[Novikov 1962] For n � 5 , there does not exist an algorithm which solves: I S S PHERE : Given a triangulated M n is it Computational complexity homeomorphic to S n ? of problems in 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. SAT: Given x ∈ F k 2 , can check all p i ( x ) = 0 S ORTED : Given a list of integers, is it sorted? in linear time. SAT: Given p 1 , . . . p n ∈ F 2 [ x 1 , . . . , x k ] is there x ∈ F k 2 with p i ( x ) = 0 for all i ? U NKNOTTED : A diagram of the unknot with c crossings can be unknotted in O ( c 11 ) U NKNOTTED : Given a planar diagram for K Reidemeister moves. [Lackenby 2013] in S 3 is K the unknot? I NVERTIBLE : Given A ∈ M n ( Z ) does it have an inverse in M n ( Z ) ? coNP: No answers can be checked in polynomial time. U NKNOTTED : Yes, assuming the GRH P: Decision problems which can be solved [Kuperberg 2011]. in polynomial time in the input size. S ORTED : O ( length of list ) n 3.5 log ( largest entry ) 1.1 � � I NVERTIBLE : O Conj: U NKNOTTED is in P .
K NOT G ENUS : Given a triangulation T , a knot Is K NOT G ENUS in P when b 1 = 0 ? K ⊂ T ( 1 ) , and a g ∈ Z � 0 , does K bound an orientable surface of genus � g ? Is the homeomorphism problem for [Agol-Hass-W.Thurston 2006] 3-manifolds in NP ? K NOT G ENUS is NP -complete. What about deciding hyperbolicity? or Conj (AHT) If b 1 ( T ) = 0 , then K NOT G ENUS being an L-space? is in coNP . Computing Khovanov homology and � HFK [AHT] K NOT A REA is NP -complete. are in EXPTIME . Just computing the Jones polynomial is #P -hard, but the Alexander [Dunfield-Hirani 2011] K NOT A REA is in P when b 1 ( T ) = 0 . polynomial can computed in poly time.
Normal surfaces meet each tetrahedra in a [Hass-Lagarias-Pippenger 1999] � exp t 2 � A fund surface has coordinates O standard way: . and correspond to lattice points in a finite [AHT 2006] K NOT G ENUS is in NP . polyhedral cone in R 7 t where t = # T : Certificate: A vector x in Z 7 t with entries with a most O ( t 2 ) digits. Check: (1) That x represents a normal surface S . (2) That χ ( S ) � 1 − 2 g . (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 surface bounding K in normal form whose need a very clever idea for (3) and (4). vector is fundamental (e.g. on a vertex ray). Hence K NOT G ENUS is decidable.
[Kuperberg 2011] Assuming GRH, “In theory, there is no difference between U NKNOTTING is in coNP . theory and practice. But, in practice, there is.” Certificate: ρ : π 1 ( S 3 − K ) → SL 2 F p –Jan L. A. van de Snepscheut � � poly ( crossings ) where log p is O . Mystery: In practice, many 3-m fl d questions Check: The following imply π 1 is not cyclic are easier than the best theoretical bounds and so K is knotted. indicate. (1) Relators for π 1 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! Here ’ s a sneak peak of joint work with Malik [Kronheimer-Mrowka 2004] When K ⊂ S 3 is nontrivial, there is a rep Obeidin, based on one natural model of π 1 ( S 3 − K ) → SU 2 with nonabelian image. random knot.
Personal best: crossings: 126 genus: 27 fi bered: No time: 7 minutes hyperbolic volume: 223.6132847441086613 tetrahedra: 243
100 genus bound from Alexander poly 80 60 40 20 0 − 20 − 50 0 50 100 150 200 250 300 350 crossings
10 5 slope=5.105 10 4 intercept=-10.037 10 3 r=0.937 alexander_magma_time 10 2 10 1 10 0 10 -1 10 -2 10 -3 10 1 10 2 10 3 crossings
2500 2000 hyperbolic volume 1500 1000 500 0 − 500 − 200 0 200 400 600 800 1000 1200 crossings
10 2 slope=2.331 10 1 intercept=-6.098 r=0.962 10 0 volume_time 10 -1 10 -2 10 -3 10 -4 10 -5 10 0 10 1 10 2 10 3 10 4 crossings
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