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Let $M$ be a geometrically finite hyperbolic surface with one cuspidal end and one funnel end so that it can be divided into $C \cup K \cup F$ where $C$ is the cusp, $F$ the funnel and $K$ the compact core.

Focus on the cusp with the following boundary defining function (due to David Borthwick: Spectral Theory of Infinite-Area Hyperbolic Surfaces)

Let $\rho (x)=e^{-r}$ where $r$ is the distance of $x$ to the compact core $K$. As a boundary defining function it needs to have a non-vanishing differential on the boundary. How can this fact be seen for this function?

If I take the look back to the upper half plane the hyperbolic distance for distinct points $a$ and $b$ located on the y-axis is given by $log(a/b)$. I suppose that's the reason for choosing $e^{-r}$. If you take a point $c$ at the boundary its distance to the compact core is infinite, so $e^{-r(c)}$ is zero, clearly. But what's with the differential? Is there a direct way of calculating it in this case with the given information? I don't understand why is it not zero at the boundary.

In a discussion with others there was the idea to show it by relating the the boundary defining function for the cusp to a (maybe given) boundary defining function for a funnel end by 'inverting', precisely:

Here $M = (0, \infty) \times S^1$ with the metric $g=(dx^2+d\theta^2)/x^2$ where the variable $x$ refers to $(0,\infty)$ and $x$ itself is a boundary defining for the funnel end. If $x$ is bdf for the funnel, is it true that then 1/x is a boundary defining function for the cusp?

My problem is the absence of a diffeomorphism from the cusp (as a point) to the funnel (which is a circle). Can this be repaired by viewing the cusp as a 'infinitely' small circle? The strategy I have in mind is: If there would be a diffeomorphism between the ends and given a boundary defining function for the funnel then the pullback of it has non-vanishing differential.

Since this is my first question on MO, I hope my explanations were precise enough and the questions aren't too basic.

Thank you for your help, Robin Neumann

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