Conjugacy search problem and the Andrews–Curtis conjecture

Dmitry Panteleev; Alexander Ushakov

doi:10.1515/gcc-2019-2005

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Conjugacy search problem and the Andrews–Curtis conjecture

Dmitry Panteleev and Alexander Ushakov

Published/Copyright: April 6, 2019

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From the journal Groups Complexity Cryptology Volume 11 Issue 1

Abstract

We develop new computational methods for studying potential counterexamples to the Andrews–Curtis conjecture, in particular, Akbulut–Kurby examples AK⁡(n). We devise a number of algorithms in an attempt to disprove the most interesting counterexample AK⁡(3). That includes an efficient implementation of the folding procedure for pseudo-conjugacy graphs, based on the original modification of a classic disjoint-set data structure. To improve metric properties of the search space (the set of balanced presentations of the trivial group), we introduce a new transformation, called an ACM-move, that generalizes the original Andrews–Curtis transformations and discuss details of a practical implementation. To reduce growth of the search space, we introduce a strong equivalence relation on balanced presentations and study the space modulo automorphisms of the underlying free group. We prove that automorphism moves can be applied to Akbulut–Kurby presentations. The improved technique allows us to enumerate balanced presentations AC-equivalent to AK⁡(3) with relations of lengths up to 20 (previous record was 17).

Keywords: Andrews–Curtis conjecture; Akbulut–Kurby presentations; trivial group; conjugacy search problem; computations

MSC 2010: 20-04; 20F05; 20E05

Funding source: National Science Foundation

Award Identifier / Grant number: DMS-1318716

Funding statement: The second author has been partially supported by NSF grant DMS-1318716.

A Used ACM-moves justification

In this section, we prove the one-relator groups identities used in Lemmas 3.4, 3.5 and 3.6 for the ACM moves. Every proof demonstrates that

c-1⁢u⁢c⁢(u′)-1=1

in 〈x,y∣v〉.

A.1 Used in Lemma 3.4

c=x⁢y⁢x, u=xk⁢Yk+1, v=x⁢y⁢x⁢Y⁢X⁢Y, u′=yk⁢XK+1:
X⁢Y⁢X⋅xk⁢(Yk+1⋅x⁢y⁢x⋅xk+1)⁢Yk=X⁢Y⁢X⁢(xk⋅x⁢y⁢x⋅YK)(Y⁢x⁢y⁢x=x⁢y)
=X⁢Y⁢X⁢x⁢y⁢x(x⁢y⁢x⁢Y=y⁢x)
=1

A.2 Used in Lemma 3.5

u=x⁢y⁢X⁢yk+1⁢x⁢Yk+2, v=x⁢y⁢X⁢y⁢x⁢Y, c=yk+1, u′=Yk+1⁢Xk:
(Yk+1)⋅x⁢y⁢X⁢yk+1⁢x⁢(Yk+2⋅yk+1)⋅xk⁢(yk+1)=(xk+1⁢y⁢X⁢yk+1)⁢x⋅Y=(xk⁢y⁢X⁢yk)⋅x⁢Y=⋯=y⁢X⁢x⁢Y=1
u=x⁢y⁢X⁢y⁢x⁢Y⁢Y⁢X⁢y, v=x⁢y⁢X⁢yk+1⁢x⁢Yk+2, c=yk+2⁢X⁢y⁢x⁢Y, u′=x⁢y⁢X⁢y⁢x⁢Y:
y⁢X⁢Y⁢x⁢Yk+2⋅x⁢y⁢X⁢y⁢x⁢Y⁢Y⁢X⁢(y⋅yk+2⁢X⁢y⁢x⁢Y⋅y⁢X⁢Y⁢x⁢Y)⁢X=y⁢X⁢Y⁢(x⁢Yk+2⁢x⁢y⁢X⁢y)⁢x⁢Y⁢Y⁢X⋅yk+2⁢X=y⁢X⁢Y⋅Yk⋅x⁢Y⁢(Y⁢X⁢yk+2⁢X)=(y⁢X)⁢Yk+1⁢(x⁢Y⋅X⁢yk+1)=Yk+1⋅yk+1=1
u=x⁢y⁢X⁢yk⁢X⁢Yk, v=x⁢y⁢X⁢y⁢x⁢Y⁢Y⁢X⁢y, c=x⁢y⁢X⁢y, u′=yk+2⁢X⁢Yk+1⁢x⁢Y⁢X:
(Y⁢x⁢Y⁢X)⋅(x⁢y⁢X⁢yk)⁢X⁢Yk⋅x⁢y⁢X⁢y⋅x⁢y⁢X⁢yk+1⁢x⁢(Yk+2)=X⁢Yk⁢x⁢y⁢(X⁢y⁢x⁢y⁢X⁢yk+1)⁢x⋅Y3=X⁢(Yk⁢x⁢y⋅y)⁢y⁢X⁢yk⋅x⁢Y3=X⋅Yk-1⁢x⁢y⁢(X⁢y⁢x⋅y⁢X⁢yk)⁢x⁢Y3=X⁢(Yk-1⁢x⁢y⋅y)⁢y⁢X⁢yk-1⋅x⁢Y3=X⁢x⁢y⋅y⁢y⁢X⋅x⁢Y3=1
u=xk⁢Yk+1, v=x⁢y⁢X⁢yk⁢X⁢Yk, c=Yk-1⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X⁢y, u′=Y⁢x⁢y⁢y⁢X⁢Y⁢x⁢Y⁢X:
(Y⁢x⁢y⁢y⁢X)⁢yk-1⁢X⁢yk-1⋅xk⁢Yk+1⋅Yk-1⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X⁢y⋅x⁢y⁢X⁢y⁢(x⁢Y⁢Y⁢X⁢y)
=(yk-1⁢X)⁢yk-1⁢xk⁢Y2⁢k⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X⁢y⁢(x⁢y⁢X⁢y)
=(yk⋅yk-1)⁢xk⁢Y2⁢k⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X⁢(y)(x⁢y⁢X⁢yk⁢X=yk)
=yk⁢x⁢(X⁢yk⋅xk)⁢Y2⁢k⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X(shift)
=yk⁢x⋅y⁢(X⁢yk⋅xk-1)⁢Y2⁢k⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X(X⁢yk⁢x=y⁢X⁢yk)
=⋯
=yk⁢x⋅yk⁢X⁢(yk⋅Y2⁢k)⁢x⁢Yk-1⁢x⁢Y⁢Y⁢X
=yk⁢x⁢(yk⁢X⋅Yk⋅x)⁢Yk-1⁢x⁢Y⁢Y⁢X
=(yk⁢x)⋅x⁢Y⋅Yk-1⁢x⁢Y⁢(Y⁢X)
=(X⁢yk⋅x)⁢Yk⁢x⁢Y
=(X⁢yk⋅x)⁢Yk⁢(x⁢Y)
=yk⋅Yk=1
u=x⁢y⁢x⁢Y⁢X⁢Y, v=xk⁢Yk+1, c=Yk, u′=yk⁢x⁢Yk⁢x⁢Y⁢X:
(yk⋅x⁢y)⁢x⁢Y⁢X⁢Y⋅Yk⋅x⁢y⁢X⁢(yk⁢X⁢Yk)=x⁢Y⁢X⁢(Y⁢Yk)⁢x⁢y⁢X⋅yk+1=x⁢Y⁢(X⋅Xk⁢x)⁢y⁢X⁢yk+1=x⁢(Y⋅Yk+1⋅y)⁢X⁢yk+1=⋯=Yk+1⁢yk+1

A.3 Used in Lemma 3.6

u=xk-1⁢y⁢Xk-1⁢y⁢X⁢Y, v=x⁢x⁢y⁢X⁢Y⁢X⁢y, c=xk-3⁢(Y⁢X)k⁢Y, u′=xk-1⁢(Y⁢X)k⁢Y:
y⁢(x⁢y)k⁢(Xk-3⋅xk-1)⁢y⁢Xk-1⁢y⁢X⁢Y⋅(xk-3⁢(Y⁢X)k⁢Y⋅y⁢(x⁢y)k⁢Xk-1)=y⁢(x⁢y)k⋅x2⁢y⁢Xk-1⁢(y⁢X⁢Y⋅X)⁢X=(y⁢x)k⁢y⁢x2⁢y⁢Xk-1⋅X⁢X⁢(Y⋅X)=⋯=y⁢x2⁢y⁢Xk-1⁢X⁢X⋅xk⁢Y⁢X=y⁢x2⁢y⁢X⁢Y⁢X=1
u=xk⁢Yk+1, v=xk-1⁢y⁢Xk-1⁢y⁢X⁢Y, c=x⁢y⁢Xk-1⁢y⁢X⁢Y⁢X⁢y, u′=Y⁢x⁢y⁢x⁢Y⁢X⁢X:
(Y⁢x⁢y⁢x⁢Y)⁢xk-1⁢Y⁢(X⋅xk)⁢Yk+1⋅x⁢y⁢Xk-1⁢y⁢X⁢Y⁢X⁢y⋅x⁢x⁢(y⁢X⁢Y⁢X⁢y)=(xk-1⁢Y⁢xk-1⁢Yk+1)⋅x⁢y⁢Xk-1⁢y⁢X⁢Y⁢X⁢y⁢(x⁢x)=⋯=Y⁢(xk-1⋅x⁢y⁢Xk-1⁢y⁢X⁢Y)⁢X⁢y=Y⋅x⋅X⁢y=1
u=x⁢y⁢x⁢Y⁢X⁢Y, v=xk⁢Yk+1, c=y⁢Xk-1⁢y⁢X⁢Y, u′=y⁢x⁢Y⁢xk-1⁢Y⁢Xk-1:
(y⁢x⁢Y⁢xk-1⁢Y)⋅x⁢y⁢x⁢Y⁢X⁢(Y⋅y)⁢Xk-1⁢y⁢X⁢Y⋅xk-1⁢(y⁢Xk-1⁢y⁢X⁢Y)=(x)⁢y⁢x⁢Y⁢Xk⁢y⁢X⁢(Y⁢xk-1)=(y)⁢x⁢Y⁢Xk⁢y⁢X⋅(yk)=(x⁢Y)⁢Xk⁢y⁢(X⋅xk)=Xk⁢y⋅yk=1

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Received: 2018-11-14

Published Online: 2019-04-06

Published in Print: 2019-05-01

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https://doi.org/10.1515/gcc-2019-2005

Keywords for this article

Andrews–Curtis conjecture; Akbulut–Kurby presentations; trivial group; conjugacy search problem; computations