# Conformal Flatness of Lorentzian 2-Manifolds

1. Jul 8, 2013

### Infrared

I have seen it stated that any Lorentzian 2-manifold is locally conformally flat; in what sense is it local? Is there a way to show this explicitly?

2. Jul 8, 2013

### WannabeNewton

As you will see shortly, the fact that a Lorentzian 2-manifold $(M,g_{ab})$ is locally conformally flat means that for any $p\in M$, there exists a neighborhood $U$ of $p$ (in the natural topology generated by the smooth structure) such that $U$ is conformally flat in the usual sense. The key part is the local existence of harmonic coordinates. In general the local existence can be guaranteed by theorems regarding specific types of PDEs but for 2-manifolds there is an easy way to prove local existence that is explicit, more instructive, and somewhat more minimal.

Let $M$ be a smooth 2-manifold, $g_{ab}$ a Lorentzian metric on $M$, and $\nabla_{a}$ the derivative operator associated with $g_{ab}$. Consider a scalar field $\alpha$ on $M$ that is harmonic meaning it satisfies $\nabla^{a}\nabla_{a}\alpha = 0$ and let $\epsilon_{ab}$ be the natural volume element on $M$ (i.e. it satisfies $\epsilon_{ab}\epsilon^{ab} = -2$). It is easy to show that the so defined natural volume element then satisfies $\epsilon_{ij}\epsilon^{ab} = -2\delta^{[a}_{i}\delta^{b]}_{j}$. Now consider another scalar field $\beta$ given by the equation $\nabla_{a}\beta = \epsilon_{ab}\nabla^{b}\alpha$.

We must show that given the 1-form $\omega_{a} = \epsilon_{ab}\nabla^{b}\alpha$, there always exists locally (in the above sense) a scalar field $\beta$ such that $\nabla_{a}\beta = \omega_{a}$. If we can show that $\omega_{a}$ is closed, meaning $\nabla_{[a}\omega_{b]} = 0$, then the Poincare Lemma guarantees that locally $\omega_{a}$ is exact, that is locally there exists a scalar field $\beta$ such that $\nabla_{a}\beta = \omega_{a}$, as desired.

Note that if we can show $\epsilon^{ab}\nabla_{a}\omega_{b} = 0$ then we are done since this would imply $\epsilon_{ij}\epsilon^{ab}\nabla_{a}\omega_{b} = \nabla_{[i}\omega_{j]} = 0$. Also note that since $\epsilon_{ab}\epsilon^{ab} = -2$, we have that $\epsilon^{ab}\nabla_{c}\epsilon_{ab} = 0$ implying $\epsilon_{ij}\epsilon^{ab}\nabla_{c}\epsilon_{ab} = \nabla_{c}\epsilon_{ij} = 0$ identically. Proceeding with this, we have $\epsilon^{ab}\nabla_{a}\omega_{b} = \epsilon^{ab}\nabla_{a}(\epsilon_{bc}\nabla^{c}\alpha) = \nabla_{a}(\epsilon^{ab}\epsilon_{bc}\nabla^{c}\alpha)$ and this reduces to $\epsilon^{ab}\nabla_{a}\omega_{b} = \nabla_{a}(\delta_{c}^{a}\nabla^{c}\alpha) = \nabla_{a}\nabla^{a}\alpha = 0$ hence $\nabla_{[a}\omega_{b]} = 0$ as desired so locally there always exists a $\beta$ such that $\nabla_{a}\beta = \epsilon_{ab}\nabla^{b}\alpha$.

Note that $\nabla^{a}\nabla_{a}\beta = \epsilon_{ab}\nabla^{a}\nabla^{b}\alpha = 0$ thus $\beta$ itself is harmonic. We wish now to make a coordinate transformation to the coordinates $(\alpha,\beta)$; these coordinates are called harmonic coordinates as all the coordinate functions are harmonic. Say our metric was originally expressed in some coordinates $(x^1,x^2)$ as $ds^{2} = g_{\mu\nu}dx^{\mu}dx^{\nu},\mu,\nu\in \{1,2\}$. We know that $g^{\mu'\nu'} = \nabla_{\mu}x^{\mu'}\nabla_{\nu}x^{\nu'}g^{\mu\nu} = \nabla_{\mu}x^{\mu'}\nabla^{\mu}x^{\nu'}$ hence $g^{\alpha\beta} = \nabla^{\mu}\alpha\nabla_{\mu}\beta = \epsilon_{\mu\gamma}\nabla^{\mu}\alpha\nabla^{\gamma}\alpha = 0$ and
$g^{\beta\beta} = \nabla_{\mu}\beta\nabla^{\mu}\beta = \epsilon_{\mu\gamma}\epsilon^{\mu\lambda}\nabla^{\gamma}\alpha \nabla_{\lambda}\alpha = -g^{\alpha\alpha} = -\Omega^{-2}$ where $\Omega^{-2} = \nabla^{a}\alpha \nabla_{a}\alpha$. Now $(g^{\mu'\nu'}) = \text{diag}(\Omega^{-2},-\Omega^{-2})$ so $(g_{\mu'\nu'}) = \text{diag}(\Omega^{2},-\Omega^{2})$ i.e. $ds^{2} = \Omega^{2}\{-d\beta^{2} + d\alpha^{2}\}$. Thus every Lorentzian 2-manifold is locally conformally flat.

3. Jul 9, 2013

### Ben Niehoff

There is a much simpler argument that any 2-manifold (Lorentzian or Riemannian) is locally conformally flat. The metric is a 2x2 symmetric matrix, which therefore has three independent entries. But coordinate transformations allow you two arbitrary functions of gauge symmetry. Therefore you can always choose a gauge in which the metric has one degree of freedom: the conformal factor.