The Scientific World Journal

Volume 2014, Article ID 504160, 3 pages

http://dx.doi.org/10.1155/2014/504160

## On Rationality of Kneading Determinants

Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

Received 30 August 2013; Accepted 10 October 2013; Published 18 February 2014

Academic Editors: J. N. Alonso Alvarez, S. Deng, and J. Hoff da Silva

Copyright © 2014 Sheng Chen and Chao Xia. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In this section, denotes a ring with identity . We study conditions under which and are left coprime or right coprime, where . As applications, we get sufficient conditions under which the Kneading determinant of a finite rank pair of operators on an infinite dimensional space is rational.

#### 1. Introduction

If is a matrix with rational coefficients, one has the well-known identity between formal power series: where denotes the trace of matrix (matrix raised to the th power) and denotes the identity matrix. This identity plays a significant role in the discussion of an important problem in dynamical systems theory. For more details see [1, 2].

We denote by the infinite dimension vector space over ; the space of linear forms on will be denoted, as usual, by , and the space of all linear endomorphisms on will be denoted by . If and is a nonnegative integer, the th iterate is defined recursively by , , for .

The subspace of whose elements are the linear endomorphism on with finite rank will be denoted by .

Let be a positive integer; we use the symbol to denote an element of ( times) and the symbol to denote an element of ; that is Given and , we define the finite rank endomorphism and the matrix by setting with the usual notation

*Definition 1 (see [3]). *A pair is said to have finite rank if .

Notice that if a pair has finite rank, then the pair also has finite rank for all , and therefore the trace of is defined.

*Definition 2 (see [3]). *For any pair with finite rank, we define the Kneading determinant of as the following invertible element of :

*Remark 3. *Kneading determinant was first studied by Milnor and Thurston in [4].

Let be the ring of the matrices whose entries lie in . If , , and , we define the matrix by

Lemma 4 (see [3]). *Let , , and such that . Denote by the identity matrix. Then,
**
holds in .*

In general, any power series can be the Kneading determinant of some pair with finite rank (see [3]). So it is interesting to study conditions under which the Kneading determinant is a rational power series.

The following is the main result of this paper.

Theorem 5. *If and are left coprime or right coprime, and with finite rank, then is a rational function.*

#### 2. Coprimeness of and

In this section, denotes a ring with identity . We discuss conditions under which and are left coprime or right coprime, where .

We say and are left coprime if there exist polynomials such that

Proposition 6. * and are left coprime if and only if there exist such that
**
for .*

*Proof. *If and are left coprime, there exist and such that

We can assume that equals . And we have

So,

If we write , then,

So,

Conversely, if there exist such that
let
and let

Then,

Now we give the definition of right coprime. We say and are right coprime if there exist polynomials such that

Proposition 7. * and are right coprime if and only if there exist such that
**
for .*

*Proof. *It is similar to the proof of Proposition 6.

#### 3. Proof of the Main Theorem

In this section, , where is an infinite dimensional vector space over , and we denote by the ring of linear transforms on .

Lemma 8. *Suppose that and are left coprime or right coprime, is of finite rank, and . Then there exists a finite dimensional space containing such that with .*

*Proof. *First we assume that if and are right coprime, then by Proposition 7 we have such that

Write ; is an invariant subspace of under . It is not very hard to check that . So is invariant under the operator .

We use induction to prove the conclusion.

If , then .

Suppose that for we have . Then for , we have . We get the conclusion in this case.

Now we assume that and are left coprime; then there exist and such that
holds. Take
Notice when , .

We use induction to prove the conclusion.

If , then .

Suppose that for we have . Then for , we have .

The proof is complete.

Now we will give the proof of Theorem 5.

*Proof. *If and are coprime, by Lemma 8, there is a finite dimensional space such that
and we have . Define

Suppose ; we denote by the identity operator. Then . For any , we have for some . For more details see [5]. So by Lemma 4, is rational; we get the conclusion.

*Example 9. *Suppose is the ring of countable infinite matrix with finite nonzero entries in each column.

Let
where and are three-dimensional row vectors. We see that . So and are left coprime. It is easy to check that

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This paper was supported by National Natural Science Foundation of China (Grant nos. 11101105 and 11001064), by the Fundamental Research Funds for the Central Universities (Grant no. HIT. NSRIF. 2014085), and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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