Question

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

Answer 1

let take the metric the following :

ds2=V(dx+4m(1−cosθ)dϕ)2+1V(dr+r2dθ2+r2sin2θdϕ2),ds2=V(dx+4m(1−cos⁡θ)dϕ)2+1V(dr+r2dθ2+r2sin2⁡θdϕ2),

where

V=1+4mr.V=1+4mr.

That is the Taub-NUT instanton.

Answer 2

This exhibits the Taub-NUT metric as the metric on the total space of a circle bundle over R2∖{0}

Let

ϑ1=V√(dx+A)        
ϑ2=1V√drϑ1=V(dx+A)        
ϑ2=1Vdr

So that we may write the metric as:⏬⏬

ds2=ϑ21+ϑ22+ϑ23+ϑ24ds2=ϑ12+ϑ22+ϑ32+ϑ42