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Computably Enumerable Semigroups, Algebras, and Groups

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## Computably Enumerable Semigroups, Algebras, and Groups

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### Computably Enumerable Semigroups, Algebras, and Groups

Bakhadyr Khoussainov

The University of Auckland

New Zealand

Research is partially supported by Marsden

Fund of New Zealand Royal Society.

Plan

- Equational presentations.
- The specification problem.
- A semigroup example.
- Algorithmically finite universal algebras.
- An algebra example.
- A group example.

Definition:

A universal algebrais a tuple

A=(A; f1,f2,…., fk, c1,…,cr),

where A is the non-empty set (domain),

each fj is a total function on A, and

each ck is a constant.

Definition (terms):

Variables and constants are terms.

Suppose that t1,…,tm are terms, f is a function

symbol, then the expression

f(t1,…,tm)

is a term.

A ground termis a term with no variables.

The set of ground terms is a universal algebra,

called the term algebra. Notation:

T = the term algebra

Fact 1: T is finitely generated and computable.

Fact 2: Every universal algebra generated by

the constants is a homomorphic image of T.

An equational presentation is a finite setS

of formulas of the type

t=q.

A quasiequational presentation is a finite set

S of formulas of the type

t1=q1…… tn=qn→ t=q.

Here t,q are terms that may contain variables.

Definition: (Specified Universal Algebras)

Let E(S) be the congruence relation generated by S.

The universal algebra TS = T/ E(S) is called

specified by S.

TS is equationally presented if S is an equational

presentation.

TS is quasi-equationally presented if S is a quasi-

equational presentation.

Examples

- Finitely presented groups and semigroups.
- Finitely presented algebras.
- Finitely presented semigroups with left-cancellation properties.

Properties of TS:

- TS satisfies S and is finitely generated.
- The equality relation E(S) of TS is computable enumerable.
- TS is universal.
- TS is unique.

The Specification Problem:

Let A be a universal algebra.

- Is Aequationally presented?
- Is A quasi-equationally presented?

Clearly, we need to assumethe following:

- A is finitely generated.
- The equality relation in A is c.e.

Definition (c.e. universal algebra)

A c.e. universal algebra is one of the form

T/E, where

(1) T is the term algebra,

(2) E is a c.e. congruence relation on T.

Examples: finitely presented groups, semi-

groups, and universal algebras.

Example:

Consider the universal algebra

( ω; x+1, 2x ).

(Bergstra and Tucker):

The universal algebra ( ω; x+1, 2x ) does

not have an equational presentation.

Important Observation

Consider the expansion:

(ω, x+1, 2x , +, x, 0).

The expanded universal algebra is now

finitely presented.

Definition: (Computable Algebra)

A universal algebra

A=(A; f1,f2,…., fk, c1,…,cr)

is computable if the set A and all functions

fj are computable.

Definition: An expansion of

A=(A; f1,f2,…., fk, c1,…,cr)

is

A’= (A; f1,f2,…., fk, g1,…,gr, c1,…,cr),

where g1,…,gr are new functions.

Theorem (Bergstra Tucker, ≈1980).

Every computable universal algebra can be

equationally presented in an expansion.

The question of Goncharov:

(early 1980s, Goncharov (also Bersgtra and Tucker))

Let A be a finitely generated computably

enumerable universal algebra:

- Does A have an equationally presented expansion?
- Does A have a quasi-equationally presented expansion?

Resemblance to Higman’s theorem

Theorem

(Kassymov, 1988; Khoussainov, 1994)

There exists a finitely generated computably

enumerable universal algebra no expansion

of which is equationally presented.

Theorem(Khoussainov, 2006)

There exists a finitely generated computably

enumerable universal algebra no expansion

of which is quasi-equationally presented.

Important comments

- The counter-examples constructed are universal algebras built specifically.
- The counter-examples do not belong to natural classes of structures such as the classes of groups, semi-groups, rings, etc.

Main Question:

Can such examples be found among the standard algebraic structures:

- Semigroups,
- Algebras (these are rings that form vector spaces over fields), and
- Groups?

Theorem A (with Hirschfeldt, 2011)

There exists a finitely generated computably

enumerable semigroup no expansion of

which is equationally presented.

Proof. Consider the free semigroup

A=({0,1}★;).

Let X be a subset of {0,1}★. We say that a

string u realizesX if u contains a substring

in X. Otherwise, we say that u avoidsX.

Define: R(X)={u | u realizes X}.

Clearly, R(X) is a subset of {0,1}★.

Define the following relation ≈X on {0,1}★.

u ≈X v if either u=v or u and v both realize X.

Lemma 1.

The relation ≈X is a congruence relation on

the free semigroup A=({0,1}★;).

Set:

A(X) = A/≈X .

Lemma 2. (Miller)

If for all k there are at most k many strings of

length ≤ k+4 in X, then R(X) is co-infinite.

Lemma 3.

There is a c.e. set X such that R(X) is simple.

Proof. Let W0, W1, …. be a standard list of

c.e. subsets of {0,1}★.

Put string y into X if for some i the string y is

the first string of length ≥i+5 appeared Wi.

The set X is a desired c.e. set.

Consider the semigroup

A(X) = A/≈X .

It is finitely generated, c.e., and infinite.

Let h1, …, hn be computable functions

compatible with ≈X. Consider the expansion

A’(X)= ( A(X); h1, …, hn)

Lemma 4 (Kassymov).

Any c.e. universal algebra whose equality relation

coincides with ≈X is residually finite. In particular,

A’(X) is residually finite.

Lemma 5 (Malcev). If a universal algebra A is

finitely presented and residually finite then the

word problem in A is decidable.

Hence, A’(X) is not equationally presented.☐

Definition (Kassymov, Khoussainov, 1986)

A finitely generated infinite c.e. universal

algebra

A = F/E

is effectively infiniteif there is an infinite

c.e. sequence u0, u1, u2 ….listing pair-wise

distinct elements of A.

If A is not effectively infinite then we call

Aalgorithmically finite (Miasnikov).

Let

A = T/E

be an algorithmically finite universal algebra.

Property 1.

Each expansion of A is algorithmically finite.

Property 2.

Each finitely generated subalgebra is

algorithmically finite.

For every term t(x) the trace

a, t(a), tt(a),…

is eventually periodic. In particular, if A is a

semigroup then every element of A is of

finite order.

Property 4.

All infinite homomorphic images of A are

also algorithmically finite.

Property 5. If A=T/Eis residually finite

then for all distinct elements x, y of A there

exists a subset S(x,y) of T such that:

- S(x,y) is E-closed and computable.
- x belongs S(x,y).
- y does belong to S(x,y).

Lemma. If A is residually finite, then all

expansions of A are also residually finite.

Proof. Let A’ be an expansion of A. Take

two distinct elements x,y of A’. Select the

separator set S(x,y) from Property 5.

Define the following binary relation ≈(x,y)

on A’:

a ≈(x,y) b

if and only if

no elements in S(x,y) and in its complement

are identified by the congruence relation on

A’ generated by the pair (a,b).

Properties of ≈(x,y) :

- ≈(x,y) is a congruence relation on A’.
- ≈(x,y) is a co-c.e. relation.
- In the quotient algebra A’/ ≈(x,y) the images of x and y are distinct.

Since A’ is algorithmically finite, A’/ ≈(x,y)

must be finite.

Theorem B.

Let A be an algorithmically finite universal

algebra.If A is residually finite then no

expansion of A has equational presentation.

Proof. If A’is an equationally presented

expansion of A, then A’is residually finite.

By Malcev’s lemma the word problem in A’

is decidable. Contradiction.

Are there algorithmically finite groups?

Motivation of the question:

Algorithmically finite groups are candidates

that have no equationally presented expansions.

Theorem (Miasnikov, Osin, 2011)

There exists an algorithmically finite group.

Miasnikov motivates the theorem from a generic

complexity view point. Algorithmically finite

groups are called Dehn monsters.

Miasnikov and Osin ask if there are

residually finite Dehn monsters.

Let K be a finite field. Consider the algebra

F=K<x1, x2,..., xm>

of polynomials in non-commuting variables.

We can represent F as the direct sum

∑Fn

where Fn is the vector space spanned over

monomials of degree n.

Let H be a set of homogeneous polynomials,

I be the ideal generated by H.

Theorem (Golod Shafarevich).

Let rn be the number of polynomials in H of

degree n, and ε be such that 0< ε < m/2 and

rn ≤ ε2(m-2 ε)n-2.

Then the algebra

A=F / I

is infinite dimensional.

Let H be a subset of {x,y}★ constructed in

Lemma 3 above. Consider the ideal I=<H>.

Theorem C. The algebra A=F/I satisfies the

following properties:

- A is effectively infinite.
- All expansions of A are residually finite.
- A has no equationally presented expansions.

Proof. It is clear that the algebra is infinite.

Write any polynomials f of F in the form

a+ h,

where a is a sum of monomials not from KH

and h is a sum of monomials in KH. So:

f=(a1+.... an)+(h1+....+hk).

Identify this sum with the set

{a1,.... an,h1,....+hk},

and call the set {a1,.... an}

the true representative of a.

Since H is simple but not hypersimple, H

has a strong array of finite sets for the

complement of H.

The identification of polynomials with finite

subsets implies that the algebra A is

effectively infinite.

Claim 1. The collection of all true

representatives is an immune set.

Claim 2. For all distinct elements x, y of A

there exists a subset S(x,y) of F such that:

- S(x,y) is I-closed and computable.
- x belongs S(x,y).
- y does belong to S(x,y).

All expansions of A are residually finite.

Proof.

Take two distinct elements x,y of an

expansion A’. Select the separator set

S(x,y) from Claim 2.

Define the following binary relation ≈(x,y):

if and only if

no elements in S(x,y) and in its complement

are identified by the congruence relation on

A’ generated by the pair (f,g).

- ≈(x,y) is a congruence relation on A’.
- ≈(x,y) is a co-c.e. relation.
- In the quotient algebra A’/ ≈(x,y) the images of x and y are distinct.

Claim 4. The collection of all true

representatives that belong to distinct

≈(x,y)-equivalence classes is a c.e. set.

Thus, the A’/ ≈(x,y) must be finite.

Hence, A’ is c.e., infinite, residually finite.

Therefore, the algebra can not be

equationally specified by Theorem B.

Theorem C. There exists an algorithmically

finite and residually finite algebra.

Proof. Consider F=K<x1, x2,..., xm>.

Construct a set H of homogeneous

polynomials by stages as follows.

Let W0, W1, …. be a list of c.e. subsets of F.

For each i, let f and g the first polynomials

occurring in Wi such that:

- f=f1+f2, g=g1+ g2, f1= g1, and
- the degrees of homogeneous polynomials occurring in both f2 and g2 are greater than i+64.

Put all homogeneous polynomials occurring

in both f2 and g2 into H.

For each n >2, in H there are at most 2n

homogeneous polynomials of degree n.

Hence, for a small ε we have 0< ε < m/2

and

rn ≤ ε2(m-2 ε)n-2

for all n.

By Golod Shafarevich theorem, we have

that the algebra A=F / I is infinite. By

construction, it is algorithmically finite.

It is well-known that Golod-Shafarevich

algebras are residually finite (Golod).

So, A is algorithmically finite and residually

finite. By Theorem B no expansion of A is

equationally presented.

Theorem D. There exists an algorithmically

finite and residually finite group G.

Proof. Consider the algebra A constructed

above. The semigroup G=G(A) generated

by the elements (1+x)/I and (1+y)/I of the

algebra A forms a group under the product

operation.

The group G is the one desired.

As a corollary we obtain the following

theorem.

Theorem E. There exists a group that has

no equationally presented expansion.

Open Question:

Are there semigroups, algebras and groups

whose all expansions are not quasi-

equationally presented?

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