What does it mean when someone writes $ds = \frac{1}{y}(dx^2+dy^2)$?

Question

What does it mean when someone writes $ds = \frac{1}{y}(dx^2+dy^2)$? (I have seen this is written in the setting of hyperbolic space.)

But essentially I have no idea as to how to interpret this information. I would be grateful if someone were to clear this up! Thanks!

Edit: All I ever see written is : consider the metric $ds = \frac{1}{y}(dx^2+dy^2)$. But of course this goes in too deep, and I can't seem to find a description of what it means.

$ds^2$ then is simply the 'first fundamental form' of the hyperbolic Riemannian metric on the plane. Set $E=G=y^{-1}$. — Dan Rust, Mar 13 '14 at 00:28
Should you edit the question (and title) to put exponent $2$ on the $ds$? — GEdgar, Mar 13 '14 at 00:46

score 2 · Answer 1 · answered Mar 13 '14 at 00:41

$\frac{1}{y}(dx^2+dy^2)$ is shorthand for $\frac{1}{y} (dx \otimes dx + dy\otimes dy)$. This is a bilinear form on the tangent space (Really it lives in a tensor bundle but this seems a bit grand for such a simple object). It eats two tangent vectors and spits out a number, and this assignment is linear with respect to each of the two tangent vectors.

Explicitly, if $\begin{bmatrix} x_1\\y_1\end{bmatrix}$ and $\begin{bmatrix} x_1\\y_1\end{bmatrix}$ are tangent vectors to the upper half plane at a point $(a,b)$, then $\frac{1}{y} (dx \otimes dx + dy\otimes dy) \left( \begin{bmatrix} x_1\\y_1\end{bmatrix},\begin{bmatrix} x_2\\y_2\end{bmatrix}\right) = \frac{1}{b} (x_1x_2+y_1y_2)$.

In other words, this object is telling you how to take a ``dot product'' in this new metric, and in this case it is the same as the regular dot product, just scaled by the reciprocal of the ordinate of the point that the tangent vectors as based at.

Once you have a dot product on tangent vectors you can do ``geometry'' locally. In particular you could talk about the length of a tiny line segment using this dot product. So when computing the length of a curve in this metric, you integrate all these tiny lengths.

BTW I think you probably meant to write $ds^2 = \frac{1}{y^2} (dx^2+dy^2)$ if you want hyperbolic metric on UHP.

score 0 · Answer 2 · answered Mar 13 '14 at 00:20

The notation $ds$ is used to denote the infinitesimal arc length. On a metric space $M$, let $\Gamma : [0, 1] → M$ be a curve on $M$, and let $g$ be the (pseudo-) metric tensor. So, $$ds=\sqrt{g_{\mu\nu}dx^\mu dx^\nu}\\ \mbox{Thus, the length of the arc is }s=\int^1_0\sqrt{\pm g(\Gamma^{'}(t),\Gamma^{'}(t))}$$ Here, the Einstein summation convention is used. Also, $x^\mu$ is a coordinate system on $M$, and $\Gamma^{'}(t)$ is the tangent vector to $\Gamma$ at $t$, i.e., $\Gamma^{'}(t)\in T_{\Gamma(t)}M$. The $\pm$ is chosen to ensure that the expression inside the $\sqrt{ }$ is a real number.

What does it mean when someone writes $ds = \frac{1}{y}(dx^2+dy^2)$?

2 Answers2