Many cusped hyperbolic 3-manifolds do not bound geometrically

Abstract

In this note, we show that there exist cusped hyperbolic 3-manifolds that embed geodesically, but cannot bound geometrically. Thus, being a geometric boundary is a non-trivial property for such manifolds. Our result complements the work by Long and Reid on geometric boundaries of compact hyperbolic 4-manifolds, and by Kolpakov, Reid and Slavich on embedding arithmetic hyperbolic manifolds.

A small cusped hyperbolic 4‐manifold

“…Concerning cusped manifolds, we found instead four commensurability classes: (we cite here the papers containing the smallest known manifolds in each class) (1)$$\chi = 1$$ both orientable and not, arithmetic [5, 12, 21, 22, 25]; (2)$$\chi = 1$$ both orientable and not, arithmetic [2, 11, 24, 26]; (3)$$\chi = 2$$ non‐orientable (so also $$\chi = 4$$ orientable), arithmetic [18, 24]; (4)$$\chi = 3$$ non‐orientable and $$\chi = 5$$ orientable, non‐arithmetic [24]. …”

Counting cusped hyperbolic 3-manifolds that bound geometrically

“…Following Long and Reid [21,22], we say that a hyperbolic n-manifold M bounds geometrically if it is isometric to ∂W , for a hyperbolic (n + 1)-manifold W with totally geodesic boundary. This class of hyperbolic manifolds has attracted significant interest, and for n = 3 some progress has been recently done -see [18,19,23,26,34,35].…”

“…As follows from [19,21], to bound geometrically is an extremely non-trivial property for a hyperbolic 3-manifold, both in the compact and non-compact setting, respectively (c.f. [19,Remark 1.4]). Despite this, in the present work we show that there are plenty of geometrically bounding cusped (i.e.…”

Embedding arithmetic hyperbolic manifolds

“…Despite the fact that "most" hyperbolic manifolds do not bound geometrically [17,20], for any n ≥ 2 there is a constant c > 0 such that the number β n (v) of ndimensional geometric boundaries of volume ≤ v is at least v cv , for v sufficiently big [10]. For n ≥ 4, the number µ n (v) of all hyperbolic n-manifolds with volume ≤ v satisfies v cv ≤ µ n (v) ≤ v dv for v large enough [5], so that β n and µ n have the same the growth rate (while usually µ n (v) = ∞ for n = 2 or 3).…”