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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?

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Your spheres should have dimension $4n-1$, not $4n+1$. – Jason DeVito Apr 19 '12 at 18:32
@JasonDeVito It should be correct now I think. – simplicity Apr 19 '12 at 18:42
I'm not really understanding what you're confused about. Is it the definition of quaternions and $\mathbb{H}$? Is it how to make the projection map? Is it about why the projection map is a fiber bundle map? – Jason DeVito Apr 20 '12 at 0:16
@JasonDeVito I think I know how to make the projection map.You would $(q_1,q_2,...,q_n) \rightarrow (q_2 q_1^{-1},....,q_n q_1^{-1})$. But, then how do you show this is a fiber bundle. – simplicity Apr 20 '12 at 0:20
I see. I don't have time to give a full answer now, but if no one else fills anything in in the next day or so, I should be able to. In the mean time, the projection should take a point in $S^7$ to a point in $\mathbb{H}P^1$, i.e., a quaternionic subspace of $\mathbb{H}^2$. The usual idea is that the projection maps sends a point $p$ on the sphere to the quaternionic subspace $p\mathbb{H}$. In terms of quaternionic coordinates, you map $(q_1,q_2)\in S^7\subseteq \mathbb{H}^2$ to $[q_1:q_2]$, where you identify $[q_1:q_2]$ with $[q_1 q: q_2 q]$ for any unit quaternion $q$. – Jason DeVito Apr 20 '12 at 0:32
up vote 2 down vote accepted

You might find this pdf note useful (but spoiler alert if you don't want all the answers). I worked through all the computations for the three Hopf bundles $\mathbb{S}^{2k-1}\hookrightarrow\mathbb{S}^{4k-1}\to\mathbb{S}^{2k}$, $k=1,2,4$, as an algebraic topology assignment last year.

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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.

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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".

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