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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?

Thanks in advance!

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# I Make a Pseudo-Riemannian Metric Conformal

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