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:

$\!\|x\|=\sum _{t=1}^{n}|x_{t}|^{2}$

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 $k^{n+1}\setminus \{0\}$ 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.