## Hopf fibration double covers circle bundle of sphere

Two days ago, I gave a seminar talk on Chern‘s proof of the generalized Gauss-Bonnet theorem. Here I record the answer to a question asked by one of my colleague during the talk. Although not directly related to the proof of the generalized Gauss-Bonnet theorem, I think it’s quite interesting itself.

Let me first give some background before stating the question.

Roughly speaking, the idea of proof goes like this: Let’s assume ${n=2}$ for simplicity. Usually the curvature form (or more appropriately, the Pfaffian of the curvature form) is not exact, for otherwise the integral of the curvature form on a closed surface is zero. However, Chern observed that the pullback of the curvature form onto the unit sphere bundle ${SM}$ is exact, and a smooth non-degenerate vector field ${X}$ on ${M}$ naturally induces a diffeomorphism from ${M'}$ onto a submanifold ${\Sigma}$ in ${SM}$. Here ${M'}$ is the open subset of ${M}$ where ${X\ne 0}$. By pulling back the curvature form onto ${\Sigma}$ and applying the Stokes theorem, we can localize the curvature integral into a sum of line integrals on small loops around the singularities of ${X}$, which turn out to give the sum of the index of the vector field. Finally by the Poincare-Hopf theorem, this would give the Euler characteristic of ${M}$.

While introducing the concept of the unit sphere bundle, I was asked by one of my colleague whether the unit sphere bundle of ${\mathbb S^2}$ is the Hopf fibration. I didn’t know the answer at that time. But then I thought about it again and found that the answer is quite obviously no. However, I found it quite interesting that the Hopf fibration is actually the double cover of ${S(\mathbb S^2)}$. I think this is a good exercise in geometry and I am recording it here.

For a Riemannian manifold ${M^n}$, the unit sphere bundle ${SM}$ is defined to be

$\displaystyle \begin{array}{rl} \displaystyle SM:=\{(x, v): x\in M, v\in T_xM, |v|=1\} \end{array}$

with projection ${\pi_S: (x, v)\mapsto x}$. This is an ${\mathbb S^{n-1}}$-bundle over ${M}$. In particular, if ${M}$ is two-dimensional, then ${SM}$ is a circle bundle.

 Proposition 1 The circle bundle ${S(\mathbb S^2)}$ is diffeomorphic to ${SO(3)}$.

Proof: We regard ${\mathbb S^2\subset \mathbb R^3}$ is the standard unit sphere, and regard ${x\in \mathbb S^2}$ as a column vector. We can define ${\Phi: S(\mathbb S^2)\rightarrow SO(3)}$ by

$\displaystyle \begin{array}{rl} \displaystyle \Phi(x, v):=\left[x, v, x\times v\right], \end{array}$

where ${x\times v}$ is the cross product of ${x}$ and ${v}$ in ${\mathbb R^3}$. Then

$\displaystyle \begin{array}{rl} \displaystyle \Phi^{-1}\left(\left[v_1\;v_2\;v_3\right]\right)=(v_1, v_2). \end{array}$

Clearly this is a diffeomorphism. $\Box$

In particular, ${S(\mathbb S^2)\cong SO(3)}$ has fundamental group ${\mathbb Z/2\mathbb Z}$, with ${\mathrm{Spin}(3)=Sp(1)}$ as its double cover (cf. here).

Recall that the Hopf fibration is given by the quotient of the action ${\mathbb S^1=\{\alpha\in \mathbb C: |\alpha|=1\}}$ on ${\mathbb S^3=\{z=(z_1,z_2)\in\mathbb C^2: |z|=1\}}$, where the action is given by ${\alpha\cdot (z_1,z_2)=(\alpha z_1, \alpha z_2)}$. It is clear that the quotient space is ${\mathbb CP^1}$, which is diffeomorphic to ${\mathbb S^2}$. The Hopf fibration is defined to be this quotient: ${\pi: \mathbb S^3\rightarrow \mathbb CP^1\cong \mathbb S^2}$.

From the above discussion, it is clear that ${\pi_S: S(\mathbb S^2)\rightarrow \mathbb S^2}$ is not the Hopf fibration. In fact, as ${\mathbb S^3}$ is simply connected, it is not diffeomorphic to ${S(\mathbb S^2)}$. However, the Hopf fibration can be regarded as the double cover of ${\pi_S: S(\mathbb S^2)\rightarrow \mathbb S^2}$, in the sense that this diagram commutes

Here ${p}$ is the double covering map from ${Sp(1)=\mathrm{Spin}(3)}$ to ${SO(3)}$. To see this, first identify ${\mathbb S^3}$ with the compact symplectic group ${Sp(1)=\{q\in \mathbb H, |q|=1\}}$ where ${\mathbb H=\{a+bi+cj+dk: a, b, c, d\in \mathbb R\}}$ is the set of quaternions, where ${i^2=j^2=k^2=-1}$. We identify ${\mathbb R^3}$ with the space of purely imaginary quaternions ${\{bi+cj+dk\}}$. Then we define ${p(q)}$ by

$\displaystyle \begin{array}{rl} \displaystyle p(q): v\mapsto q^* v q, \textrm{ where } v \textrm{ is a purely imaginary quaternion}. \end{array}$

It is clear that ${p(q)\in SO(3)}$. Under this identification, then ${\pi_S(p(q))=q^* i q}$ as we identify ${i}$ with ${e_1}$. Let ${q=a+bi+cj+dk=z_0+z_1j\in Sp(1)}$, where ${z_i\in \mathbb C}$. Suppose ${z_0=a+bi}$, ${z_1=c+di}$. Then by a direct calculation (done by Mathematica here)

$\displaystyle \begin{array}{rl} \displaystyle \pi_S(p(q))=q^* i q =(a^2+b^2-c^2-d^2)i+2(bc-ad)j+2(ac+bd)k. \ \ \ \ \ (1)\end{array}$

On the other hand, for ${q=a+bi+cj+dk=z_0+z_1j\in Sp(1)}$, ${\pi}$ can be defined to be

$\displaystyle \begin{array}{rl} \displaystyle \pi(q)=\pi(z_0, z_1):= (|z_0|^2-|z_1|^2, 2\mathrm{Im}(z_0z_1^*), 2\mathrm{Re}(z_0z_1^*)). \end{array}$

Expanding the above, we have

$\displaystyle \begin{array}{rl} \displaystyle (|z_0|^2-|z_1|^2, 2z_0z_1^*)=(a^2+b^2-c^2-d^2, 2(bc-ad), 2(ac+bd)). \end{array}$

Comparing with (1), we have proved the commutativity.