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Math Help - An isometry between riemannian metric spaces

  1. #1
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    An isometry between riemannian metric spaces

    I can do all the questions apart from part iv). After some work, I got df_{z}(1,0)=\frac{|z|^2-2z^2}{|z|^4}. Unfortunately, the image of this under the metric is not 1/im(z)^2 as it ought to be.
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  2. #2
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    Re: An isometry between riemannian metric spaces

    let v=df(1,0)=\frac{(y^2-x^2, -2xy)}{(x^2+y^2)^2}, w=df(0,1)=\frac{(-2xy,x^2-y^2)}{(x^2+y^2)^2}
    g(v,v)=\frac{v \cdot v}{y^2/(x^2+y^2)^2}=\frac{1}{(x^2+y^2)^2}\cdot\frac{(x^2+  y^2)^2}{y^2}=1/y^2
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    Re: An isometry between riemannian metric spaces

    If you take the real part of what I got above, you get Re(\frac{|z|^2-2z^2}{|z|^4})=\frac{x^2+y^2-2(x^2-y^2)}{(x^2+y^2)^2}=\frac{3y^2-x^2}{(x^2+y^2)^2} which disagrees with your answer due to the 3y^2 instead of y^2.
    Last edited by Plato13; May 11th 2013 at 11:25 AM.
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    Re: An isometry between riemannian metric spaces

    so you may need to check your calculation
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    Re: An isometry between riemannian metric spaces

    Ok I've had another go. Can you tell me where I go wrong. Think of R^2 as C and let z=x+iy. Then, a(t)=z+t is a generating curve for e=(1,0).

    f(z)=\frac{z}{|z|^2} so f(a(t))=\frac{z+t}{|z+t|^2} whose derivative is \frac{|z+t|^2-(z+t)(2+x)t}{|z+t|^4}so that evaluating at t=0 gives \frac{1}{|z|^2} which appears even further from the truth!. I think it hinges on what the derivative of |z+t|^2 is. My reasoning was |z+t|^2=|z|^2+t^2+2Re(tz)=|z|^2+t^2+2tx which differentaing w.r.t t gives 2t+2x. Thanks
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    Re: An isometry between riemannian metric spaces

    Sorted it now thanks
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