How to glue hyperbolic manifolds?

Question

I want to understand the details of gluing for hyperbolic manifolds. By gluing I mean something along the lines of the statement that "a Riemann surface is glued together on the overlap of local charts via holomorphic transition functions" etc.

I am having a hard time finding an equivalent description for hyperbolic manifolds. For example, I often come across the statement that "one can clearly glue to hyperbolic trousers together if their boundary components have the same length" but nowhere (that I have seen) is it described why two boundaries having the same length is enough.

That is, it seems entirely possible that the two hyperbolic metrics could fail to be compatible in a neighborhood of the two boundaries.

Answer 1

Here I sketch how to glue two hexagons in order to make a trouser.

Let $P_1$, $P_2$ be two two-dimensional real manifolds, with boundary and corners. That is, around each point, there is a neighborhood that is diffeomorphic to $\mathbb{R}^2$, or a half-plane of it, or a quarter-of-a-plane of it. Let us assume that $P_1$ and $P_2$ have only finitely many corners.

Let us now assume that $P_1$ and $P_2$ are endowed with hyperbolic structures, that is, there is an atlas such that the domains of the charts are open subsets of hyperbolic polygonals, and such that the change-of-chart-maps (I don't know the word) are (restrictions of) hyperbolic isometries.

Let us now assume that $s_1$ is a side of $P_1$, and $s_2$ is a side of $P_2$ (a side is a connected component of the boundary with the corners removed) that have equal length (length is well-defined and can be defined on charts). Let $f: s_1 \rightarrow s_2$ be an isometry. We are going to endow $P_1 \cup P_2 / f$ with a hyperbolic structure such that the inclusion are isomorphisms of the hyperbolic structure. The set $P_1 \cup P_2/f$ is the disjoint union of (copies of) $P_1$ and $P_2$, quotiented by the relation $x \sim y$ if $f(x) = y$.

For points in the interiors of $P_1$ of $P_2$, we just take the charts we already have; we are now going to define charts on points in the interior of $s$, $s$ being the image of $s_1$ (and $s_2$) in the quotient space $P_1 \cup P_2/f$.

Let $V_1$ be a neighborhood of $s_1$ in $P_1$, and $V_2$ be a neighborhood of $s_2$ in $P_2$. Consider (hyperbolic) charts $\phi_1$ and $\phi_2$ going from $V_1$ and $V_2$ to the hyperbolic plane. Up to composing with isometries of the hyperbolic plane, one can assume that (since the group of isometries act transitively on segments of fixed length), for every $x \in s_1$, $\phi_1(x) = \phi_2(f(x))$, and that $\phi_1(V_1)$ and $\phi_2(V_2)$ lie on different sides of the hyperbolic line passing through $\phi_1(s_1)$ (which is equal to $\phi_2(s_2)$).

Let $x$ be a point in the interior of $s$ and $V$ be a neighborhood of $V$. Then the map $\phi : y \mapsto \{\phi_1(y) \mbox{ if }y\in P_1,\ \phi_2(y)\mbox{ if }y\in P_2\}$ maps $V$ diffeomorphically on an open subset of the hyperbolic plane.

You now just have to check that change-of-chart-maps are hyperbolic isometries, but it should be fine!

In order to glue two hyperbolic trousers along circles that have the same length, you can do the same thing: mark two antipodal points on each circle (so each circle is the union of two segments sharing vertices), you can glue the segments separately by using the arguments above. What is left to check is to prove that you can build charts on the two marked points. Both of these points are at the intersection of four corners, so a necessary condition should be that the sum of the angles is $2\pi$. And the condition is also sufficient: you can then map the four corners around a point in hyperbolic plane and build the chart this way.

I confess having never tried to write everything in detail!

Answer 2

The (sufficiently small) tubular neighborhood of the boundary curve is the quotient of the region between a geodesic and an equidistant curve by a translation along the geodesic by $\ell(\partial).$ The hyperbolic plane is homogeneous enough that the geometry of this tubular neighborhood is entirely determined by the translation distance, so locally when you are gluing the two pairs of pants the neighborhood of the geodesic will be the quotient of the two sided neighborhood of the geodesic by the self-same length.

If you want to avoid group actions, a pair of pants is the double of a right-angled hexagon, so you are asking whether you can glue two such hexagons along a side of equal length. The answer is "yes", and the two right angles add up to $\pi.$ on each side (sorry, don't have a picture).

To summarize, this is just simple hyperbolic geometry. A similar argument works in higher dimensions - you can glue two hyperbolic manifolds with totally geodesic boundaries if the boundaries are isometric (this is the base of the famous Gromov-Piatetski-Shapiro construction of non-arithmetic manifolds).