Fibered product of Hopf Fibrations

Question

I am wondering: what is the vector bundle associated to the Hopf-fibration $S^{3}\to S^{2}$?

What is the fibered product of two Hopf-fibrations?

Thanks.

Answer 1

$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Modulo orientation, the complex line bundle coming from the Hopf map $\pi:S^{3} \to S^{2}$ is the tautological line bundle $\mathcal{O}_{\Proj^{1}}(-1)$, whose total space is the blow-up of $\Cpx^{2}$ at the origin. (The same real $2$-plane bundle with the opposite orientation is the hyperplane bundle $\mathcal{O}_{\Proj^{1}}(1)$, whose total space is the complement of a point in $\Cpx\Proj^{2}$.)

The fibre product of the Hopf map with itself (i.e., the pullback of the Hopf bundle by the Hopf map) is the torus bundle over $S^{2}$ whose fibre over a point $p$ is $\pi^{-1}(p) \times \pi^{-1}(p)$. The total space is a circle bundle over $S^{3}$ in two ways, by projecting away the respective factors.

---

Identify the circle $S^{1}$ with the multiplicative group of unit complex numbers. If $\pi:P \to M$ is a principal $S^{1}$-bundle over a smooth manifold $M$ and $\rho$ is an effective linear representation of $S^{1}$ on $\Cpx$ (i.e., $\rho(e^{it})$ is multiplication by $e^{it}$, or by $e^{-it}$), then there is a complex line bundle $$ E = P \times_{S^{1}} \Cpx \to M, $$ in which the action of $S^{1}$ on $P$ is the principal action, and the action on $\Cpx$ is via $\rho$.

The usual norm on $\Cpx$ induces an Hermitian metric in $E$, for which the total space of $P$ is the unit circle bundle.

The $2$-sphere may be identified with $\Cpx\Proj^{1}$, the space of complex lines through the origin in $\Cpx^{2}$, and the $3$-sphere may be identified with the set of unit vectors in $\Cpx^{2}$, whereupon the Hopf bundle $P = S^{3} \to S^{2}$ is identified with the projection sending a unit vector $x$ to the (complex) line through $x$, viewed as an element of $S^{2}$.

The action $S^{1} \times S^{3} \to S^{3}$ defined by $(e^{it}, x) \mapsto e^{it}x$ endows $S^{3}$ with the structure of a principal circle bundle over $S^{2}$. If $\rho$ is an effective representation of the circle on $\Cpx$, the associated complex line bundle is obtained from the restriction of the circle action $$ S^{1} \times (\Cpx^{2} \times \Cpx) \to \Cpx^{2} \times \Cpx,\qquad (e^{it}, x, v) \mapsto (e^{it} x, e^{\pm it}v). \tag{1} $$

If $S^{1}$ acts by conjugate multiplication on $\Cpx$, the associated complex line bundle is obtained from the restriction of the circle action $$ S^{1} \times (\Cpx^{2} \times \Cpx) \to \Cpx^{2} \times \Cpx,\qquad (e^{it}, x, v) \mapsto (e^{it} x, e^{-it}v). $$ The product $(x, v) \mapsto vx$ is invariant under the action, so we may identify the second factor $\Cpx$ with the line through $x$, subject to the understanding that "each line has its own origin/zero vector". The latter is precisely the total space of the blow-up of $\Cpx^{2}$ at the origin. The induced complex line bundle is $\mathcal{O}_{\Proj^{1}}(-1)$.

The action of $S^{1}$ by ordinary multiplication also has a familiar interpretation. Equation (1) becomes the action of $S^{1}$ on $\Cpx^{3}$ by scalar multiplication, whose quotient is $\Cpx\Proj^{2}$. Since $x$ is point of $S^{3}$, i.e., a unit vector in $\Cpx^{2}$, the point $[0:0:1]$ is not in the image; that is, the total space of this line bundle is the complement of a point in $\Cpx\Proj^{2}$. The projective line $S^{2}$ sits "at infinity", and the bundle projection is radial projection from the origin. In algebraic geometry, this complex (in fact, holomorphic) line bundle is the "hyperplane bundle" $\mathcal{O}_{\Proj^{1}}(1)$.

The underlying (unoriented) real $2$-plane bundles are equivalent under fibrewise conjugation.