TITLE: What does $ds^2$ mean and how does it specify a metric?
QUESTION [13 upvotes]: Let $H$ be the upper half-plane in $\mathbb{R}^2$. How does the following expression $$ds^2= \frac{dx^2+dy^2}{y^2}$$ specify a Riemannian metric on $H$?
I don't understand what the expression means. If the $y^2$ wasn't there, then $g=dx^2+dy^2=dx\otimes dx+dy\otimes dy$ is the standard metric. This does make sense to me: at every point it associates the standard inner product. But the above expression is not even symmetric in $x,y$ so it isn't meant to be read the same way, but I don't know how to make sense of it.

REPLY [7 votes]: A "fundamental form"  $$ds^2=E(x,y)\>dx^2+2F(x,y)\>dx\>dy+G(x,y)\>dy^2\ ,$$
in your case
$$ds^2={dx^2+dy^2\over y^2}\ ,$$ tells you how to measure the length of a curve $\gamma$ given in the form
$$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$
The rule is: Write formally 
$$dx=\dot x(t)\ dt,\quad dy=\dot y(t)\ dt$$
and compute
$$L(\gamma)=\int_a^b\sqrt{E\bigl(x(t),y(t)\bigr)\dot x^2(t)+2F\bigl(x(t),y(t)\bigr)\dot x(t)\dot y(t)+G\bigl(x(t),y(t)\bigr)\dot y^2(t)}\ dt\ ,$$
in your case
$$L(\gamma)=\int_a^b{\sqrt{\dot x^2(t)+\dot y^2(t)}\over y(t)}\ dt\ .$$
The intuitive idea is the following: For reasons particular to the problem at hand the length of an "infinitesimal segment" connecting the points $(x,y)$ and $(x+dx,y+dy)$ should not be the usual pythagorean $\sqrt{dx^2+dy^2}$; instead it should depend in a still "natural", but more complicated way on the vector $(dx,dy)$. In your case the length of this "infinitesimal segment" should be its euclidean length, divided by the $y$-coordinate of the point where it is located.