associated disk bundle of a sphere bundle over $S^4$ We are given an $R^4$ bundle $\xi$ over $S^4$, whose total space is E, and we know that the associated sphere bundle is the Hopf fibration $S^3 \rightarrow S^7 \rightarrow S^4$. How can we show that the total space E' of the associated disk bundle is $HP^2-$ an open 8-cell?
I know that the Thom space $T$ of $\xi$ is obtained from $E'$ by gluing a cone over $\partial E'\approx S^7$ and cell decomposition of $T$ is the same as that of  $HP^2$. But I cannot combine these..
 A: Define an $\mathbb{H}$-bundle
$$\mathbb{H} \hookrightarrow \mathbb{H}P^2 - \{[0:0:1]\} \xrightarrow{~\pi~} \mathbb{H}P^2,$$
$$\pi([u:v:w]) = [u:v] \in \mathbb{H}P^1$$
as follows. $\mathbb{H}P^1$ decomposes as the union of two disks
$$D_1 = \{[u:1] : \|u\| \leq 1\},$$
$$D_2 = \{[1:v] : \|v\| \leq 1\}$$
glued along their common boundary via the map $[u:1] \mapsto [1:u^{-1}]$. Then take the local trivializations of our bundle to be
$$\psi_1: D_1 \times \mathbb{H} \longrightarrow \pi^{-1}(D_1),$$
$$\psi_1([u:1],w) = [u:1:w],$$
$$\psi_2: D_2 \times \mathbb{H} \longrightarrow \pi^{-1}(D_2),$$
$$\psi_2([1:v],w) = [1:v:w].$$
Making the identifications $\mathbb{H} \cong \mathbb{R}^4$ and $\mathbb{H}P^1 \cong S^4$, this defines an $\mathbb{R}^4$-bundle over $S^4$. The transition function on the equatorial $S^3$ is
$$\psi_2^{-1} \psi_1([u:1],w) = \psi_2^{-1}([u:1:w]) = \psi_2^{-1}([1:u^{-1}:u^{-1}w]) = ([1:v],vw).$$
Now from the total space $\mathbb{H}P^2 - \{[0:0:1]\}$ of our bundle, remove the open $8$-disk
$$D = \{[u:v:1] : \|u\|^2 + \|v\|^2 < 1\}$$
centered at $[0:0:1]$. For every $[u:v] \in \mathbb{H}P^1$, this restricts the fiber over $[u:v]$ to
$$\{[u:v:w] : \|w\|^2 \leq \|u\|^2 + \|v\|^2\} \cong D^4.$$
Therefore we have a disk bundle
$$D^4 \hookrightarrow \mathbb{H}P^2 - D \xrightarrow{~\pi~} S^4$$
with transition function
$$\psi_2^{-1} \psi_1([u:1], w) = ([1:v], vw) \tag{$\ast$}$$
on the equatorial $S^3$.
$S^3$-bundles over $S^4$ are classified by elements of
$$\pi_3(\mathrm{SO}(4)) \cong \mathbb{Z} \oplus \mathbb{Z}.$$
An explicit isomorphism identifies the pair $(h,j) \in \mathbb{Z} \oplus \mathbb{Z}$ with the $S^3$-bundle over $S^4$ with transition function
$$f_{hj}: S^3 \longrightarrow \mathrm{SO}(4),$$
$$f_{hj}(u) \cdot v = u^h v u^j$$
on the equatorial $S^3$, where here we consider $u \in S^3$ and $v \in \mathbb{R}^4$ as quaternions, i.e. the expression $u^h v u^j$ is understood as quaternion multiplication. From $(\ast)$, we see that the bundle
$$D^4 \hookrightarrow \mathbb{H}P^2 - D \xrightarrow{~\pi~} S^4$$
has the transition function $f_{10}$. You can easily check that the Hopf bundle
$$S^3 \hookrightarrow S^7 \longrightarrow S^4$$
has transition function $f_{10}$ as well, so by the above identification of $S^3$-bundles over $S^4$, its associated disk bundle must be isomorphic to
$$D^4 \hookrightarrow \mathbb{H}P^2 - D \xrightarrow{~\pi~} S^4.$$
Therefore the total space of the disk bundle associated to the Hopf bundle over $S^4$ is $\mathbb{H}P^2$ with an open $8$-disk removed.
