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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?

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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.

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There is a general theorem of differential topology that says that every vector field tangent to a (topological) sphere must be zero in at least one point.

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  • $\begingroup$ How do you know that the manifold is a topological sphere? $\endgroup$
    – auxsvr
    Commented Apr 24, 2014 at 16:24
  • $\begingroup$ @auxsvr: you don't know that the manifold (and the whole manifold will almost always not be ) is a topological sphere, but part of the assumptions of asymptotic flatness is that you're doing analysis on a "sphere at infinity" $\endgroup$
    – Zo the Relativist
    Commented Apr 24, 2014 at 16:36
  • $\begingroup$ 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? $\endgroup$
    – auxsvr
    Commented Apr 24, 2014 at 17:12
  • $\begingroup$ @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. $\endgroup$
    – Zo the Relativist
    Commented Apr 24, 2014 at 17:16
  • $\begingroup$ Oh, I mean to write about the spatial part, compactification of $\mathbb{R}^3$. Now, we're in agreement, right? $\endgroup$
    – auxsvr
    Commented Apr 24, 2014 at 18:37

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