Hopf map fiber bundle help I need help on constructing the fiber bundles $S^3 \rightarrow S^{7} \rightarrow \mathbb{H}P^1$ I heard that you just use the same idea as the hopf map to $\mathbb{C}P^n$.
So I guess you say $S^{7}$ are the vectors in $\mathbb{H}^{2}$. But, then what does would that mean. Also, do I then take a quotient $q: S^{7} \rightarrow \mathbb{H}P^1$.
Also, I'm a bit confused on showoing $q^-1(U_a) \cong U_a \times S^3$. I assume I have to prove this and then give the manifolds for it to prove that it's a fiber bundle. 
Anyone know a decent paper that explain this Hopf map?
 A: Thinking of $\mathbb{H}P^1$ as the one point compactification of $\mathbb{R}^4$, the Hopf map is given by $p \colon S^7\rightarrow \mathbb{R}^4\cup \infty$, $(v, w)\mapsto vw^{-1}$, if $w\neq 0$, $(v, w) \mapsto \infty$, if $w = 0$.
What is a good candidate for a trivialization on $U :=\mathbb{R}^4\cup\infty-\{\infty\}$? Well, remember that you can multiply two quaternions, so you might try $\phi\colon U\times S^3 \rightarrow p^{-1}(U)$, $(v, z)\mapsto (vz, z)$.
However, the element on the right hand side does not have  norm equal to 1 anymore. But $(v, z)\mapsto \sqrt{|v|^2+1}^{-1} ( vz, z)$ does the job.
You have to check that this a homeomorphism. To see that $p\circ \phi  = pr_1$, note that $w^{-1} = w^*/|w|^2$, where $w^*$ is the conjugate quaternion.
The trivialization over $\mathbb{R}^4\cup \infty- \{0\}$ is similar.
A: One way of solving this is to define normed vector spaces over the field $K$ which is one of  the real numbers, complex numbers, or quaternions. Because $K$ might then be a skew-field, i.e. in the last case, it is better to take right vector spaces, i.e. scalar multiplication on the right, to make the matrix theory work well. In this way  you can get the cell structure of projective spaces, all three cases at once, and all the Hopf maps. See Section 5.3 of my book Topology and Groupoids. 
