[Novikov 1962] For n > 5, there does not exist an algorithm which solves: n Computational complexity ISSPHERE: Given a triangulated M is it n of problems in homeomorphic to S ? 3-dimensional topology Thm (Geometrization + many results) There is an algorithm to decide if two compact 3-mflds are homeomorphic.
Nathan Dunfield University of Illinois
slides at: http://dunfield.info/preprints/ Today: How hard are these 3-manifold questions? How quickly can we solve them? NP: Yes answers have proofs that can be Decision Problems: Yes or no answer. checked in polynomial time. k x F pi(x) = 0 SORTED: Given a list of integers, is it sorted? SAT: Given ∈ 2 , can check all in linear time. SAT: Given p1,... pn F2[x1,..., xk] is k ∈ x F pi(x) = 0 i UNKNOTTED: A diagram of the unknot with there ∈ 2 with for all ? 11 c crossings can be unknotted in O(c ) UNKNOTTED: Given a planar diagram for K 3 Reidemeister moves. [Lackenby 2013] in S is K the unknot?
A Mn(Z) INVERTIBLE: Given ∈ does it have an inverse in Mn(Z)? coNP: No answers can be checked in polynomial time. UNKNOTTED: Yes, assuming the GRH P: Decision problems which can be solved [Kuperberg 2011]. in polynomial time in the input size. SORTED: O(length of list) 3.5 1.1 INVERTIBLE: O n log(largest entry) NKNOTTED is in . Conj: U P T KNOTGENUS: Given a triangulation , a knot b1 = 0 (1) Is KNOTGENUS in P when ? K T g Z 0 K ⊂ , and a ∈ > , does bound an orientable surface of genus 6 g? Is the homeomorphism problem for [Agol-Hass-W.Thurston 2006] 3-manifolds in NP? KNOTGENUS is NP-complete. What about deciding hyperbolicity? or Conj (AHT) If b1(T) = 0, then KNOTGENUS being an L-space? is in coNP.
Computing Khovanov homology and HFK[ [AHT] KNOTAREA is NP-complete. are in EXPTIME. Just computing the Jones polynomial is #P-hard, but the Alexander [Dunfield-Hirani 2011] KNOTAREA is in P when b1(T) = 0. polynomial can computed in poly time. Normal surfaces meet each tetrahedra in a [Hass-Lagarias-Pippenger 1999] 2 standard way: A fund surface has coordinates O exp t . and correspond to lattice points in a finite [AHT 2006] KNOTGENUS is in NP. 7t polyhedral cone in R where t = #T: 7t Certificate: A vector x in Z with entries 2 with a most O(t ) digits.
Check: (1) That x represents a normal surface S. (2) That χ(S) 6 1 − 2g. (3) That S connected and orientable. (4) That ∂S is as advertised. [Haken 1961] There is a minimal genus All can be done in time polynomial in t but K surface bounding in normal form whose need a very clever idea for (3) and (4). vector is fundamental (e.g. on a vertex ray). Hence KNOTGENUS is decidable. Assuming GRH, [Kuperberg 2011] “In theory, there is no difference between NKNOTTING is in . U coNP theory and practice. But, in practice, there is.” 3 –Jan L. A. van de Snepscheut ρ: π1(S − K) SL2Fp Certificate: → where log p is O poly(crossings) . Mystery: In practice, many 3-mfld questions Check: The following imply π1 is not cyclic are easier than the best theoretical bounds and so K is knotted. indicate. (1) Relators for π1 hold, so ρ is a rep. Q: How big a knot can we compute the genus (2) A pair of generators have for? noncommuting images. Q: Where do we even get big knots from? Proof that such a rep exists uses algebraic There are more 100 crossing prime knots than geometry/number theory and: there are atoms in the Earth!
[Kronheimer-Mrowka 2004] Here’s a sneak peak of joint work with Malik K S3 Obeidin, based on one natural model of When ⊂ is nontrivial, there is a rep π (S3 − K) SU random knot. 1 → 2 with nonabelian image. 0 erhda 243 tetrahedra: volume: hyperbolic minutes 7 time: fi 27 genus: 126 crossings: best: Personal ee:No bered: 223.6132847441086613 100
80
60
40
20
0 genus bound from Alexander poly
−20 −50 0 50 100 150 200 250 300 350
crossings 105
slope=5.105
104 intercept=-10.037
3 10 r=0.937
102
101
100 alexander_magma_time
10-1
10-2
10-3 101 102 103 crossings 2500
2000
1500
1000
0
−500 −200 0 200 400 600 800 1000 1200
crossings 102
slope=2.331
101 intercept=-6.098
r=0.962 100
10-1
10-2 volume_time
10-3
10-4
10-5 100 101 102 103 104 crossings