# An isometry between riemannian metric spaces

• May 11th 2013, 03:16 AM
Plato13
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.
• May 11th 2013, 08:42 AM
xxp9
Re: An isometry between riemannian metric spaces
let $v=df(1,0)=\frac{(y^2-x^2, -2xy)}{(x^2+y^2)^2}$, $w=df(0,1)=\frac{(-2xy,x^2-y^2)}{(x^2+y^2)^2}$
$g(v,v)=\frac{v \cdot v}{y^2/(x^2+y^2)^2}=\frac{1}{(x^2+y^2)^2}\cdot\frac{(x^2+ y^2)^2}{y^2}=1/y^2$
• May 11th 2013, 10:41 AM
Plato13
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.
• May 11th 2013, 02:00 PM
xxp9
Re: An isometry between riemannian metric spaces
so you may need to check your calculation
• May 12th 2013, 08:00 AM
Plato13
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
• May 12th 2013, 08:29 AM
Plato13
Re: An isometry between riemannian metric spaces
Sorted it now thanks