As I was reading Jost's Compact Riemann Surfaces, I came across the definition of a (conformal) Riemannian metric:
Definition 2.3.1 A conformal Riemannian metric on a Riemann surface $\Sigma$ is given in local coordinates by $$\lambda^2(z)dzd\bar z,\quad \lambda(z)>0$$
As i was trying to make sense of this definition, I found another one on the Internet (Ben Andrews's lecture notes on differential geometry):
Definition A Riemannian metric $g$ on a smooth manifold $M$ is a smoothly chosen inner product $g_x:T_xM\times T_xM\to\mathbb{R}$ on each of the tangent spaces $T_xM$ of $M$.
The second definition seems to me easier to understand, so I am trying to understand how they are equivalent (putting aside the conformal part for now, and assuming that $M$ in the second definition is also a Riemann surface).
My questions:
(1) In the first definition, what is $z$? Is it a point in $\Sigma$ or what? If it is a point in $\Sigma$, then how come it says $\lambda(z)dzd\bar z$ is the metric given in local coordinates?
(2) What does $dzd\bar z$ mean and what does it do? As I understand, according to the second definition, $\lambda(z)dzd\bar z$ takes a pair $(u,v)$ of tangent vectors as an input and outputs a real number, so the reasonable understanding is that
$$dzd\bar z=dz\wedge d\bar z=-2i\ dx\wedge dy\\
\implies dzd\bar z(u,v)=-2i\ dx\wedge dy(u,v)=-2i\left|
\begin{matrix}dx(u)&dx(v)\\dy(u)&dy(v)\end{matrix}\right|$$
But in this case $\lambda(z)dzd\bar z(u,v)$ is not a real number.
(3)Please excuse me if my questions seem absurd for those who have learned differential geometry. It would be great if you can provide me some introductory reference books that can get me quickly started on the subject.
Best Answer
$z$ is a conformal/holomorphic coordinate chart on $\Sigma,$ i.e. a conformal map $z = x+iy : \Sigma \supset U \to \mathbb C$ that is a diffeomorphism onto its image. Think about this in the same way you think about real coordinate charts $x^i : M \to \mathbb R^n.$
$\def\bdz{\overline{dz}}$ The notation $dz\;\bdz$ denotes the symmetric tensor product, not the wedge product. Thus $\lambda^2\;dz\;\bdz=\lambda^2\;(dx+i\;dy)(dx - i\;dy) = \lambda^2(dx^2+dy^2),$ which is real positive-definite symmetric and thus a (smoothly varying) inner product as desired.