Why are Euclidean and hyperbolic lengths proportional to first order?

Question

In his book “Three-Dimensional Geometry and Topology”, Thurston constructs a Riemannian metric for the Poincare disk model and begins as follows. He says that, given any (hyperbolic) line segment $s$ in the disk, one may choose a hyperbolic line $L$ as well as a Euclidean circle $C$ such that $s$ and $L$ as well as $s$ and $C$ meet perpendicularly. He then uses a geometric argument to show that the hyperbolic length of $s$ depends only on the angle between $C$ and $L$.

He then considers the length $l$ of such a line segment as a function of the angle $\alpha$ and concludes that “this function has a finite derivative at $\alpha=0$, because the Euclidean length $l_E$ of any particular transversal segment is roughly proportional to $\alpha$ for $\alpha$ small, and Euclidean and hyperbolic lengths should be proportional to first order”.

I assume that rough proportionality means that $l_E(\alpha)=c\cdot\alpha+O(\alpha^2)$ as $\alpha\to 0$; a property I can prove. But I don’t see any geometric reason why the Euclidean and hyperbolic lengths should be proportional to first order. Again, I suppose it means that $l(\alpha)=c\cdot l_E(\alpha)+O(\alpha^2)$ as $\alpha \to 0$.

Answer 1

The above interpretation of being proportional to first order is questionable, as this notion should not depend on Thurston's angle construction. It seems more plausible to demand for two metrics $g,h$ that there exists a constant such that $\|\exp_g(t\cdot v)'\|_g = c\cdot \|\exp_h(t\cdot v)'\|_h + O(t^2)$. This condition is satisfied if the metrics are conformally equivalent.

It therefore remains to give a geometric explanation for conformal equivalence. Let $\rho\colon D^2\to D^2$ denote the rotation by $\pi/2$ around the origin. This is a hyperbolic isometry. Clearly $T\rho\colon TD^2\to TD^2$ has the property that it maps a nowhere vanishing vector field $X\in\Gamma(TD^2)$ to a linearly independent vector field. Thus $(X,T\rho(X))$ is a global orthogonal frame with respect to the Euclidean metric and with respect to the hyperbolic metric Thurston goes on constructing. This shows conformal equivalence as $\|X\|=\|T\rho\circ X\|$ with respect to both metrics.

Answer 2

For infinitesimal lengths, hyperbolic and Euclidean metrics are the same. (This follows from the fact that the ratio between squared lengths and Gaussian curvature becomes zero.) So perhaps the author meant the following argument: for infinitesimal lengths, the hyperbolic metric and the Euclidean one agree, and the latter is proportional to first order, so the former must be proportional to first order as well.