First Fundamental Form and the Metric On the upper half-plane with standard $x,y$ coordinates we have the hyperbolic metric given by $g_{11}=g_{22}=1/y^2$ and $g_{12}=0$.
In terms of the first fundamental form, we write
$$ds^2=\frac{dx^2+dy^2}{y^2}=
\frac{-4\,dz\,d\bar{z}}{(z-\bar{z})^2}.$$
I can interpret $dz$ and $d\bar{z}$ as (complex-valued) $1$-forms on the upper half plane.
Suppose I now want to show that a fractional linear transformation $z\mapsto z'$ where $z'=\frac{az+b}{cz+d}$ and $ad-bc=1$ gives an isometry.
By definition of an isometry, I would want to show that the pushforward of two tangent vectors in the domain yield the same result evaluated under the metric.
But when I see other people approach the problem, what they do is calculate $dz'$ and $d\bar{z'}$, and then prove that
$$\frac{-4\,dz'\,d\bar{z'}}{(z'-\bar{z'})^2}=\frac{-4\,dz\,d\bar{z}}{(z-\bar{z})^2}.$$
I'm a little confused by what this calculation is actually doing. I would think that what I need to do is evaluate the same fundamental form (without the 's in the numerator) at the point $z'$ and $\bar{z'}$ on the pushforward of some tangent vectors in the domain.  I am not sure why it is valid to simply substitute in $dz'$ and $\bar{dz'}$ and compute "the same thing", and along the way I seem to have lost any intuition of what $dz'$ and $d\bar{z'}$ really mean as $1$-forms.
 A: As you know, the metric $g_p$ at each point $p$ of the space $M$ is just a bilinear form on tangent vectors at the point $p$, that is, a function that associates to each pair of vectors $X_p, Y_p$ their "generalized scalar product" $g_p(X_p, Y_p)$.
To prove that a map $T:M \rightarrow M$ is an isometry, you can do what you said and show that for each point $p$ and each pair $X_p, Y_p$ of tangent vectors at $p$, we have following identity:
$g_p(X_p, Y_p) = g_{T(p)}({T_{*,p}(X_p), T_{*,p}(Y_p)})$
where $T_{*,p}$ is the pushforward, that is, the map on tangent vectors induced by $T$. In terms of matrices of coordinates, what you are proving is that
$V^t G_p W = (AV)^t G_{T(p)} AW$
where $V, W$ are the column vectors formed with the coordinates of $X_p, Y_p$, $A$ is the matrix of the pushforward map $T_{*,p}$, and $G_q$ is the matrix of the metric at each point $q$.
Modifying this equation we find that is equivalent to
$V^t G_p W = V^tA^t G_{T(p)} AW = V^t(A^tG_{T(p)}A)W = V^tG'_p W$
And we just need to show the identity between the two matrices $G_p$ and $G_p' = A^t G_{T(p)} A$, which represent two metrics at the point $p$. One is $g_p$, our first fundamental form, and the other one is $g_p'$, defined by:
$g_p'(X_p,Y_p) = (AV)^t G_{T(p)} (AW) = g_{T(p)}({T_{*,p}(X_p), T_{*,p}(Y_p)})$
this metric is also called the pullback of $g$ by $T$, and denoted by $T^*g$. As it should be obvious by now, $T$ is an isometry if and only if $g = T^*g$ at each point, which is what you are encountering, because the pullback metric is obtained from the coordinate expression of $g$ simply by replacing each coordinate $z, \overline{z}$ (or $x, y$) with the same coordinate transformed by $T$, $z' = z\circ T, \overline{z}' = \overline{z}\circ T$ (or $x' = x\circ T, y' = y\circ T$).
If you think about this, you can discover that every multilinear form on tangent vectors is transformed by each map $T$ via some definition of pullback. For example, the area form $\Omega$, which associates to every pair of vectors the (oriented) area of the parallelogram which they span, can be pulled back the same way:
$(T^*\Omega)_p(X_p, Y_p) = \Omega_{T(p)}(T_{*,p}(X_p), T_{*,p}(Y_p))$
And the same is true for linear functions of more arguments (also called covariant tensors), like the (purely covariant) curvature tensor. The name "covariant" comes precisely from the way they "vary", that is, the way they transform under smooth maps. In general, tensors represent geometric concepts, and smooth maps such as $T$ don't transform points and vectors only: they transform angles, distances, volumes, curvatures and everything tensorial. Isometries are just those maps that leave the metric tensor invariant.
