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

+ 4 like - 0 dislike
2066 views

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.

This post imported from StackExchange Physics at 2014-06-27 11:28 (UCT), posted by SE-user silvrfück
asked Jun 25, 2014 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]
Have you tried computing $\det g$ for $\theta = \pi$?

This post imported from StackExchange Physics at 2014-06-27 11:28 (UCT), posted by SE-user Robin Ekman
@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

This post imported from StackExchange Physics at 2014-06-27 11:28 (UCT), posted by SE-user silvrfück
@RobinEkman maybe you are interested in the edit I just made in the question

This post imported from StackExchange Physics at 2014-06-27 11:28 (UCT), posted by SE-user silvrfück

1 Answer

+ 4 like - 0 dislike

Rewrite the metric as follows:

$$ds^2 = V (dx + A)^2 + \frac1V (dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2)$$

This exhibits the Taub-NUT metric as the metric on the total space of a circle bundle over $\mathbb{R}^2\setminus\{0\}$.  Let

$$\vartheta_1 = \sqrt{V} (dx + A) ~~~~~~~~ \vartheta_2 = \frac1{\sqrt V} dr$$

$$\vartheta_3 = \frac1{\sqrt V} r d\theta ~~~~~~~~ \vartheta_4 = \frac1{\sqrt V} r \sin\theta d\phi$$

so that we may write the metric as 

$$ ds^2 = \vartheta_1^2 + \vartheta_2^2 + \vartheta_3^2 + \vartheta_4^2$$

and the volume form is

$$d\mathrm{vol} = \vartheta_1 \wedge \vartheta_2 \wedge \vartheta_3 \wedge \vartheta_4 = \frac{r^2\sin\theta}{V} dx \wedge dr \wedge d\theta \wedge d\phi$$

If $\theta = \pi$, then $\sin\theta = 0$ and hence $d\mathrm{vol}=0$, whence it is singular.

answered Jun 28, 2014 by José Figueroa-O'Farrill (2,135 points) [ revision history ]
edited Jun 28, 2014 by José Figueroa-O'Farrill

Good answer, but there seems to be a minor typo in the line where you rewrote the metric (\(\sin^\theta\) should be \(\sin\theta\)). Sorry for nitpicking though!

straight to the key point. Thanks!

Yes, sorry -- I left a '2' out :)  Should be corrected now.

There's something I don't see. Following your argument the volume form does also vanish at $\theta=0$ so this should also be a singilarity. Nonetheless, in the paper I linked it is explicitly stated that $\theta=0$ is not a singularity. Why is this so?

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