GEOMETRY OF GROUPS AND COMPUTATIONAL COMPLEXITY · Hence the co-compact lattices in semi-simple Lie...

GEOMETRY OF GROUPS AND

COMPUTATIONAL COMPLEXITY

Mark V. Sapir

Prague, June 17, 2010

The word problem

Let G = 〈X | R〉 be a finitely presented group; G = F (X )/N.

The word problem


The word problem: Given a word w ∈ F (X ), decide if w ∈ N,

The word problem


The word problem: Given a word w ∈ F (X ), decide if w ∈ N,i.e. if w = 1 in G ,

The word problem


The word problem: Given a word w ∈ F (X ), decide if w ∈ N,i.e. if w = 1 in G ,i.e. w =

∏m

i=1si ri s

−1

iwhere si ∈ F (X ), ri ∈ R .

The word problem


The word problem: Given a word w ∈ F (X ), decide if w ∈ N,i.e. if w = 1 in G ,i.e. w =

∏m

i=1si ri s

−1

iwhere si ∈ F (X ), ri ∈ R .

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q q q

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r1

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Vam Kampen diagrams and tilings

After cancelation, we get a planar graph with boundary label w :



Thus the word problem is a tiling problem



Thus the word problem is a tiling problemThe direct part of van Kampen lemma: If w = 1 in G , thenthere is a van Kampen diagram ∆ over the presentation of G withboundary label w .

An elementary problem and non-elementary solution

Let P be the standard 8× 8 chess board with two opposite squaresremoved. Prove that P cannot be tiled by the standard 2× 1dominos.

An elementary problem and non-elementary solution

Let P be the standard 8× 8 chess board with two opposite squaresremoved. Prove that P cannot be tiled by the standard 2× 1dominos.

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a a a a a a a

a

a a

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ab bb b

b bb

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The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab.

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1.

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1. Consider the group G with the presentation〈a, b | ab2a−1b−2 = 1, a2ba−2b−1 = 1〉 (the Conway tiling group).

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1. Consider the group G with the presentation〈a, b | ab2a−1b−2 = 1, a2ba−2b−1 = 1〉 (the Conway tiling group).The word W is equal to 1 in G .

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1. Consider the group G with the presentation〈a, b | ab2a−1b−2 = 1, a2ba−2b−1 = 1〉 (the Conway tiling group).The word W is equal to 1 in G . Consider the 6-element symmetricgroup S3 and two permutations α = (1, 2), β = (2, 3) in it.

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1. Consider the group G with the presentation〈a, b | ab2a−1b−2 = 1, a2ba−2b−1 = 1〉 (the Conway tiling group).The word W is equal to 1 in G . Consider the 6-element symmetricgroup S3 and two permutations α = (1, 2), β = (2, 3) in it. Themap a 7→ α, b 7→ β extends to a homomorphism G → S3.

The solution

The (counterclockwise) boundary of P has labelW = a7b7a−1ba−7b−7ab. Every domino can be placed eithervertically or horizontally. In the first case its boundary label isab2a−1b−2, and in the second case its boundary label isa2ba−2b−1. Consider the group G with the presentation〈a, b | ab2a−1b−2 = 1, a2ba−2b−1 = 1〉 (the Conway tiling group).The word W is equal to 1 in G . Consider the 6-element symmetricgroup S3 and two permutations α = (1, 2), β = (2, 3) in it. Themap a 7→ α, b 7→ β extends to a homomorphism G → S3.W (α, β) = (αβ)4 = αβ = (1, 3, 2) which is not trivial. Hence W

is not equal to 1 in G , a contradiction.

The converse part of the van Kampen lemma

Lemma (A. Olshanskii, S.) Let ∆ be a van Kampen diagramover a presentation 〈X | R〉.


Lemma (A. Olshanskii, S.) Let ∆ be a van Kampen diagramover a presentation 〈X | R〉. Let w be the boundary label of ∆.Then w is equal in the free group to a word of the formu1r1u2r2...umrdum+1 where:


Lemma (A. Olshanskii, S.) Let ∆ be a van Kampen diagramover a presentation 〈X | R〉. Let w be the boundary label of ∆.Then w is equal in the free group to a word of the formu1r1u2r2...umrdum+1 where:1. ri ∈ R ; u1u2...um+1 = 1 in the free group;2.

∑m+1

i=1|ui | ≤ 4e where e is the number of edges of ∆.

In particular, w = 1 in G .



∑m+1



Thus the size of a van Kampen diagram is approximately equal tothe number of cells plus the length of W .



∑m+1




The non-deterministic complexity of the word problem in G

is at most the Dehn function of G , i.e. the function f (n)which is the maximal area of van Kampen diagram withboundary length at most n



∑m+1




The non-deterministic complexity of the word problem in G

is at most the Dehn function of G , i.e. the function f (n)which is the maximal area of van Kampen diagram withboundary length at most n

In particular the word problem is decidable if and only if the Dehnfunction is bounded by a recursive function.

The Baumslag-Solitar group, its Dehn function.

BS(1, 2) = 〈a, b | b−1ab = a2〉

The Baumslag-Solitar group, its Dehn function.

BS(1, 2) = 〈a, b | b−1ab = a2〉

- -

- -- -

R R� �K Kb b

a a

a

6K

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�b

b b b

ba

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bn bn

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cup

cap

A characterization of groups with word problem in NP

If G < H, then we can tile discs with boundary labels from G byrelations of H.


If G < H, then we can tile discs with boundary labels from G byrelations of H. So if H has polynomial Dehn function, then theword problem in G is in NP.


If G < H, then we can tile discs with boundary labels from G byrelations of H. So if H has polynomial Dehn function, then theword problem in G is in NP.

Theorem (Birget, Olshanskii, Rips, S., Ann. of Math. 2002)The word problem of a finitely generated group G is in NP if andonly if G is embedded into a finitely presented group withpolynomial Dehn function.

A connection with the classical isoperimetric problem for

manifolds

Theorem (Gromov, Bridson): If a group G acts properlyco-compactly on a simply connected Riemannian manifold withisoperimetric function f , then the Dehn function of G is equivalentto f .


manifolds

Theorem (Gromov, Bridson): If a group G acts properlyco-compactly on a simply connected Riemannian manifold withisoperimetric function f , then the Dehn function of G is equivalentto f .Hence the co-compact lattices in semi-simple Lie groups of rank> 1 have quadratic Dehn function.


manifolds


Non-uniform lattices are harder to deal with.


manifolds


Non-uniform lattices are harder to deal with.

Theorem (R. Young, solving a conjecture of Thurston): Thegroup SL(n,Z) has quadratic Dehn function if n ≥ 5.

Nilpotent groups

Theorem (Gromov, Gersten+Riley): The Dehn function of anilpotent group of class c is at most nc+1.

Nilpotent groups


Theorem (Alcock, Olshanskii+S., R. Young): The Dehnfunctions of higher dimensional Heisenberg groups are quadratic.

Nilpotent groups


Theorem (Alcock, Olshanskii+S., R. Young): The Dehnfunctions of higher dimensional Heisenberg groups are quadratic.

Theorem (S. Wenger): There are nilpotent groups with Dehnfunctions not of the form nk for any k (bigger than n2 and smallerthan n2 log n.

Connections with asymptotic cones

Finitely generated groups are metric spaces. dist(a, b) = |a−1b|.



If (X , dist) is a metric space, o ∈ X , dn → ∞, and ω is anultrafilter, we can consider the limit Conω(X , (dn), o)




Examples. The asymptotic cone of Zn is Rn with the l1-metric.




Examples. The asymptotic cone of Zn is Rn with the l1-metric.The asymptotic cone of a nilpotent group is a nilpotent Lie group.




Examples. The asymptotic cone of Zn is Rn with the l1-metric.The asymptotic cone of a nilpotent group is a nilpotent Lie group.The asymptotic cone of a (non-elementary) hyperbolic group is areal tree of degree continuum at each point




Examples. The asymptotic cone of Zn is Rn with the l1-metric.The asymptotic cone of a nilpotent group is a nilpotent Lie group.The asymptotic cone of a (non-elementary) hyperbolic group is areal tree of degree continuum at each point The asymptotic coneof a co-compact lattice in a Lie group is a building, etc.





Theorem (Gromov): If all asymptotic cones of G are simplyconnected, then G is finitely presented and its Dehn function ispolynomial.





Theorem (Gromov): If all asymptotic cones of G are simplyconnected, then G is finitely presented and its Dehn function ispolynomial.

Question: Is it true that every NP-group is inside a group withsimply connected asymptotic cones?

Quadratic Dehn functions

Theorem (Gromov, Papasoglu). Quadratic Dehn functionimplies simply connected asymptotic cones.

Quadratic Dehn functions

Theorem (Gromov, Papasoglu). Quadratic Dehn functionimplies simply connected asymptotic cones.

Examples: SLn(Z), n ≥ 5, the CAT(0)-groups, automatic groups,the R. Thompson group, etc.

The set of Dehn functions

Theorem (Gromov, Olshanskii, Papasoglu, Bowditch,

Wenger) If the Dehn function is smaller than n2

4π, then the group

is hyperbolic and its Dehn function is linear.




4π, then the group


Theorem (Bridson, Brady, etc.): There are Dehn functions ofthe form na for some transcendental a.




4π, then the group



Theorem (S., Birget, Rips): For every superadditive timefunction T of a non-deterministic Turing machine, there exists afinitely presented group with Dehn function equivalent to T 4.




4π, then the group




Theorem (S., Birget, Rips): Let α be a number > 4 computablein time at most 22

n

. Then there exists a finitely presented groupwith Dehn function nα.




4π, then the group





n

. Then there exists a finitely presented groupwith Dehn function nα. If nα is equivalent to the Dehn function ofa finitely presented group, then α is computable in time at most

222α

.




4π, then the group





n

. Then there exists a finitely presented groupwith Dehn function nα. If nα is equivalent to the Dehn function ofa finitely presented group, then α is computable in time at most

222α

.

Examples. π + e, any algebraic number > 4, etc.

Some weird Dehn functions

Theorem (Olshanskii, S.) There exists a finitely presented groupwith non-recursive word problem and almost quadratic Dehnfunction.

Thank you!

GEOMETRY OF GROUPS AND COMPUTATIONAL COMPLEXITY · Hence the co-compact lattices in semi-simple Lie...

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