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.

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