# 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.

• Have you tried computing $\det g$ for $\theta = \pi$? – Robin Ekman Jun 25 '14 at 10:22
• @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 – Yossarian Jun 25 '14 at 10:46
• @RobinEkman maybe you are interested in the edit I just made in the question – Yossarian Jun 25 '14 at 10:52
• 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. – Dilaton Jun 28 '14 at 6:45
• @Dilaton admittedly overflow has been a great idea – Yossarian Jun 28 '14 at 15:41