \mapsto (f^*\sigma ,\kappa ([f])). \end{aligned}$$

  is an isomorphism of abelian groups.

Proof

Clearly we have $[\eta \circ \omega ] \in \ker \Psi$. For (a) it is enough to check that $\kappa ([\eta \circ \omega ])$ is non-zero in $\pi _1^S$. We choose an odd degree map $\omega :M \rightarrow S^{n+1}$ as follows: Let $\{p_1,\ldots ,p_l\}$ be the preimage of a regular value $y_0$ and choose open sets $U_1,\ldots ,U_l \subset M$ as well as $V\subset S^{n+1}$ such that for all $i=1,\ldots ,l$

• (a) $U_i$ and V are contractible,

• (b) $p_i \in U_i$ and $y_0 \in V$,

• (c) there are charts $\psi _i:U_i \rightarrow {\mathbb {R}}^{n+1}$, $\psi :V \rightarrow {\mathbb {R}}^{n+1}$,

• (d) $\omega _i :=\omega |_{U_i}$ is an orientation preserving diffeomorphism onto V.

Since $\omega$ has odd degree, l has to be an odd number (such maps exists e.g. using the Pontryagin-Thom construction). Furthermore let $x_0 \in S^n$ be a regular value of $\eta$ and $S_0 =\eta ^{-1}(x_0)$. We may assume that $S_0$ is connected (e.g. see [15, Theorem C]) and $S_0 \subset V$. Let $\varphi _0$ be the framing of $\nu (S_0)$ induced by $\eta$, then $0 \ne [S_0,\varphi _0]\in \Omega _1^{\text {fr}}(S^{n+1})\cong \pi _{n+1}(S^n) \cong {\mathbb {Z}}_2$ and therefore by definition we have ${\text {ind}}(S_0,\varphi _0) \ne 0$.

Denote by $S_i := \omega _i^{-1}(S_0)$ and frame $\nu (S_i)$ by $\varphi _0$ and $d\omega _i$. Then $C= S_1\cup \ldots \cup S_l$ together with the framings $\varphi _i$ is a Pontryagin manifold for $\eta \circ \omega$ to the regular value $x_0$. Note that $w(S_i)=0$ for $i=1,\ldots ,l$, since they are contained in a chart of M. By Proposition 3.11 this means that their indices do not depend on the spin structure of M. Clearly we deduce ${\text {ind}}(S_i,\varphi _i) = {\text {ind}}(S_0,\varphi _0)\ne 0$ for all $i=1,\ldots ,l$ and from that we infer

$$\begin{aligned} \kappa ([\eta \circ \omega ]) = \sum _{i=1}^{l} {\text {ind}}(S_i,\varphi _i) =l \cdot {\text {ind}}(S_0,\varphi _0) \ne 0 \end{aligned}$$

since l is odd, which proves (a).

Part (b) follows directly from part (a). $\square$

Corollary 4.2

Suppose M is simply connected, then, up to homotopy, there are exactly two maps $M \rightarrow S^n$ and one of them is the constant map. The homotopy class of the non-trivial map is represented by $\eta \circ \omega :M \rightarrow S^n$, see Theorem 4.1.

Finally we would like to show that $\kappa$ is natural with respect to maps between manifolds which preserve the spin structure

Proposition 4.3

Suppose $\Phi :M_1 \rightarrow M_2$ is a map between two closed and connected spin manifolds of dimension $(n+1)$. We assume that the spin structure of $M_1$ coincides with the pull-back spin structure by $\Phi$ of $M_2$. Then for the natural homomorphism $\Phi ^\# :\pi ^n(M_2) \rightarrow \pi ^n(M_1)$, $f\mapsto \Phi \circ f$ we have

$$\begin{aligned} \kappa \left( \Phi ^\#(f) \right) = \deg _2 \Phi \cdot \kappa (f). \end{aligned}$$

where $\deg _2 \Phi$ is the mod 2 degree of $\Phi$. Therefore using the isomorphism

$$\begin{aligned} \pi ^n(M) \cong H^n(M;{\mathbb {Z}}) \oplus \pi _{n+1}(S^n) \end{aligned}$$

we have

$$\begin{aligned} \Phi ^\# :\pi ^n(M_2) \rightarrow \pi ^n(M_1),\quad (\alpha ,\nu )\mapsto (\Phi ^*(\alpha ),\deg _2 \Phi \cdot \nu ) \end{aligned}$$

Proof

First note that $\Phi ^\#$ is well-defined on the homotopy class of $\Phi$. For $f \in \pi ^n(M_2)$ there is a decomposition $f=f_\alpha + f_\nu$ with $\kappa (f_\alpha )=0$, $f_\alpha ^*(\sigma )=\alpha$ and $\kappa (f_\nu )=\nu$ as well as $f_\nu ^*(\sigma )=0$.

Let us show first $\Phi ^\#(f_\alpha )=f_{\Phi ^*(\alpha )}$. Clearly we have $\Phi ^\#(f_\alpha )(\sigma )=\Phi ^*(\alpha )$ thus it remains to show $\kappa (\Phi ^\#(f_\alpha ))=0$. Let $C_2$ be the preimage of a regular value of $f_\alpha$ with a normal framing $\varphi _0$ such that $\kappa ([C_2,\varphi _0])=0$. Moreover we may choose $f_\alpha$ such that each framed circle of $(C_2,\varphi _0)$ has index 0. Deform $\Phi$ to be transversal to $C_2$, thus $C_1:=\Phi ^{-1}(C_2)$ is a closed 1-dimensional submanifold of $M_1$. The normal bundle to $C_1$ is isomorphic to the pull back of the normal bundle of $C_2$ by $\Phi$. This induces a framing on $C_2$ such that every framed circle thereof has index 0 (note that the spin structure of $M_1$ is the pulled back by $\Phi$ from $M_2$) which is also the framing induced by the map $\Phi \circ f_\alpha$. But this means $\kappa (\Phi ^\#(f_\alpha ))=0$.

On the other hand we may assume a preimage of a regular point in $S^n$ under $f_\nu$ is a contractible circle $S_2$ in $M_2$ with normal framing $\varphi$ such that the index of the framed circle $(S_2,\varphi )$ is $\nu \in \pi _{n+1}(S^n)$. Then making again $\Phi$ transverse to $S_2$ we obtain a normally framed submanifold $(C_1,\varphi )$ such that the index of each framed circle in $C_1$ has index $\nu$. As in the proof of Theorem 4.1 the degree of $(C_1,\varphi )$ is just $\deg _2\Phi \cdot \nu$. Therefore $\Phi ^\#(f_\nu ) = f_{\deg _2\Phi \cdot \nu }$ and the proposition follows. $\square$

Corollary 4.4

Let $f :M \rightarrow S^n$ and $x_0 \in S^n$ a regular value. Write $S_1 \cup \ldots \cup S_k = f^{-1}(x_0)$ such that $S_i$ is a connected component of $f^{-1}(x_0)$ and denote by $\varphi _i$ the induced framing from f. Then the number

$$\begin{aligned} \# \{ i : \kappa ([S_i,\varphi _i])\ne 0 \} \mod 2 \end{aligned}$$

does not depend on $x_0$ and is a homotopy invariant. Using the notation of Lemma 3.14we can rewrite the above number as

$$\begin{aligned} \# \{ i : \overline{[A_i]}\ne 0 \} \mod 2 \end{aligned}$$

where $[A_i]$ is the homotopy class of matrices corresponding to $[S_i,\varphi _i]$.

5 Application to vector bundles

In this section $\pi :E \rightarrow M$ should denote an oriented vector bundle of rank n endowed with a spin structure. Let $s :M \rightarrow E$ be a section. If not otherwise stated, we say s is transversal if s is transversal to the zero section $0_E$ of E. For a transversal section s the zero locus C is a smooth 1-dimensional closed submanifold of M. The differential $ds :TM \rightarrow TE$ restricted to $\nu (C)$ is an isomorphism of the vector bundles $\nu (C) \rightarrow E|_C$. Since E possesses a spin structure, by Lemma 3.1$E|_C$ has a framing and with ds this endows $\nu (C)$ with the framing $\varphi$ of $E|_C$. Note that the homology class $[C] \in H_1(M;{\mathbb {Z}})$ is the Poincaré dual of the Euler class of E.

Proposition 5.1

The class $[C,\varphi ] \in \Omega _1^{\text {fr}}(M)$ does not depend on the section s.

Proof

Let $s' :M \rightarrow E$ be another transversal section and denote the corresponding normally framed zero locus by $(C',\varphi ')$. Let $s^*:M \times I \rightarrow \text {pr}^*(E)$ be a section of $\text {pr}^*(E) \rightarrow M \times I$ (where $\text {pr} :M \times I \rightarrow M$) such that $s^*|_{M \times 0} = s$ and $s^*|_{M \times 1} = s'$. We may deform $s^*$ to a section ${\hat{s}}$ which is transverse to the zero section of $\text {pr}^*(E) \rightarrow M\times I$ and agrees with s and $s'$ on the boundary of $M \times I$. The zero locus of ${\hat{s}}$, call it $\Sigma \subset M \times I$ is a bordism between C and $C'$ by construction. Moreover by Lemma 3.1$T(M\times I)|_\Sigma$ inherits a framing from the spin structure of M as well as $\nu (\Sigma )$ from $d{\hat{s}}$ and the spin structure of $\text {pr}^*(E)|_\Sigma$. Thus $\Sigma$ is a normally framed bordism between $(C,\varphi )$ and $(C',\varphi ')$. $\square$

Definition 5.2

The bordism class $[C,\varphi ] \in \Omega _1^{\text {fr}}(M)$ constructed above is called the framed divisor of $E\rightarrow M$. Furthermore we define the degree $\kappa (E)$ of E as $\kappa ([C,\varphi ])$

For $[C,\varphi ]\in \Omega _1^{\text {fr}}(M)$ we denoted by $w(C) \in H^{n}(M;{\mathbb {Z}}_2)$ the Poincaré dual of the ${\mathbb {Z}}_2$ fundamental class $[C] \in H_1(M;{\mathbb {Z}}_2)$. If $[C,\varphi ]$ is the framed divisor of $E \rightarrow M$ then w(C) is the n-th Stiefel-Whitney class $w_{n}(E)$ (the zero locus of a generic section of E represents the Poincaré dual of the Euler class of E, cf. [4, Proposition 12.8]. Therefore, since n-th Stiefel-Whitney class of E is the Euler class modulo 2, we have that [C] is Poincaré dual to $w_n(E)$)). Moreover if one changes the spin structure of E by an element $\alpha \in H^1(M;{\mathbb {Z}}_2)$ then for the framed divisor it is the same as keeping the spin structure of E and changing that of M by $\alpha$. Hence if $w_{n}(E)=0$ then the degree of E does not depend on the spin structure (see Lemma 3.1 and Proposition 3.11).

Proposition 5.3

If $w_n(E)=0$ then the framed divisor is independent of the spin structures on M and E.

For the next theorem we will need a technical Lemma. Let $D^{m}$ denote the closed unit ball in ${\mathbb {R}}^m$ and consider a smooth map $f :D^{n+k+1} \rightarrow {\mathbb {R}}^{n+1}$. Assume that $0 \in {\mathbb {R}}^{n+1}$ is a regular value for f and $\Sigma _f^k := f^{-1}(0)$ does not intersect the boundary of $D^{n+k+1}$. Denote by $\varphi _f$ the induced framing on $\nu (\Sigma _f^k)$. Since $\Sigma _f^k$ is a submanifold of ${\mathbb {R}}^{n+k+1}$ the trivialization $\varphi _f$ defines a stable tangential framing of $\Sigma _f^k$ thus the pair $(\Sigma _f^k,\varphi _f)$ defines an element in $\Omega _k^{\text {fr}}$. On the other side, consider

$$\begin{aligned} g :S^{n+k}= \partial D^{n+k+1} \rightarrow S^n,\quad g(x):=\frac{f(x)}{|f(x)|} \end{aligned}$$

and choose a regular value $y \in S^{n}$. Denote by $(\Sigma _g^k,\varphi _g)$ the induced stably framed manifold.

Lemma 5.4

With the notation above we have that $(\Sigma _f^k,\varphi _f)$ and $(\Sigma _g^k,\varphi _g)$ are stably framed bordant, thus they define the same element in $\Omega _k^{\text {fr}}$.

Proof

There is an $\varepsilon >0$ such that the closed ball $D_\varepsilon$ centered in $0 \in {\mathbb {R}}^{n+1}$ with radius $\varepsilon$ contains only regular values of f. The preimage of $D_\varepsilon$ under f is a disc bundle $D(\Sigma _f^k)$ of the normal bundle $\nu (\Sigma _f^k \hookrightarrow {\mathbb {R}}^{n+k+1})$. Denote by $S(\Sigma _f^k)$ its sphere bundle. Then $f|_{S(\Sigma _f^k)}$ has image $S_\varepsilon = \partial D_\varepsilon$. Thus for $y' \in S_\varepsilon$, $\Sigma _{y'} = \left( f|_{S(\Sigma _f^k)} \right) ^{-1}(y')$ lies completely in $S(\Sigma _f^k)$. Moreover the Pontryagin manifold $(\Sigma _{y'},\varphi _{y'})$ is framed bordant to $(\Sigma _f^k,\varphi _f)$. Thus we would like to show that $(\Sigma _{y'},\varphi _{y'})$ represents the same element in $\Omega _{k}^{\text {fr}}$ as $(\Sigma _g^k,\varphi _g)$. Since the normal bundle of $S(\Sigma _f^k)$ is trivial the framing $\varphi _{y'}$ induces a framing $\varphi _{y'}'$ on $\nu (\Sigma _{y'}\hookrightarrow S(\Sigma _{f}^k))$ such that $(\Sigma _{y'},\varphi _{y'})$ is stably framed bordant to $(\Sigma _{y'},\varphi _{y'}')$. But the latter normally framed manifold is the Pontryagin manifold to the map $f|_{S(\Sigma _f^k)} :S(\Sigma _f^k) \rightarrow S_\varepsilon$ at the point $y ' \in S_\varepsilon$.

Let N be the complement of the interior of $D(\Sigma _f^k)$ in $D^{n+k+1}$. Then N is a framed cobordism between $S^{n+k}=\partial D^{n+k+1}$ and $S(\Sigma _f^k)$. The restriction of the map

$$\begin{aligned} F :N \rightarrow S^n,\quad F(x):=\frac{f(x)}{|f(x)|} \end{aligned}$$

to $S^{n+k}$ is equal to g and F restricted to $S(\Sigma ^k)$ is equal to $\varepsilon ^{-1}{\hat{f}}$. Hence F defines a framed bordism between $(\Sigma _g^k,\varphi _g)$ and $(\Sigma _{y'},\varphi '_{y'})$ which proves the lemma. $\square$

Theorem 5.5

Let $E \rightarrow M$ be an oriented vector bundle of rank n with $w_2(E)=0$ over a closed spin manifold M of dimension $n+1$. Then E admits a nowhere vanishing section if and only if the Euler class is zero and $\kappa (E)=0$.

Proof

Suppose there is a nowhere vanishing section of E then clearly this section is transverse and has an empty framed divisor. Thus from Theorem 3.13 we have that the Euler class must be zero and $\kappa (E)=0$.

Assume now that $e(E)=0$ and $\kappa (E)=0$. Consider the fibration

$$\begin{aligned} S^{n-1} \longrightarrow B \text {SO}(n-1) \longrightarrow B \text {SO}(n). \end{aligned}$$

where $B \text {SO}(k)$ denotes the classifying space to the special orthogonal group $\text {SO}(k)$. Consider the classifying map $g :M \rightarrow B \text {SO}(n)$ for $E \rightarrow M$. There exists a nowhere vanishing section if and only if there is a lift ${\hat{g}}:M \rightarrow B \text {SO}(n-1)$ of g up to homotopy.

First we put a CW-structure on M (e.g. induced by a Morse function) then over the $(n-1)$-skeleton of M there exists such a lift ${\hat{g}}$ of g. The obstruction to extend the lift over the n-skeleton lies in $H^n(M;\pi _{n-1}(S^{n-1}))=H^n(M;{\mathbb {Z}})$ which is given by the Euler class e(E). Since this is assumed to be zero ${\hat{g}}$ extends over the n-skeleton of M. The obstruction to extend ${\hat{g}}$ over the top cell of M lies in $H^{n+1}(M;\pi _{n}(S^{n-1}))\cong \pi _{n}(S^{n-1})\cong {\mathbb {Z}}_2$. Let $e_{n+1}$ be the top cell of M and $\psi :\partial e_{n+1}\cong S^{n} \rightarrow M$ the corresponding attaching map. The bundle $E|_{e_{n+1}}$ is canonical isomorphic to $e_{n+1} \times {\mathbb {R}}^n$. Let $\sigma :M \rightarrow E$ be a section which has no zeroes over the n-skeleton of M and which is transverse to the zero section of E. Then consider the map

$$\begin{aligned} g :\partial e_{n+1}\cong S^n \rightarrow S^{n-1},\quad g(x):=\frac{\sigma \circ \psi (x)}{|\sigma \circ \psi (x)|} \end{aligned}$$

(where the norm is taken with respect to a Euclidean bundle metric on E). The homotopy class of g in $\pi _{n}(S^{n-1})$ is the obstruction to extend a no where vanishing section over the n-skeleton to the $(n+1)$-skeleton of M. Since $\pi _n(S^{n-1})$ is isomorphic to the stable homotopy group $\pi _1^S$ we consider the homotopy class of g as an element therein.

From Lemma 5.4 we infer that $[g] \in \pi _1^S\cong \Omega _1^{\text {fr}}$ is equal to the framed divisor $\kappa (E)$ of E defined by $\sigma$, thus E admits a nowhere vanishing section in case $e(E)=0$ and $\kappa (E)=0$. $\square$

Example 5.6

As an application of our theory we will reprove the following fact due to Whitehead [23] and Eckmann [6]: The number of linear independent vector fields on $S^{4k+1}$ is equal to 1 (see also [1] and in [22]).

Denote by $\langle \cdot , \cdot \rangle$ the standard Euclidean product in ${\mathbb {R}}^{4k+2}$. The vector field

$$\begin{aligned} v :{\mathbb {R}}^{4k+2}\rightarrow {\mathbb {R}}^{4k+2},\quad v(x_1,x_2,\ldots ,x_{4k+2})=(-x_2,x_1,\ldots ,-x_{4k+2},x_{4k+1}) \end{aligned}$$

defines a nowhere vanishing vector field on $S^{4k+1}$ since $\langle v(x), x \rangle =0$ for $x \in S^{4k+1}$. Let E be the subbundle of $TS^{4k+1}$ orthogonal to the line bundle spanned by v. For any vector field on $S^{4k+1}$ which is in every point linear independent to v there is a nowhere vanishing section of E². Since the Euler class of E vanishes, it suffices to show that $\kappa (E)$ is not zero by Theorem 5.5 (note that the spin structures of $S^{4k+1}$ and that of E are unique up to homotopy).

Consider now the vector field

$$\begin{aligned} w :{\mathbb {R}}^{4k+2} \rightarrow {\mathbb {R}}^{4k+2},\quad w(x) = (0,0,-x_5,x_6,x_3,-x_4,-x_9,x_{10},x_7,-x_8,\ldots ) \end{aligned}$$

Since $\langle w(x), x \rangle = \langle w(x), v(x) \rangle =0$ we have that w is a section of E. Furthermore w is transverse to the zero section of E and the zero locus is given by

$$\begin{aligned} S=\{(x_1,x_2,0,\ldots ,0) \in S^{4k+1} : x_1^2+x_2^2 =1\}. \end{aligned}$$

In Example 3.10 we saw that $TS^{4k+1}|_S$ inherits the standard framing from the spin structure. But the induced framing on $E|_S$ cannot be the standard framing. To see this assume it inherits the standard framing and let $\tau _1,\ldots ,\tau _n$ be a trivialization of $E|_S$, then, since the spin structure on E is induced by $TS^{4k+1}$ and v, the map $S \rightarrow \mathrm {SO}(4k+2)$, $x\mapsto (x,v(x),\tau _1(x),\ldots ,\tau _n(x))$ has to be nullhomotopic cf. Example 3.10 (note that $v|_S$ is tangent to S) which is a contradiction. Thus from Example 3.10 we deduce that the index of the framed divisor is not zero, hence $\kappa (E)=1$ and therefore E does not admit a nowhere vanishing section from Theorem 5.5.

Remark 5.7

In [7, Theorem 1.6] the authors show that for any n-dimensional CW-complex of dimension X and any k-dimensional integral cohomology class $a \in H^k(X;{\mathbb {Z}})$ there exists an oriented vector bundle over X whose Euler class equals $2 \cdot N(n,k) \cdot a$.

Suppose $\dim X = 2k+1$. By Steenrod’s exact sequence (ST) it follows that the Hurewicz map $\pi ^n(X) \rightarrow H^n(X;{\mathbb {Z}})$ is surjective. Then for every $a \in H^n(X;{\mathbb {Z}})$ there is a map $f_a \in \pi ^n(X)$ such that $f_a^*(\sigma )=a$, where $\sigma \in H^n(S;{\mathbb {Z}})$ denotes the generator such that $2\sigma$ equals to the Euler class of the tangent bundle $TS^n$ of $S^n$. Clearly the vector bundle $f_a^*(TS^n)$ has Euler class $2 \cdot a$ and therefore $N(2k,2k+1)=1$ in the notation of [7].

Note that any vector bundle over $S^n$ for $n\ne 2,4,8$ has an Euler class divisible by 2, cf. [2, 13]. In the cases $n=2,4,8$ there are real vector bundles whose Euler class is a generator of $H^n(S^n;{\mathbb {Z}})$, namely the associated bundles to the Hopf fibrations $S^{2n-1} \rightarrow S^n$. We deduce

Proposition 5.8

Suppose $n=4$ or $n=8$ and let M be a $(n+1)$-dimensional closed spin manifold. Denote by $\mathrm {Vect}_n(M)$ the set oriented vector bundles over M of rank n up to isomorphism. Let $E_0 \rightarrow S^{n}$ denote the oriented rank n vector bundle such that the Euler class of $E_0$ is a generator of $H^n(S^n;{\mathbb {Z}})$. Then the map

$$\begin{aligned} \pi ^n(M) \rightarrow \mathrm {Vect}_n(M), \quad f\mapsto f^*(E_0) \end{aligned}$$

is injective.

Proof

We consider $f_1,f_2 \in \pi ^n(M)$ such that $E_1:=f_1^*(E_0)\cong f_2^*(E_0)=:E_2$ since they represent the Euler class the respective bundles. This implies $f_1^*(\sigma )=f_2^*(\sigma )$ for a generator in $H^n(S^n;{\mathbb {Z}})$. Thus it remains to show that $\kappa (f_1) = \kappa (f_2)$. Let $x_i \in S^n$ be a regular value for $f_i$ for $i=1,2$. There is a section $\sigma _{0,i} :S^n \rightarrow E_0$ which is transverse to the zero section with an isolated zero in $x_i$ (note that the Poincaré dual of $x_i$ in $S^n$ represents the Euler class of $E_0$. Therefore $\sigma _{0,i}$ can only exist since if the Euler class is a generator, since the index of transverse sections is always $\pm 1$). Then $\sigma _i:=f^*(\sigma _{0,i})$ is a transverse section of $E_i$. Note that from the Pontryagin-Thom construction we may assume that $f_i^{-1}(x_i)$ is connected, hence the zero locus of $\sigma _i$ coincides with $f_i^{-1}(x_i)$. Moreover the framed divisor of $E_i$ coincides with the degree of $f_i$ (cf. Definitions 3.8 and 5.2). Since $E_1\cong E_2$ we have $\kappa (E_1) \cong \kappa (E_2)$ by construction of the framed divisor and Proposition 5.1. From $f_1^{*}(\sigma ) = f_2^{*}(\sigma )$ and $\kappa (f_1)=\kappa (E_1)=\kappa (E_2)=\kappa (f_2)$ it follows from Theorem 4.1 that $f_1$ is homotopic to $f_2$. $\square$