Proof. With $n\geq 5$ any map $S^{1}\rightarrow M$ is homotopic to a smooth embedding, and any two homotopic embeddings $S^{1}\subset M$ are isotopic (by a result of Whitney). Furthermore, since both $S^{1}$ and M are oriented, the normal disc bundle $N(S^{1})$ of an embedding $S^{1}\subset M$ has the induced orientation, hence has a trivialisation $N(S^{1})\cong S^{1}\times D^{n-1}$ , showing that $\gamma $ surjects.

As the group of trivialisations of $N(S^{1})$ is $\pi _{1}(SO(n-1))=\mathbb{Z}_{2}$ , which is generated by $\alpha $ , the pre-image $\gamma ^{-1}[S^{1}]$ of any $S^{1}\subset M$ contains at most two elements: if one of them is $[S^{1},F]$ , the other must be $[S^{1},F\widetilde{\alpha }]$ .

Proof. The first assertion, together with the properties (i)–(iii), have been shown by Goldstein and Linger in [ Reference Goldstein and Lininger13 , theorem 1].

Let $S^{1}\subset $ M be an oriented circle on M. Since M is 1-connected with dimension $n\geq 5$ there is an embedded disc $D^{2}\subset M$ with $\partial D^{2}=S^{1}\subset M$ . Taking a tubular neighbourhood $D^{n}$ of the disc $D^{2}$ one gets a frame F of the circle $S^{1}$ such that $\text{Im}\;F\subset \overset{\circ }{D^{n}}$ . That is, the manifold M has a decomposition $M=M\#S^{n}$ so that $(S^{1},F) $ is a framed circle on the sphere $S^{n}$ .

For $M=S^{n}$ we have $[M,S^{n-1}]=\mathbb{Z}_{2}$ , while the map $\varepsilon $ is one to one, implying that $S^{n}$ possesses the two decompositions stated in (2·1).

Proof. Since the generator $\alpha \in \pi _{1}(SO(n-1))$ is of order 2, the relation (2·3) follows from the definition of the tunnel sum in (2·2).

Since M is 1-connected, we can assume by Lemma 2·2 that $M=S^{n}$ , and take a decomposition of $S^{n}$ relative to the index $\varepsilon (S^{1},F)$ as in (2·1). If $\varepsilon (S^{1},F)=\delta (S^{1},F^{\prime })=0$ (resp. if $\varepsilon (S^{1},F)=1$ but $\delta (S^{1},F^{\prime })=0$ ) we get (2·4) directly from the definition of $\Sigma _{i}:$

\begin{align*}&(S^{n},F)\circ (S^{1}\times N,F^{\prime })=D^{{\small 2}}\times S^{{\small n-1}}\cup _{id}(S^{{\small 1}}\times (N-\overset{\circ }{D^{n-1}}))=\Sigma_{0}N\\ &(\text{resp.} =D^{{\small 2}}\times S^{{\small n-1}}\cup _{\tau }(S^{{\small 1}}\times (N-\overset{\circ }{D^{n-1}})=\Sigma _{1}N).\end{align*}

For the remaining case $\varepsilon (S^{1},F)=\delta(S^{1},F^{\prime })=1$ (resp. $\varepsilon (S^{1},F)=0$ but $\delta(S^{1},F^{\prime })=1$ ), we can compose both F and $F^{\prime }$ with the diffeomorphism $\widetilde{\alpha }$ of $S^{{\small 1}}\times D^{n-1}$ by (2·3), then apply (i) of Lemma 2·2 and Lemma 2·4 to reduce the current case to the previous one.

Finally, the relation (2·5) has been shown by (iii) of Lemma 2·2.

Proof of Theorem B. Let $p\;:\;E\rightarrow B$ an oriented circle bundle over a manifold $B\in \mathcal{T}_{n-1}$ , where the total space E is 1-connected with $\dim E=n\geq 5$ . Take an orientation preserving embedding $i\;:\;D^{n-1}\subset B$ and set $i(0)=b\in B$ . Then the inclusion $F_{b}\;:\;E\mid D^{n-1}\subset E$ defines a frame $(p^{-1}(b),F_{b})$ of the oriented circle $p^{-1}(b)\subset E$ , to be called a canonical frame of $p^{-1}(b)$ . Recall from [ Reference Goldstein and Lininger13 , theorem 8] that the function $\varepsilon \;:\;B\rightarrow \{0,1\}$ by $\varepsilon(b)\;:\!=\;\varepsilon (p^{-1}(b),F_{b})$ is constant, and satisfies that

(2·6) \begin{equation}\varepsilon (p^{-1}(b),F_{b})=0\ \text{or}\ 1\ \text{in accordance to}\ w_{2}(B)\neq 0\ \text{or}\ w_{2}(B)=0.\end{equation}

On the other hand, let $p^{\prime }\;:\;S^{1}\times N\rightarrow N$ be the trivial $S^{1}$ -bundle on $N\in \mathcal{T}_{n-1}$ . Take an orientation preserving embedding $i\;:\;D^{n-1}\subset N$ and set $x_{0}\;:\!=\;i(0)\in N$ . Since the inclusion $F_{x_{0}}\;:\;p^{\prime }\mid D^{n-1}\subset S^{1}\times N$ is precisely the canonical frame $(p^{\prime -1}(x_{0}),F_{x_{0}})$ of the oriented circle $p^{\prime -1}(x_{0})=S^{1}\times x_{0}$ on $S^{1}\times N$ we get by the definition of the index $\delta $ that

(2·7) \begin{equation}\delta (p^{\prime -1}(x_{0}),F_{x_{0}})=0.\end{equation}

Consider now the projection $f_{N}\;:\;B\#N\rightarrow B$ ( $f_{N}(N-\overset{\circ }{D}^{n-1})=b$ ), and let $p_{N}\;:\;E_{N}\rightarrow $ $B\#N$ be the induced bundle $f_{N}^{\ast }E$ . Then the total space $E_{N}$ admits the decomposition into the tunnel sum

\begin{equation*}E_{N}=(E,F_{b})\circ (S^{1}\times N,F_{x_{0}}),\end{equation*}

where $b\in B$ , $x_{0}\in N$ , and where both $(p^{-1}(b),F_{b})$ and $(S^{1}\times \{x_{0}\},F_{x_{0}})$ are canonical. In view of (2·6) and (2·7), the desired formulae (1·4) and (1·5) of Theorem B have been shown by the relations (2·4) and (2·5) in Lemma 2·5, respectively.

Proof of Theorem A. Let $N\in \mathcal{T}_{n}$ be a manifold with base point $x_{0}\in N$ , and let $F_{0}$ be the canonical frame of the oriented circle $S^{1}\times \{x_{0}\}$ on $S^{1}\times N$ . A map $\beta \;:\;S^{1}\rightarrow Diff(N,x_{0})$ gives rise to a diffeomorphism of $S^{1}\times N$

\begin{equation*}\widetilde{\beta }\;:\;S^{1}\times N\rightarrow S^{1}\times N \;\text{by}\; \widetilde{\beta }(z,x)=(z,\beta (z)\cdot x)\end{equation*}

that fixes the circle $S^{1}\times x_{0}$ pointwisely. In particular, in addition to $F_{0}$ the composition $\widetilde{\beta }\circ F_{0}$ is a second frame of the circle $S^{1}\times \{x_{0}\}$ .

Assume that the manifold N is $\Sigma $ -stable. Then there exists a $\beta$ such that $e_{N}\circ \beta =\alpha $ , implying $\widetilde{\beta }\circ F_{0}=F_{0}\widetilde{\alpha }$ . Thus, surgeries along $S^{1}\times x_{0}$ with respect to the two frames $F_{0}$ and $\widetilde{\beta }\circ F_{0}$ , we obtain from $\widetilde{\beta }$ a desired diffeomorphism $\Sigma_{0}N\rightarrow \Sigma _{1}N$ .

Proof. Let $\{e_{1},\ldots ,e_{n+1}\}$ be the standard orthonormal basis of $\mathbb{R}^{n+1}$ and take $e_{n+1}\in S^{n}$ as the base point. The generator $\alpha \in \pi_{1}(SO(n))$ gives rise to the loop on $Diff(S^{n},x_{0})$

\begin{equation*}\beta \;:\;S^{1}\rightarrow Diff(S^{n},x_{0}), \beta (z)(x)=\alpha (z)x, z\in S^{1}, x\in S^{n},\end{equation*}

that satisfies $e_{S^{n}}(\beta )=\alpha $ , implying that $S^{n}$ is $\Sigma $ -stable. We get $\Sigma _{0}S^{n}=\Sigma _{1}S^{n}$ by Theorem A. Moreover, from $W_{S^{n}}=D^{n+2}$ we get by (3·1) that

\begin{equation*}\Sigma _{0}S^{n}=\partial W_{S^{n}}=S^{n+1},\end{equation*}

showing the first statement of the proposition.

For (3·2) we get from (3·1) that (e.g. [ Reference Suciu25 , lemma 1·3], [ Reference Gitler and Medrano11 , lemma 2])

\begin{equation*}\Sigma _{0}(S^{p}\times S^{q})=\partial W_{S^{p}\times S^{q}}=S^{p}\times S^{q+1}\#S^{p+1}\times S^{q}.\end{equation*}

Since $S^{p}\times S^{q}$ is $\Sigma $ -stable by Corollary 3·1, we get (3·2) by Theorem A.

Proof. Set $r\;:\!=\;\dim _{\mathbb{R}}\mathbb{F}$ . In view of $\partial D(\lambda _{\mathbb{F}}^{n-1})=S^{rn-1}$ the manifold $\mathbb{F}P^{n}$ has the decomposition $\mathbb{F}P^{n}=D(\lambda _{\mathbb{F}}^{n-1})\cup _{id}D^{rn}$ , implying

\begin{equation*}W_{\mathbb{F}P^{n}}=D^{2}\times (\mathbb{F}P^{n}-\overset{\circ }{D^{rn}})=D^{2}\times D(\lambda _{\mathbb{F}}^{n-1})=D(\lambda _{\mathbb{F}}^{n-1}\oplus \varepsilon ^{2}).\end{equation*}

One obtains by (3·1) that

\begin{equation*}\Sigma _{0}\mathbb{F}P^{n}=\partial W_{\mathbb{F}P^{n}}=S(\lambda _{\mathbb{F}}^{n-1}\oplus \varepsilon ^{2}).\end{equation*}

Since $\Sigma _{0}\mathbb{F}P^{n}$ is the total space of a d-dimensional sphere bundle over $\mathbb{F}P^{n-1}$ which has a section, we obtain (3·3) by the homotopy exact sequence of the fiber bundle $\Sigma _{0}\mathbb{F}P^{n}\rightarrow \mathbb{F}P^{n-1}$ .

Example 3·4. Let $\Omega _{n}^{Spin}$ be the n-dimensional Spin bordism group. Then $\Sigma _{0}\mathbb{H}P^{2}=0$ in $\Omega _{9}^{Spin}$ by $\Sigma _{0}\mathbb{H}P^{2}=\partial W_{\mathbb{H}P^{2}}$ . On the other hand, there exists a surgery along the circle $S^{1}\times \{x_{0}\}\subset S^{1}\times \mathbb{H}P^{2}$ such that the resulting manifold M satisfies $M\neq 0$ in $\Omega _{9}^{Spin}$ (e.g. [ Reference Anderson, Brown and Peterson1 , p.258], [ Reference Crowley and Yang6 ]). That is

\begin{equation*}M\neq \Sigma _{0}\mathbb{H}P^{2} \;\text{but}\; M=\Sigma _{1}\mathbb{H}P^{2}.\end{equation*}

In particular, the manifold $\mathbb{H}P^{2}$ fails to be $\Sigma $ -stable by Theorem A.

Example 3·5. Formula (1·10) in Section 1 suggests that Theorem B may be applicable to derive the homotopy groups of certain manifolds. Combining (1·6) with (3·3) we get further evidence:

\begin{eqnarray*}\pi _{r}(\mathbb{C}P^{2n}\#\mathbb{R}P^{4n})&=&\left\{\begin{array}{c}\mathbb{Z}_{2}\text{ if }r=1\text{,} \\[5pt] \pi _{r}(S^{4n-1})\oplus \pi _{r}(S^{2})\text{ if }r\neq 1\end{array}\right.\!\!;\\[10pt]\pi _{r}(\mathbb{C}P^{2n}\#\mathbb{C}P^{2n})&=&\left\{\begin{array}{c}\mathbb{Z}\oplus \mathbb{Z}\text{ if }r=2 \\[5pt] \pi _{r}(S^{4n-1})\oplus \pi _{r}(S^{3})\text{ if }r\neq 2\end{array}\right.\!\!;\\[10pt]\pi _{r}(\mathbb{C}P^{2n}\#\mathbb{H}P^{n})&=&\left\{\begin{array}{c}\mathbb{Z}\text{ if }r=2,4 \\[5pt] \pi _{r}(S^{4n-1})\oplus \pi _{r}(S^{5})\text{ if }r\neq 2,4\end{array}\right.\!\! .\end{eqnarray*}

Example 3·6. For $n=2$ we have $\pi _{1}(SO(2))=\mathbb{Z}$ with generator $\alpha $ . Therefore, the operation $\Sigma _{i}$ applies also to the orientable surfaces to yield the correspondences $\Sigma _{i}\;:\;\mathcal{T}_{2}\rightarrow \mathcal{T}_{3}$ , $i=0,1$ .

Let $M_{g}\in \mathcal{T}_{2}$ be the orientable surface with genus $g\geq 1$ , and let $D^{2}\subset M_{g}$ be an embedded disc. Then the fundamental group $\pi _{1}(M_{g}-\overset{\circ }{D^{2}})$ is the free group generated by the standard elements

\begin{equation*}a_{1},b_{1},\ldots a_{g},b_{g}\in \pi _{1}(M_{g}-\overset{\circ }{D^{2}}).\end{equation*}

Applying the Van Kampen Theorem to (1·2) one gets the presentations of the fundamental groups $\pi _{1}(\Sigma _{i}M_{g})$ by the generators and relations [ Reference Freedman10 ]

\begin{eqnarray*}\pi _{1}(\Sigma _{0}M_{g})&=&\left\langle a_{1},b_{1},\cdots a_{g},b_{g}\right\rangle ;\\[5pt] \pi _{1}(\Sigma _{1}M_{g})&=&\Big\langle a_{1},b_{1},\cdots a_{g},b_{g},z\mid a_{i}z=za_{i},b_{i}z=zb_{i}, \\[5pt] && z^{-1}=a_{1}b_{1}a_{1}^{-1}b_{1}^{-1}\cdots a_{n}b_{n}a_{n}^{-1}b_{n}^{-1}\Big\rangle .\end{eqnarray*}

Precisely, $\Sigma _{0}M_{g}=S^{1}\times S^{2}\#\cdots \#S^{1}\times S^{2}$ (2g-copies), while $\Sigma _{1}M_{g}$ is the total space of the oriented circle bundle over $M_{g}$ whose Euler class is the orientation class on $M_{g}$ (e.g. Freedman [ Reference Freedman10 ]). This shows that $\Sigma _{0}M_{g}\neq \Sigma _{1}M_{g}$ .

Proof. It suffices to show (4·1) for the case of $i=1$ . The remaining case $i=0$ follows from the same idea.

By the definition of the space $E_{N}$ in (1·3) we have $E_{N\#M}=(E_{N})_{M} $ . In addition, the Van–Kampen Theorem [ Reference Hatcher14 , p.43] implies that, if $N\in \mathcal{T}_{n-1}$ is 1-connected, then

\begin{equation*}\pi _{1}(\Sigma _{1}N)=\pi _{1}(N)=0, i=0,1.\end{equation*}

Since $\Sigma _{1}N_{1},\cdots ,\Sigma _{1}N_{k}\in \mathcal{T}_{n-1}$ are all 1-connected, Theorem B can be applied repeatedly to show that, if $w_{2}(B)=0$ , then

\begin{equation*}E_{N_{1}\#\cdots \#N_{k}}=E\#\Sigma _{1}N_{1}\#\cdots \#\Sigma _{1}N_{k}.\end{equation*}

In particular, taking p to be the circle bundle $S^{3}\times S^{n-2}\rightarrow S^{2}\times S^{n-2}$ in Example 1·5, we get

\begin{equation*}S^{3}\times S^{n-2}\#\Sigma _{1}(N_{1}\#\cdots \#N_{k})=S^{3}\times S^{n-2}\#\Sigma _{1}N_{1}\#\cdots \#\Sigma _{1}N_{k}.\end{equation*}

Since the surgery on the 3-sphere $S^{3}\times \{\ast \}\subset $ $S^{3}\times S^{n-3}$ with trivial framing yields $S^{n}$ , applying surgery on both sides to kill $S^{3}\times \{\ast \}$ shows (4·1).

Proof. It suffices to show formulae (4·3) and (4·4). In view of (1·2) we set $A\;:\!=\;S^{1}\times $ $(N-\overset{\circ }{D^{n}})$ and $B\;:\!=\;D^{2}\times S^{n-1}$ to get

\begin{equation*}\Sigma _{i}N=A\cup B, A\cap B=S^{1}\times S^{n-1},\end{equation*}

as well as the Mayer–Vietoris sequence

\begin{equation*}\cdots \rightarrow H^{r-1}(A\cap B)\overset{\delta }{\rightarrow }H^{r}(\Sigma _{i}N)\overset{i_{1}^{\ast }\oplus i_{2}^{\ast }}{\rightarrow }H^{r}(A)\oplus H^{r}(B)\overset{j_{1}^{\ast }-j_{2}^{\ast }}{\rightarrow }H^{r}(A\cap B)\rightarrow \cdots \end{equation*}

For a $M\in \mathcal{T}_{n}$ let $\omega _{M}\in H^{n}(M)$ be the orientation class. Then the graded group $H^{\ast }(A\cap B)$ is torsion free with basis

\begin{equation*}\{1,\omega _{S^{1}}\times 1,1\times \omega _{S^{n-1}},\omega _{S^{1}\times S^{n-1}}\},\end{equation*}

where $\omega _{S^{1}}\times 1\in \text{Im}\;j_{1}^{\ast }$ , $1\times \omega _{S^{n-1}}\in \text{Im}\;j_{2}^{\ast }$ . Moreover, the Mayer–Vietoris sequence can be summarised as the short exact sequence

(4·5) \begin{equation}0\rightarrow \mathbb{Z}\{\omega _{S^{1}\times S^{n-1}}\}\overset{\delta }{\rightarrow }H^{\ast }(\Sigma _{i}N)\overset{i_{1}^{\ast }}{\rightarrow }\frac{H^{\ast }(S^{1}\times (N-\left\{ x_{0}\right\}\!))}{\mathbb{Z}\{\omega _{S^{1}}\times 1\}}\rightarrow 0,\end{equation}

where

1. (a) $\mathbb{Z}\{\omega \}$ denotes the cyclic group with generator $\omega $ ;

2. (b) $N-\overset{\circ }{D^{n}}$ has been replaced by $N-\left\{ x_{0}\right\}$ , and

3. (c) the map $\delta $ is an isomorphism onto $H^{n+1}(\Sigma _{i}N)=\mathbb{Z}$ .

In particular, the sequence (4·5) is splittable to yields (4·3), under the convention that the orientation class $\omega _{\Sigma_{i}N}=\delta (\omega _{S^{1}\times S^{n-1}})$ agrees with the top degree generator $s\wedge \omega _{N}$ of $\widetilde{H^{n+1}}(SN)=\mathbb{Z}$ .

Finally, let TM be the tangent bundle of a smooth manifold M. Then the inclusion $i_{1}\;:\;$ $S^{1}\times (N-\overset{\circ }{D^{n}})\rightarrow\Sigma _{i}N$ satisfies that

\begin{equation*}i_{1}^{\ast }T(\Sigma _{i}N)=\varepsilon ^{1}\oplus T(N-\left\{x_{0}\right\}\!),\end{equation*}

where $\varepsilon ^{1}$ is the 1-dimensional trivial bundle on $S^{1}\times (N-\overset{\circ}{D^{n}})$ consisting of the tangent directions of the first factor $S^{1}$ . The formulae (4·4) follows now from the naturality of the characteristic classes, together with the decomposition (4·3).

Proof of Theorem C. For a 1-connected 6-manifold M with a regular circle action let $\pi \;:\;$ $M\rightarrow N\;:\!=\;M/S^{1} $ be the quotient map. Since the quotient N is then an 1-connected and 5-dimensional manifold which is not diffeomorphic to $S^{5}$ , formula (5·2) is applicable to decompose it as a connected sum of the basic ones. Moreover, since the Euler class $\omega \in H^{2}(N)$ of $\pi $ is primitive by Lemma 5·1, we can take a decomposition of N relative to $\omega $ as that stated in (5·3), to get by the formula (1·4) in Theorem B the decomposition

\begin{equation*}M=S^{3}\times S^{3}\#\Sigma _{i}N^{\prime }, N^{\prime }\in \mathcal{T}_{5},\end{equation*}

where $i=0$ or 1 in accordance to $\omega \equiv w_{2}(N)\;\mathrm{mod}\;2$ , or $\omega \neq w_{2}(N)\;\mathrm{mod}\;2$ . Decomposing $N^{\prime }$ in the form of (5·2) and using (4·1) to expand $\Sigma _{i}N^{\prime }$ , we obtain formula (1·7) in Theorem C, in which

\begin{equation*}\Sigma _{0}(S^{2}\times S^{3})=\Sigma _{1}(S^{2}\times S^{3})=S^{3}\times S^{3}\#S^{2}\times S^{4}\; \textrm{({by};(3${\cdot}$2)).}\end{equation*}

This completes the proof.

Example 6·2. For $k\geq 1$ the manifold $Q_{k}$ constructed in (1·9) satisfies that

(6·1) \begin{equation}P_{t}(Q_{k})=1+c_{1}\big(t^{3}+t^{k+1}\big)+c_{2}\big(t^{4}+t^{k}\big)\cdots +c_{r}\big(t^{r+2}+t^{k-r-2}\big), r=\left[ \frac{k}{2}\right], \end{equation}

where the sequence of integers $(c_{1},\ldots ,c_{r})$ is determined by the integer k by

\begin{eqnarray*}c_{i}&\;:\!=\;&(k-1)\binom{k-1}{i}-\binom{k-1}{i+1}+(k-1)\binom{k-1}{i-1}-\binom{k-1}{i-2}, 1\leq i\,{\lt}\,r\\[5pt] c_{r}&\;:\!=\;&\frac{1}{2}\bigg((k-1)\binom{k-1}{r}-\binom{k-1}{r+1}+(k-1)\binom{k-1}{r-1}-\binom{k-1}{r-2}\bigg).\end{eqnarray*}

The proof of Theorem C (see Section 1) relies on the following two lemmas, whose proofs will be postponed.

Proof of Theorem D. Let M be an 1-connected 4-manifold with $\beta _{2}(M)=k$ . According to [ Reference Duan and Liang8 , corollary 3] there is a primitive element $e\in H^{2}(M)$ so that the total space of the circle bundle $\pi \;:\;M(e)\rightarrow M$ with Euler class e is

\begin{equation*}M(e)=\#_{1\leq i\leq k-1}(S^{2}\times S^{3}) (=N\; \text{in the notation of Lemma 6$\cdot$4}).\end{equation*}

Moreover, in term of the Gysin sequence of $\pi $ [ Reference Milnor and Stasheff20 , p.143], one finds a basis $\left\{ \alpha _{1},\ldots ,\alpha _{k}\right\} $ of $H^{2}(M)$ such that

\begin{equation*}\alpha _{1}=e, \pi ^{\ast }(\alpha _{i})=\omega _{i-1}, 2\leq i\leq k,\end{equation*}

and that $\{\omega _{1},\ldots ,\omega _{k-1}\}$ is the basis of $H^{2}(M(e))$ specified in Lemma 6·4. That is, in the notation of (1·8),

\begin{equation*}M(\alpha _{1},\ldots ,\alpha _{k})= M(e)(\omega _{1},\ldots ,\omega_{k-1})=N(\omega _{1},\ldots ,\omega _{k-1}).\end{equation*}

We get $M(\alpha _{1},\ldots ,\alpha _{k})=Q_{k}$ by Lemma 6·4

Finally, if $\left\{ \alpha _{1}^{\prime },\cdots ,\alpha _{k}^{\prime}\right\} $ is an arbitrary basis of $H^{2}(M)$ , we have by i) of Lemma 6·3 a diffeomorphism

\begin{equation*}M(\alpha _{1}^{\prime },\ldots ,\alpha _{k}^{\prime })\cong M(\alpha_{1},\ldots ,\alpha _{k}).\end{equation*}

This completes the proof of Theorem D.

Proof of Lemma 6·3. Let $\xi _{k}$ be the universal principal $T^{k}$ -bundle on $BT^{k}$ , and let M be a manifold whose second cohomology $H^{2}(M)$ is torsion free. For two bases $\left\{ \alpha_{1},\ldots ,\alpha _{k}\right\} $ and $\left\{ \alpha _{1}^{\prime },\ldots,\alpha _{k}^{\prime }\right\} $ of $H^{2}(M)$ with classifying maps f and $f^{\prime }\;:\;M\rightarrow BT^{k}$ , respectively, we have by (1·8) that

(6·2) \begin{equation}M(\alpha _{1},\ldots ,\alpha _{k})=f^{\ast }\xi _{k}, M\big(\alpha _{1}^{\prime },\ldots ,\alpha _{k}^{\prime }\big)=f^{\prime \ast }\xi_{k}.\end{equation}

On the other hand, expressing the basis elements $\alpha _{i}$ ’s in the $\alpha _{i}^{\prime }$ ’s we get a $k\times k$ integral invertible matrix $C=(c_{ij})_{k\times k}$ such that

\begin{equation*}\left\{ \alpha _{1},\ldots ,\alpha _{k}\right\} =\left\{ \alpha_{1}^{\prime },\ldots ,\alpha _{k}^{\prime }\right\} \cdot C.\end{equation*}

Regard $S^{1}$ as the group of the unit complexes $z\in \mathbb{C}$ , define in term of C the group isomorphism

\begin{equation*}g\;:\;T^{k}\rightarrow T^{k}, g(z_{1},\ldots ,z_{k})=\big(z_{1}^{c_{11}}\cdots z_{k}^{c_{1k}},\ldots ,z_{1}^{c_{k1}}\cdots z_{k}^{c_{kk}}\big),\end{equation*}

and let $B_{g}\;:\;BT^{k}\rightarrow BT^{k}$ be the induced map of g. From $f^{\ast }=f^{\prime \ast }B_{g}^{\ast }$ we get $f\simeq B_{g}\circ f^{\prime }$ . By the homotopy invariance of the induced bundles, together with the fact that the map $B_{g}$ has a homotopy inverse, we obtain by (6·2) a diffeomorphism $M(\alpha _{1},\ldots ,\alpha _{k})\cong M(\alpha_{1}^{\prime },\ldots ,\alpha _{k}^{\prime })$ , showing part (i) of Lemma 6·3.

For ii) we can assume by $M\in \mathcal{S}_{n}$ and $n\geq 5$ that

(6·3) \begin{equation}M=\#_{k}\big(S_{i}^{2}\times S_{i}^{n-2}\big)\#_{1\leq j\leq r}\big(S_{j}^{t_{j}}\times S_{j}^{n-t_{j}}\big), t_{j}\leq n-t_{j}, t_{j}\neq 2,\end{equation}

where $S_{t}^{p}\times S_{t}^{n-p}=S^{p}\times S^{n-p}$ . For each $1\leq i\leq k$ let $q_{i}\;:\;M\rightarrow S_{i}^{2}\times S_{i}^{n-2}$ be the quotient of M by the summands other than the component $S_{i}^{2}\times S_{i}^{n-2}$ , $p_{1}\;:\;$ $S_{i}^{2}\times S_{i}^{n-2}\rightarrow S^{2}$ be the projection onto the first factor, and set

\begin{equation*}g_{i}\;=\!:\;p_{1}\circ q_{i}\;:\;M\rightarrow S_{i}^{2}\times S_{i}^{n-2}\rightarrow S^{2}, \omega _{i}\;:\!=\;g_{i}^{\ast }(\omega )\in H^{2}(M),\end{equation*}

where $\omega \in H^{2}(S^{2})=\mathbb{Z}$ is the orientation class of $S^{2}$ . In terms of (i) it suffices to show $M(\omega _{1},\ldots,\omega _{k})\in \mathcal{S}_{n+k}$ . This can be done by induction on k.

Let $\pi _{1}$ be the circle bundle $M(\omega _{1})\rightarrow M$ over M with Euler class $\omega _{1}\in H^{2}(M)$ . Then

\begin{equation*}M(\omega _{1},\ldots ,\omega _{k})=M(\omega _{1})(\pi _{1}^{\ast }(\omega_{2}),\ldots ,\pi _{1}^{\ast }(\omega _{k-1})),\end{equation*}

where, by the Gysin sequence [ Reference Milnor and Stasheff20 , p.143] of spherical fibrations, the set $\{\pi _{1}^{\ast }(\omega _{2}),\ldots ,$ $\pi_{1}^{\ast }(\omega _{k-1})\}$ is a basis of $H^{2}(M(\omega _{1}))$ . However, it follows from Example 1·5, together with formulae (6·3) and (3·2), that

\begin{equation*}M(\omega _{1})=\big(S^{3}\times S^{n-2}\big)\#_{k-1}\Sigma _{1}\big(S_{i}^{2}\times S_{i}^{n-2}\big)\#_{r}\Sigma _{1}\big(S_{j}^{k_{j}}\times S_{j}^{n-k_{j}}\big)\in\mathcal{S}_{n+1}.\end{equation*}

This completes the inductive procedure of showing part (ii).

Proof of Lemma 6·4. Suppose that $N=\#_{1\leq i\leq k-1}(S_{i}^{2}\times S_{i}^{3})$ , and that $\{\omega _{1},\ldots ,\omega_{k-1}\}$ is the basis of $H^{2}(N)$ specified as that in Lemma 6·4. Then the cohomology $H^{\ast }(N)$ has the basis

(6·4) \begin{equation}\left\{ 1,\omega _{1},\ldots ,\omega _{k-1},y_{1},\ldots,y_{k-1},z\right\} , \deg \omega _{i}=2, \deg y_{i}=3, \deg z=5,\end{equation}

that is subject to the multiplicative relations

(6·5) \begin{equation}\omega _{i}\cdot \omega _{j}=0;\quad y_{i}\cdot y_{j}=0;\quad \omega _{i}\cdot y_{j}=\delta _{ij}\cdot z, 1\leq i,j\leq k-1, \end{equation}

where $\delta _{ij}$ is the Kronecker symbol. By (ii) of Lemma 6·3, for the principal $T^{k-1}$ -bundle

\begin{equation*}\pi \;:\;N(\omega _{1},\ldots ,\omega _{k-1})\rightarrow N\end{equation*}

over N we have $N(\omega _{1},\cdots ,\omega _{k-1})\in S_{4+k}$ . According to Lemma 6·1 it suffices to derive that

(6·6) \begin{equation}P_{t}(N(\omega _{1},\ldots ,\omega _{k-1}))=P_{t}(Q_{k})\; \textrm{(see ;(6${\cdot}$1))}.\end{equation}

To this end we compute with the Leray–Serre spectral sequence $\{E_{r}^{\ast ,\ast },d_{r}\}$ of the bundle $\pi $ [ Reference Whitehead30 , p.645].

Since both cohomologies $H^{\ast }(N)$ and $H^{\ast }(T^{k-1})$ are torsion free, while the base manifold N is 1-connected, the Leray–Serre theorem tells that

\begin{equation*}E_{2}^{\ast ,\ast }=H^{\ast }(N)\otimes H^{\ast }(T^{k-1}).\end{equation*}

In addition, there are unique elements $t_{1},\ldots ,t_{k-1}\in H^{1}(T^{k-1})$ such that

\begin{equation*}H^{\ast }(T^{k-1})=\Lambda (t_{1},\ldots ,t_{k-1});\; d_{2}(1\otimes t_{i})=\omega _{i}\otimes 1,\end{equation*}

implying

(6·7) \begin{equation}d_{2}(x\otimes t_{i})=\omega _{i}\cdot x\otimes 1, d_{2}(x\otimes 1)=0, x\in H^{\ast }(N).\end{equation}

For a multi-index $I\subseteq \{1,\ldots ,k-1\}$ we set $t_{I}\;:\!=\;\Pi _{i\in I}t_{i}$ , and let $B_{i}\subset E_{2}^{\ast ,\ast }$ , $1\leq i\leq 4$ , be the subgroups with the basis:

\begin{align*}&B_{1}\;:\;\{1\otimes t_{I}\mid I\subseteq \{1,\ldots ,k-1\}\}; \\[5pt] &B_{2}\;:\;\{1,\omega _{i}\otimes 1,\omega _{i}\otimes t_{I}\mid \quad i\in\{1,\ldots ,k-1\},I\subseteq \{1,\ldots ,k-1\}\};\\[5pt] &B_{3}\;:\;\{y_{i}\otimes 1,y_{i}\otimes t_{I}\mid i\in \{1,\ldots,k-1\},I\subseteq \{1,\ldots ,k-1\}\}; \\[5pt] &B_{4}\;:\;\{z\otimes 1,z\otimes t_{I}\mid I\subseteq \{1,\ldots ,k-1\}\},\end{align*}

respectively. Then $E_{2}^{\ast ,\ast }=B_{1}\oplus B_{2}\oplus B_{3}\oplus B_{4}$ by (6·4) and (as is clear):

\begin{align*}P_{t}(B_{1})&=\underset{1\leq i\leq k-1}{\Sigma }\binom{k-1}{i}t^{i};\\[5pt] P_{t}(B_{2})&=1+(k-1)\underset{0\leq i\leq k-1}{\Sigma }\binom{k-1}{i}t^{i+2} ;\\[5pt] P_{t}(B_{3})&=(k-1)\underset{0\leq i\leq k-1}{\Sigma }\binom{k-1}{i}t^{i+3};\\[5pt] P_{t}(B_{4})&=\underset{0\leq i\leq k-1}{\Sigma }\binom{k-1}{i}t^{i+5}. \end{align*}

Moreover, the $d_{2}$ -action on the basis elements $x\in B_{i}$ has been decided by (6·5) and (6·7) as

\begin{equation*}d_{2}(x)=\left\{\begin{array}{l}0\; \text{if}\; x=\omega _{i}\otimes 1,y_{i}\otimes 1,z\otimes t_{I},\omega_{i}\otimes t_{I}, y_{j}\otimes t_{J}\; \text{if}\; j\notin J; \\[5pt] \underset{1\leq s\leq t}{\Sigma }(\!-\!1)^{s-1}\omega _{i_{s}}\otimes t_{I_{s}}\;\text{if} \;x=1\otimes t_{I} \;\text{with ;} I=\{i_{1},\cdots ,i_{t}\}; \\[5pt] (\!-\!1)^{s-1}z\otimes t_{I_{s}} \;\text{if}\; x=y_{i}\otimes t_{I} \;\text{with}\; i=i_{s}\in I=\{i_{1},\cdots ,i_{t}\},\end{array}\right.\end{equation*}

where $I_{s}$ denotes the multi-index obtained from I by deleting the sth coordinate $i_{s}$ . It follows that:

1. (a) $d_{2}(B_{2})=0$ , $d_{2}$ maps $B_{1}$ monomorphically into $B_{2}$ ;

2. (b) $d_{2}(B_{4})=0$ , $d_{2}(B_{3})\subset B_{4}$ with $B_{4}/d_{2}(B_{3})=\mathbb{Z}$ generated by $z\otimes t.$

As results,

(6·8) \begin{equation}E_{3}^{\ast ,\ast }=B_{2}/d_{2}(B_{1})\oplus B_{4}/d_{2}(B_{3})\oplus \overline{B}_{3} \;\text{with}\; \overline{B}_{3}\;:\!=\;\ker[d_{2}\;:\;B_{3}\rightarrow B_{4}];\end{equation}

(6·9) \begin{equation}E_{3}^{\ast ,\ast }=E_{\infty }^{\ast ,\ast }=H^{\ast}(N(\omega _{1},\ldots ,\omega _{k-1})).\end{equation}

Since

\begin{align*}&P_{t}(B_{2}/d_{2}(B_{1}))=P_{t}(B_{2})-t\cdot P_{t}(B_{1}) \text{by a});\\[5pt] &P_{t}(B_{4}/d_{2}(B_{3}))=t^{k+4} \text{by b}),\end{align*}

and since

\begin{equation*}P_{t}(\overline{B}_{3})=P_{t}(B_{3})-t^{-1}(P_{t}(B_{4})-P_{t}(B_{4}/d_{2}(B_{3})),\end{equation*}

by the exactness of the sequence of the free $\mathbb{Z}$ -modules

\begin{equation*} 0\rightarrow \overline{B}_{3}\rightarrow B_{3}\overset{d_{2}}{\rightarrow }B_{4}\rightarrow B_{4}/d_{2}(B_{3})(=\mathbb{Z})\rightarrow 0,\end{equation*}

we get, in the order of (6·9), (6·8) and (6·1), the equalities

\begin{align*}&P_{t}(N(\omega _{1},\ldots ,\omega _{k-1}))=P_{t}(E_{3}^{\ast ,\ast })\\[5pt] &=P_{t}(B_{2})-t\cdot P_{t}(B_{1})+P_{t}(B_{3})-t^{-1}\cdot (P_{t}(B_{4})-t^{k+4})+t^{k+4}\\[5pt] &=P_{t}(Q_{k}).\end{align*}

This verifies the relation (6·6), completing the proof of Lemma 6·4.