Fiber bundle of sphere over projective space

General version

Statement of general version

Suppose $k=\mathbb{R},\mathbb{C},\mathbb{H}$, i.e., $k$ is either the real numbers, or the complex numbers, or the Hamiltonian quaternions. Let $|\cdot |$ denote the absolute value/modulus operation in $k$. For $x=(x_1,x_2,\dots ,x_n)$ an element of $k^n$, we define:

$\left|x\right|=\sum _{t=1}^n|x_t{|}^2$

Now define the sphere:

$S^n(k)=\{x\in k^{n+1}\mid |x|=1\}$

with the subspace topology from the topology on $k^{n+1}$ arising from the product topology on $k^{n+1}$ from the usual Euclidean topology on $k$.

Note that $S^0(k)$ is a group, because it is the kernel of the modulus homomorphism from $k\setminus \{0\}$ to the multiplicative group of nonzero reals.

Define the projective space:

${\mathbb{P}}^n(k)=(k^{n+1}\setminus \{0\})/(k\setminus \{0\})$

where the quotient is via the diagonal left multiplication action. We put the quotient topology from the subspace topology on $k^{n+1}\setminus \{0\}$ arising from the product topology on $k^{n+1}$.

There is a fiber bundle $S^n(k)\to {\mathbb{P}}^n(k)$ with fiber $S^0(k)$. The map composes the inclusion of $S^n(k)in<math>k^{n+1}\setminus \{0\}$ with the quotient map to ${\mathbb{P}}^n(k)$.

Interpretation in the three special cases

Interpretation for arbitrary $n$:

$k$	$k^{n+1}$ becomes ...	$S^n(k)$ becomes ...	$S^0(k)$ becomes	Conclusion about fiber bundle
$\mathbb{R}$	${\mathbb{R}}^{n+1}$	$S^n$	$S^0$	$S^n\to {\mathbb{P}}^n(\mathbb{R})$ with fiber $S^0$. In other words, $S^n$ is a covering space of ${\mathbb{P}}^n(\mathbb{R})$, or more precisely a double cover. Since $S^n$ is simply connected, ${\mathbb{P}}^n(\mathbb{R})$ has fundamental group $\mathbb{Z}/2\mathbb{Z}$.
$\mathbb{C}$	${\mathbb{R}}^{2n+2}$	$S^{2n+1}$	$S^1$ -- the circle	$S^{2n+1}\to {\mathbb{P}}^n(\mathbb{C})$ with fiber $S^1$.
$\mathbb{H}$	${\mathbb{R}}^{4n+4}$	$S^{4n+3}$	$S^3$ -- the 3-sphere	Failed to parse (unknown function "\nathbb"): {\displaystyle S^{4n + 3} \to \nathbb{P}^n(\mathbb{H})} with fiber $S^3$.

Interpretation for $n=1$: In this case, ${\mathbb{P}}^n(k)$ itself becomes a sphere. We get some very special fiber bundles:

$k$	${\mathbb{P}}^1(k)$ is the sphere ...	$n=1$ gives the fiber bundle of spheres ...
$\mathbb{R}$	$S^1$, i.e., the circle	$S^1\to S^1$ with fiber $S^0$, i.e., the circle as a double cover of itself.
$\mathbb{C}$	$S^2$, i.e., the 2-sphere	$S^3\to S^2$ with fiber $S^1$. This map is termed the [{Hopf fibration]].
$\mathbb{H}$	$S^4$, i.e., the 4-sphere	$S^7\to S^4$ with fiber $S^3$.

In fact, these are the only fibrations where the base space, total space, and fiber space are all spheres.