This map is an isometry (in the Riemannian sense) of the hyperbolic plane. Why is the following a proof of it?

Question

I'm making my way through a textbook on elementary undergraduate geometry. The author has defined the notion of an isometry between two subsets of $\mathbb{R}^2$ equipped with a Riemannian metric. It is a diffeomorphism $\phi:\tilde{V} \to V$ such that for all $P \in \tilde{V}$ and all $x,y \in \mathbb{R}^2$, $\langle x, y \rangle^{^\sim} _P = \langle (d\phi)_P(x), (d\phi)_P(y) \rangle _{\phi (P)}$.

In a subsequent chapter we encounter the upper half-plane model for the hyperbolic plane, which is the set $H = \{(x,y)\in\mathbb{R}^2 | y > 0\}$ equipped with the Riemannian metric $(dx^2 + dy^2)/y^2$. Identifying $\mathbb{R}^2$ and $\mathbb{C}$ in the usual way the metric becomes $|dz|^2/Im(z)^2$.

We then go on to consider the map $w:H\to H$ where $w(z) = -1/z$. The author shows that $|dw|^2/Im(w)^2 = |dz|^2/Im(z)^2$ and states that this shows that $w$ is an isometry. He's not explicitly defined an isometry of $H$, so I assume he means in the sense defined above, but it's not clear to me why $|dw|^2/Im(w)^2 = |dz|^2/Im(z)^2$ implies that $w$ is an isometry in that sense.

I have been able to explicitly show that $w$ is an isometry in the above sense, but I assume that there's a way to do it using $|dw|^2/Im(w)^2 = |dz|^2/Im(z)^2$. My question is why does $|dw|^2/Im(w)^2 = |dz|^2/Im(z)^2$ imply that $w$ is an isometry as defined above?

Answer 1

It is more a matter of notation.

Take $z\in H$ and $x,y\in\mathbb R^2$. Then $$ \langle d_zw\, x,d_zw \,y\rangle_{w(z)}=:\frac{|dw|^2}{Im(w)^2}(x,y)=\frac{|dz|^2}{Im(z)^2}(x,y):=\langle x,y\rangle_z. $$

Added:

More explicitly, write $z=z_1+iz_2$, $x=(x_1,x_2)\equiv x_1+ix_2$ and $y=(y_1,y_2)\equiv y_1+iy_2$. Then, by definition, $$ \langle x,y\rangle_z:=\frac{|dz|^2}{Im(z)^2}(x,y)=\frac{x_1y_1+x_2y_2}{z_2^2}. $$ Now take $w(z)=-1/z=w_1+iw_2$ and define $$ x'=d_zw\,x=\frac{x_1+ix_2}{z^2}\equiv x_1'+ix_2', \quad y'=d_zw\,y=\frac{y_1+iy_2}{z^2}\equiv y_1'+iy_2'. $$ Then $$ \langle d_zw\, x,d_zw \,y\rangle_{w(z)}=\frac{x_1'y_1'+x_2'y_2'}{w_2^2}=:\frac{|dw|^2}{Im(w)^2}(x,y). $$ Really, their notation is not the best.