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.

This entry was posted in Algebra, Differential geometry, Group theory. Bookmark the permalink.

One Response to Hopf fibration double covers circle bundle of sphere

  1. ricanry says:

    The commutativity part is interesting.

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