# Lorentz surfaces, conformal metrics and eigenvalues

From what I understand of Lorentz surfaces (spacetimes of dimension 2), it seems that, according to Kulkarni's theorem, two reasonable enough Lorentz surfaces (I am only interested in surfaces with topology $\Bbb R^2$) are conformally equivalent, that is, $g_1 = \Omega^2 g_2$. This includes Minkowski space, meaning that they must all be conformally flat.

To find the equivalent conformally flat metric, I assumed that since they are conformal, the metric's eigenvalues should be $-\Omega^2$ and $\Omega^2$. This would then mean that, given a real symmetric $2\times 2$ matrix with negative determinant, the eigenvalues should always be inverses of each other.

From some calculations, this seems not to be the case. Did I misunderstand Kulkani's theorem or is the method I tried incorrect for such a task?

• $\uparrow$ Which calculations? Which method did you try? Jul 2, 2016 at 13:53

The mistake here is to forget that the only coordinate-independent property of the eigenvalues of the metric is the sign. Consider the Minkowski metric in usual Cartesian coordinates $$d s^2 = -dt^2 + dx^2$$ Or in matrix form $$\left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$$ I.e., the eigenvalues are always in a ratio $\lambda_1/\lambda_2=-1$.
Now let us use a new time coordinate coordinate $\tau$ such that $$t=\frac{2}{3} \tau^{3/2}$$ which is restricted to $t \in (0,\infty) \to \tau \in (0,\infty)$. Now the metric transforms to $$ds^2 = -\tau d\tau^2 + dx^2$$ or in matrix form $$\left( \begin{array}{cc} -\tau & 0 \\ 0 & 1 \end{array} \right)$$ Now we see that the eigenvalue ratio is $\lambda_1/\lambda_2=-\tau$ and as we go through $\tau \in (0,\infty)$, the ratio goes through all the possible values admissible for a non-degenerate matrix of signature $(-+)$.