Triangulated 3-manifolds: from Haken to Thurston Feng Luo Rutgers University May, 19, 2010 Barrett Memorial Lecture University of Tennessee
Part of this is a joint work with Stephan Tillmann.
Recall of surfaces Uniformization Thm. Given a Riemann surface S, there exists a conformal constant curvature metric g on S. Question. How to compute g? Input: triangulated surface (S, T) with some structure. Output: a constant curvature metric g on (S, T). The work: Thurston on circle packing, Colin de Verdiere, Stephenson, Bobenko-Springborn, Gu-Yau, Leibon, Rivin, Braegger, Chow-Luo, and many others
The work of Colin de Verdiere, 1990 He constructed a variational principle for circle packings. 1.wrote down the energy function 2.Introduced Euclidean angle structures (EAS) on (S, T): assign each corner a positive real number so that sum in each triangle is π, sum at each vertex is 2π..
THM(Colin de Verdiere). For any (S, T), then (a) either there are no EAS on (S, T), or (b) the energy function has a max point on the space of all radii. max point of energy → a metric g on (S,T), the topological obstruction →modify triangulation T.
3-manifolds: the geometrization thm of Perelman-Thurston M closed, oriented. 1.Cut M open along S 2 ’s to obtain M = N 1 # N 2 # … # N k where N i cannot be decomposed further (irreducible) 2. M irreducible, cut open along (essential) tori T 2 ’s to obtain W 1, W 2, …, W n. W i cannot be decomposed further (atoriodal) THM(Perelman-Thurston) W atoriodal, then either W is hyperbolic or W admits an S 1 -action. By Mostow, the hypebolic metric is unique.
Similarity between 2-d and 3-d There are prime knots with < 17 crossing, all but 32 of them are hyperbolic.
Problem Given triangulated (M 3, T), compute the spheres, the tori and the hyperbolic metrics. There are softwares available: Snapea etc, mainly for compact manifolds M with torus boundary. Work of Haken, Jaco-Rubinstein, Li, Weeks and others We propose a variational principle to approach it.
1. collection of oriented tetra A 1, …, A k 2. Identify faces in pairs by affine orientation reversing homeomorphisms. The result is an oriented closed pseudo 3- manifold with a triangulation (M, T). Triangulation of 3-manifolds
Thurston’s example of
Normal quads and triangles in tetra
properties of quads 1. Each quad = pair of opposite edges 2. q 1, q 2, q 3 are 3 quads in an oriented tetra, then cyclic order q 1 ->q 2 ->q 3 -> q 1 depends only on the orientation.
Haken’s theory of Normal surfaces A surface S in (M, T) is normal if for each tetra K, S ∩ K is a collection of triangles and quads:
Haken’s question Given a collection of t’s and q’s in T, when can you produce a normal surface out of these?
Haken’s equation S normal, its coordinate x is in R ∆ X R □ x(t)= # copies of t in S x(q) = # copies of q in S Q. Is x in R ∆ X R □ a coordinate? Ans. It satisfies Haken’s equation x(t)+ x(q) = x(t’) + x(q’) (*) Def. a 2-quad-type solution x of (*): exist two quads q 1 ≠q 2 so that x(q)=0 for all other q’s and x(q 1 )≠0.
Thurston’s way to produce geometry from T Make each tetra ideal hyperbolic tetra Glue by isometries Match them nicely at edges
Thurston’s parametrization Shape of an ideal tetra in H 3 is given by z є C-{0,1}. Opposite edges have the same parameter, i.e. defined on quads.
Thurston’s equation on (M, T) Shape parameter z in (C-{0,1}) □ so that, 1.For q->q’, z(q’) = 1/(1-z(q)). 2. For each edge e, ∏ q ~e z(q) = 1. If the right-hand-side is ±1, it is the generalized Thurston’s equation.
Remark Thurston used a solution of his equation to produce the hyperbolic metric on the figure-8 knot complement in 1978.
Theorem 1. (M, T) closed oriented pseudo 3-manifold, then either (1) there exists a solution to generalized Thurston’s equation, or (2) there exists a 2-quad-type solution to Haken’s equation.
A variational principle in 3-D Def. S 1 -angle structure (SAS) on (M, T) assigns each quad q, x(q) є S 1 so that (1)If x 1, x 2, x 3 assigned to 3 quads in a tetra, then x 1 x 2 x 3 =-1 (2)For each edge e, ∏ q ~e x(q) = 1. SAS(T)= the set of all SAS’s on T. It is non-empty and is a smooth closed manifold. RM. This generalizes Casson, Rivin’s earlier definition.
Volume of SAS Lobachevsky formula for volume of tetra Def. The volume of an SAS x is Vol(x) =sum of volume of tetra
The Lobachevsky function is periodic of period π and is continuous, non-smooth at Zπ. So vol: SAS(T) -> R is continuous, non-smooth. Vol has a maximum point p. Thm 1’. If p is a smooth point, then (1) holds. If p is non-smooth, then (2) holds. Conj. For a minimal triangulated closed 3-manifolds, if all max points are smooth, then there is a solution to Thurston’s equation. (Thanks to Burton and Segerman).
Sketch of proof
A triangulation of M is minimal if it has the smallest number of tetra among all triangulations of M. Thm 2(L-Tillmann). If (M,T) minimal triangulated closed 3-manifold and vol: SAS(T) -> R has a non-smooth max point, then M is either (1) reducible, or (2) toriodal, or (3) admitting an S 1 -action, or (4) contains the connected sum of the Kleinbottle and the projective plane.
A very recent development I was informed by Tillmann 3 weeks ago that he and Segerman can prove. Thm(Segerman-Tillmann) If (M, T) a closed 3- manifold supports a solution to Thurston’s equation, then each edge in T is homotopically essential. This thm and the conjecture (all smooth max -> sol to Th. eq) give a new proof of the Poincare conjecture (without using Ricci flow).
the argument Suppose M is a closed 3-manifold homotopic to S 3. By Kneser (1929), may assume M irreducible. Take T to be a minimal triangulation of M. By Jaco-Rubinstein (2001), T has only one vertex. Since each edge is null homotopic, by Segerman Tillmann, T has no solution to Thurston’s eq. By the conjecture (all smooth max-> sol. to Th), then T has a non-smooth max point. By L-Tillmann and irreducibility of M, thus M is S 3.
Conclusion Volume maximization on circle-valued angle structures on (M, T). 1.non-smooth max point links to Haken’s normal surfaces, 2.smooth max points links to Thurston’s equation which produces hyperbolic structures. Ref. F. Luo: F. Luo:
Thank you