Here is the following problem:

Let $g_0$ be the Euclidean metric on $\mathbb C=\mathbb R^2$.

Let $M=\{z \in \mathbb C| \ |z|<1 \}$ and equip it with the Riemannian metric $g=\frac{1}{(1-|z|^2)^2}g_0.$

Let $N=\{z \in \mathbb C| \ \text{Im} \ z>0 \}$ and equip it with the Riemannian metric $\tilde{g}=\frac{1}{(\text{Im} \ z)^2}g_0.$

Show that $f:M\rightarrow N$ is an isometry if $f$ is given by

$$z\mapsto i\frac{1-z}{1+z}$$

So I am trying to show that $f^*\tilde{g}=g$. It seems like an easy calculation to check at first, but when I wrote down everything in real $(x,y)$ coordinates, I get some horribly long calculations. So I am either missing some trick which would simplify them, or I took a wrong approach, can anyone help?

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    $\begingroup$ You may try complex coordinate then.. $\endgroup$ – user99914 Dec 3 '15 at 19:45
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    $\begingroup$ I am 95% sure that to get an isometry, you'll need a $4$ in the coefficient of $g$. But I concur with @JohnMa: You should do it in complex coordinates. $dx\otimes dx+dy\otimes dy = \frac12(dz\otimes d\bar z + d\bar z\otimes dz) = |dz|^2$. $\endgroup$ – Ted Shifrin Dec 3 '15 at 19:54

Computing pullbacks is always a bit of a hairy task. The key point to notice here is that if we write $z\mapsto w = i\frac{1-z}{1+z}$ express the non-constant constant portion of the metric $g_N$ at $w$ in terms of $z$, we find

$$ \frac{1}{\text{Im}(w)^2} = \frac{(1+z)^2(1+\bar{z})^2}{(1-|z|^2)^2}$$


\begin{align*} \text{Im}(w) &= \text{Im}\left(i\frac{1-z}{1+z}\right)\\ &= \frac{1}{2i}\left[i\frac{1-z}{1+z} - \overline{\left(i\frac{1-z}{1+z}\right)}\right]\\ &= \frac{1}{2i}\left[i\frac{1-z}{1+z} + i\frac{1-\bar{z}}{1+\bar{z}}\right]\\ &= \frac{1}{2}\left[\frac{(1-z)(1+\bar{z}) + (1-\bar{z})(1+z)}{(1+z)(1+\bar{z})}\right]\\ &= \frac{1}{2}\left[\frac{1 -z + \bar{z} - z\bar{z} + 1- \bar{z} + z - z\bar{z} }{(1+z)(1+\bar{z})}\right]\\ &= \frac{1- |z|^2}{(1+z)(1+\bar{z})} \end{align*}

This factor of $(1-|z|^2)$ is the first indication that we are on the right track! So, when formally computing the pullback of your metric you'll begin to see

\begin{align*} f^*(g_N)_z &= f^*\left(\frac{1}{Im(z)^2}dz\otimes d\bar{z}\right)\\ &:= (g_N)_{f(z)}\\ &= \frac{1}{\text{Im}(w)}dw \otimes d\bar{w} \\ &= \frac{(1+z)^2(1+\bar{z})^2}{(1-|z|^2)^2}dw \otimes d\bar{w}\\ \end{align*}

It remains to compute $|dw|^2 = dw\otimes d\bar{w}$ which we can only hope will cancel with those nasty copies of $(1+z)(1+\bar{z})$ in the numerator. This will look something like

$$dw = d\left(i\frac{1-z}{1+z}\right) = i\frac{-(1+z)-(1-z)}{(1+z)^2}dz = -\frac{2i}{(1+z)^2}dz$$ and similarly $$d\bar{w} = d\left(\overline{i\frac{1-z}{1+z}}\right) = -i d\left(\frac{1-\bar{z}}{1+\bar{z}}\right) = \frac{2i}{(1+\bar{z})^2}d\bar{z}.$$

Combining these yields the pullback as

$$f^*(g_N)_z = \frac{4}{(1-|z|^2)^2}dz\otimes d\bar{z} = 4g_M.$$

Geometrically this is classical and well recognized conformal isometry is between the Poincaré disk and the upper-half plane!

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    $\begingroup$ Indeed :) I think one will find that I was right about the factor of $4$ being needed, too. :) Also, you don't want $dw^2$ or $d\bar w^2$. You want $|dw|^2$. Keep going! $\endgroup$ – Ted Shifrin Dec 3 '15 at 23:10
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    $\begingroup$ As I said in my comment way up there ^^^, this is the shorthand for writing out $\frac12(dz\otimes d\bar z+d\bar z\otimes dz)$. Note that this is the standard notation used when the tensor symbol is dropped: $|dz|^2 = dz\bullet d\bar z = (dx+i\,dy)\bullet (dx-i\,dy) = dx^2+dy^2$, the $\bullet$ denoting symmetric product. $\endgroup$ – Ted Shifrin Dec 3 '15 at 23:18
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    $\begingroup$ I used to call myself one of those too. :) $\endgroup$ – Ted Shifrin Dec 3 '15 at 23:22
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    $\begingroup$ Yeah, so I believe you've vindicated my comments about the original statement needing a factor of $4$. It's a standard computation that to get $K=-1$ one needs that $4$, and the metric on the upper half-plane is "well known" to have $K=-1$. $\endgroup$ – Ted Shifrin Dec 3 '15 at 23:24
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    $\begingroup$ Yup, conformally equivalent. $\endgroup$ – Ted Shifrin Dec 3 '15 at 23:30

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