Consider the vector field $\textbf{B} = \frac{1}{\rho^2}\textbf{e}_\rho$ on $\mathbb{R}^3 - \{0\}$ where $(\rho, \theta, \phi)$ are the usual spherical coordinates and $\textbf{e}_\rho = \frac{\partial \rho}{|\partial \rho|}$ is the unit tangent vector in the $\rho$ direction. $\textbf{B}$ does not have a global vector potential on $\mathbb{R}^3 - \{0\}$ by Stokes' theorem, but does have local vector potentials on $\mathbb{R}^3 - \{z_+\}$ and $\mathbb{R}^3 - \{z_-\}$ (the complement of the positive $z$-axis and negative $z$-axis, respectively). If you pull these local vector potentials back to the $2$-sphere $S^2$ and invoke the usual identification between vector fields and $1$-forms then you get $\mu_+ = (1 - \cos \phi)d\theta$ and $\mu_- = (1 + \cos \phi)d\theta$ defined on $S_+^2$ and $S_-^2$.

Now, the Hopf bundle $S^3 \to S^2$ trivializes over $S_+^2$ and $S_-^2$, yielding sections $S_+^2 \to S^3$ and $S_-^2 \to S^3$. One can check by direct calculation that if you pull back the $1$-form $$\omega = -x_2 dx_1 + x_1 dx_2 - x_4 dx_3 + x_3 dx_4$$ on $\mathbb{R}^4$ to $S_+^2$ and $S_-^2$ along the composition of the sections above with the inclusion $S^3 \to \mathbb{R}^4$ then you get $\mu_+$ and $\mu_-$. Thus the Hopf bundle is home to a sort of global vector potential for the magnetic monopole.

My question is: what is the geometric meaning of $\omega$? Its pullback to $S^3$ has a nice interpretation: it is a natural connection $1$-form on the Hopf bundle (maybe up to a factor of $i$). But what about $\omega$ itself?

  • $\begingroup$ If you think of $\mathbb R^4 - \{0\}$ as $\mathbb C^2 - \{0\}$ then it seems likely (I haven't done any computations) that $\omega$ is a connection form on the principal $\mathbb C^\times$ bundle $\mathbb C^2 - \{0\} \to \mathbb C P^1 \simeq S^2$. $\endgroup$ – Eric O. Korman Apr 19 '14 at 2:39

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