In "Exercises and Solutions in Mathematics", Ta-Tsien, 2nd Edition, exercise 3343.
Statement of the exercise
Let $(\mathbb{H}, g)$ be the two-dimensional hyperbolic space, where
\begin{equation} \mathbb{H} = \{(x, y) \in \mathbb{R}^2 : y > 0\} \end{equation}
is the upper half plane of $\mathbb{R}^2 = \mathbb{C}$ and the metric $g$ is given by
\begin{equation} g = \frac{dx^2 + dy^2}{y^2} \end{equation}
Suppose $a$, $b$, $c$ and $d$ are real numbers such that $ad - bc = 1$. Define
\begin{equation} \phi(z) = \frac{az + b}{cz + d} \end{equation}
for any $z = x + \sqrt{-1} y$. Prove that $\phi$ is an isometry for $(\mathbb{H}^2, g)$.
Statement of the answer
To prove that $\phi$ is an isometry, the authors compute:
\begin{equation} d\phi = \frac{a(dz)(cz+d) - c(dz)(az+b)}{(cz + d)^2} \end{equation}
and after some computations, concludes that since:
\begin{equation} \Vert d\phi(z) \Vert^2 = \frac{d\phi(z) d\overline{\phi(z)}}{[Im \phi(z)]^2} = \frac{dx^2 + dy^2}{y^2} = \Vert dz \Vert^2 \end{equation}
then $\phi$ is an isometry.
My question
What is the mathematical nature of the operator $d$ in this context ? It seems to me that in order to prove that $\phi$ is an isometry, one has to prove that:
$g = \phi^* g$
i.e. that the pullback of $g$ by $\phi$ is $g$. In the context of differential geometry, I have only seen $d\phi$ standing for the exterior derivative of $\phi$ or for the differential map associated to $\phi$.
In a more "intuitive" manner, considering small variations of $\phi(z)$ and $z$, I understand that an isomorphism maps a small increment $dz$ to a small increment $d\phi(z)$. But I would like to understand the differential geometry meaning.
Is $dz$ a differential form in the context of this exercise ? Then what is the precise meaning of $\Vert dz \Vert^2$ ? Is $d\phi(z)$ of the same nature than $dz$ ? Is it a vector-valued differential form ?
$\newcommand{\vv}{\mathbf{v}}\newcommand{\Cpx}{\mathbf{C}}\DeclareMathOperator{\Im}{Im}$Since $\phi:\Cpx \to \Cpx$ is meromorphic, $d\phi$ may be viewed as the ordinary derivative of a complex-valued function, as the exterior derivative of a complex-valued $0$-form, or as the total differential of a mapping (surely among other interpretations).
To bridge the classical differential calculus and modern language, let $z$ and $w$ denote complex coordinates, and write $w = \phi(z)$. The coordinate $1$-forms satisfy $dw(\vv) = dz(\vv) = \vv$ for every tangent vector $\vv$.
The "classical" chain rule $dw = \phi'(z)\, dz$ has the modern expression $$ \phi^{*}(dw) = \phi'(z)\, dz. $$ Indeed, for every tangent vector $\vv$, $$ \phi^{*}(dw)(\vv) = dw(\phi_{*}\vv) = \phi_{*}\vv = d\phi(z)(\vv) = \phi'(z)\, dz(\vv). $$
If we write $w = u + iv$, then $dw = du + i\, dv$. The product $$ dw\, d\bar{w} = du^{2} + dv^{2} $$ refers to the quadratic form on $\Cpx$ that sends a tangent vector $\vv$ to $$ dw(\vv)\, d\bar{w}(\vv) = \vv\bar{\vv} = \|\vv\|^{2}. $$ (Compare the tensor product $$ dz \otimes d\bar{z} = (du \otimes du + dv \otimes dv) - 2i\, du \wedge dv, $$ which accepts two vectors as input, has non-zero imaginary part, etc.)
For the mapping at hand, $w = \phi(z) = \dfrac{az + b}{cz + d}$, and therefore $\phi^{*}(dw) = \dfrac{dz}{(cz + d)^{2}}$. Since $\|dw\|^{2} := g = \dfrac{dw\, d\bar{w}}{(\Im w)^{2}}$ as a quadratic form, the final line of the author's computation asserts $$ \phi^{*}g = \|\phi^{*}(dw)\|^{2} = \frac{\phi^{*}(dw)\, \phi^{*}(d\bar{w})}{(\Im w)^{2}} = \frac{\|\phi'(z)\|^{2}\, dz\, d\bar{z}}{(\Im \phi(z))^{2}} = \frac{dz\, d\bar{z}}{(\Im z)^{2}} = \|dz\|^{2} = g. $$