# Standard (special) spines and hyperbolic structure on 3-manifolds

My question relates to constructing angled triangulations or hyperbolic triangulations for $3$--manifolds. Briefly, an angle triangulation can be considered as an assignment of a real number (called an angle) to each edge of a tetrahedron such that when we glue the tetrahedron up to obtain the $3$--manifold, around each edge the sum of the angles are $2\pi$, and around each vertex the sum of angles is $\pi$ (we also require that opposite edges of each tetrahedron have the same angle). In short, an angle structure corresponds to the linear part of Thurston's gluing equations -- it corresponds to a weak hyperbolic structure in that the induced metric on the manifold may not be complete.

Now I have assumed the topological conditions required to obtain a hyperbolic structure on the manifold via Thurston's hyperbolization theorem (i.e., the manifolds I am looking at are compact, irreducible, atoroidal with torus boundary) and I am able to explicitly construct special spines for these. Dual to these special spines are ideal triangulations (this comes from the work of Matveev).

My question is whether there are well known conditions for a special spine to be 'geometric'. That is given a special spine, we can look at the special spine as a special polyhedron onto which the manifold collapses. Does it follow that if I look at the special spine as a hyperbolic polyhedron, and I know that the special spine when thickened gives me a $3$--manifold, does this imply that the $3$--manifold is a hyperbolic $3$--manifold. Any discussion, questions and ideas will be appreciated. This is my first time posting here so apologies in advance if I have not followed protocol.

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Maybe you could define what you mean by "hyperbolic polyhedron"? Presumably you want the faces to be hyperbolic polygons, glued isometrically along the edges (this is satisfied e.g. by the Ford domain special spine). But what condition do you want for the angles at each vertex? – Ian Agol Jul 23 '13 at 13:58
Yes, I want the faces to be hyperbolic polygons glued isometrically together along the edges. So I guess I am looking at is a decomposition of a 2-sphere with hyperbolic polygons glued together by isometries. Each vertex is either trivalent or quadrivalent. To give a bit more detail -- the special spine is actually obtained by decomposing the Haken manifold along a hierarchy such as that used by Waldhausen and Johannson. – Don Shanil Jul 24 '13 at 5:32
(continued) As you suggest another way of asking the question is 'what are the combinatorial conditions needed to get a special Ford spine?' – Don Shanil Jul 24 '13 at 5:59

@user37434: there is a sort of dual angle structure on the special spine. Each corner of a polygon dual to an ideal edge will be "dual" to a dihedral corner of a tetrahedron, with $\alpha$. One places the angle $\pi-\alpha$ at the corner of the polygon. Then this angle structure has the property that the sum of the angles over every 3-cycle at a vertex is $2\pi$, and the sum of the "exterior angles" at a polygon is $2\pi$, corresponding to a Euclidean polygon. These sorts of angle structures in the 2-complex context were introduced by Gersten: ams.org/mathscinet-getitem?mr=919828 – Ian Agol Jul 23 '13 at 14:06