Asymptotic flatness implies existence of rotation axis

Suppose we have an asymptotically flat, globally hyperbolic spacetime $M$ endowed with two one-parameter isometry groups $\sigma_t$ and $\chi_{\phi}$ which commute (i.e. $\sigma_t \circ \chi_{\phi}= \chi_{\phi} \circ \sigma_t.$)

Assume moreover that the orbits of $\sigma_t$ are timelike curves generated by the Killing vector field $\xi^a.$ The orbits of $\chi_{\phi}$ are closed spacelike curves generated by the Killing field $\psi^a$.

In chapter 7.1, p 165 of Wald's GR text, he states that the asymptotic flatness of the spacetime implies that "there must be a rotation axis on which $\psi^a$ vanishes." Why must this be the case?

Asymptotically, the spacetime metric approaches the Minkowski metric, which has the property that for $\theta = 0$, $\theta=\pi$ (on the rotation axis), $\psi^a \psi_a =0$, hence, since $\psi^a$ is spacelike, it vanishes there. These two points are enough to satisfy the first hypothesis of theorem 7.1.1.
• I only know few things about topology; isn't $\mathbb{R}^4$ not homeomorphic to $S^3$? Wouldn't we need to perform first one-point compactification on $\mathbb{R}^4$, then apply the theorem? Apr 24 '14 at 17:12
• @auxsvr: read what I said above. When you try to look at the asymptotic flatness of, say Schwarzchild (which is homeomorphic to $\mathbb(R}^{4}$ with strips removed), you take a constant time slice, do a conformal transformation to bring infinity to a finite distance, and then do analysis on the sphere that is the boundary of this conformal manifold. Part of the assumptions of asymptotic flatness is that this boundary will be a sphere. Apr 24 '14 at 17:16
• Oh, I mean to write about the spatial part, compactification of $\mathbb{R}^3$. Now, we're in agreement, right? Apr 24 '14 at 18:37