# Choosing a coordinate transformation

I was reading the following paper

http://scitation.aip.org/docserver/fulltext/aip/journal/jmp/4/7/1.1704018.pdf?expires=1460721373&id=id&accname=2112043&checksum=607EDBF852A9E209F384DB6DA228B416

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

$\zeta'=g(\zeta)$

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

• 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 $Y_2 = \frac{1}{\sqrt{2}}+\frac{M}{2\sqrt{2}}((y^{3})^{2}+(y^{4})^{2})$. – Igor Khavkine Apr 15 '16 at 14:11
• @IgorKhavkine could you please explain your first sentence? How do $(Y_{2},Y_{1})$ form local coordinates? thank you – GregVoit Apr 15 '16 at 17:49
• @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 – GregVoit Apr 15 '16 at 17:52
• @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. – Willie Wong May 5 '16 at 19:24

## 1 Answer

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 $Y_2 = \frac{1}{\sqrt{2}} + \frac{M}{2\sqrt{2}} ((y^3)^2+(y^4)^2)$.

• 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 – GregVoit Apr 18 '16 at 5:24
• 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. – Igor Khavkine Apr 18 '16 at 12:13
• 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 – GregVoit Apr 23 '16 at 8:03
• @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. – Igor Khavkine Apr 23 '16 at 15:42
• @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. – Willie Wong May 6 '16 at 17:39