Quaternionic clutching bundles ${\xi }_{h,j}$ over $S^4$

Identify the oriented fibre ${\mathbb{R}}^4$ with the quaternions $\mathbb{H}$ in the ordered basis $(1,i,j,k)$, so that $N(a)=1$ defines the unit sphere $S^3\subseteq \mathbb{H}$ (The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$, Euclidean spheres and closed balls as subspaces of ${\mathbb{R}}^n$). For a unit quaternion $a$ and an integer $m$, let $a^m$ denote the integer power in the group $\mathbb{H}\setminus \{0\}$ ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$); on $S^3$ this is $a^m$ for $m\ge 0$ and $(\bar{a}{)}^{-m}$ for $m<0$.

Write $S^4=D_{+}^4{\cup }_{S^3}{D}_{-}^4$ for the union of the two closed four-disks glued along their common boundary sphere. We fix the following orientations. The disk $D_{+}^4$ and the disk $D_{-}^4$ carry the standard orientation of ${\mathbb{R}}^4$ in the basis $(1,i,j,k)$; the equator $S^3=\mathrm{\partial }{D}_{+}^4$ carries the induced boundary orientation; the fibre ${\mathbb{R}}^4$ carries the orientation of the ordered basis $(1,i,j,k)$; and $S^4$ is oriented so that the standard coordinate orientation on $D_{-}^4$ agrees with its manifold orientation, while that on $D_{+}^4$ is opposite. Thus the common coordinate sphere $a\in S^3$, oriented as $\mathrm{\partial }{D}_{+}^4$ in its standard coordinates, is also the positively oriented coordinate boundary of $D_{-}^4$. The calibration lemma proves that this convention gives Euler number $+1$ to both basic multiplication maps. Let $u\in H^4(S^4;\mathbb{Z})$ be the resulting positive generator.

For integers $h,j$, let $g_{h,j}:S^3\to G{L}_4(\mathbb{R})$ be the clutching map $g_{h,j}(a)v=a^{h}v{a}^j,a\in S^3,v\in \mathbb{H},$ so that $g_{h,j}(a)$ is the composite of the $\mathbb{R}$-linear maps of left multiplication by $a^h$ and right multiplication by $a^j$. Then ${\xi }_{h,j}$ is the oriented real rank-four bundle over $S^4$ obtained by the clutching construction from $g_{h,j}$, with the upper-to-lower identification $(a,v{)}_{+}\sim (a,g_{h,j}(a)v{)}_{-},(a,v)\in S^3\times \mathbb{H},$ as in Clutching construction for bundles over a suspension.

Well-definedness of the clutching data. The multiplication of $\mathbb{H}$ is given by a polynomial formula in the coordinates, so it is smooth in each variable and, for fixed $a$, the maps $L_a(v)=av$ and $R_a(v)=va$ are $\mathbb{R}$-linear endomorphisms of $\mathbb{H}={\mathbb{R}}^4$ (The quaternions $\mathbb{H}$: real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1,i,j,k$, $\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$). Writing $a=(a_0,a_1,a_2,a_3)$ and $v=(x_0,x_1,x_2,x_3)$, the product formula gives $L_{a}v=M(a)v$ with $M(a)=\left(\begin{array}{cccc}{a}_0& -a_1& -a_2& -a_3\\ a_1& a_0& -a_3& a_2\\ a_2& a_3& a_0& -a_1\\ a_3& -a_2& a_1& a_0\end{array}\right)\mathrm{.}$ Expanding the sixteen entries of $M(a{)}^{\mathsf{T}}M(a)$ shows that its rows are pairwise orthogonal and each has squared length $N(a)=a_0^2+a_1^2+a_2^2+a_3^2$, so $M(a{)}^{\mathsf{T}}M(a)=N(a)I$; this is the coordinate form of the classical four-square identity and gives $N(av)=N(a)N(v)\text{for all }a,v\in \mathbb{H}\mathrm{.}$ The same expansion applied to the matrix of right multiplication $v\mapsto va$, whose columns are $a,ia,ja,ka$, gives $N(va)=N(v)N(a)$ as well. Consequently, for $N(a)=1$ and all $m$, $L_{a^m}$ and $R_{a^m}$ are norm-preserving, hence orthogonal. Their determinants are $1$: the maps $a\mapsto \det L_a$ and $a\mapsto \det R_a$ are continuous on the path-connected sphere $S^3$ (For $n\ge 2$, the sphere $S^{n-1}$ is path-connected and connected, Every path-connected space is connected, and every path component lies inside a component) with values in $\{\pm 1\}$, and at $a=1$ both are the identity; hence $L_a,R_a\in SO(4)$ for every unit $a$.

Consequences for the clutching construction. For fixed $(h,j)$ the map $g_{h,j}:S^3\to SO(4)$ is continuous, being the restriction to $S^3$ of a polynomial map; equivalently, on $S^3$ every negative power is the polynomial map $a\mapsto {\bar{a}}^{\mathrm{\mid }m\mathrm{\mid }}$, because $a\bar{a}=1$ there ($\mathbb{H}$ is a division ring that is not commutative, hence not a field: $q^{-1}=\bar{q}\mathrm{/}N(q)$ for $q\mathrm{\ne }0$, while $ij=k$ and $ji=-k$). Thus $g_{h,j}$ is a continuous map from the compact based space $S^3$ into $G{L}_4(\mathbb{R})$ taking values in $SO(4)$, and the clutching construction applies to produce the rank-four bundle ${\xi }_{h,j}\to S^4$ with the stated upper-to-lower transition. The calibration of the two basic maps $g_{1,0}$ and $g_{0,1}$ to degree $+1$ is the content of the following local lemma; the bundle orientation uses that calibration and the fibre orientation $(1,i,j,k)$.