Quaternionic projective space and homotopy groups of spheres Let $\mathbb{H}$ be the $\mathbb{C}$ with formal basis {1,j} and define multiplication by distributively extending $j^2 =-1$ and $ji =-ij$:
$$(a+bj)(u+vj):= au-b\overline{v} +(av+b \overline{u})j.$$ It is easy to see that $(\mathbb{H},+,.)$ is a skew field with $$||a+bj||= \sqrt{(a\overline{a}+b\overline{b})}, ||pq||= ||p||||q||.$$
I have the following three questions:


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*How can I define a free action of $\mathbb{S}^3$ on $\mathbb{S}^{4n+3}$ and obtain a fibration $$q:\mathbb{S}^{4n+3} \rightarrow \mathbb{S}^{4n+3}/\mathbb{S}^3 =:  \mathbb{H}P^n.$$ The base space is quaternion projective $n$-space. 

*How is $\mathbb{H}P^1$  homeomorphic to $\mathbb{S}^4$. This looks obvious from the quotient defined above but I cant seem to provide an argument/proof.

*Lastly, from the Long exact sequence, how can one show that for $k \ge1$, $$\pi_k\mathbb{S}^4 \cong \pi_k\mathbb{S}^7 \oplus \pi_{k-1}\mathbb{S}^3.$$
Thanks for you kind help and explanations.
 A: Suppose $\mathbb{K}$ is one of $\mathbb{R},\mathbb{C},\mathbb{H}$ and $d=\dim_{\mathbb{R}}(\mathbb{K})$ (so one of $1,2,4$).
The elements of $\mathbb{K}^n$ with unit norm form the sphere $S^{dn-1}$.
$\mathbb{K}^n$ is a right vector space over $\mathbb{K}$ of dimension $n$, and a real inner product space of dimension $dn$ with inner product $\langle x,y\rangle=\overline{x_1}y_1+\cdots+\overline{x_n}y_n$ (using the physicist convention of linearity in the second argument, so they can write things like $x^{\dagger}y$ or $\langle x\mid y\rangle$). I have scalars multiplied from the right so that matrices may be applied from the left and yield $\mathbb{K}$-linear transformations.
To form the projective space $\mathbb{KP}^{n-1}$, we put an equivalence relation $\sim$ on nonzero elements of $\mathbb{K}^n$ in which $x\sim x\lambda$ for all nonzero vectors $x\in\mathbb{K}^n\setminus0$ and nonzero scalars $\lambda\in\mathbb{K}^\times$, then create the quotient space $(\mathbb{K}^n\setminus0)/\!\sim$. The equivalence class of $(x_1,\cdots,x_n)$ is denoted $[x_1:\cdots:x_n]$.
Every unit vector of $\mathbb{K}^n$ defines a $\mathbb{K}$-line, i.e. a $1$-dimensional subspace as a right vector space over the (possibly skew) field $\mathbb{K}$. In other words, we have a map $S^{dn-1}\to\mathbb{KP}^{n-1}$. As every $\mathbb{K}$-line contains a unit vector, this map is surjective.
A fiber of this map will be all unit vectors in a given $\mathbb{K}$-line. Since every $\mathbb{K}$-line looks like $\mathbb{K}^1$, all of the fibers will be $S^{d-1}$. Thus, we have a fiber bundle $S^{d-1}\to S^{dn-1}\to\mathbb{KP}^{n-1}$.
Moreover, the group $S^{d-1}\subset\mathbb{K}^\times$ acts regularly on each fiber (the $\mathbb{K}$-lines are equivalence classes under scalar multiplication, and the intersection of a $\mathbb{K}$-line with the sphere of unit vectors must be a copy of $S^{d-1}\subseteq\mathbb{K}^1$), making $S^{dn-1}\to\mathbb{KP}^{n-1}$ a principal $S^{d-1}$-bundle.
The projective line $\mathbb{KP}^1$ can be identified with $S^d$. One way is as follows. Every vector $(x,y)$ can normalized to have the form $(xy^{-1},1)$, unless $y=0$ in which case it may be normalized  to have the form $(1,0)$. Thus, $\mathbb{KP}^1$ minus a point may be identified with $\mathbb{K}$ (in the first coordinate). If we extend $\mathbb{K}$ by adjoining the symbol $\infty$ to it, call the result $\widehat{\mathbb{K}}=\mathbb{K}\cup\{\infty\}$, topologize it as the one-point compactification of $\mathbb{K}$, then we can extend this to a homeomorphism $\mathbb{KP}^1\to\widehat{\mathbb{K}}$. In the case of $\mathbb{K}=\mathbb{C}$, this is called the Riemann sphere.
This is also the origin of Mobius transformations (aka linear fractional transformations). $\mathrm{GL}(2,\mathbb{K})$ acts on $\mathbb{K}^n$ and commutes with the right action by scalars in $\mathbb{K}^\times$, hence descends to an action on $\mathbb{KP}^1\simeq\widehat{\mathbb{K}}$. Since $[\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}][\begin{smallmatrix} x \\ y \end{smallmatrix}]=[\begin{smallmatrix} ax+by \\ cx+dy\end{smallmatrix}]\mapsto (ax+by)(cx+dy)^{-1}=(axy^{-1}+b)(cxy^{-1}+d)^{-1}$, the transported action of $\mathrm{GL}(2,\mathbb{K})$ on $\widehat{\mathbb{K}}$ is $[\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}]z=(az+b)(cz+d)^{-1}$.
To see $\widehat{\mathbb{K}}$ is a sphere, we may use stereographic projection. Given any real inner product space (which $\mathbb{K}$ is an example of), invoke coordinates so it looks like $\mathbb{R}^d$. Any line through the "north pole" $(0,\cdots,0,1)\in S^d\subset \mathbb{R}^{d+1}$ and a point $x\in\mathbb{R}^d$ (viewed as a subspace of $\mathbb{R}^{d+1}$ in the obvious way) intersects the sphere $S^d$ in a unique point, and conversely, which sets up a bijection between $\mathbb{R}^d$ and ($S^d$ minus the north pole), in which case $S^d$ is the one-point compactification of $\mathbb{R}^d$.
Here's another trick, which IIRC Baez attributes to Penrose. Given any $\mathbb{K}$-line $L$ in $\mathbb{K}^n$ (equivalently, point in $\mathbb{KP}^{n-1}$) there is an orthogonal projection onto $L$. This has the property that it acts as the identity on $L$ and as the zero map on $L^\perp$ (the orthogonal complement with respect to the real inner product). Concretely, if we represent $L=x\mathbb{K}$ with $x$ a unit vector, then $\langle x,-\rangle$ siphons out the "$x$-component" of vectors, so the projection is $x\langle x,-\rangle$, which may be represented as left-multiplication by the matrix $xx^{\dagger}$ (where $x$ is treated as a column vector and $x^{\dagger}$ denotes conjugate transpose). This is an element of $\mathfrak{h}_n(\mathbb{K})$ ($n\times n$ hermitian matrices over $\mathbb{K}$, i.e. matrices $A$ for which $A^{\dagger}=A$) with trace $\mathrm{tr}(xx^{\dagger})=|x|^2=1$.
In the case of $n=2$, you can subtract $\frac{1}{2}I_2$ from $xx^{\dagger}$ and you get an element of a sphere centered at the origin in the space of traceless $n\times n$ hermitian matrices (equipped with the Frobenius norm), and thus establish a bijection $\mathbb{KP}^1\to S^d$. I give more detail here, and relate it to stereographic projection and exceptional isomorphisms.
Given a fiber bundle $F\to E\to B$, there is a long homotopy exact sequence
$$ \cdots\to\pi_{k+1}(B)\to ~~~ \pi_k(F)\to \pi_k(E)\to \pi_k(B) ~~~ \to \pi_{k-1}(F)\to\cdots $$
If all inclusions $F\to E$ are null-homotopic, then $\pi_k(F)\to\pi_k(E)$ is the zero map, in which case the long homotopy exact sequence becomes a collection of short exact sequences
$$ 0\to \pi_k(E)\to \pi_k(B)\to\pi_{k-1}(F)\to0. $$
Moreover, we have projections $\pi_k(B)\to \pi_k(E)$, so a split exact sequence
$$ 0\to \pi_k(E)\rightleftarrows \pi_k(B)\to \pi_{k-1}(F)\to0 $$
which implies $\pi_k(B)\cong\pi_k(E)\oplus\pi_{k-1}(F)$.
In the case of $S^{d-1}\to S^{2d-1}\to S^d$ (with $n=2$ of our discussion above) this gives the desired result that $\pi_k(S^{2d-1})=\pi_k(S^{d-1})\oplus \pi_{k-1}(S^d)$. To see that $S^{d-1}\to S^{2d-1}$ is null-homotopic, notice it is certainly not surjective, so there is a point outside the image (of a fixed choice of inclusion) we can use for stereographic projection, getting an embedding $S^{d-1}\to \mathbb{R}^{2d-1}$ which can be contracted to the trivial based map by rescaling embeddings by $t$ with $0\le t\le 1$.
Note that by sufficiently changing the definition of projective space (using equivalent algebraic characterizations), we can define a projective line $\mathbb{OP}^1$ (and even projective plane $\mathbb{OP}^2$) over the octonions $\mathbb{O}$, in which case the above results still hold true.
