Consider the following metric


where $$V=1+\frac{4m}{r}.$$

That is the Taub-NUT instanton. I have been told that it is singular at $\theta=\pi$ but I don't really see anything wrong with it. So, why is it singular at $\theta=\pi$?

EDIT:: I have just found in this paper that the metric is singular "since the $(1-\cos\theta)$ term in the metric means that a small loop about this axis does not shrink to zero at lenght at $\theta=\pi$" but this is still too obscure for me, any clarification would be much appreciated.

  • $\begingroup$ Have you tried computing $\det g$ for $\theta = \pi$? $\endgroup$ Jun 25, 2014 at 10:22
  • $\begingroup$ @RobinEkman just did, if I have't messed up the calculations it is $\frac{r^2(-256m^2V^2)}{V^2}$, nothin weird I'd say $\endgroup$
    – Yossarian
    Jun 25, 2014 at 10:46
  • $\begingroup$ @RobinEkman maybe you are interested in the edit I just made in the question $\endgroup$
    – Yossarian
    Jun 25, 2014 at 10:52
  • 1
    $\begingroup$ As nicely explained here, after rewriting the metric it can easily be shown that the volume form vanishes for $\theta = \pi$. This is the singularity. $\endgroup$
    – Dilaton
    Jun 28, 2014 at 6:45
  • 2
    $\begingroup$ @Dilaton admittedly overflow has been a great idea $\endgroup$
    – Yossarian
    Jun 28, 2014 at 15:41


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