Exponent matrices and topological equivalence of maps
Conjugate classes of continuous maps of the interval [0,1] into itself, whose iterations form a finite group are described. For each of possible groups of iterations one to one correspondence between conjugate classes of maps and equivalent classes of (0,1)-exponent matrices of special form is const...
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Інститут прикладної математики і механіки НАН України
2007
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| Цитувати: | Exponent matrices and topological equivalence of maps / V. Fedorenko, V. Kirichenko, M. Plakhotnyk // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 4. — С. 45–58. — Бібліогр.: 5 назв. — англ. |
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irk-123456789-1523812019-06-11T01:25:35Z Exponent matrices and topological equivalence of maps Fedorenko, V. Kirichenko, V. Plakhotnyk, M. Conjugate classes of continuous maps of the interval [0,1] into itself, whose iterations form a finite group are described. For each of possible groups of iterations one to one correspondence between conjugate classes of maps and equivalent classes of (0,1)-exponent matrices of special form is constructed. Easy way of finding the quiver of the map in terms of the set of its extrema is found. 2007 Article Exponent matrices and topological equivalence of maps / V. Fedorenko, V. Kirichenko, M. Plakhotnyk // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 4. — С. 45–58. — Бібліогр.: 5 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:05С50, 37C15, 37C25. http://dspace.nbuv.gov.ua/handle/123456789/152381 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Conjugate classes of continuous maps of the interval [0,1] into itself, whose iterations form a finite group are described. For each of possible groups of iterations one to one correspondence between conjugate classes of maps and equivalent classes of (0,1)-exponent matrices of special form is constructed. Easy way of finding the quiver of the map in terms of the set of its extrema is found. |
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Fedorenko, V. Kirichenko, V. Plakhotnyk, M. |
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Fedorenko, V. Kirichenko, V. Plakhotnyk, M. Exponent matrices and topological equivalence of maps Algebra and Discrete Mathematics |
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Fedorenko, V. Kirichenko, V. Plakhotnyk, M. |
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Fedorenko, V. |
| title |
Exponent matrices and topological equivalence of maps |
| title_short |
Exponent matrices and topological equivalence of maps |
| title_full |
Exponent matrices and topological equivalence of maps |
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Exponent matrices and topological equivalence of maps |
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Exponent matrices and topological equivalence of maps |
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exponent matrices and topological equivalence of maps |
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Інститут прикладної математики і механіки НАН України |
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2007 |
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Exponent matrices and topological equivalence of maps / V. Fedorenko, V. Kirichenko, M. Plakhotnyk
// Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 4. — С. 45–58. — Бібліогр.: 5 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT fedorenkov exponentmatricesandtopologicalequivalenceofmaps AT kirichenkov exponentmatricesandtopologicalequivalenceofmaps AT plakhotnykm exponentmatricesandtopologicalequivalenceofmaps |
| first_indexed |
2025-07-13T02:57:41Z |
| last_indexed |
2025-07-13T02:57:41Z |
| _version_ |
1837498859801018368 |
| fulltext |
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.
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 4. (2007). pp. 45 – 58
c© Journal “Algebra and Discrete Mathematics”
Exponent matrices and topological equivalence of
maps
Volodymyr Fedorenko, Volodymyr Kirichenko
and Makar Plakhotnyk
Abstract. Conjugate classes of continuous maps of the in-
terval [0, 1] into itself, whose iterations form a finite group are
described. For each of possible groups of iterations one to one
correspondence between conjugate classes of maps and equivalent
classes of (0, 1)-exponent matrices of special form is constructed.
Easy way of finding the quiver of the map in terms of the set of its
extrema is found.
Introduction. Recall that a right Noetherian semiperfect semiprime
ring A is semimaximal if the endomorphism ring of every indecompos-
able finitely generated projective A-module is a discrete valuation ring.
It is known [2, chapter 14] that semiperfect semiprime ring Λ is semi-
maximal if and only if it is isomorphic to the finite direct product of
prime semimaximal rings. Every semimaximal ring is semidistributive
two-sided Noetherian ring with nonzero Jacobson radical. Prime Noethe-
rian semiperfect semidistributive ring with nonzero Jacobson radical is
called tiled order.
Every tiled order Λ has the classical ring of fractions Mn(D) which is
the ring of all square matrices of order n over the division ring D. Denote
by eij the matrix units of Mn(D) where 1 6 i, j 6 n.
It is known [2, Chapter 14] that every tiled order is isomorphic to a
tiled order of the following form Λ =
n∑
i,j=1
eijπ
αijO where O is a discrete
valuation ring with the prime element π, and E = (αij) is integer matrix of
order n > 1 with zeros on the diagonal, for whose elements the inequality
αij + αjk − αik > 0 takes place for all i, j, k, 1 6 i, j, k 6 n. These
2000 Mathematics Subject Classification: 05С50, 37C15, 37C25.
Key words and phrases: exponent matrix, finite orbits, topological equivalence.
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.46 Exponent matrices and topological equivalence of maps
inequalities are called ring inequalities and matrix E is called exponent
matrix.
A tiled order can be considered as a pair (O, E). Semiperfect ring Λ
with Jacobson radical R is called reduced if the quotient ring Λ/R is a
direct product of division rings [2]. Tiled order is reduced if and only if
its exponent matrix has no symmetrical zeros. Such exponent matrix is
called reduced.
A tiled order Λ is called a Gorenstein tiled order if inj.dimΛΛΛ = 1.
A reduced tiled order Λ = (O, E(Λ) = (αij)) is Gorenstein if and
only if there exists a permutation σ without fixed points such that for
any numbers i, j the equality αij + αjσ(i) = aiσ(i) take place.
A reduced exponent matrix such that for all i, j these equalities take
place is called Gorenstein matrix.
Theorem 1. [3, Theorem 7.1.1] Let Λ = {O, E(Λ)} be a reduced prime
Noetherian SPSD-ring Λ with the exponent matrix E(Λ) = (αij) ∈ Z.
Then inj.dimΛΛΛ = 1 if and only if matrix E(Λ) is Gorenstein. In this
case inj.dimΛΛΛ = 1.
It is known that there in one to one correspondence between finite
posets and (0, 1)-reduced exponent matrices of order n. This equivalence
can be realized in the next way. Let Θ = (θ1, . . . , θn) be a finite poset
of n elements. The equality αij = 0 should be equivalent to inequality
θi > θj . Ring inequalities for all indices give that this correspondence
will be really one to one correspondence.
Let F = {f1, . . . , fn} be a finite linearly ordered set of n elements.
Let ψ be real function defined on F . This function ψ induces one more
relation 6 on the set F by the rule fi 6 fj if and only if ψ(fi) 6 ψ(fj).
The set F with this relation will be called doubly ordered set. For the
doubly ordered set F correspond exponent matrix can be defined by the
same rule as for a poset.
Let Un be matrix of order n all whose elements are equal to 1 and
En be identity matrix. Then for any order n exponent (0, 1) matrix E ,
consider the matrix Ẽ = Un −E −En. This formula defines an adjacency
matrix of the diagram of the finite partially ordered set Θ [3, p. 277].
Note that for any poset its correspond matrix Ẽ should not be neces-
sarily an exponent matrix.
Example 1. Consider the poset of 3 elements, whose diagram is the next
1
3
OO
2
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 47
According to the definition, its exponent matrix is E =
0 1 1
1 0 1
0 1 0
and Ẽ =
0 0 0
0 0 0
1 0 0
. Matrix Ẽ is not exponent matrix because the
inequality α̃32 + α̃21 ≥ α̃31 does not take place.
Example 2. Consider the doubly ordered set of 3 elements, whose dia-
gram is the next
1
3
OO
2
According to the definition, its exponent matrix is E =
0 0 0
1 0 0
1 0 0
and Ẽ =
0 1 1
0 0 1
0 1 0
. Matrix Ẽ is an exponent matrix.
Example 3. Consider the doubly ordered set of 3 elements, whose dia-
gram is the next
1 2
3
OO
According to the definition, its exponent matrix is E =
0 0 0
0 0 0
1 1 0
and Ẽ =
0 1 1
1 0 1
0 0 0
. Matrix Ẽ is an exponent matrix.
In our work we prove that for any doubly ordered set its correspond
matrix Ẽ is always an exponent matrix.
Maps f and g, which map interval [0, 1] into itself, are named topolo-
gical equivalent if they are conjugate with invertible map of this interval
into itself.
Iterations of each each map f form a semigroup. It is known [5, 4] that
if this semigroup is finite group then it is either trivial or is the group
C2, i.e. the group of order two. More then that, there exist numbers
p, q, 0 6 p 6 q 6 1, dependent on f , such that for every x ∈ [0, 1] the
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.48 Exponent matrices and topological equivalence of maps
inclusion f(x) ∈ [p, q] takes place. Iterations group is trivial if and only
if for any x ∈ [p, q] the equality f(x) = x takes place and this group is C2
if and only if f decrease on [p, q] and this interval is invariant under the
action of this map. Up to the end of our paper all maps of interval will
be considered as continuous whose iteration semigroup is finite group.
In this paper for the map f and its iterations group G we construct an
(0, 1)-exponent matrix, which is a matrix of some doubly ordered set and
has some additional properties, which can be easily described as in terms
of matrices as in terms of doubly ordered sets. Any class of topological
equivalent maps corresponds to the pair of matrices, each of them is a
matrix of some doubly ordered set and one of them can be obtained
from another by an easy rule. And conversely for any pair of matrices of
doubly ordered sets such that one of them can be obtained from another
by the rule mentioned above and such that satisfy some easily formulated
restrictions, the conjugate class of the map f with iterations group G can
be uniquely restored.
We also determine more explicit definition of the quiver of exponent
matrix if it is given that it is a matrix of the equivalence class of contin-
uous map.
Remind the notion of equivalence of exponent matrices, which ap-
peared in their algebraic introduction.
Exponent matrices A,B are called equivalent, if one may be obtained
from another by the following transformations:
1) adding an integer to all elements of some row with simultaneous
subtracting it from the elements of the column with the same number.
2) simultaneous interchanging of two rows and equally numbered
columns,
or by compositions of such transformations.
Lemma 1. Under the transformations of the first type Gorenstein matrix
goes to Gorenstein one with the same correspond permutation [1].
If one consider (0, 1)-matrices Hn = (hij) such that hij = 1 if and
only if i > j and
G2m =
(
Hm H
(1)
m
H
(1)
m Hm
)
where H
(1)
m = E + Hm, then it will be easy to see, that Hn is Goren-
stein matrix with permutation σ(Hn) = (n, n − 1, . . . 2, 1), and G2m is
Gorenstein one with permutation σ(G2m) =
m∏
i=1
(i,m+ i).
Lemma 2. Gorenstein (0, 1) matrix is equivalent to either matrix Hn or
to G2m [1].
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 49
In our work we will show that Gorenstein matrix which is a matrix
of some doubly order, not all whose elements are equal, is equivalent to
Hn.
1. Doubly orders. For every doubly order {f1, . . . , fn} construct a
(0, 1)-matrix E = (αij) of order n in the next way:
αij =
{
0 fi > fj ,
1 fi < fj .
Matrix E is exponent matrix without symmetrical 1. Really as it is (0, 1)-
matrix, then inequality αij + αjk > αik can be violated only in the case
when αik = 1 and in the same time αij = αjk = 0. From the construction
of the matrix, equalities αij = αjk = 0 mean that fi > fj and fj >
fk, which is fi > fk, whence αik = 0, but not αik = 1. Absence of
symmetrical 1 is obvious.
Lemma 3. Considering exponent matrix without symmetrical 1, con-
structed by the doubly order {f1, f2} it is possible to restore unambigu-
ously the correspondence “bigger-smaller” between symbols f1 and f2.
Proof. Exponent (0, 1)-matrices of order 2 which have no symmetrical 1
can be found explicitly, and so we get
E1 =
(
0 0
0 0
)
; E2 =
(
0 0
1 0
)
; E3 =
(
0 1
0 0
)
.
For the matrix E1 obtain f1 > f2 and f2 > f1, whence f1 = f2.
For the matrix E2 obtain f1 > f2 and f2 < f1, whence f2 < f1.
For the matrix E3 obtain f1 < f2 and f2 > f1, whence f1 < f2.
Using lemma 3, considering any exponent matrix E , which has no
symmetrical 1, we can construct a map A from the pair sets (i, j) to the
set of letter expressions {“fi < fj”, “fi = fj“, “fi > fj”}, which will
compare symbols fi and fj , with using the matrix
(
0 αij
αji 0
)
,
where αij and αji are elements of the matrix E . Determine some proper-
ties of the map A(i, j).
1. If A(m, t) = “fm = ft” and in the same time A(t, k) = “ft = fk”,
then A(m, k) = “fm = fk”.
In other words, if αmt = αtm = 0 together with αtk = αkt = 0, then
αmk = αkm = 0. From the inequality αmt + αtk > αmk and equalities
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.50 Exponent matrices and topological equivalence of maps
αmt = αtk = 0 obtain that αmk = 0. From the inequality αkt+αtm > αkm
and equalities αkt = αtm = 0 obtain that αkm = 0, which is necessary.
2. If A(m, t) = “fm = ft” and in the same time A(t, k) = “ft < fk”,
then A(m, k) = “fm < fk”.
In other words, if αmt = αtm = 0 and αtk = 1, then αmk = 1. consider
the ring inequality αtm + αmk > αtk. As αtm = 0, and αtk = 1, then
αmk = 1, which is necessary.
3. If A(m, t) = “fm = ft” and in the same time A(t, k) = “ft > fk”,
then A(m, k) = “fm > fk”.
In other words, if αmt = αtm = 0 and αkt = 1, then αkm = 1.
Consider ting inequality αkm +αmt > αkt. As αmt = 0 and αkt = 1, then
αkm = 1.
4. If A(m, t) = “fm > ft” and in the same time A(t, k) = “ft > fk”,
then A(m, k) = “fm > fk”.
In other words, if αtm = αkt = 1, then αkm = 1. As matrix E has
no symmetrical 1, then the equality αkt = 1 holds αtk = 0. Consider the
inequality αtk + αkm > αtm. As αtm = 1 and αtk = 0, then αkm = 1.
Lemma 4. Considering exponent matrix without symmetrical 1, con-
structed by the doubly order {f1, . . . fn} it is possible to restore unambigu-
ously the correspondence “bigger-smaller” between elements of this doubly
order.
Proof. Prove this lemma by induction for n. Induction base for n = 2
holds from lemma 3.
Now, let symbols f1, . . . , fn−1 be renumbered in such a way that
fi1 6 fi2 6 . . . 6 fin−1
. By the map A(n, ik), k = 1, 2, . . . compare
consequently fn with fi1 , fi2 etc. till get the inequality fn < fi1 , or
quality fn = fik , or pair of inequalities fik < fn < fik+1, or inequality
fn > fin−1
. Doing in such a way, we will find the place of the symbol fn
in the set of symbols fi1 , . . . , fin−1
. Unambiguousness of ordering holds
from the construction and consistence holds from the properties 1, . . . , 4
of the map A.
Lemma 5. For any exponent (0, 1)-matrix E = (αij) without symmetri-
cal 1, matrix Ẽ = (α̃ij) is reduced exponent (0, 1)-matrix.
Proof. Ẽ is reduced because of definitions. Check ring inequalities for
this matrix.
As inequality α̃ij + α̃jk > α̃ik can be violated only in the case when
α̃ij = α̃jk = 0 and α̃ik = 1. As according to lemma 4 there is the
unique doubly ordered set {f1, . . . fn} with Ẽ as correspond matrix then
inequalities α̃ij = α̃jk = 0 means that fi 6 fj 6 fk, whence fi 6 fk and
α̃ik = 0.
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 51
Theorem 2. Let E = (αij) be such reduced exponent (0, 1)-matrix that
matrix Ẽ is also exponent. There is one to one correspondence between
such matrices and doubly orders.
Let h be invertible map, p1, p2, q1 and q2 be numbers such that
f([0, 1]) = [p1, q1], and g([0, 1]) = [p2, q2]. Let a1, . . . , an and b1, . . . bm
be extrema of maps f and g correspondingly. Points 0, 1, ends of each
maximal interval of constantness of the map and each end of maximal
interval of fixed points also consider as extrema.
2. Increase conjugate map for idempotent maps. Let f and
g be idempotent increasing maps. Everywhere instead situations where
this has to be proven consider the equality g = h−1(f(h)) to take place.
Lemma 6. The equalities h(p2) = p1 and h(q2) = q1 take place.
Proof. Let x0 ∈ Fix(g). If h(x0) > q1 then f(h(x0)) < h(x0), because
f(h(x0)) ∈ f([0, 1]) = [p1, q1], and h(x0) > q1. As the map h increase,
then h−1 also increase, whence h−1(f(h(x0))) < h−1(h(x0)). But the last
equality means that g(x0) < x0, which contradicts to fact that x0 is the
fixed point of g. The case when h(x0) < p1 should be considered in the
same way.
So, the image of the interval [p2, q2] under acting of h is contained in
[p1, q1]. As the equality g = h−1(f(h)) holds f = h(g(h−1)) then in the
same way we can prove that image of the interval [p1, q1] under the action
of h−1 is contained in [p2, q2]. The last together with invertibleness of h
proves lemma.
Lemma 7. m = n and for any i ∈ [1, n] the equality h(bi) = ai takes
place. Also for every r, s ∈ [1, n] the inequality g(br) 6 g(bs) is equivalent
to f(ar) 6 f(as).
Proof. Really, this lemma holds from the fact that composition of in-
creasing monotone functions is still increasing function.
Fix arbitrary number i ∈ [1, n− 1] and consider the monotonicity in-
terval [ai, ai+1] of the map f . Let f increase at this interval. Show that
in this case the map g increase at [h−1(ai), h
−1(ai+1)]. For arbitrary
x1, x2 ∈ [a1, ai+1], x1 < x2, obtain h(x1) < h(x2); as the map f in-
crease at [ai, ai+1] then f(h(x1)) < f(h(x2)), whence we get an inequality
h−1(f(h(x1))) < h−1(f(h(x2))), which shows that [h−1(ai), h
−1(ai+1)] is
increasing interval of g. If f decrease at [ai, ai+1], then proving of de-
creasing of g on [h−1(ai), h
−1(ai+1)] is almost similar.
Consider points bi = h−1(ai) for i = 1, . . . , n. Proving above gives
that these points are extrema of the map g, and for any i = 1, . . . , n the
extremum character of bi is the same as one of ai.
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.52 Exponent matrices and topological equivalence of maps
Let for some r, s ∈ [1, n] there is an inequality g(br) 6 g(bs) i.e.
h−1(f(h(br))) 6 h−1(f(h(bs))). Taking h of both sides, using increasing
of h, obtain f(h(br)) 6 f(h(bs)). Using equalities br = h−1(ar) and
bs = h−1(as) obtain f(ar) 6 f(as) which is necessary.
Lemma 8. Let m = n, for any i ∈ [1, n] the equality f(ai) 6 f(aj) is
equivalent to g(bi) 6 g(bj) and p1 and p2 have the same places in sets of
extrema of these maps. Then maps f and g are conjugate.
Proof. Let us construct the map h such that equality g = h−1(f(h)) take
place. Let the graph of h contains points (g(bi), f(ai)) for all i. Note that
as f(ai) 6 f(aj) is equivalent to g(bi) 6 g(bj) then the restriction for h
does not contradict to its monotonicity. More then that as h(p2) = p1
and h(q2) = q1 that this restriction is localized at [p2, q2].
Let the map h be arbitrary increasing map on the interval [p2, q2]
whose graph contained reminded points.
Consider any number i < n such that interval [ai, ai+1] not to be
interval [p1, q1]. Consider an arbitrary number x0 ∈ [bi, bi+1]. Then the
condition g(x0) = h−1(f(h(x0))) is equivalent to h(g(x0)) = f(h(x0)).
As map h is already defined on the interval [p2, q2] then the number
h(g(x0)) can be found. As x0 ∈ [bi, bi+1], then h(x0) ∈ [ai, ai+1]. As the
map f is monotone on the interval [ai, ai+1], then it has locally inverse
fi, which is defined on the interval (f(ai), f(ai+1)), or on the interval
(f(ai+1), f(ai)), dependent on increasing or decreasing of f on [ai, ai+1].
So, for the equality h(g(x0)) = f(h(x0)) to take place it is enough to have
h(x0) = f−1
i (h(g(x0))).
For any i = 1, . . . , n define the map h on the interval [bi, bi+1] by the
formula h = f−1
i (h(g)). From the constructing obtain that the map h
defines conjugation of maps f and g.
Using f construct the matrix Ẽ = (α̃ij) of order n by the rule
α̃ij =
{
1 f(ai) > f(aj) and i 6= j;
0 f(ai) < f(aj) or i = j.
Denote it Ẽ(f(a1), . . . , f(an)). It is (0, 1)-reduced exponent matrix and
the matrix E = Un − Ẽ − En is an exponent matrix.
As f is idempotent map then there should exist a number i such that
f(ai) = min
k
f(ak) and f(ai+1) = max
k
f(ak). This condition can be easily
reformulated in terms of matrix Ẽ .
Using the theorem 2, we can formulate the next theorem.
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 53
Theorem 3. Consider the set of pairs (E , i0), where E is exponent ma-
trix and Ẽ is reduced exponent such that conditions necessary condition
mentioned above is satisfied. There is one to one correspondence between
such pairs and equivalent classes of maps (conjugated by increasing map),
whose iterations group is trivial.
3. Increase conjugate map for generators of C2. Let maps f
and g be generators of the group C2 and h be increasing map.
Lemma 9. Let g = h−1(f(h)). Then h(p2) = p1 and h(q2) = q1.
Proof. The inequality g = h−1(f(h)) holds g2 = h−1(f2(h)). Now lemma
is a corollary of lemma 6.
Lemma 10. Let f and g be two decreasing maps of interval such that
f2 = id and g2 = id. Then they are conjugate by increasing map.
Proof. Let xf
0 and xg
0 be fixed points of maps f and g correspondingly.
Construct new maps f1 and g1 by the next rule: f1(x) = f(x) for x 6 xf
0
and f1(x) = x for x > xf
0 ; also g1(x) = g(x) for x 6 xg
0 and g1(x) = x for
x > xg
0.
Construct increasing map h1, which realizes conjugateness of f1 and
g1. Let the graph of h contains the point (xg
0, x
f
0) and be defined arbitrary
on the interval [xg
0, 1].
Then for an arbitrary x ∈ [xg
0, 1] we have h(x) ∈ [xf
0 , 1], whence
f1(h(x)) = h(x), which means that h−1(f1(h(x))) = h−1(h(x)) = x, and
so g1 = h−1(f1(h)) on the interval [xf
0 , 1]. As g1 is monotone on the
interval [0, xg
0], then it is enough to take h = f−1
1 (h(g1)) on the interval
[0, xg
0], for conjugateness of f1 and g1. Here map h in the righthand side
is already defined, because g−1
1 ([0, xg
0]) = [xg
0, 1].
Let us show that map h which has been constructed in such a way
really realizes the conjugateness of map f and g. the fact that equal-
ity g(x) = h−1(f(h(x))) takes place for every x ∈ [0, xg
0] holds from
construction. As there are compositions of monotone functions at the
righthand side of the equality g1(x) = h−1(f1(h(x))) then it is possible
to write the equality of inverse functions and get the equality g−1
1 (x) =
h−1(f−1
1 (h(x))) for x ∈ [0, xg
0]. Considering equalities f2 = id and
g2 = id we have g−1
1 (x) = g(x) and f−1
1 (h(x)) = f(h(x)) for x ∈ [xg
0, 1],
which means that the equality g = h−1(f(h)) s correct at the whole
interval [0, 1]. The last finishes the proof of the lemma.
Lemma 11. Let g = h−1(f(h)). Then for arbitrary i ∈ [1, n] the equality
h(bi) = ai take place. Also the inequality g(br) 6 g(bs) is equivalent to
f(ar) 6 f(as) for all r, s ∈ [1, n].
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.54 Exponent matrices and topological equivalence of maps
Proof. The proof of this lemma is exactly the same as of lemma 7.
For any functions w with extrema v1, . . . , vn denote
A′(w) = Ẽ(w(v1), w
2(v1), . . . , w(vn), w2(vn)).
Lemma 12. Let the equality g = h−1(f(h)) to take place for maps f and
g. Then matrices A′(f) and A′(g) coincide.
Proof. Plug the value x = g(bi) into equality g(x) = h−1(f(h(x))) and
get g2(bi) = h−1(f(h(g(bi)))). As according to lemma 8 the graph of
the map h contains points (g(bi), f(ai)) for all i ∈ [1, n], then the ob-
tained equality is equivalent to g2(bi) = h−1(f(f(ai))). Plugging left
hand and righthand side of obtained equality into function h get an equal-
ity h(g2(bi)) = f2(ai), which means that the graph of function h contains
points (g2(bi), f
2(ai)) for all i ∈ [1, n]. From this fact, from increasing
of h and from the fact that points (g(bi), f(ai)) for all i ∈ [1, n] belong
to the graph of h, obtain the correctness of lemma.
Lemma 13. Maps f and g are conjugate if and only if m = n, numbers
of extrema which are ends of intervals f([0, 1]) and g([0, 1]), and matrices
A′(f) and A′(g) coincide.
Proof. It is proven in the previous lemma that matrices of conjugated
maps are equal. So the only think we have to prove is that inverse propo-
sition is still correct.
Let for maps f and g matrices A′(f) and A′(g) coincide. Let’s show
that these maps are conjugate by increasing function.
From he equality A′(f) = A′(g) obtain that there exists continuous
map h1 : [p2, q2] −→ [p1, q1] which realizes conjugateness of maps f and
g on their periodic points and whose graph contains points (g(bi), f(ai))
and (g2(bi), f
2(ai)) for all i ∈ [1, . . . n]. This map is constructed analo-
gously by one from lemma 10.
The map h such that g = h−1(f(h)) on the set [0, 1] \ [p2, q2] should
be found in the same way as in the proof of the lemma 8.
As for the map f the equality f = f3 take place then there should
exist a number i such that f(ai) = f2(ai), f(ai−1) = max
k
f(ak) and
ai+1 = min
k
f(ak). Also f(ai−1) = f2(ai+1), f(ai+1) = f2(ai−1) and for
all i, j such that f(ai) 6 f(aj) the inequality f2(ai) 6 f2(aj) takes place.
This condition can be easily reformulated in terms of matrix E ′(f), which
gives us the next theorem.
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 55
Theorem 4. Consider the set of pairs (E , i), where E is exponent matrix
such that matrix Ẽ is reduced exponent such that the necessary condition
mentioned above is satisfied. There is one to one correspondence between
such pairs and equivalent classes of maps (conjugated by increasing map),
whose iterations group is nontrivial.
4. Decreasing conjugate map. Let us investigate the changing of
exponent matrix of the map under conjugating by decreasing map.
As any arbitrary decreasing map h can be presented in the form
h(x) = 1 − h1(x) where h1(x) = 1 − h(x) in increasing map, then conju-
gating by the map h is a composition of conjugating by the map 1 − x
and increasing one. As it is proven that increasing map does not change
the matrix of the map then is is enough to investigate this changing if we
have conjugation by the map 1 − x.
The action of conjugating by the map 1 − x on the graph f we can
understand as consequent symmetrical reflecting in the line x = 1/2 and
symmetrical reflecting in the line y = 1/2. The first means acting by
permutation (1, n)(2, n−1) · · · or (1, n−1)(2, n)(3, n−3)(4, n−2) · · · on
lines and columns of the matrix E ′ dependently on if the iterations group
is trivial or nontrivial.
Symmetrical reflecting of the graph in the line y = 1/2 moves each
inequality fαi(ai) 6 fαj (aj), where αi, αj ∈ {1, 2} to fαj (aj) 6 fαi(ai).
For the exponent matrices it means that this mappings A −→ B is defined
in the next way:
1) for any numbers i, j equality aij = aji = 1, holds bij = bji = 1;
2) for any different i, j the equality aij = 0 holds bij = 1 and bji = 0;
3) bii = 0 for all i.
Here for both trivial and nontrivial iterations group case the number
i0 from theorems 3 and 4 goes to n− i0 + 1.
So, any class of conjugate maps corresponds to pair of exponent ma-
trices of doubly ordered set each of them are image of another in the
self invertible action, mentioned just above. These matrices should have
obvious additional properties, dependent on the iteration group to make
correspond doubly ordered set to be a set of extrema of necessary map.
5. The quiver of a map with finite group of iterations. Let
matrix E = Ẽ(b1, . . . , bm) = (α̃ij) be constructed by the rule
α̃ij =
{
1 bi > bj and i 6= j;
0 bi < bj or i = j.
E is reduced exponent matrix. Remind how to construct the quiver of
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.56 Exponent matrices and topological equivalence of maps
such matrix [1]. First, construct the matrix E(1) = (β̃ij) = E + E, i.e.
β̃ij =
{
1 bi > bj ;
0 bi < bj .
Considering matrix E(1) construct the matrix E(2) = (γ̃ij) defined by the
rule γ̃ij = min
k
(β̃ik + β̃kj). Consider matrix Q = E(2) − E(1) as adjacency
matrix of a quiver, which we name a quiver, which corresponds to the
map f .
Let us understand, when the quiver has a loop at the vertex i. As
β̃ii = 1 then for the loop equalities β̃ik = β̃ki = 1 have to take place for
all k. It is possible but only in the case when b1 = . . . = bm. It is obvious
that in this case all the vertexes of the quiver have one loop each and
there is exactly one arrow from each vertex to each.
Lemma 14. γ̃ij = 0 if and only if there exists k, such that bi < bk < bj.
Proof. According to definition of γ̃ij we get that γ̃ij = 0 if and only if
there exists a k such that β̃ik = β̃kj = 0. From the construction of E(1)
it means that bi < bk < bj , which is necessary.
Its obvious that if bi < bk < bj for some k, then γ̃ij = 0.
Lemma 15. γ̃ij = 2, if and only if bi = max
k
bk, and bj = min
k
bk.
Proof. If γ̃ij = 2, then β̃ik = β̃kj = 1 for all k, in other words bi > bk > bj
for all k whence bi = max
k
bk, and bj = min
k
bk.
It is obvious that if bi = max
k
bk, and bj = min
k
bk then γ̃ij = 2.
Lemma 16. γ̃ij = 1 and β̃ij = 0 if and only if bj > bi and there is no
any bk between them.
Proof. Let γ̃ij = 1 and β̃ij = 0. The equality γ̃ij = 1 holds that if
for some k the equality β̃kj = 0 takes place then β̃ik = 1. From the
constructing of the matrix E(1) obtain that for all k such that bk < bj the
inequality bk 6 bi takes place. This means that bj 6 bi, or bj > bi, but
there is no bk between them.
As β̃ij = 0 then bi < bj which contradicts to the case when bj 6 bi.
Let bj > bi, but there is no any bk between them. Show that in
this case γ̃ij = 1 and β̃ij = 0. According to lemmas 14 and 15 obtain
that equivalent conditions for bj and bi to the quality γ̃ij = 0 or γ̃ij = 2
are braked. That is why γ̃ij = 1. As bi < bj then β̃ij = 0 which is
necessary.
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.V. Fedorenko, V. Kirichenko, M. Plakhotnyk 57
Using proven lemmas we can rewrite the definition the quiver of f :
qij = 1 if and only if bi = max
k
bk, and bj = min
k
bk, or bj > bi but there is
no any bk between them, where max and min are considered in not strict
meaning.
Describe the quiver constructed above. Let bi1 < bi2 < . . . < bik be
such numbers that {bi1 , . . . , bik} = {b1, . . . , bm}. From the construction
we get that the quiver has cycle bi1 → bi2 → . . .→ bik → bi1 .
Let for some numbers j, t the equality bj = bit takes place. From the
construction of the quiver obtain that vertexes bj are bit connected with
all others arrows in the same way. More then that, there is an arrow
which connects them if and only if b1 = . . . = bm. So, we have got that
the quiver of the map f is inflation of the cycle, which s defined in the
next way:
Definition 1. 1. The quiver Γ2 is an inflation of the vertex v of the
quiver Γ1, if the former is obtained from the last by adding the vertex ṽ,
which is connected with all another vertexes in the seme way as it was for
v in Γ1. Vertexes v and ṽ are connected with arrow if and only if there
was a loop at v in Γ1. In this case ṽ also has a loop.
2. The quiver Γ2 is a inflation of Γ1 if it can be obtained from the
last by some number of sequent inflations of vertexes.
6. Gorenstein matrix of the map. Like in previous section let
matrix Ẽ = Ẽ(b1, . . . , bm) = (α̃ij) be constructed by the rule
α̃ij =
{
1 bi > bj and i 6= j;
0 bi < bj or i = j.
As it was mentioned matrix Ẽ is reduced exponent matrix. Let us
study conditions under which this matrix will be Gorenstein matrix.
Lemma 17. If matrix Ẽ is Gorenstein with permutation σ then all num-
bers b1, . . . , bm are different.
Proof. Let for different numbers i, j, the equality bi = bj takes place.
Then α̃ij = 1. Next, as bi = bj then independently on the permutation
σ, the inequality bi < bσ(i) and bj < bσ(i) take place or no simultaneously,
which means that α̃ij + α̃jσ(i) = α̃iσ(i) + 1, and means that matrix Ẽ is
not Gorenstein.
Theorem 5. If all numbers b1, . . . , bm are different then matrix Ẽ is
Gorenstein.
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.58 Exponent matrices and topological equivalence of maps
Proof. Using lemma 1 we may assume that b1 < b2 < . . . bm. Consider the
permutation σ = (m, m−1, . . . , 1) and show that matrix Ẽ is Gorenstein
with σ as correspond permutation.
Consider the equality α̃ij + α̃jσ(i) = α̃iσ(i). Let i > 1. then this
equality can be rewritten as α̃ij + α̃j,i−1 = α̃i,i−1. As bi > bi−1 then
α̃i,i−1 = 1. If j 6= i and j 6= i− 1 then, obviously, α̃ij + α̃j,i−1 = 1, which
is necessary. Otherwise if j = i then α̃ij = 0 and α̃j,i−1 = α̃i,i−1 and if
j = i− 1 then α̃j,i−1 = 0 and α̃i,j = α̃i,i−1 which is necessary.
Now let i = 1. Then for any j the equality α̃i,j = α̃i,σ(i) = 0 takes
place because b1 is minimal element of the set {b1, . . . , bm}. As σ(1) = m
then for any j the equality α̃j,σ(i) = 0 takes place, which means that
matrix Ẽ is really Gorenstein matrix.
Note that according to the proved theorem and lemma 1 if the matrix
of a doubly ordered set is Gorenstein matrix then this matrix is Hn.
References
[1] Dokuchaev M.A., Kirichenko V.V., Zelensky A.V. and Zhuravlev V.N., Goren-
stein Matrices, Algebra and discrete mathematics. – 2005 No 1. pp. 8–29.
[2] M. Hazewinkel, N. Gubareni and V.V. Kirichenko, Algebras, Rings and Modules,
MIA, 575, Vol. 1, Kluwer Academic Publishers, 2004, 380p.
[3] M. Hazewinkel, N. Gubareni and V.V. Kirichenko, Algebras, Rings and Modules,
MIA, 586, Vol. 2, Springer, 2007, 400p.
[4] Pelukh G.P., Sharkovskii A.N., Introduction to the theory of functional equations,
Kyiv: Naukova Dumka, 1974, 119 p.
[5] Plakhotnyk M.V., Fedorenko V.V. and Fedorenko Y.V. One-dimensional dy-
namical systems, whose cardinalities of orbits are uniformly bounded, Bulletin of
the University of Kyiv, No 4. pp. 119–128.
Contact information
V. Fedorenko Department of dynamical systems of the
Mathematical institute NASU Tereshchen-
kivska str., 3, Kyiv, Ukraine.
V. Kirichenko,
M. Plakhotnyk
Department of Mechanics and Mathemat-
ics, Kyiv National Taras Shevchenko Univ.,
Volodymyrska str., 64, 01033 Kyiv, Ukraine.
Received by the editors: 25.02.2008
and in final form 10.04.2008.
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