I was reading the following paper


about the Taub-NUT metric, and I don't understand the origin of equation (2.55). Here is the (more or less self-contained) description of the issue:

the authors have constructed an orthonormal tetrad $(l,n,m,\overline{m})$, in the coordinate system $(x^{1},r,x^{3},x^{4})$. At some point the vector field $m$ has the form

$m=\overline{\rho}\hskip 1pt(Y_{1}\: , \: 0 \: , \: Y_{2} \: , \: iY_{2})$

where $\rho$ is a function of $r$, and $Y_{2}=Y_{2}(x^{3},x^{4})$ is a real function. The coordinate freedom still available is the coordinate transformation of the form


where $\zeta:=x^{3}+ix^{4}$.

QUESTION: the authors say, "it is possible to choose such a coordinate transformation $g$ so that $Y_{2}=\frac{1}{\sqrt{2}}+\frac{M}{2\sqrt{2}}((x^{3})^{2}+(x^{4})^{2})$", where $M$ is some constant. How do they know that? How could I see that? In general, what kind of calculation do I have to do for $Y_{2}$ to see what kind of $g$ I could pick?

Thank you for any hints

  • $\begingroup$ Pick a second function $Y_1(x^3,x^4)$ such that $(Y_2,Y_1)$ form local coordinates on the $(x^3,x^4)$-plane. Then, pretend that $Y_2 = \frac{1}{2} + \frac{M}{2\sqrt{2}} r^2$ and $Y_1 = \theta$ in "polar" coordinates $(r,\theta)$. Then obtain "cartesian" coordinates by the usual formula $(y^3,y^4) = (r\cos\theta, r\sin\theta)$, so that $Y_2 = \frac{1}{\sqrt{2}}+\frac{M}{2\sqrt{2}}((y^{3})^{2}+(y^{4})^{2})$. $\endgroup$ Apr 15, 2016 at 14:11
  • $\begingroup$ @IgorKhavkine could you please explain your first sentence? How do $(Y_{2},Y_{1})$ form local coordinates? thank you $\endgroup$
    – GregVoit
    Apr 15, 2016 at 17:49
  • $\begingroup$ @IgorKhavkine I also don't understand when you say "pretend $Y_{2}=...$", since that's exactly what I'm trying to show. How do I know it can have such a form? I'd be very grateful for clarifications $\endgroup$
    – GregVoit
    Apr 15, 2016 at 17:52
  • $\begingroup$ @GregVoit: can you edit your question to include a non-expiring link? Your current link doesn't go anywhere and we don't know what paper you are linking to. $\endgroup$ May 5, 2016 at 19:24

1 Answer 1


I will add here some more details to expand my comment. Any two functions $Y_1(x^3,x^4)$ and $Y_2(x^3,x^4)$ give local coordinates on any open domain of the $(x^3,x^4)$-plane where their Jacobian determinant is non-vanishing, $\frac{\partial(Y_1,Y_2)}{\partial(x^3,x^4)}$. Moreover, if you already have the function $Y_2$ given to you, you can always complete it by another function $Y_1$ to a local coordinate system, at least on some open subset that contains no critical points of $Y_2$.

Next, let $r = \sqrt{\frac{2\sqrt{2}}{M} Y_2 - \frac{2}{M}}$ and $\theta = Y_1$. This transformation from $(Y_1,Y_2)$ coordinates to $(r,\theta)$ coordinates is a local diffeomorphism on any open domain that avoids the values $Y_2 \le 1/\sqrt{2}$.

Finally, let $(y^3,y^4) = (r\cos\theta, r\sin\theta)$, which is again a local diffeomorphism on any open domain that avoids the values $r \le 0$. Then, as desired, $(y^3,y^4)$ form a local coordinate system on some open subset of $Y_2 > 1/\sqrt{2}$ and $Y_2 = \frac{1}{\sqrt{2}} + \frac{M}{2\sqrt{2}} ((y^3)^2+(y^4)^2)$.

  • $\begingroup$ Thank you for the explanation. I was wondering one more thing: with this reasoning, can I also set $Y_{2}$ to a constant? Or is there an underlying motivation to set it equal to this particular function? @IgorKhavkine $\endgroup$
    – GregVoit
    Apr 18, 2016 at 5:24
  • $\begingroup$ I can't say anything about motivation, since I haven't looked at the paper you referenced. A function is constant iff every point is a critical point (which is a coordinate invariant property). So you can't have it both ways. $\endgroup$ Apr 18, 2016 at 12:13
  • $\begingroup$ I went now further on with reading the paper, and the authors pick then three choices for the value of $M$: -1/2, 0 and 1/2. But with the reasoning above, how can $M$ be allowed to be zero? @IgorKhavkine $\endgroup$
    – GregVoit
    Apr 23, 2016 at 8:03
  • $\begingroup$ @GregVoit, it can't. If what the paper says is correct, then something must be different in the context of the statement. I can't say either way. $\endgroup$ Apr 23, 2016 at 15:42
  • 1
    $\begingroup$ @GregVoit: Every constant curvature surface is locally isometric to the plane, the sphere, or the hyperbolic plane. For the plane (M = 0), the standard Euclidean coordinates exhibits the desired form. For the hyperbolic plane (M<0), you are talking of the the Poincare disc model (Sorry, I misspoke earlier when I referred to geodesic normal coordinates; that only agrees up to order 2). For the sphere (M>0) this is the metric as presented in a stereographically projected chart. $\endgroup$ May 6, 2016 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.