Monday, 24 December 2018

differential geometry - Why is the Taub-NUT instanton singular at $theta=pi$?


Consider the following metric


$$ds^2=V(dx+4m(1-\cos\theta)d\phi)^2+\frac{1}{V}(dr+r^2d\theta^2+r^2\sin^2\theta{}d\phi^2),$$


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.




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