D. Tong's notes on string theory (PDF), subsection 5.1.1, feature the following in introducing the symmetries used in the Faddeev-Popov method:

We have two gauge symmetries: diffeomorphisms and Weyl transformations. We will schematically denote both of these by $\zeta$. The change of the metric under a general gauge transformation is $g \rightarrow g^{\zeta}$. This is shorthand for, \begin{equation} g_{\alpha \beta} (\sigma) \rightarrow g_{\alpha \beta}^{\zeta} (\sigma)= e^{2\omega (\sigma)} \frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}} g_{\gamma \delta} (\sigma)\tag{p.111} \end{equation} In two dimensions these gauge symmetries allow us to put the metric into any form that we like — say, $\hat{g}$. This is called the fiducial metric and will represent our choice of gauge fixing.

Is there a proof available that shows that the combined symmetry allows us to "put the metric in any form (locally, of course)?" Is this property restricted to two dimensions?


1 Answer 1


Theorem. Every 2D pseudo-Riemannian manifold $(M,g)$ is conformally flat, i.e. there locally exist isothermal coordinates.$^1$

Sketched proof of theorem: Given a point $p\in M$. Consider local coordinates $u,v$ in a neighborhood of $p$.

  1. Generic case $g_{uu}(p)\neq 0$: Do a Weyl scaling such that $g_{uu}\equiv 1$. Then $$\begin{align}g~=~&\mathrm{d}u\odot \mathrm{d}u + 2g_{uv}\mathrm{d}u\odot \mathrm{d}v + 2g_{vv}\mathrm{d}v\odot \mathrm{d}v\cr ~\stackrel{(2)}{=}~&\omega_+\odot \omega_-,\end{align}\tag{1}$$ where we have defined 2 non-vanishing one-forms $$ \omega_{\pm}~:=~\mathrm{d}u + [g_{uv} \pm \sqrt{-\det(g)}]\mathrm{d}v. \tag{2}$$
  • Minkowskian case $\det(g)<0$: Then the one-forms $\omega_{\pm}$ are real. There exist locally 2 real integrating factors $\lambda_{\pm}\neq 0$ such that $$\omega_{\pm} ~=~\lambda_{\pm}\mathrm{d}x^{\pm}.\tag{3}$$ Then the metric tensor (1) reads $$g~\stackrel{(1)+(3)}{=}~\lambda_+\lambda_-\mathrm{d}x^+\odot \mathrm{d}x^-.\tag{4} $$ A Weyl scaling brings the metric on light-cone form. $\Box$

  • Euclidean case $\det(g)>0$: Then the one-form $\omega_{\pm}~=~\omega^{\ast}_{\mp}$ is complex. There exists locally a complex integrating factor $\lambda_+\neq 0$ such that $$\begin{align}\omega_+ ~=~&\lambda_+\mathrm{d}z, \cr z ~=~&x+iy, \cr \omega_- ~=~&\omega^{\ast}_+ ~=~\lambda^{\ast}_+\mathrm{d}z^{\ast}.\end{align}\tag{5}$$
    Then the metric tensor (1) reads $$\begin{align}g~\stackrel{(1)+(5)}{=}&~|\lambda_+|^2\mathrm{d}z\odot \mathrm{d}z^{\ast}\cr ~=~&|\lambda_+|^2[\mathrm{d}x\odot \mathrm{d}x+\mathrm{d}y\odot \mathrm{d}y],\end{align}\tag{6} $$ which is again manifestly real. A Weyl scaling brings the metric on standard Euclidean form. $\Box$

  1. Special case $g_{uu}(p)=0$:
  • Euclidean case $\det(g)>0$: Impossible. $\Box$

  • Minkowskian case $\det(g)<0$: Then $g_{uv}(p)\neq 0$. In the Gauss elimination procedure (if we were to bring $g$ on diagonal form) this corresponds to a case of vanishing diagonal element. It is possible to perform an affine coordinate transformation $(u,v) \to (u^{\prime},v^{\prime})$ so that $g_{u^{\prime}u^{\prime}}(p)\neq 0$. Now use the generic case. $\Box$


  1. M. Nakahara, Geometry, Topology and Physics, 1989; Example 7.32.

  2. M. Nakahara, Geometry, Topology and Physics, 2003; Example 7.9.


$^1$ And yes, the theorem is not true in higher dimensions, cf. e.g. this related Phys.SE post.

  • $\begingroup$ Notes for later. 1. Given a non-vanishing real one-form $\omega = f\mathrm{d}x+g\mathrm{d}y$. Existence of real integrating factor $\lambda$ such that $\mathrm{d}(\lambda\omega)=0$. $\quad (g\partial_x-f\partial_y)\ln\lambda=\partial_yf-\partial_xg$ first-order linear PDE. 2. Given a non-vanishing complex one-form $\omega = f\mathrm{d}z+g\mathrm{d}\bar{z}$. Existence of complex integrating factor $\lambda$ such that $\mathrm{d}(\lambda\omega)=0$. View $z$ and $\bar{z}$ as independent complex variables, and essentially proceed as in the real case. $\endgroup$
    – Qmechanic
    Aug 5, 2019 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.