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Let $\phi:S^2\to S^2$ be a Möbius transformation. I would like to show that $$\frac1{\sqrt2}|d\phi(z)|=|\phi'(z)|\frac{1+|z|^2}{1+|\phi(z)|^2}.$$ Here I am using the Fubini–Study metric on both domain and codomain. I am pretty unfamiliar with the Fubini–Study metric, but I know it is given by $(dx^2+dy^2)/(1+r^2)^2$.

I tried doing this in local coordinates, but somehow I ended up with the expression $d\phi(z)(v)$ being a real number instead of a tangent vector. I think any explanation would help, but I would really love one in local coordinates so that I can figure out what I was doing wrong. Thanks!

boink
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  • You missed a square on $(1+r^2)$. It is $ds^2=(dx^2+dy^2)/(1+r^2)^2$ (where of course $r^2=x^2+y^2$). – user10354138 Oct 07 '23 at 16:49
  • Thanks, just edited! – boink Oct 07 '23 at 18:02
  • Sorry I guess I don't quite see it. I am not even sure what $|d\phi(z)|$ means here. I guess it should be something like $|d\phi(z)|^2=(dx^2(d\phi(z),d\phi(z))+dy^2(d\phi(z),d\phi(z)))/(1+r^2)^2$ but I am not sure what it would mean to take $dx(d\phi(z))$. Is there some kind of operator norm-type thing going on? I'm pretty new to Riemannian metrics in general unfortunately – boink Oct 07 '23 at 18:44

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