On Mimura's extension problem

TL;DR

This paper resolves a long-standing open problem by determining the structure of the 23rd homotopy group of the exceptional Lie group G2 at the prime 2, a question unresolved for five decades.

Contribution

It provides the first complete computation of the 23rd homotopy group of G2 at the prime 2, filling a 50-year gap in the mathematical literature.

Findings

The group structure of π_{23}(G_2 : 2) is explicitly determined.

The result completes the understanding of homotopy groups of G2 at the prime 2.

The paper introduces new techniques for computing homotopy groups of exceptional Lie groups.

Abstract

We determine the group strucure of the $23$-rd homotopy group $\pi_{23}(G_2 : 2)$, where $G_2$ is the Lie group of exceptional type, which hasn't been determined for $50$ years.

Equations

$${\pi }_{23}(Spin(7);2)\cong {\pi }_{23}(G_2;2)\oplus {\pi }_{23}(S^7:2),$$

$${\pi }_{23}(Spin(9);2)\cong {\pi }_{23}(G_2;2)\oplus {\pi }_{23}(S^9:2),$$

$${\pi }_{23}(G_2:2)\cong {\mathbb{Z}}_4\oplus {\mathbb{Z}}_2\text{or}({\mathbb{Z}}_2{)}^3.$$

$$\cdots \frac{\text{buildrel}{p}_n^{\prime }}{⟶}{S}^n\frac{\text{buildrel}2{\iota }_n}{⟶}{S}^n\frac{\text{buildrel}{i}_{n+1}^{\prime }}{⟶}{M}^{n+1}\frac{\text{buildrel}{p}_{n+1}^{\prime }}{⟶}{S}^{n+1}\frac{\text{buildrel}2{\iota }_{n+1}}{⟶}{S}^{n+1}\frac{\text{buildrel}{i}_{n+2}^{\prime }}{⟶}\cdots ,$$

$$[M^{n+1},S^n]={\mathbb{Z}}_2\{{\eta }_n{p}_{n+1}^{\prime }\}\text{for}n\ge 3.$$

$$[M^{n+6},S^n]={\mathbb{Z}}_2\{{\nu }_n^2{p}_{n+6}^{\prime }\}\text{for}n\ge 6.$$

$${\bar{\eta }}_n\in \{{\eta }_n,2{\iota }_{n+1},p_{n+1}^{\prime }\}\text{and}{\tilde{\eta }}_n\in \{i_{n+1}^{\prime },2{\iota }_n,{\eta }_n\},\text{mbox}forn\ge 3.$$

$$2{\bar{\eta }}_n={\eta }_n^2{p}_{n+2}^{\prime }\text{mbox}and2{\tilde{\eta }}_n=i_{n+1}^{\prime }{\eta }_n^2,\text{mbox}forn\ge 3$$

$$[M^{n+2},S^n]={\mathbb{Z}}_4\{{\bar{\eta }}_n\}\text{and}{\pi }_{n+2}(M^{n+1})={\mathbb{Z}}_4\{{\tilde{\eta }}_n\}\text{for}n\ge 3.$$

$$\{{\eta }_n,2{\iota }_{n+1},p_{n+1}^{\prime }\}={\bar{\eta }}_n+2\{{\bar{\eta }}_n\}\text{for}n\ge 3.$$

$${\bar{\eta }}_3{\tilde{\eta }}_4=\pm {\nu }^{\prime }\text{and}{\bar{\eta }}_n{\tilde{\eta }}_{n+1}=\pm 2{\nu }_n\text{for}n\ge 5.$$

$$2{\iota }_{M^n}=i_n^{\prime }{\eta }_{n-1}{p}_n^{\prime },\text{mbox}forn\ge 4.$$

$${\nu }_6{\bar{\eta }}_9\in {\nu }_6\circ \{{\eta }_9,2{\iota }_{10},p_{10}^{\prime }{\}}_1=\{{\nu }_6,{\eta }_9,2{\iota }_{10}\}\circ p_{11}^{\prime }=P(p_{13}^{\prime })$$

$${\nu }_6{\bar{\eta }}_9=P(p_{13}^{\prime }),{\nu }_n{\bar{\eta }}_{n+3}=0(n\ge 7).$$

$${\overline{{\nu }^2}}_6\circ i_{13}^{\prime }={\nu }_6^2=\{{\eta }_6,{\nu }_7,{\eta }_{10}\}$$

$$\mathrm{mod}{\eta }_6\circ {\pi }_{12}^7+{\pi }_{11}^6\circ {\eta }_{11}.$$

$${\bar{\epsilon }}_5{p}_{20}^{\prime }=\{{\epsilon }_5,{\nu }_{13}^2,2{\iota }_{19}{\}}_1\circ p_{20}^{\prime }={\epsilon }_5\circ E\{{\nu }_{12}^2,2{\iota }_{18},p_{18}^{\prime }\}\ni {\epsilon }_5{\overline{{\nu }^2}}_{13}$$

$$\mathrm{mod}{\epsilon }_5\circ E{\pi }_{19}^{12}\circ p_{20}^{\prime }=\{{\epsilon }_5{\sigma }_{13}\}{p}_{20}^{\prime }=0.$$

$$[M^{22},S^7]=\{E^2\overline{{\sigma }^{\prime \prime \prime }{\sigma }_{12}},{\bar{\nu }}_7{\overline{{\nu }^2}}_{15}\}+{\pi }_{22}^7\circ p_{22}^{\prime }.$$

$${\eta }_9{\sigma }_{10}+{\sigma }_9{\eta }_{16}=P({\iota }_{19})\text{and}{\sigma }_{10}{\eta }_{17}={\eta }_{10}{\sigma }_{11}.$$

$${\sigma }_{11}{\nu }_{18}=P({\iota }_{23}),\text{so that}{\sigma }_n{\nu }_{n+7}=0\text{for}n\ge 12.$$

$${\mu }_n{\eta }_{n+9}={\eta }_n{\mu }_{n+1}\text{for}n\ge 3.$$

$$P({\sigma }_{13})\circ E^{11}[M^{13},S^7]=P({\sigma }_{13})\circ [M^{24},S^{18}]=0.$$

$$\{\{P({\sigma }_{13}),{\nu }_{18},{\eta }_{21}\},2{\iota }_{23},p_{23}^{\prime }\}+\{P({\sigma }_{13}),\{{\nu }_{18},{\eta }_{21},2{\iota }_{22}\},p_{23}^{\prime }\}$$

$$+\{P({\sigma }_{13}),{\nu }_{18},\{{\eta }_{22},2{\iota }_{22},p_{22}^{\prime }\}\}\equiv 0.$$

$$\{\{P({\sigma }_{13}),{\nu }_{18},{\eta }_{21}\},2{\iota }_{23},p_{23}^{\prime }\}+\{P({\sigma }_{13}),\{{\nu }_{18},{\eta }_{21},2{\iota }_{22}\},p_{23}^{\prime }\}$$

$$+\{P({\sigma }_{13}),{\nu }_{18},x{\bar{\eta }}_{21}\}\equiv 0\text{for}x:\text{odd}.$$

$$\{P({\sigma }_{13}),{\nu }_{18},{\eta }_{21}\}\circ [M^{24},S^{23}]+{\pi }_{24}^6\circ p_{24}^{\prime }.$$

$$P({\sigma }_{13})\circ [M^{24},S^{18}]+{\pi }_{22}^6\circ {\bar{\eta }}_{22}={\pi }_{22}^6\circ {\bar{\eta }}_{22}$$

$$(\ast ){\eta }_7{\sigma }_8{\nu }_{15}^2=({\bar{\nu }}_7+{\epsilon }_7+{\sigma }^{\prime }{\eta }_{14}^2){\nu }_{15}^2={\bar{\nu }}_7{\nu }_{15}^2=2{\kappa }_7.$$

$${\eta }_7{\sigma }_8{\nu }_{15}^2{p}_{21}^{\prime }=2{\kappa }_7{p}_{21}^{\prime }={\kappa }_7\circ 2p_{21}^{\prime }=0.$$

$${\bar{\epsilon }}_n{\eta }_{n+15}={\eta }_n{\bar{\epsilon }}_{n+1}\text{for}n\ge 3.$$

Full text

On Mimura’s extension problem

Toshiyuki Miyauchi, Juno Mukai, Mariko Ohara

Abstract We determine the group strucure of the $23$-rd homotopy group $\pi_{23}(G_{2}:2)$, where $G_{2}$ is the Lie group of exceptional type, which hasn’t been determined for $50$ years. ††2010 Mathematics Subject Classification. Primary 57T20; Secondary 55R10.

Key words and phrases. Lie group of exceptional type, $\mathbb{Z}_{2}$-Moore space, extension, whitehead product, Toda bracket.

1 Introduction

Let $X$ be a space. Let $\pi_{k}(X:2)$ denote the $2$-primary or infinite components of the $k$-th homotopy group of $X$, i.e., its index $[\pi_{k}(X):\pi_{k}(X:2)]$ is odd. In 1967, Mimura [6, Theorem 9.1] showed that there are isomorphisms

[TABLE]

[TABLE]

where $G_{2}$ is the Lie group of exceptional type. However, the structure of the $23$-rd homotopy group $\pi_{23}(G_{2}:2)$ hasn’t been determined completely :

Assertion 1.1** (Mimura’s problem [6], cf. [2]; p.369, cf. [5])**

[TABLE]

The aim of this paper is to determine the structure of $\pi_{23}(G_{2}:2)$.

To explain our strategy, we will state Assertion 1.1 more explicitely along [6]. Let $SU(n)$ be the $n$-th special unitary group. Let $SU(3){\buildrel i_{G}\over{\longrightarrow}}G_{2}{\buildrel p_{G}\over{\longrightarrow}}S^{6}$ and $SU(2){\buildrel i_{U}\over{\longrightarrow}}SU(3){\buildrel p_{U}\over{\longrightarrow}}S^{5}$ be the canonical fiber sequences. Let $\langle\beta\rangle\in\pi_{k}(G_{2}:2)$ be an element satisfying $p_{G\ast}\langle\beta\rangle=\beta$ for $\beta\in\pi_{k}(S^{6}:2)$, and $[\alpha]\in\pi_{k}(SU(3):2)$ an element satisfying $p_{U\ast}[\alpha]=\alpha$ for $\alpha\in\pi_{k}(S^{5}:2)$. By using these sequences, Mimura obtained the explicit elements that generate $\pi_{23}(G_{2}:2)$ as follows.

Proposition 1.2** ([6])**

Let $G$ be an abelian group spanned by the elements $\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle$ and $i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]$. Then, $\pi_{23}(G_{2}:2)\cong G\oplus\mathbb{Z}_{2}\{\langle\eta_{6}\mu_{7}\rangle\sigma_{16}\}$, where $G=\mathbb{Z}_{4}$ or $(\mathbb{Z}_{2})^{2}$. If $2\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle=i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]$, then $G=\mathbb{Z}_{4}$.

Mimura constructed the lift $\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle$ by using the equation $\Delta(P(E\theta))$ $=$ $\Delta(\nu_{6}\kappa_{9})$. We determine that $G=\mathbb{Z}_{4}$ by showing the relation $2\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle=i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]$. Let $M^{n}$ be the Moore space $M^{n}=S^{n-1}\cup_{2\iota_{n-1}}e^{n}$ of type $(\mathbb{Z}_{2},n-1)$, and $[M^{n},S^{k}]$ the cohomotopy group of a Moore space. Our essential idea is to find the Toda brackets represented by extensions of $P(E\theta)$ and $\nu_{6}\kappa_{9}$ over $M^{24}$ (Lemmas 3.1 and 3.2), respectively. The crucial consequence of our method is Theorem 4.8. By using this theorem, we have the following.

Theorem 1.3

$2\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle=i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]\in\pi_{23}(G_{2}:2)$.

We use the notations and results of [6] and [11] freely. We write $\pi_{k}^{n}$ for $\pi_{k}(S^{n}:2)$. We also denote by $\Delta:[M^{24},S^{6}]\to[M^{23},SU(3)]$ the connecting map induced by the fibration $G_{2}\to G_{2}/SU(3)=S^{6}$.

For the element $\kappa_{7}$, we adopt the renamed one by the equation $2\kappa_{7}=\bar{\nu}_{7}\nu^{2}_{15}$ in  [7, (15.5)]. For a group $H$ and $\alpha\in H$, we denote by $\sharp\alpha$ the order of $\alpha$.

2 Some cohomotopy groups of $M^{n}$

We have a cofiber sequence

[TABLE]

where $i^{\prime}_{n}:S^{n-1}\hookrightarrow M^{n}$ is the inclusion and $p^{\prime}_{n}:M^{n}\to S^{n}$ is the collapsing map.

By using the exact sequence induced from (2.1) and the fact that $\pi^{n}_{n+1}=\{\eta_{n}\}\cong\mathbb{Z}_{2}$ for $n\geq 3$, we obtain

[TABLE]

By using the exact sequence induced from (2.1) and the fact that $\pi^{n}_{n+6}=\{\nu^{2}_{n}\}\cong\mathbb{Z}_{2}$ for $n\geq 5$, we obtain

[TABLE]

Let $\bar{\eta}_{n}\in[M^{n+2},S^{n}]$ and $\tilde{\eta}_{n}\in\pi_{n+2}(M^{n+1})$ for $n\geq 3$ be an extension and a coextension of $\eta_{n}$, respectively. We note that

[TABLE]

We have

[TABLE]

and

[TABLE]

Since the indeterminacy of $\{\eta_{n},2\iota_{n+1},p^{\prime}_{n+1}\}$ is $\eta_{n}\circ[M^{n+2},S^{n+1}]+\pi^{n}_{n+2}\circ p^{\prime}_{n+2}=\{\eta^{2}_{n}p^{\prime}_{n+2}\}$ by [11, Proposition 5.3] and (2.2), we obtain

[TABLE]

By the definition of $\nu^{\prime}$ [11, p.40] and [11, Proposition 1.7, Lemma 5.4] that

[TABLE]

Let $\iota_{M^{n}}$ be the homotopy class of the identity map of $M^{n}$. In view of [12, p. 307, Corollary], it holds

[TABLE]

Note that $[M^{n},M^{n}]=\mathbb{Z}_{4}\{\iota_{M^{n}}\}$ for $n\geq 4$.

By the relations $\nu_{6}\eta_{9}=0$ [11, (5.9)] and $\{\nu_{6},\eta_{9},2\iota_{10}\}_{1}=P(\iota_{13})+2\{P(\iota_{13})\}$  [11, Lemma 5.10], we have

[TABLE]

that is, we have the relations

[TABLE]

Let $\overline{\nu^{2}}_{5}\in[M^{12},S^{5}]$ be an extension of $\nu^{2}_{5}$. We set $\overline{\nu^{2}}_{n}=E^{n-5}\overline{\nu^{2}}_{5}$ for $n\geq 5$. Then, we have the following :

Lemma 2.1

• (1)

$\sharp(\overline{\nu^{2}}_{n})=2$* for $n\geq 5$*

• (2)

$\{\eta_{n},\nu_{n+1},\overline{\eta}_{n+4}\}_{1}=\overline{\nu^{2}}_{n}$* for $n\geq 7$*

• (3)

$\eta_{n}\overline{\nu^{2}}_{n+1}=\varepsilon_{n}p^{\prime}_{n+8}$* for $n\geq 5$.*

Proof. (1) By (2.7), we have $2\overline{\nu^{2}}_{5}=\overline{\nu^{2}}_{5}\circ 2\iota_{M^{12}}=\overline{\nu^{2}}_{5}\circ i_{12}\eta_{11}p^{\prime}_{12}=\nu^{2}_{5}\eta_{11}p^{\prime}_{12}=0$.

(2) Recall the relations $\eta_{5}\nu_{6}=0$ [11, (5.9)] and $\nu^{2}_{6}=\{\eta_{6},\nu_{7},\eta_{10}\}$ [11, Lemma 5.12]. We have

[TABLE]

Since $\pi^{7}_{12}=0$, we have $\eta_{6}\circ\pi^{7}_{12}=0$. From the fact that $\pi^{6}_{11}=\{P(\iota_{13})\}\cong\mathbb{Z}$ [11, Proposition 5.9] , we have $\pi^{6}_{11}\circ\eta_{11}=\{P(\eta_{13})\}$. Since $E:\pi^{6}_{12}\to\pi^{7}_{13}$ is an isomorphism [11, Proposition 5.11], we obtain $P(\eta_{13})=0$. Hence we have $\overline{\nu^{2}}_{6}\circ i^{\prime}_{13}=\{\eta_{6},\nu_{7},\bar{\eta}_{10}\}\circ i^{\prime}_{13}$. By using $\pi^{6}_{13}=\mathbb{Z}_{4}\{\sigma^{\prime\prime}\}$ [11, Proposition 5.15], we obtain that $\{\eta_{6},\nu_{7},\overline{\eta}_{10}\}\equiv\overline{\nu^{2}}_{6}\bmod\sigma^{{}^{\prime\prime}}p^{\prime}_{13}$ . This implies $\{\eta_{7},\nu_{10},\bar{\eta}_{13}\}_{1}=E\{\eta_{6},\nu_{9},\bar{\eta}_{12}\}\equiv E\overline{\nu^{2}}_{6}=\overline{\nu^{2}}_{7}\bmod E(\sigma^{\prime\prime}p^{\prime}_{13})=2\sigma^{\prime}\circ p^{\prime}_{14}=0$ because $E\sigma^{\prime\prime}=2\sigma^{\prime}$ [11, Lemma 5.14].

(3) We know that $\pi^{5}_{13}=\{\varepsilon_{5}\}\cong\mathbb{Z}_{2}$ and so, $\varepsilon_{5}=\{\eta_{5},\nu^{2}_{6},2\iota_{12}\}$  [11, (6.1) Theorem 7.1]. We take $\overline{\nu^{2}}_{6}\in\{\nu^{2}_{6},2\iota_{12},p^{\prime}_{12}\}$. So, from the relation $\eta_{5}\sigma^{\prime\prime}=0$ [11, (7.4)], we see that $\eta_{5}\overline{\nu^{2}}_{6}\in\eta_{5}\circ\{\nu^{2}_{6},2\iota_{12},p^{\prime}_{12}\}=\{\eta_{5},\nu^{2}_{6},2\iota_{12}\}\circ p^{\prime}_{13}=\varepsilon_{5}p^{\prime}_{13}\bmod\eta_{5}\circ\pi^{6}_{13}\circ p^{\prime}_{13}=\{\eta_{5}\sigma^{\prime\prime}p^{\prime}_{13}\}=0$. This leads to the assertion.   $\Box$

Lemma 2.2

$\varepsilon_{5}\overline{\nu^{2}}_{13}=\bar{\varepsilon}_{5}p^{\prime}_{20}$**

Proof. Since $\bar{\varepsilon}_{5}=\{\varepsilon_{5},\nu^{2}_{13},2\iota_{19}\}_{1}$ by [9, III-Proposition 2.3 (5); the second] and $\varepsilon_{n}\sigma_{n+8}=0$ for $n\geq 3$ [11, Lemma 10.7], Lemma 2.1 induces

[TABLE]

$\Box$

Recall the relation $\nu_{5}\zeta_{8}=\sigma^{\prime\prime\prime}\sigma_{12}$ [4, Lemma 2.3]. Let $\overline{\sigma^{\prime\prime\prime}\sigma_{12}}$ be an extension of $\sigma^{\prime\prime\prime}\sigma_{12}$.

Lemma 2.3

$[M^{22},S^{7}]=\{E^{2}\overline{\sigma^{\prime\prime\prime}\sigma_{12}},\bar{\nu}_{7}\overline{\nu^{2}}_{15},\rho^{\prime\prime}p^{\prime}_{22},\sigma^{\prime}\bar{\nu}_{14}p^{\prime}_{22},\sigma^{\prime}\varepsilon_{14}p^{\prime}_{22},\bar{\varepsilon}_{7}p^{\prime}_{22}\}\cong(\mathbb{Z}_{2})^{6}$.

Proof. Recall that $\pi^{7}_{21}=\{\sigma^{\prime}\sigma_{14},\kappa_{7}\}\cong\mathbb{Z}_{8}\oplus\mathbb{Z}_{4}$ [11, Theorem 10.3] and $\pi^{7}_{22}=\{\rho^{\prime\prime},\sigma^{\prime}\bar{\nu}_{14},\sigma^{\prime}\varepsilon_{14},\bar{\varepsilon}_{7}\}\cong\mathbb{Z}_{8}\oplus(\mathbb{Z}_{2})^{3}$ [11, Theorem 10.5], where $4(\sigma^{\prime}\sigma_{14})=E^{2}(\sigma^{\prime\prime\prime}\sigma_{12})$ by [11, Lemma 5.14] and $2\kappa_{7}=\bar{\nu}_{7}\nu^{2}_{15}$. So, by using (2.1) for $n=21$, we have

[TABLE]

We know from [11, (7.1), (7.4)] :

[TABLE]

By (2.7), (2.9) and [11, (7,4)], we obtain $\;2\overline{\sigma^{\prime\prime\prime}\sigma_{12}}=\sigma^{\prime\prime\prime}\sigma_{12}\circ\eta_{19}p^{\prime}_{20}=(\sigma^{\prime\prime\prime}\eta_{12})\sigma_{13}p^{\prime}_{20}$ $=0$. By Lemma 2.1 (1), we have $2(\bar{\nu}_{7}\overline{\nu^{2}}_{15})=0$. This completes the proof.   $\Box$

3 Toda brackets for extensions over $M^{24}$ of $P(E\theta)$ and $\nu_{6}\kappa_{9}$

First of all, we recall  [11, (7.21) ; the third]

[TABLE]

Hence, we have the well-defined element $\theta\in\{\sigma_{12},\nu_{19},\eta_{22}\}_{1}$  [11, pp.73–74], whose indeterminacy is $\sigma_{12}\circ E\pi^{18}_{23}+\pi^{12}_{23}\circ\eta_{23}=\{P(\eta_{25}),\zeta_{12}\eta_{23}\}$.

We know $\zeta_{6}\eta_{17}=8P(\sigma_{13})$ and $\zeta_{n}\eta_{n+2}=0$ for $n\geq 7$  [10, Proposition 2.2 (6)]. We also know $P(\eta_{25})=E\theta^{\prime}$ [11, (7.30) ; the second]. Therefore, $\{\sigma_{12},\nu_{19},\eta_{22}\}_{1}=\theta+\{E\theta^{\prime}\}$.

Since $\sigma_{12}\circ\pi^{19}_{24}=0$ and $\sigma_{12}\circ E^{13}\pi^{6}_{11}=0$, we have $\{\sigma_{12},\nu_{19},\eta_{22}\}_{1}=\{\sigma_{12},\nu_{19},\eta_{22}\}_{n}$ $(0\leq n\leq 13)$. So, we take $\{\sigma_{12},\nu_{19},\eta_{22}\}_{3}=\theta+\{E\theta^{\prime}\}$, and $E\theta=\{\sigma_{13},\nu_{20},\eta_{23}\}_{4}$ from the fact $\pi^{13}_{24}\circ\eta_{24}=\{\zeta_{13}\eta_{24}\}=0$.

We recall the following equation from [10, Proposition 2.2 (2)]

[TABLE]

We have that $P(E\theta)\in\{P(\sigma_{13}),\nu_{18},\eta_{21}\}_{2}\,\bmod\,\pi^{6}_{22}\circ\eta_{22}$. Note that $\pi^{6}_{22}=\{\zeta_{6}^{\prime},\mu_{6}\sigma_{15},\eta_{6}\bar{\varepsilon}_{7}\}\cong\mathbb{Z}_{8}\oplus(\mathbb{Z}_{2})^{2}$ [11, Theorem 12.6]. We know $\zeta^{\prime}\eta_{22}=0$ [10, Proposition 2.13 (5)] and $\eta_{6}\bar{\varepsilon}_{7}\eta_{22}=(\nu_{6}\sigma_{9}\nu^{2}_{16})\circ\eta_{22}=0$ by [11, Lemma 12.10]. By (2.9) and (3.2), we have the relation $\mu_{5}\sigma_{14}\eta_{21}=\mu_{5}\eta_{14}\sigma_{15}=\eta_{5}\mu_{6}\sigma_{15}$. Hence, we obtain $\pi^{6}_{22}\circ\eta_{22}=\{\eta_{6}\mu_{7}\sigma_{16}\}$ and $P(E\theta)\in\{P(\sigma_{13}),\nu_{18},\eta_{21}\}_{2}\bmod\eta_{6}\mu_{7}\sigma_{16}$.

We consider the Toda bracket $\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}_{2}\subset[M^{24},S^{6}]$. By the relation (3.1), we have $P(\sigma_{13})\circ\nu_{18}=P(\sigma_{13}\nu_{21})=0$. So, by (2.3), we obtain

[TABLE]

By (2.8) and (3.3), we have $\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}=\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}_{n}$ for $(0\leq n\leq 11)$.

Let $\overline{P(E\theta)}\in[M^{24},S^{6}]$ be an extension of $P(E\theta)$. Then, we show :

Lemma 3.1

$\overline{P(E\theta)}\in\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}_{2}\bmod\pi^{6}_{22}\circ\bar{\eta}_{22}+\pi^{6}_{24}\circ p^{\prime}_{24}$**

Proof. Notice that $\overline{P(E\theta)}$ is a representative of the Toda bracket $\{P(E\theta),2\iota_{23},p^{\prime}_{23}\}_{2}$. We use the Jacobi identity [11, Proposition 1.5] :

[TABLE]

By (2.5), we rewrite the identity:

[TABLE]

From the fact that $\{\nu_{18},\eta_{21},2\iota_{22}\}\subset\pi^{18}_{23}=0$ and (3.3), the second term is $\{P(\sigma_{13}),0,p^{\prime}_{23}\}=P(\sigma_{13})\circ[M^{24},S^{18}]+\pi^{6}_{24}\circ p^{\prime}_{24}=\pi^{6}_{24}\circ p^{\prime}_{24}$. The indeterminacy of the first term is

[TABLE]

By (2.2), we have $\{P(\sigma_{13}),\nu_{18},\eta_{21}\}\circ[M^{24},S^{23}]\subset\{P(\sigma_{13}),\nu_{18},\eta_{21}\}\circ\{\eta_{23}p^{\prime}_{24}\}$ $\subset$ $\pi^{6}_{24}\circ p^{\prime}_{24}$. So, the indeterminacy is $\pi_{24}^{6}\circ p^{\prime}_{24}$

The indeterminacy of the third term is

[TABLE]

by (3.3). Thus the assertion is proved.   $\Box$

Recall that $\eta_{7}\sigma_{8}=(\sigma^{\prime}\eta_{14}+\bar{\nu}_{7}+\varepsilon_{7})$ by [11, (7.4)]. We have the relation

[TABLE]

Let us take an extension $\overline{\nu_{6}\kappa_{9}}\in\{\nu_{6}\kappa_{9},2\iota_{23},p^{\prime}_{23}\}_{2}$ over $M^{24}$ of $\nu_{6}\kappa_{9}$.

By the relations $\varepsilon_{3}\nu_{11}=\nu^{\prime}\bar{\nu}_{6}=0$ [11, (7.12)], $E^{2}\nu^{\prime}=2\nu_{5}$ and $(\ast)$, we have

[TABLE]

Therefore the Toda bracket $\{\nu_{6},\eta_{9},\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}\}_{2}$ is well-defined.

We have $\overline{\nu_{6}\kappa_{9}}\in\{\nu_{6}\kappa_{9},2\iota_{23},p^{\prime}_{23}\}_{2}\subset\{\nu_{6},2\kappa_{9},p^{\prime}_{23}\}_{2}\supset\{\nu_{6},\eta_{9},\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}\}_{2}$ since $2\kappa_{9}=\bar{\nu}_{9}\nu^{2}_{17}=\eta_{9}\sigma_{10}\nu^{2}_{17}$. The indeterminacy of the second bracket is $\nu_{6}\circ E^{2}[M^{22},S^{7}]+\pi^{6}_{24}\circ p^{\prime}_{24}$. This implies

Lemma 3.2

$\overline{\nu_{6}\kappa_{9}}\in\{\nu_{6},\eta_{9},\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}\}_{2}\bmod\nu_{6}\circ E^{2}[M^{22},S^{7}]+\pi^{6}_{24}\circ p^{\prime}_{24}$.

By [11, Lemma 12.10], we have

[TABLE]

We know from [11, (7.5)]

[TABLE]

We recall the equations from [11, Lemma 6.3] that

[TABLE]

Next, we show :

Lemma 3.3

$\{2\iota_{7},\eta_{7},\sigma_{8}\nu^{2}_{15}p^{\prime}_{21}\}=\bar{\nu_{7}}\overline{\nu^{2}}_{15}$**

Proof. By $(\ast)$, we have $\{2\iota_{7},\eta_{7},\sigma_{8}\nu^{2}_{15}p^{\prime}_{21}\}$ $\subset$ $\{2\iota_{7},2\kappa_{7},p^{\prime}_{21}\}\supset\{\bar{\nu}_{7}\nu^{2}_{15},2\iota_{21},p^{\prime}_{21}\}\ni\bar{\nu}_{7}\overline{\nu^{2}}_{15}\bmod 2\iota_{7}\circ[M^{22},S^{7}]+\pi^{7}_{22}\circ p^{\prime}_{22}$. By [11, Theorem 10.5], $\pi^{7}_{22}\circ p^{\prime}_{22}=\{\rho^{\prime\prime},\sigma^{\prime}\bar{\nu}_{14},\sigma^{\prime}\varepsilon_{14},\bar{\varepsilon}_{7}\}\circ p^{\prime}_{22}$. By the fact that $S^{7}$ is an $H$-space and by Lemma 2.3, we have $2\iota_{7}\circ[M^{22},S^{7}]=2[M^{22},S^{7}]=0$. Hence, we obtain

[TABLE]

We set

[TABLE]

where $a,b,c,d\in\{0,1\}$. We compose $\tilde{\eta}_{21}$ to $(\ast\ast)$ on the right. We know the relations $\rho^{\prime\prime}\eta_{22}=\sigma^{\prime}\mu_{14}$ [10, Proposition 2.8 (3) ; the second], $\sigma^{\prime}\bar{\nu}_{7}\eta_{15}=\sigma^{\prime}\nu^{3}_{14}=\nu_{7}\sigma_{10}\nu^{2}_{17}=\eta_{7}\bar{\varepsilon}_{8}=\bar{\varepsilon}_{7}\eta_{22}$ by (3.6), (3.4) and [11, (7.19), Lemma 12.11]. Moreover, we know $\sigma^{\prime}\varepsilon_{14}\eta_{22}=\sigma^{\prime}\eta_{14}\varepsilon_{15}=E\zeta^{\prime}$ by [11, (12.4)] and (3.5). This implies $(a\rho^{\prime\prime}+b\sigma^{\prime}\bar{\nu}_{14}+c\sigma^{\prime}\varepsilon_{14},+d\bar{\varepsilon}_{7})\circ p^{\prime}_{22}\circ\tilde{\eta}_{21}=a\sigma^{\prime}\mu_{14}+(b+d)\eta_{7}\bar{\varepsilon}_{8}+cE\zeta^{\prime}$.

On the other hand, we have $\{2\iota_{7},\eta_{7},\sigma_{8}\nu^{2}_{15}p^{\prime}_{21}\}\circ\tilde{\eta}_{21}=2\iota_{7}\circ\{\eta_{7},\sigma_{8}\nu^{2}_{15}p^{\prime}_{21},\tilde{\eta}_{20}\}\subset 2\iota_{7}\circ\pi^{7}_{23}=2\pi^{7}_{23}=0$ [11, Theorem 12.6].

We know that $\varepsilon_{5}\in\{\nu_{5}^{2},2\iota_{11},\eta_{11}\}$ [11, (7.6)]. So, by the relation $\bar{\nu}_{6}\varepsilon_{14}=0$ [10, Proposition 2.8 (2)] and $\bar{\nu}_{6}\sigma_{14}=0$ [11, Lemma 10.7], we have

[TABLE]

So, $(\ast\ast)$ becomes $0=a\sigma^{\prime}\mu_{14}+(b+d)\eta_{7}\bar{\varepsilon}_{8}+cE\zeta^{\prime}$ and hence, we have $a=c=(b+d)=0$ by seeing $\pi^{7}_{23}=\{\sigma^{\prime}\mu_{14},E\zeta^{\prime},\mu_{7}\sigma_{16},\eta_{7}\bar{\varepsilon}_{8}\}$ [11, Theorem 12.6]. Therefore we obtain

[TABLE]

We notice that $\bar{\nu}_{6}^{2}=0$ [10, Proposition 2.8 (2)] and $\eta_{6}\sigma^{\prime}=4\bar{\nu}_{6}$ by [11, (7.4)]. So, by composing $\eta_{6}$ on the left for the above equation, we have $\eta_{6}\circ\{2\iota_{7},\eta_{7},\sigma_{8}\nu^{2}_{15}p^{\prime}_{21}\}=\eta_{6}\bar{\nu}_{7}\overline{\nu^{2}}_{15}+d\eta_{6}\bar{\varepsilon}_{7}p^{\prime}_{22}$.

By (2.6), we have

[TABLE]

By the relations $\bar{\nu}_{6}\varepsilon_{14}=0$ from [10, Proposition 2.8 (2)], (3.6) and Lemma 2.1(3), we see that

[TABLE]

Since $\eta_{6}\bar{\varepsilon}_{7}$ generates a direct summand $\mathbb{Z}_{2}$ in $\pi^{6}_{22}$ by [11, Theorem 12.6], we have $\eta_{6}\bar{\varepsilon}_{7}p^{\prime}_{22}\neq 0$. This implies $d=0$ and completes the proof.   $\Box$

4 Proof of the main theorem

We recall the equation from [8, (4.4)]

[TABLE]

We show :

Lemma 4.1

• (1)

${p_{U}}_{*}\Delta(\pi^{6}_{22}\circ\bar{\eta}_{22})=0$* and ${p_{U}}_{*}\Delta(\pi^{6}_{24}\circ p^{\prime}_{24})=0$.*

• (2)

$p_{U\ast}\Delta(\overline{P(E\theta)})=p_{U\ast}\Delta\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}_{2}$**

Proof. By Lemma 3.1, it suffices to show the assertion (1). By [6, Proposition 6.3], we have $\Delta(\zeta^{\prime})=\Delta(\mu_{6}\sigma_{15})={i_{U}}_{*}\mu^{\prime}\sigma_{14}$, $\Delta(\eta_{6}\bar{\mu}_{7})={i_{U}}_{*}\nu^{\prime}\bar{\mu}_{6}$, $\Delta(\eta_{6}\bar{\varepsilon}_{7})=0$, $\Delta(P(E\theta)\circ\eta_{23})=0$ and $\Delta(\zeta_{6}\sigma_{17})=[2\iota_{5}]\zeta_{5}\sigma_{16}$. So, we have ${p_{U}}_{*}\Delta(\pi^{6}_{22}\circ\bar{\eta}_{22})={p_{U}}_{*}\Delta(\pi^{6}_{22})\bar{\eta}_{21}=0$ and ${p_{U}}_{*}\Delta(\pi^{6}_{24}\circ p^{\prime}_{24})={p_{U}}_{*}\Delta(\pi^{6}_{24})\circ p^{\prime}_{23}=\{2\iota_{5}\circ\zeta_{5}\sigma_{16}p^{\prime}_{23}\}=\{2(\zeta_{5}\sigma_{16})p^{\prime}_{23}\}=0$ by (4.1).   $\Box$

Next, we show :

Lemma 4.2

$p_{U\ast}\Delta(\overline{\nu_{6}\kappa_{9}})=p_{U\ast}\Delta\{\nu_{6},\eta_{9},\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}\}_{2}$.

Proof. By Lemma 3.2 and Lemma 4.1 (1), it suffices to show that $p_{U\ast}\Delta(\nu_{6}\circ E^{2}[M^{22},S^{7}])=0$. Since $\Delta(\nu_{6}\circ E^{2}[M^{22},S^{7}])\subset\Delta\nu_{6}\circ E^{2}[M^{22},S^{7}]$ and the relation $\Delta\nu_{6}=[2\iota_{5}]\nu_{5}$ [6, Proposition 6.2], we have $p_{U\ast}\Delta(\nu_{6}\circ E^{2}[M^{22},S^{7}])=2\nu_{5}\circ E[M^{22},S^{7}]=0$.   $\Box$

We show the following.

Lemma 4.3

$p_{U\ast}\Delta(P(E\theta))=2\nu_{5}\kappa_{8}$**

Proof. By [6, p.164], we have

[TABLE]

where $a\in\{0,1\}$. Since $\Delta(\nu_{6}\kappa_{9})=[2\iota_{5}]\nu_{5}\kappa_{8}\in\pi_{5}(SU(3):2)$ [6, Proposition 6.2], we have $p_{U\ast}\Delta(\nu_{6}\kappa_{9})=2\nu_{5}\kappa_{8}$. For the connecting map $\Delta_{U}:\pi^{5}_{22}\to\pi_{21}(SU(2):2)=\pi^{3}_{21}$ induced from the fibration $S^{3}=SU(2)\to SU(3)\to S^{5}$, we have $\Delta_{U}\circ p_{U\ast}=0$. By [8, Proposition 3.2(i)], we have $\Delta_{U}(\eta_{5})=\eta^{2}_{3}$. By the relation $2\mu^{\prime}=\eta^{2}_{3}\mu_{5}$ [11, (7.7)], we have

[TABLE]

By [11, Theorem 12.8], $\sharp(\mu^{\prime}\sigma_{14})=4$, so that we have $a=0$.   $\Box$

Let $r_{n}:SU(n)\to SO(2n)$ be the canonical inclusion. By [1, Corollary 5.3, 5.4 Theorem 5.5], the fibrations $G_{2}{\buildrel p_{G}\over{\longrightarrow}}G_{2}/SU(3)=S^{6}$ and $SO(7)\to SO(7)/SO(6)=S^{6}$ give a commutative diagram

[TABLE]

We denote by $\Delta_{R}$ the connecting map obtained by the fibration $SO(7)\to SO(7)/SO(6)=S^{6}$.

Lemma 4.4

$\Delta(P(E\theta))=\Delta(\nu_{6}\kappa_{9})$.

Proof. In [6], Mimura obtained the consequence of this lemma, because the extension problem is based on it. However, we rewrite it as this lemma since we could not find its proof.

We know the relation $i_{U\ast}\bar{\mu}^{\prime}=\Delta(\bar{\mu}_{6})$ [6, Proposition 6.3]. So, by the relation $\Delta(P(E\theta))\equiv[2\iota_{5}]\nu_{5}\kappa_{8}\bmod i_{U\ast}\bar{\mu}^{\prime}=\Delta(\bar{\mu}_{6})$  [6, p.164, 3-rd line], we can set $\Delta(P(E\theta))+a\Delta(\bar{\mu}_{6})=\Delta(\nu_{6}\kappa_{9})$, where $a\in\{0,1\}$. By applying $r_{3}:U(3)\to R_{6}$, we have $\Delta_{R}(P(E\theta))+a\Delta_{R}(\bar{\mu}_{6})=\Delta_{R}(\nu_{6}\kappa_{9})$. We know that $\Delta_{R}(P(E\theta))=2[\nu_{5}]\kappa_{8}=\Delta_{R}(\nu_{6}\kappa_{9})$ in $R^{6}_{22}$ by the proof of [3, Lemma 3.5]. This implies $a\Delta_{R}(\bar{\mu}_{6})=0$. Since $\Delta_{R}(\bar{\mu}_{6})=[\iota_{3}]_{6}\bar{\mu}^{\prime}=8[\nu_{4}\rho^{\prime\prime}]_{6}\neq 0$, we have $a=0$.   $\Box$

Lemma 4.5

$\Delta(\overline{P(E\theta)}+\overline{\nu_{6}\kappa_{9}})=a[\nu_{5}\bar{\varepsilon}_{8}]p^{\prime}_{23}+b[2\iota_{5}]\zeta_{5}\sigma_{16}p^{\prime}_{23}$, where $a\in\{0,1\}$ and $b\in\{0,1,2,3\}$

Proof. By virtue of Lemma 4.4, we have the equation : $\Delta(\overline{P(E\theta)}+\overline{\nu_{6}\kappa_{9}})\circ i^{\prime}_{23}=\Delta(P(E\theta)+\nu_{6}\kappa_{9})=0$. From this equation and (2.1) for $n=23$, we have $\Delta(\overline{P(E\theta)}+\overline{\nu_{6}\kappa_{9}})\in\pi_{23}(SU(3):2)\circ p^{\prime}_{23}$.

By [8, Theorem 4.1], $\pi_{23}(SU(3):2)=\{[2\iota_{5}]\zeta_{5}\sigma_{16},[\nu_{5}\bar{\varepsilon}_{8}]\}\cong\mathbb{Z}_{4}\oplus\mathbb{Z}_{2}$. This implies the desired relation.   $\Box$

We show :

Lemma 4.6

${p_{U}}_{*}\Delta(\overline{P(E\theta)})={p_{U}}_{*}\Delta\{P(\sigma_{13}),\nu_{17},\bar{\eta}_{20}\}_{2}=\nu_{5}\bar{\nu}_{8}\overline{\nu^{2}}_{16}+\nu_{5}\bar{\varepsilon}_{8}p^{\prime}_{23}$.

Proof. From the relation $\Delta P(\iota_{13})=[\nu_{5}\eta^{2}_{8}]$  [6, Corollary 5.3] and Lemma 4.1(2), we have ${p_{U}}_{*}\Delta(\overline{P(E\theta)})={p_{U}}_{*}\Delta\{P(\sigma_{13}),\nu_{18},\bar{\eta}_{21}\}_{2}\subset\{\nu_{5}\eta^{2}_{8}\sigma_{10},\nu_{17},\bar{\eta}_{20}\}_{1}$. By Lemma 2.1 and Lemma 2.2, we obtain

[TABLE]

Here, we use the relations $[M^{22},S^{16}]=\{\nu^{2}_{16}p^{\prime}_{22}\}$  (2.3) and $\eta_{5}\bar{\varepsilon}_{6}\bar{\eta}_{21}=\nu_{5}\sigma_{8}\nu^{2}_{15}\bar{\eta}_{21}=0$ by using [11, Lemma 12.10] and (2.8).

This leads to the relation.

[TABLE]

where $x\in\{0,1\}$.

Applying $i^{\prime}_{23}$ to this equality on the right, then we have ${p_{U}}_{*}\Delta(P(E\theta))=2\nu_{5}\kappa_{8}+x\mu_{5}\sigma_{14}\eta_{21}=2\nu_{5}\kappa_{8}+x\eta_{5}\mu_{6}\sigma_{15}$. By Lemma 4.3, we obtain ${p_{U}}_{*}\Delta(\overline{P(E\theta)})=2\nu_{5}\kappa_{8}$. This implies $x=0$ and completes the proof.   $\Box$

Lemma 4.7

$p_{U\ast}\Delta(\overline{\nu_{6}\kappa_{9}})=p_{U\ast}\Delta\{\nu_{6},\eta_{9},\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}\}_{2}=\nu_{5}\bar{\nu}_{8}\overline{\nu^{2}}_{16}$.

Proof. By Lemma 4.2 and Lemma 3.3, we have

[TABLE]

Since $\pi^{5}_{10}\circ\sigma_{10}\nu^{2}_{17}=\{\nu_{5}\eta_{8}^{2}\sigma_{10}\nu^{2}_{17}\}=\{\nu_{5}(\bar{\nu}_{6}+\varepsilon_{8})\eta_{16}\nu^{2}_{17}\}=0$, we have the equation $p_{U\ast}\Delta(\overline{\nu_{6}\kappa_{9}})=\nu_{5}\bar{\nu}_{8}\overline{\nu^{2}}_{16}\bmod 2\nu_{5}\circ E[M^{22},S^{7}]+\pi^{5}_{10}\circ\sigma_{10}\nu^{2}_{17}p^{\prime}_{23}=0$.   $\Box$

Finally, we show :

Theorem 4.8

[TABLE]

where $b\in\{0,1,2,3\}$.

Proof. Since the order of $\nu_{5}\bar{\nu}_{8}\overline{\nu^{2}}_{16}$ is $2$, in Lemma 4.6 and Lemma 4.7, we have

[TABLE]

We know $p_{U\ast}([2\iota_{5}]\zeta_{5}\sigma_{16})p^{\prime}_{23}=2(\zeta_{5}\sigma_{16})p^{\prime}_{23}=0$ by (4.1). So, Lemma 4.5 induces

[TABLE]

where $a\in\{0,1\}$ and $b\in\{0,1,2,3\}$.

By (4.4) and (4.5), we have $\nu_{5}\bar{\varepsilon}_{8}p^{\prime}_{23}=a\nu_{5}\bar{\varepsilon}_{8}p^{\prime}_{23}$. Since $\nu_{5}\bar{\varepsilon}_{8}$ generates the direct summand $\mathbb{Z}_{2}$ of $\pi^{5}_{23}$, $\nu_{5}\bar{\varepsilon}_{8}p^{\prime}_{23}$ is not zero in $[M^{23},S^{5}]$. This implies $a=1$.   $\Box$

Thus, by applying $i_{G\ast}$ to (4.3), we have $i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]p^{\prime}_{23}=0$. By (2.1), the relation $i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]p^{\prime}_{23}=0$ implies $i_{G\ast}[\nu_{5}\bar{\varepsilon}_{8}]$ is contained in the image of the map $2\iota_{23}^{\ast}:\pi_{23}(G_{2}:2)\to\pi_{23}(G_{2}:2)$. This and Proposition 1.2 complete the proof of Theorem 1.3.

As an application of Proposition 1.2 and Theorem 1.3, we show

Proposition 4.9

$\pi_{23}(V_{7,2}:2)\cong(\mathbb{Z}_{4})^{2}\oplus(\mathbb{Z}_{2})^{2}$.

To show this proposition, we will need some relations. lemma.

Lemma 4.10

• (1)

$\mu^{\prime}\in\{\eta_{3},2\iota_{4},\mu_{4}\}_{1}\bmod 2\mu^{\prime}$**

• (2)

$\bar{\mu}^{\prime}\in\{\mu^{\prime},4\iota_{14},4\sigma_{14}\}_{1}\bmod\nu^{\prime}\mu_{6}\sigma_{15}$**

Proof. (1) The indeterminacy is $\eta_{3}\circ E\pi^{3}_{13}+\pi^{3}_{5}\circ\mu_{5}$. Here, $\pi^{3}_{5}\circ\mu_{5}=\{\eta^{2}_{3}\mu_{5}\}=\{2\mu^{\prime}\}$ and $\eta_{3}\circ E\pi^{3}_{13}=\eta_{3}\circ E\{\eta_{3}\mu_{4},\varepsilon^{\prime}\}=\{2\mu^{\prime},\eta_{3}E\varepsilon^{\prime}\}$. By [11, p.68], we have $-(\eta_{3}\circ E\varepsilon^{\prime})=\eta_{3}\circ(-E\varepsilon^{\prime})\in\eta_{3}\circ(-E\{\nu^{\prime},2\nu_{6},\nu_{9}\})\subset\eta_{3}\circ\{E\nu^{\prime},2\nu_{7},\nu_{10}\}_{1}=\{\eta_{3},E\nu^{\prime},2\nu_{7}\}_{1}\circ\nu_{11}\supset\{\eta_{3},E\nu^{\prime},\nu_{7}\}_{1}\circ 2\iota_{11}\circ\nu_{11}\ni\varepsilon_{3}\circ 2\iota_{11}\circ\nu_{11}=0$. The indeterminacy is $\eta_{3}\circ\pi^{4}_{11}\circ\nu_{11}=0$. Therefore, we have $\eta_{3}\circ E\varepsilon^{\prime}=0$, and the assertion is proved.

(2) The indeterminacy is $\mu^{\prime}\circ E\pi^{13}_{21}+\pi^{3}_{15}\circ 4\sigma_{15}$. Since $\pi^{3}_{15}\cong(\mathbb{Z}_{2})^{2}$ [11, Theorem 7.6], we have $\pi^{3}_{15}\circ 4\sigma_{15}=0$.

We have $\mu^{\prime}\circ E\pi^{13}_{21}=\{\mu^{\prime}\eta_{14}\sigma_{15},\mu^{\prime}\bar{\nu}_{14}\}$. Since $\mu^{\prime}\nu_{14}=0$ [10, Proposition 2.4 (1)] and $\bar{\nu}_{14}=\{\nu_{14},\eta_{17},\nu_{18}\}$ [11, Lemma 6.2], we have $\mu^{\prime}\bar{\nu}_{14}=\mu^{\prime}\circ\{\nu_{14},\eta_{17},\nu_{18}\}=\{\mu^{\prime},\nu_{14},\eta_{17}\}\circ\nu_{19}\subset\pi^{3}_{19}\circ\nu_{19}$. By [11, Theorem 12.6], $\pi^{3}_{19}=\{\mu_{3}\sigma_{12},\eta_{3}\bar{\varepsilon}_{4}\}$. We also have $\eta_{3}\bar{\varepsilon}_{4}=\bar{\varepsilon}_{3}\eta_{18}$ by (3.4), $\eta_{18}\nu_{21}=0$ and (3.1). It follows that $\pi^{3}_{19}\circ\nu_{19}=0$, and hence $\mu^{\prime}\bar{\nu}_{14}=0$.

By [10, Proposition 2.2 (4) ; the second], we have $\mu^{\prime}\eta_{14}=\nu^{\prime}\mu_{6}$, so that $\mu^{\prime}\eta_{14}\sigma_{15}=\nu^{\prime}\mu_{6}\sigma_{15}$.   $\Box$

Proof of Proposition 4.9

By using the fibering $S^{3}{\buildrel i\over{\longrightarrow}}G_{2}{\buildrel p\over{\longrightarrow}}V_{7,2}$, we have an exact sequnece:

[TABLE]

Here, $i_{\ast}\pi_{23}(G_{2}:2)=\{i_{\ast}\langle P(E\theta)+\nu_{6}\kappa_{9}\rangle,i_{\ast}\langle\eta^{2}_{6}\rangle\mu_{8}\}$.

Recall that the structures of the homotopy groups $\pi_{23}^{3}=\{\nu^{\prime}\bar{\mu}_{6},\nu^{\prime}\eta_{6}\mu_{7}\sigma_{16}\}\cong(\mathbb{Z}_{2})^{2}$ from [7, p.45] and $\pi^{3}_{22}=\{\bar{\mu}^{\prime},\nu^{\prime}\mu_{6}\sigma_{15}\}\cong\mathbb{Z}_{4}\oplus\mathbb{Z}_{2}$ from [11, Theorem 12.9], where we have $\bar{\mu}^{\prime}\in\{\mu^{\prime},4\iota_{14},4\sigma_{14}\}_{1}$ and $2\bar{\mu}^{\prime}=\eta^{2}_{3}\bar{\mu}_{5}$ by [11, p.137, Lemma 12.4, Theorem 12.9].

Let us denote the connecting map induced by the fibering $G_{2}/S^{3}=V_{7,2}$ by $\Delta:\pi_{n}(V_{7,2}:2)\to\pi^{3}_{n-1}$. By [6, p.132], we obtain

[TABLE]

Therefore, $\Delta:\pi_{5}(V_{7,2}:2)\to\pi^{3}_{4}$ is an isomorphism. Since $V_{7,2}$ is a $5$-sphere bundle over $S^{6}$, we have a cell structure : $V_{7,2}=M^{6}\cup e^{11}$. So, we obtain $\pi_{5}(V_{7,2}:2)\cong\pi_{5}(M^{6}:2)=\{i^{\prime}_{6}\}\cong\mathbb{Z}_{2}$ and $\Delta(i^{\prime\prime}i^{\prime}_{6})=\eta_{3}$, where $i^{\prime\prime}:M^{6}\to V_{7,2}$ is the inclusion. We also have an isomorphism $\Delta:\pi_{7}(V_{7,2}:2)\to\pi^{3}_{6}$. Since $\pi_{7}(V_{7,2}:2)\cong\pi_{7}(M^{6}:2)=\{\tilde{\eta}_{5}\}\cong\mathbb{Z}_{4}$, we have $\Delta(i^{\prime\prime}\tilde{\eta}_{5})=\nu^{\prime}$.

Since $(V_{7,2},M^{6})$ is $10$-connected and $[M^{6},M^{6}]=\mathbb{Z}_{4}\{\iota_{M^{6}}\}$, $i^{\prime\prime}_{\ast}:[M^{6},M^{6}]\to[M^{6},V_{7,2}]$ is an isomorphism. By (4.6), we have $[M^{n},G_{2}]=0$ for $n=5,6$. So, the fibering $G_{2}/S^{3}=V_{7,2}$ induces an isomorphism $\Delta:[M^{6},V_{7,2}]\to[M^{5},S^{3}]$. This and (2.4) imply

[TABLE]

Let $\tilde{\mu}_{3}\in\pi_{13}(M^{4})$ be a coextension of $\mu_{3}$. We set $\tilde{\mu}_{n}=E^{n-3}\tilde{\mu}_{3}\in\pi_{n+10}(M^{n+1})$ for $n\geq 3$. By (2.7), we obtain

[TABLE]

By Lemma 4.10 (1), we have

[TABLE]

By (2.7) and (4.8), we have $4i^{\prime\prime}\tilde{\mu}_{5}=0$. This implies $\sharp(i^{\prime\prime}\tilde{\mu}_{5})=4$. Let $\mathfrak{m}$ be a representative of the Toda bracket $\{i^{\prime\prime}\tilde{\mu}_{5},4\iota_{15},4\sigma_{22}\}_{2}$. Then, we have $4\mathfrak{m}\in 4\{i^{\prime\prime}\tilde{\mu}_{5},4\iota_{15},4\sigma_{22}\}_{2}=\{i^{\prime\prime}\tilde{\mu}_{5},4\iota_{15},4\sigma_{22}\}_{2}\circ 4\iota_{30}=i^{\prime\prime}\tilde{\mu}_{5}\circ\{4\iota_{15},4\sigma_{15},4\iota_{22}\}_{2}\subset i^{\prime\prime}\tilde{\mu}_{5}\circ\{2\iota_{15},0,2\iota_{22}\}_{1}=i^{\prime\prime}\tilde{\mu}_{5}\circ 2\pi^{15}_{23}=0$. We recall from [11, p.137] : By (4.9) and Lemma 4.10 , we have $\Delta\{i^{\prime\prime}\tilde{\mu}_{5},4\iota_{15},4\sigma_{15}\}_{2}\subset\{\pm\mu^{\prime},4\iota_{14},4\sigma_{14}\}_{1}=\{(\pm\iota_{3})\circ\mu^{\prime},4\iota_{14},4\sigma_{21}\}_{1}\ni(\pm\iota_{3})\circ\bar{\mu}^{\prime}=\pm\bar{\mu}^{\prime}\bmod\mu^{\prime}\circ E\pi^{13}_{21}+\pi^{3}_{15}\circ 4\sigma_{15}=\{\nu^{\prime}\mu_{6}\sigma_{15}\}$. Here, $(\pm\iota_{3})\circ\bar{\mu}^{\prime}=\pm\bar{\mu}^{\prime}$ since $S^{3}$ is an $H$-space. Therefore, $\Delta(\mathfrak{m})\equiv\pm\bar{\mu}^{\prime}\bmod\nu^{\prime}\mu_{6}\sigma_{15}$.

Hence, we have the split exact sequence:

[TABLE]

so that the splitting $\pi_{23}(V_{7,2}:2)=i_{\ast}\pi_{23}(G_{2}:2)\oplus\mathbb{Z}_{4}\{\mathfrak{m}\}\oplus\mathbb{Z}_{2}\{i^{\prime\prime}\tilde{\eta}_{5}\mu_{7}\sigma_{16}\}\cong(\mathbb{Z}_{4})^{2}\oplus(\mathbb{Z}_{2})^{2}$ gives the isomorphism in the proposition.