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I Make a Pseudo-Riemannian Metric Conformal

  1. Oct 9, 2016 #1
    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?

    Thanks in advance!
     
  2. jcsd
  3. Oct 10, 2016 #2

    Ben Niehoff

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    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##.
     
  4. Oct 10, 2016 #3

    Ben Niehoff

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    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)##.
     
  5. Oct 10, 2016 #4
    Thanks for your help first!!!

    I noticed that too! But then I felt hard to solve it. In the Riemannian case the complex number help me to separate the equation, and now I don't know how to have further result...

    I'll keep trying~ thank you!
     
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