# I Make a Pseudo-Riemannian Metric Conformal

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1. Oct 9, 2016

### tommyxu3

Hello everyone:

I studied in differential geometry recently and have seen a statement with its proof:
Suppose there is a Riemannian metric: $dl^2=Edx^2+Fdxdy+Gdy^2,$ with $E, F, G$ are real-valued analytic functions of the real variables $x,y.$ Then there exist new local coordinates $u,v$ for the surface in terms of which the induced metric takes the conformal form
$$dl^2=f(u,v)(du^2+dv^2).$$

The proof of this applied the complex method to construct a function $\lambda(x,y)$ such that
$$\lambda (\sqrt{E}dx+\frac{F+i\sqrt{g}}{\sqrt{E}}dy)=du+idv,$$
$$\lambda (\sqrt{E}dx+\frac{F-i\sqrt{g}}{\sqrt{E}}dy)=du-idv,$$
where $g=EG-F^2,$ the determinant of the first fundamental form, and find such $\lambda$ then $f=\frac{1}{|\lambda|^2}$ is what we want.

What confused me was that is there a corresponding statement for a pseudo-Riemannian metric, at least in $\mathbb{R}^2_1?$ That is, I want to prove:

A pseudo-Riemannian metric $dl^2=Edx^2+Fdxdy+Gdy^2$ (of type $(1,1)$) with real analytic coefficients, takes the form
$$dl^2=f(t,x)(dt^2-dx^2)$$
after a suitable coordinate change.

I try to prove it in the similar way but it seems not feasible for I just can factorize $dt^2-dx^2=(dt+dx)(dt-dx),$ without $i$ in the factorization. Directly decomposing it as real function, I still cannot find an answer, so I ask for help here >< Could anyone give me any advice?

2. Oct 10, 2016

### Ben Niehoff

Not sure if this helps, but remember in the Lorentzian-signature case that your determinant $g$ is negative, hence your $\sqrt{g}$ will spit out an extra factor of $i$.

3. Oct 10, 2016

### Ben Niehoff

Ah, a further comment. In the Riemannian case, there is no requirement that $\lambda$ be a real function. You can have (complex) $\lambda$ on one line and $\bar \lambda$ on the other, such that $f(u,v) = 1/|\lambda|^2$ appears in the metric. I have a feeling this might be necessary, actually.

Then, in the Lorentzian case, you actually have not just one integrating factor $\lambda$, but two independent (real) ones, which you can call $\lambda_1, \lambda_2$. Then your conformal factor will be $f(t,x) = 1/(\lambda_1 \lambda_2)$.

4. Oct 10, 2016