On the dimension of Kirichenko space
We introduce a notion of the Kirichenko space which is connected with the notion of Gorenstein matrix (see [2], ch. 14). Every element of Kirichenko space is an n × n matrix, whose elements are solutions of the equations ai,j +aj,σ(i) = ai,σ(i) ; a₁,i = 0 for i, j = 1, . . . , n determined by a...
Saved in:
| Published in: | Algebra and Discrete Mathematics |
|---|---|
| Date: | 2006 |
| Main Author: | |
| Format: | Article |
| Language: | English |
| Published: |
Інститут прикладної математики і механіки НАН України
2006
|
| Online Access: | https://nasplib.isofts.kiev.ua/handle/123456789/157355 |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| Journal Title: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Cite this: | On the dimension of Kirichenko space / M. Plakhotnyk // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 2. — С. 87–126. — Бібліогр.: 8 назв. — англ. |
Institution
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
nasplib_isofts_kiev_ua-123456789-157355 |
|---|---|
| record_format |
dspace |
| spelling |
Plakhotnyk, M. 2019-06-20T02:46:01Z 2019-06-20T02:46:01Z 2006 On the dimension of Kirichenko space / M. Plakhotnyk // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 2. — С. 87–126. — Бібліогр.: 8 назв. — англ. 1726-3255 https://nasplib.isofts.kiev.ua/handle/123456789/157355 We introduce a notion of the Kirichenko space which is connected with the notion of Gorenstein matrix (see [2], ch. 14). Every element of Kirichenko space is an n × n matrix, whose elements are solutions of the equations ai,j +aj,σ(i) = ai,σ(i) ; a₁,i = 0 for i, j = 1, . . . , n determined by a permutation σ which has no cycles of the length 1. We give a formula for the dimension of this space in terms of the cyclic type of σ. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On the dimension of Kirichenko space Article published earlier |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| title |
On the dimension of Kirichenko space |
| spellingShingle |
On the dimension of Kirichenko space Plakhotnyk, M. |
| title_short |
On the dimension of Kirichenko space |
| title_full |
On the dimension of Kirichenko space |
| title_fullStr |
On the dimension of Kirichenko space |
| title_full_unstemmed |
On the dimension of Kirichenko space |
| title_sort |
on the dimension of kirichenko space |
| author |
Plakhotnyk, M. |
| author_facet |
Plakhotnyk, M. |
| publishDate |
2006 |
| language |
English |
| container_title |
Algebra and Discrete Mathematics |
| publisher |
Інститут прикладної математики і механіки НАН України |
| format |
Article |
| description |
We introduce a notion of the Kirichenko space
which is connected with the notion of Gorenstein matrix (see [2],
ch. 14). Every element of Kirichenko space is an n × n matrix,
whose elements are solutions of the equations ai,j +aj,σ(i) = ai,σ(i)
;
a₁,i = 0 for i, j = 1, . . . , n determined by a permutation σ which
has no cycles of the length 1. We give a formula for the dimension
of this space in terms of the cyclic type of σ.
|
| issn |
1726-3255 |
| url |
https://nasplib.isofts.kiev.ua/handle/123456789/157355 |
| citation_txt |
On the dimension of Kirichenko space / M. Plakhotnyk // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 2. — С. 87–126. — Бібліогр.: 8 назв. — англ. |
| work_keys_str_mv |
AT plakhotnykm onthedimensionofkirichenkospace |
| first_indexed |
2025-11-25T12:56:41Z |
| last_indexed |
2025-11-25T12:56:41Z |
| _version_ |
1850512637367943168 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2006). pp. 87 – 126
c© Journal “Algebra and Discrete Mathematics”
On the dimension of Kirichenko space
Makar Plakhotnyk
Communicated by M. Ya. Komarnytskyj
Dedicated to the memory of V. M. Usenko
Abstract. We introduce a notion of the Kirichenko space
which is connected with the notion of Gorenstein matrix (see [2],
ch. 14). Every element of Kirichenko space is an n × n matrix,
whose elements are solutions of the equations ai,j +aj,σ(i) = ai,σ(i);
a1,i = 0 for i, j = 1, . . . , n determined by a permutation σ which
has no cycles of the length 1. We give a formula for the dimension
of this space in terms of the cyclic type of σ.
Introduction
We remind some definitions and notions from [2], ch 14. Denote by
Mn(B) the ring of all n× n matrices over the ring B. Let Z be a ring of
all integers.
An integer matrix E = (aij) is called
an exponent matrix, if aij + ajk ≥ aik for all i, j, k;
an reduced exponent, if aij + aji > 0 for all i, j.
A reduced exponent matrix E will be called Gorenstein, if there exists
a permutation σ of {1, . . . , n} such that ai,k + ak,σ(i) = ai,σ(i).
We will denote σ by σ(E). As usual, we will call σ(E) the Kirichenko
permutation. Note that σ(E) of Gorenstein matrix has no cycles of length
1. We will name relations of the type ai,k + ak,σ(i) = ai,σ(i) Gorenstein
relations. Exponent matrices are widely used in the theory of tiled
orders over a discrete valuation ring. Many properties of tiled orders and
their quivers are fully determined by their exponent matrices. Gorenstein
matrices appeared at the first time in [5].
Type your thanks here..
Key words and phrases: Gorenstein matrix, Gorenstein tiled order.
88 On the dimension of Kirichenko space
It is easy to see, that from the condition ai,k + ai,σ(k) = ai,σ(i) for
i = k we obtain ak,k = 0 for every k, 1 ≤ k ≤ n.
Theorem. (see [1], p. 17) Let permutation σ of the set {1, 2, . . . n} be
arbitrary permutation without fixed elements. Then there exists Goren-
stein matrix A with σ as Kirichenko permutation, such that all elements
of A belong to the set {0, 1, 2}.
Proof. Matrix A, which has σ as correspondent Kirichenko permutation
may be constructed in direct way:
ai,i = 0 and ai,σ(i) = 2 for i = 1, . . . , n;
ai,j = 1 for all i, j ∈ [1, n], such that i 6= j and j 6= σ(i).
It is obvious that such matrix will be Gorenstein and σ will be
Kirichenko permutation.
Exponent matrices A, B are called equivalent, if one may be ob-
tained 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.
Theorem (see [1], p. 15): Under the transformations of the first
type Gorenstein matrix goes to Gorenstein one with the same Kirichenko
permutation.
Note that transformations of the first type define free commutative
group with n generators.
If one consider matrices
Hn =
0 0 0 · · · 0 0
1 0 0 · · · 0 0
1 1 0 · · · 0 0
· · · · · · · · · · · · · · · · · ·
1 1 1 · · · 0 0
1 1 1 · · · 1 0
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 Gorenstein
matrix with Kirichenko permutation σ(Hn) = (n, n−1, . . . 2, 1), and G2m
is Gorenstein one with Kirichenko permutation σ(G2m) =
m∏
i=1
(i, m + i).
A matrix B = (bi,j) is called (0, 1) matrix, if bi,j is either zero or one.
Theorem (see [1], p. 15): Gorenstein (0, 1) matrix is equivalent
either Hn or G2m.
M. Plakhotnyk 89
Theorem (see [1], p. 15): If A is Gorenstein matrix with Kirichenko
permutation σ, and B is obtained from it by transformation of the second
type, defined by transposition τ , then B is Gorenstein matrix with the
permutation τστ .
According to the last theorem, without bounding of generality we may
suppose that for the cyclic index {l1, . . . , lq} of correspond Kirichenko
permutation the conditions 2 ≤ l1 ≤ . . . ≤ lq and l1 + . . . + lq = n take
place. Denote ni =
i∑
k=1
lk for every 1 ≤ i ≤ q. Then σ will look as
σ = (1, 2 . . . n1)(n1 + 1, n1 + 2 . . . , n2) · · · (nq−1 + 1, nq−1 + 2, . . . , nq).
Further we will consider only Gorenstein matrices and under equiv-
alence of matrices we will consider possibility of obtaining one from
another only by transformations of the first type.
One may come to a question of describing of Gorenstein matrices in
inverse way to the one, presented in the definition. In the set of square
matrices of order n over the field K of characteristic 0 consider linear
space K(n, σ) of all matrices A = (ai,j) such that Gorenstein relations
take place for A.
When a problem of describing of such matrices will be solved, it will
become possible to talk about restrictions, generated by inequalities from
the definition of reduced exponent matrix.
If A ∈ K(n, σ), B is equivalent to it, then B ∈ K(n, σ), and ev-
ery equivalence class contains the unique matrix with zeros at the first
line, which is canonical representor of the equivalence class. Call ma-
trix is called Kirichenko matrix. More over, it’s easy to see, that for
Kirichenko matrix from the condition ai,k + ak,σ(i) = ai,σ(i) for i = 1
one may obtain ak,σ(1) = 0 for all k. It is easy to prove that the set
of Kirichenko matrices is linear space and we will name it Kirichenko
space.
We need the next notations. Let A = (ai,j) be Kirichenko matrix of
order n with Kirichenko permutation σ, and σ is the product of q cycles
of length l1, l2, . . . , lq, respectively 2 ≤ li ≤ li+1, 1 ≤ i ≤ q − 1 and
l1 + . . . + lq = n.
The main result of this work is calculating the dimension of Kirichen-
ko space. To do this, some elements of Kirichenko matrix were considered
as parameters, and represent all another elements of matrix in terms of
these parameters. The relations which represent the elements of matrix
through parameters we name element relations, and relations between
parameters we name count relations. We were succeed to find the set
of parameters, element and count relations in the form such that it is
90 On the dimension of Kirichenko space
possible to prove the equivalence of the set of Gorenstein relations to the
set of element relations and count relations. After this, we were succeed
to find the formula for the defect of the system of count relations i.e.
the dimension of Kirichenko space may be represented in terms of cyclic
index of Kirichenko permutation σ as
2 − 2q +
q∑
r=1
[
lr
2
]
+
1
2
∑
r 6=s
(ls, lr),
where (a, b) denotes the greatest common divisor of numbers a and b, and
[x] denotes integer part of a number x which is the greatest integer, not
larger then x.
Definition 1. Denote xk = ak,1 for every k, 2 ≤ k ≤ n. For arbitrary
r, 0 < r < q, and for every k, nr + 2 ≤ k ≤ n, denote as well zk,r =
= anr+1,k. Variables xk and zk,r we name parameters.
Definition 2. Element relation is formula which is corollary of Goren-
stein relations and expresses some element of Gorenstein matrix as linear
function of parameters. Full set of element relations is set of relations
which contains the formula for every element of matrix A.
Example 1. Relations
ak,σ(k) = xk, 2 ≤ k ≤ n (1)
are element.
Really, if we substitutes value i = 1 to the equality ak,i + ai,σ(k) =
ak,σ(k) and take into account the fact that the first row of A is zero, we
obtain the necessary relation.
Example 2. Relations
ak,2 = 0, 1 ≤ k ≤ n
ak,k = 0, 1 ≤ k ≤ n
a1,k = 0, 1 ≤ k ≤ n
(2)
and definition of parameters are element relations.
Element relations from two previous examples we name trivial.
M. Plakhotnyk 91
1. Element relations are corollaries of Gorenstein rela-
tions
Proposition 1. Equalities (1)-(11) are full set of element relations,
where
a2,k =
{
xns+1
, if k = ns + 1 for some s
xk−1, if k 6= ns + 1 for any s
(3)
For every m, 3 ≤ m ≤ n1, every s, 1 ≤ s ≤ q − 1 and every k, ns +
1,≤ k ≤ ns+1 or m + 1 ≤ k ≤ n1 write out
ak,m =
m−1∑
i=2
xi −
m−3∑
i=0
xns+1−i, if k = ns + 1;
m−1∑
i=2
xi −
l−1∑
i=1
xk−i −
m−l−2∑
i=0
xns+1−i, if k = ns + l, 2 ≤ l ≤ m −2;
m−1∑
i=2
xi −
m−2∑
i=1
xk−i, if ns + m − 1 ≤ k ≤ ns+1, or m < k ≤ n1
(4)
and
am,k =
m−2∑
i=0
xns+1−i −
m−1∑
i=2
xi, if k = ns + 1,
l−1∑
i=1
xk−i +
m−l−1∑
i=0
xns+1−i −
m−1∑
i=2
xi, if k = ns + l, 2 ≤ l < m
m−1∑
i=1
xk−i −
m−1∑
i=2
xi, if ns + m ≤ k ≤ ns+1, or m < k ≤ n1
(5)
Writing out element relations for other elements of matrix A, we will
write indices of its elements as ak,m and am,k, where k > m for nr +
1 ≤ m ≤ nr+1 and some r, 1 ≤ r ≤ q − 1. in this case for every
s, r ≤ s ≤ q − 1, and k, ns + 1 ≤ k ≤ ns+1 relations (6)-(11) look like:
ak,nr+1 =
{
xk − zns+1,r, if k = ns+1, s > r
xk − zk+1,r, if ns + 1 ≤ k < ns+1, s > r
(6)
ak,nr+1 =
{
xk, if k = nr+1.
xk − zk+1,r, if nr + 1 < k < nr+1
(7)
ak,nr+2 = xnr+1 − zk,r, if k > nr + 2 (8)
anr+2,k =
{
xns+1
− xnr+1 + zns+1,r, if k = ns + 1, s > r
xk−1 − xnr+1 + zk−1,r, if ns + 2 ≤ k ≤ ns+1, k > nr + 2
(9)
92 On the dimension of Kirichenko space
ak,nr+m =
−zns+1−m+3,r +
m−1∑
i=1
xnr+i −
m−3∑
i=0
xns+1−i,
if k = ns + 1 > nr + m;
−zns+1−m+l+2,r +
m−1∑
i=1
xnr+i −
m−l−2∑
i=0
xns+1−i −
l−1∑
i=1
xk−i,
if k = ns + l > nr + m 2 ≤ l ≤ m − 2;
−zk−m+2,r +
m−1∑
i=1
xnr+i −
m−2∑
i=1
xk−i,
if ns + m − 1 ≤ k ≤ ns+1, k > nr + m;
(10)
anr+m,k =
zns+1−m+2,r −
m−1∑
i=1
xnr+i +
m−2∑
i=0
xns+1−i,
if k = ns + 1 > nr + m;
zns+1−m+l+1,r −
m−1∑
i=1
xnr+i +
l−1∑
i=1
xk−i +
m−l−1∑
i=0
xns+1−i,
if k = ns + l > nr + m, 2 ≤ l ≤ m − 1;
zk−m+1,r −
m−1∑
i=1
xnr+i +
m−1∑
i=1
xk−i,
if ns + m ≤ k ≤ ns+1, k > nr + m.
(11)
For proving this proposition we need the following lemmas.
Lemma 1. If 3 ≤ m ≤ n1, max(ns + 1, m + 1) ≤ k ≤ ns+1, then
relations (4) and (5) are corollaries of Gorenstein ones.
Proof. Consider equality a2,k + ak,3 = x2, whence, using (3) obtain, that
ak,3 = x2−xns+1
, if k = ns+1, and ak,3 = x2−xk−1 if ns+2 ≤ k ≤ ns+1,
which is equality (4) for m = 3.
Consider ak,3+a3,k+1 = xk, k 6= ns+1. For use (4), consider situations
k = ns +1 and k 6= ns +1. For k = ns +1, obtain (x2−xns+1
)+a3,ns+2 =
xns+1, whence a3,p = xp−1+xns+1
−x2, for p = ns+2. Consider situation,
when ns + 2 ≤ k < ns+1. In this case (x2 − xk−1) + a3,k+1 = xk, whence
a3,k+1 = xk + xk−1 − x2, that is a3,p = xp−1 + xp−2 − x2 for ns + 3 ≤ p ≤
ns+1. Now, substituting k = ns+1 to an equality ak,3 + a3,σ(k) = xk, we
obtain equality (5) for m = 3.
Let us prove the validity of relations (4) and (5) for 3 ≤ m ≤ n1 by
induction. The induction base is just proved. Suppose these relations to
be valid for some m, 3 ≤ m < n1,. Let us proof their validity for some
m + 1 too.
M. Plakhotnyk 93
Consider equality am,k + ak,m+1 = xm, m < n1, whence ak,m+1 =
xm − am,k.
For k = ns + 1 we obtain ak,m+1 = xm −
(
m−2∑
i=0
xns+1−i −
m−1∑
i=2
xi
)
=
(m+1)−1∑
i=2
xi −
(m+1)−3∑
i=0
xns+1−i, which is necessary.
Fix l ∈ [2, m−1]. Then for k = ns+l we obtain ak,m+1 = xm−am,k =
xm −
(
l−1∑
i=1
xk−i +
m−l−1∑
i=0
xns+1−i −
m−1∑
i=2
xi
)
=
(m+1)−1∑
i=2
xi −
l−1∑
i=1
xk−i−
−
(m+1)−l−2∑
i=0
xns+1−i for k = ns+l, 2 ≤ l ≤ (m+1)−2, which is necessary.
Consider equality am,k + ak,σ(m) = xm for k, ns + m ≤ k ≤ ns+1 and
obtain ak,m+1 = xm − am,k = xm −
(
m−1∑
i=1
xk−i −
m−1∑
i=2
xi
)
=
(m+1)−1∑
i=2
xi −
(m+1)−2∑
i=1
xk−i, ns + (m + 1) − 1 ≤ k ≤ ns+1.
Thus we have proved that ak,m+1 =
=
(m+1)−1∑
i=2
xi −
(m+1)−3∑
i=0
xns+1−i, ifk = ns + 1,
(m+1)−1∑
i=2
xi −
l−1∑
i=1
xk−i −
(m+1)−2−l∑
i=0
xns+1−i, ifk = ns + l, 2 ≤ l < m
(m+1)−1∑
i=2
xi −
(m+1)−2∑
i=1
xk−i, ifk > ns + m − 1
.
Consider equality ak,m+1 + am+1,k+1 = xk, k 6= ns.
Let k = ns + 1. Then am+1,ns+2 = xns+1 − ans+1,m+1 = xns+1 −(
m∑
i=2
xi −
m−2∑
i=0
xns+1−i
)
= xp−1 −
(m+1)−1∑
i=2
xi +
(m+1)−3∑
i=0
xns+1−i = am+1,p ,
for p = ns + l, l = 2.
Let k = ns + l, 2 ≤ l ≤ m − 1. Then am+1,ns+(l+1) = xns+l −
−ans+l,m+1 = xns+l−
(
m∑
i=2
xi −
l−1∑
i=1
xk−i −
m−l−1∑
i=0
xns+1−i
)
= −
(m+1)−1∑
i=2
xi+
(l+1)−1∑
i=1
x(k+1)−i +
(m+1)−(l+1)−1∑
i=0
xns+1−i, which is the same as am+1,ns+l =
= −
(m+1)−1∑
i=2
xi +
l−1∑
i=1
xk−i +
(m+1)−l−1∑
i=0
xns+1−i, for 3 ≤ l ≤ (m + 1) − 1.
94 On the dimension of Kirichenko space
Let ns + m ≤ k ≤ ns+1. Then am+1,k+1 = xk − ak,m+1 = xk −
−
(
m∑
i=2
xi −
m−1∑
i=1
xk−i
)
= −
m∑
i=2
xi +
m∑
i=1
xk+1−i, for ns + m ≤ k ≤ ns+1,
which is the same as am+1,k = −
(m+1)−1∑
i=2
xi +
(m+1)−1∑
i=1
xk−i for ns + (m +
1) ≤ k ≤ ns+1.
Consider equality ak,m+1 + am+1,σ(k) = xk for k = ns+1, and ob-
tain am+1,ns+1 = xns+1
− ans+1,m+1 = xns+1
−
(
m∑
i=2
xi −
m−1∑
i=1
xns+1−i
)
=
(m+1)−2∑
i=0
xns+1−i −
(m+1)−1∑
i=2
xi = am+1,p , where p = ns + 1.
thus we have proved that am+1,k =
=
(m+1)−2∑
i=0
xns+1−i −
(m+1)−1∑
i=2
xi, if k = ns + 1,
l−1∑
i=1
xk−i +
(m+1)−l−1∑
i=0
xns+1−i −
(m+1)−1∑
i=2
xi, if k = ns + l, 2 ≤ l < m + 1
(m+1)−1∑
i=1
xk−i −
(m+1)−1∑
i=2
xi, if ns + (m + 1) ≤ k ≤ ns+1
,
and it finishes the proof of relations (4) and (5).
Fix arbitrary r, 1 ≤ r ≤ q−1, arbitrary s, r ≤ s ≤ q−1, and consider
k, ns + 1 ≤ k ≤ ns+1. For convenience instead of zk,r we will write zk.
Lemma 2. For r, s and k from the intervals noted above, equalities (6)-
(9) take place.
Proof. Consider equality ak,nr+1 + anr+1,σ(k) = xk. For k 6= ns+1 obtain
ak,nr+1 + zk+1 = xk, whence ak,nr+1 = xk − zk+1.
Considering k = nr+1 obtain anr+1,nr+1 +anr+1,nr+1 = xnr+1
, whence
ak,nr+1 = xk, that is
ak,nr+1 =
{
xk, if k = nr+1.
xk − zk+1, if nr + 1 < k < nr+1
,
which coincides with (7).
For k = ns+1, s > r obtain ans+1,nr+1 + anr+1,ns+1 = xns+1
, whence
ak,nr+1 = xk − zns+1, that is
ak,nr+1 =
{
xk − zns+1, if k = ns+1.
xk − zk+1, if ns + 1 ≤ k < ns+1
,
which coincides with (6).
M. Plakhotnyk 95
Consider equality anr+1,k + ak,nr+2 = xnr+1, whence, according to
denotations, ak,nr+2 = xnr+1 − anr+1,k = xnr+1 − zk, which is (8).
If lr > 2, then consider equality ak,nr+2+anr+2,k+1 = xk for k 6= ns+1.
Whence anr+2,k+1 = xk − ak,nr+2 = xk − xnr+1 + zk, that is
anr+2,p = xp−1 − xnr+1 + zp−1 for p 6= ns + 1.
For k = ns+1 obtain ans+1,nr+2 + anr+2,ns+1 = xns+1
, and so
anr+2,k = xns+1
− ans+1,nr+2 = xns+1
− (xnr+1 − zns+1
)
for k = ns + 1, whence
anr+2,p =
{
xns+1
− xn1+1 + zns+1
, if k = ns + 1, which is (9).
xk−1 − xn1+1 + zk−1, if ns + 2 < k ≤ ns+1
Leaving in force denotations for boundaries for r, s and k, consider
arbitrary m, 3 ≤ m ≤ lr.
Lemma 3. For every r, s, k and m from noted intervals, relations (10)
and (11) are corollaries of Gorenstein relations.
Proof. Proof by induction for m.
Consider the equality anr+2,k + ak,nr+3 = xnr+2, whence
ak,nr+3 = xnr+2 − anr+2,k, i.e.
ak,nr+3 =
{
xnr+1 + xnr+2 − xns+1
− zns+1
, if k = ns + 1.
xnr+1 + xnr+2 − xk−1 − zk−1, if k 6= ns + 1
, and so
obtain induction base for (10).
Consider equality ak,nr+3 + anr+3,k+1 = xk for k 6= ns+1, whence
anr+3,p = xp−1 − ap−1,nr+3, for ns + 2 ≤ p ≤ ns+1.
For k 6= ns + 2 then the last equality is equivalent to anr+3,k =
xk−1−(xnr+1+xnr+2−xk−2−zk−2) = xk−1+xk−2−xnr+1−xnr+2+zk−2.
For k = ns+2 obtain anr+3,k = xk−1−(xnr+1+xnr+2−xns+1
−zns+1
) =
xk−1 − xnr+1 − xnr+2 + xns+1
+ zns+1
Substitute k = ns+1 to equality ak,nr+3 + anr+3,σ(k) = xk, and obtain
anr+3,ns+1 = xns+1
− ans+1,nr+3 = xns+1
− (xnr+1 + xnr+2 − xns+1−1 −
zns+1−1) = xns+1
+ xns+1−1 − xnr+1 − xnr+2 + zns+1−1. Whence obtain
anr+3,k =
xns+1
+ xns+1−1 − xnr+1 − xnr+2 + zns+1−1; if k = ns + 1,
xk−1 − xnr+1 − xnr+2 + xns+1
+ zns+1
; if k = ns + 2,
xk−1 + xk−2 − xnr+1 − xnr+2 + zk−2, if ns + 3 ≤ k ≤ ns+1,
which gives induction base for (11).
96 On the dimension of Kirichenko space
Assume that relations (10) and (11) are valid for some 3 ≤ m ≤ lr−1,
and proof, that in this case they will be valid for m + 1 also.
Consider equality anr+m,k +ak,nr+m+1 = xnr+m, whence ak,nr+m+1 =
xnr+m − anr+m,k, that is
ak,nr+m+1 = xnr+m −
(
zns+1−m+2 −
m−1∑
i=1
xnr+i +
m−2∑
i=0
xns+1−i
)
if k = ns + 1; For the case k = ns + l, 2 ≤ l ≤ m − 1 the formula for
ak,nr+m+1 will be ak,nr+m+1 =
= xnr+m −
(
zns+1−m+l+1 −
m−1∑
i=1
xnr+i +
l−1∑
i=1
xk−i +
m−l−1∑
i=0
xns+1−i
)
,
and if ns + m ≤ k ≤ ns+1, then
ak,nr+m+1 = xnr+m −
(
zk−m+1 −
m−1∑
i=1
xnr+i +
m−1∑
i=1
xk−i
)
,
which is necessary.
Consider equality ak,nr+m+1+anr+m+1,σ(k) = xk, whence for k 6= ns+1
obtain anr+m+1,k+1 = xk − ak,nr+m+1.
Substitute k = ns + 1 to the last equality and obtain
anr+m+1,ns+2 = xns+1 − ans+1,nr+m+1 = xns+1 −
(
m∑
i=1
xnr+i−
−
m−2∑
i=0
xns+1−i − zns+1−m+2
)
= zns+1−m+2 + xns+1 −
m∑
i=1
xnr+i+
+
m−2∑
i=0
xns+1−i, that is xnr+m+1,p = zns+1−(m+1)+3 + xp−1−
−
(m+1)−1∑
i=1
xnr+i +
(m+1)−3∑
i=0
xns+1−i for p = ns + 2, which is necessary.
Substitute k = ns + l, 2 ≤ l ≤ m− 1 to equality anr+m+1,k+1 = xk −
ak,nr+m+1, and obtain anr+m+1,k+1 = xk−
(
−zns+1−m+l+1 +
m∑
i=1
xnr+i−
−
l−1∑
i=1
xk−i −
m−1−l∑
i=0
xns+1−i
)
= zns+1−m+l+1 −
m∑
i=1
xnr+i +
l∑
i=1
xk+1−i +
M. Plakhotnyk 97
m−l−1∑
i=0
xns+1−i, that is anr+m+1,p =
= zns+1−(m+1)+l+2 −
(m+1)−1∑
i=1
xnr+i +
l−1∑
i=1
xp−i +
m−l−1∑
i=0
xns+1−i
for p = ns + l, 3 ≤ l ≤ (m + 1) − 1, which is necessary. Substitute
ns + m ≤ k ≤ ns+1 − 1 to an equality ak,nr+m+1 + anr+m+1,σ(k) = xk,
whence, according to assumption of induction, obtain anr+m+1,k+1 =
= xk − ak,nr+m+1 = xk −
(
m∑
i=1
xnr+i − zk−m+1 −
m−1∑
i=1
xk−i
)
=
= zk−m+1 −
m∑
i=1
xnr+i +
m−1∑
i=0
xk−i,
that is
anr+m+1,p = zp−(m+1)+2 −
(m+1)−1∑
i=1
xnr+i +
(m+1)−1∑
i=1
xp−i
for ns + (m + 1) ≤ p ≤ ns+1, which is necessary.
Substitute k = ns+1 to an equality anr+m+1,σ(k) = xk − ak,nr+m+1,
and obtain anr+m+1,ns+1 = xns+1
− ans+1,nr+m+1 = xns+1
−
−
(
m∑
i=1
xnr+i − zns+1−m+1 −
m−1∑
i=1
xns+1−i
)
= zns+1−(m+1)+2−
−
(m+1)−1∑
i=1
xnr+i +
(m+1)−2∑
i=0
xns+1−i, which is necessary to prove.
2. Count relations are corollaries of Gorenstein relations
Definition 3. Untrivial relation between parameters is called count re-
lation, if it is provided by Gorenstein relations.
Proposition 2. Relations (12)-(18) are count relations.
If n1 = 2, then for every s, 1 ≤ s ≤ q − 1, equalities
{
xk−1 + xk = x2, if ns + 2 ≤ k ≤ ns+1,
xns+1
+ xk = x2, if k = ns + 1
(12)
98 On the dimension of Kirichenko space
take place.
For every m, 1 ≤ m ≤ n1 − 1, equalities
xm+1 = xn1−m+1 (13)
take place.
If n1 > 2, then for arbitrary s, 1 ≤ s ≤ q − 1, equalities
xns+1 +
n1−2∑
i=0
xns+1−i =
n1∑
i=2
xi, (14)
{
xns+p = xns+1−n1+p, if 1 ≤ p ≤ n1
xns+p = xns+p−n1
, if n1 < p ≤ ls
(15)
take place.
For every r, 1 ≤ r ≤ q−1, and m, 2 ≤ m ≤ lr−1, equalities (16)-(18)
znr+2,r = xnr+1, (16)
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i = znr+1−m+2,r + znr+m+1,r, (17)
zns+1 − zns+1−(lr−1) =
ns+1∑
i=ns+1−(lr−1)
xi −
nr+1∑
i=nr+1
xi =: f0
zns+p+1 − zns+1−lr+p+1 =
=
p−1∑
i=0
x(ns+p)−i +
lr−p−1∑
i=0
xns+1−i −
nr+1∑
i=nr+1
xi =: fp
for 1 ≤ p ≤ lr − 1
zns+p+1 − zns+p−(lr−1) =
lr−1∑
i=0
xns+p−i −
nr+1∑
i=nr+1
xi =: fp
for lr ≤ p ≤ ls − 1
(18)
take place.
The proof of this proposition is broken to some natural parts which
we will formulate as lemmas. For convenience of further calculations we
will write zk instead of zk,r, if it well be obvious r, 0 < r < q, which is
under consideration.
Lemma 4. If n1 = 2, then relation (12) takes place for every s, 1 ≤ s ≤
q − 1 .
Proof. Assume that n1 = 2 and consider the equality a2,k + ak,1 = a2,1.
For ns +2 ≤ k ≤ ns+1, according to (3) obtain xk−1 +xk = x2, which
coincides with the first line of (12).
According to (3) for k = ns + 1 we obtain xns+1
+ xns
= x2, which
coincides with second line of (12).
M. Plakhotnyk 99
Lemma 5. The equality (13) takes place for every m, 1 ≤ m ≤ n1 − 1.
Proof. We will prove this lemma by induction for m.
Consider equality ak,2 + a2,σ(k) = xk for k = n1, and obtain an1,2 +
a2,1 = xn1
. According to definition and according to (2) we have an1,2 = 0
and a2,1 = x2, whence x2 = xn1
, that is (13) for m = 1 and gives the
induction base.
Let for some m, 1 < m ≤ n1 − 1, the equality (13) takes place for
every i < m. Show, that in this case it well take place for i = m also.
Substitute k = n1 into equality ak,m+1 + am+1,σ(k) = ak,σ(k), and
obtain an1,m+1 +xm+1 = xn1
. To express an1,m+1 we use (4) substituting
respectively n1 and m+1 instead of k and m, whence using the third line
of (4) we obtain
(
m∑
i=2
xi −
m−1∑
i=1
xn1−i
)
+ xm+1 = xn1
, that is
m+1∑
i=2
xi =
m−1∑
i=0
xn1−i, whence, using induction base obtain that xm+1 = xn1−m+1,
which is necessary.
Lemma 6. If n1 > 2, then for arbitrary s, 1 ≤ s ≤ q−1 the equality (14)
takes place.
Proof. For arbitrary s, 1 ≤ s ≤ q−1 consider equality an1,ns+1+ans+1,1 =
an1,1. Then according to (5) for m := n1 and k = ns + 1 (for calculating
an1,ns+1, according to the first line of this relation), and notation, obtain
n1−2∑
i=0
xns+1−i −
n1−1∑
i=2
xi + xns+1 = xn1
, that is
n1−2∑
i=0
xns+1−i + xns+1 =
n1∑
i=2
xi,
which is (14).
Lemma 7. For every n1 < p ≤ ls the equalities, which are second line
of (15), are corollaries of element relations and Gorenstein ones.
Proof. Fix arbitrary value of p, n1 < p ≤ ls, and consider equality
an1,k + ak,1 = an1,1 for k = ns + p − 1 and k = ns + p, whence obtain
n1−1∑
i=0
xns+p−1−i =
n1∑
i=2
xi, that is
n1∑
i=1
xns+p−i =
n1∑
i=2
xi and
n1−1∑
i=0
xns+p−i =
n1∑
i=2
xi, whence, subtracting the last equality from previous one, obtain
xns+p = xns+p−n1
,
that proves lemma.
100 On the dimension of Kirichenko space
Lemma 8. For every 2 ≤ s ≤ q − 1 and p, 1 ≤ p ≤ n1 the equalities,
defined by the first line of (15) are corollaries of Gorenstein relations and
element relations.
Proof. We prove this lemma by induction for p.
According to (2,5), for k = ns + p, 1 < p < n1 equality an1,ns+p +
ans+p,1 = an1,1 may be changed to
p−1∑
i=1
xk−i +
n1−p−1∑
i=0
xns+1−i −
n1−1∑
i=2
xi +
xns+p = xn1
, whence
p−1∑
i=0
xns+p−i +
n1−p−1∑
i=0
xns+1−i =
n1∑
i=2
xi (19)
Consider equality an1,ns+1 + ans+1,1 = an1,1, whence, according to (5),
obtain
(
n1−2∑
i=0
xns+1−i −
n1−1∑
i=2
xi
)
+ xns+1 = xn1
whence (19) is valid for
every p, 1 ≤ p < n1.
Using (19) for p = 2 obtain xns+1 + xns+2 +
n1−3∑
i=0
xns+1−i =
n1∑
i=2
xi.
Subtracting from it equality for p = 1, obtain
xns+2 = xns+1−(n1−2),
and so, the induction base, that is the first line of (15) for p = 2 is proved.
Consider arbitrary arbitrary p, from the interval [3, n1−1]. Substitute
k = ns + p − 1 and k = ns + p into equality an1,k + ak,1 = an1,1, and
using (5) for m = n1 obtain
(p−1)−1∑
i=1
xns+p−1−i +
n1−(p−1)−1∑
i=0
xns+1−i −
n1−1∑
i=2
xi
+ xns+p−1 = xn1
,
that is
p−1∑
i=1
xns+p−1−i +
n1−(p−1)−1∑
i=0
xns+1−i =
n1∑
i=2
xi
and
p−1∑
i=0
xns+p−i +
n1−p−1∑
i=0
xns+1−i =
n1∑
i=2
xi,
whence, subtracting the last equality from previous one, obtain xns+p =
xns+1−(n1−p), which proves the validity of lemma for 3 ≤ p ≤ n1 − 1.
For k = ns + n1, equality an1,k + ak,1 = an1,1 according to (5) one
may change to the form
n1−1∑
i=1
xns+n1−i −
n1−1∑
i=2
xi + xns+n1
= xn1
, that is
M. Plakhotnyk 101
n1−1∑
i=0
xns+n1−i =
n1∑
i=2
xi. Subtracting this equality from (19) for p = n1 −1
obtain (15) for p = n1.
Let’s return to equality an1,k + ak,1 = xn1
. According to (5), it
may be rewritten in the form
(
n1−1∑
i=1
xk−i −
n1−1∑
i=2
xi
)
+ xk = xn1
, that is
n1−1∑
i=0
xk−i =
n1∑
i=2
xi. From the second line of equality (15) (which is proved
in previous lemma) obtain, that for arbitrary s, 2 ≤ s ≤ q − 1 equalities
xns+1 = xns+n1+1, xns+2 = xns+n1+2 and so on up to xns+1−n1
= xns+1
take place, so sequence {xns+1, . . . , xns+1
} is periodic with period n1. It
means that
n1−1∑
i=0
xk−i is sum of n1 elements of this sequence, which are
equal to it’s last elements, whence obtain equality
n1−1∑
i=0
xk−i =
n1∑
i=2
xi,
which according to (14) one may change to the form
n1−1∑
i=0
xns+1−i =
n1−2∑
i=0
xns+1−i+xns+1, whence xns+1−n1+1 = xns+1, which is (15) for p = 1,
and finishes the proof of equalities (15) in general.
Substitute k = nr + 1 into equality ak,nr+1 + anr+1,σ(k) = xk, and
obtain znr+2 = xnr+1, which is (16).
Lemma 9. For every 2 ≤ m ≤ lr − 1 equalities (17) are corollaries of
element relations and Gorenstein ones.
Proof. Proof by induction for m.
Substitute k = nr+1 into equality ak,nr+2+anr+2,σ(k) = xk and obtain
anr+1,nr+2+anr+2,nr+1 = xnr+1
, whence, according to (8), and (9), obtain
(xnr+1 − znr+1
) + (xnr+2 − znr+3) = xnr+1
, that is
znr+1
+ znr+3 = xnr+1 + xnr+2 − xnr+1
,
which is the equality (17) for m = 2 and so induction base is proved.
For every m, 3 ≤ m ≤ lr − 1, substitute k = nr+1 into equal-
ity ak,nr+m + anr+m,σ(k) = xk, and obtain anr+1,nr+m + anr+m,nr+1 =
xnr+1, whence, according to (7) and (10) obtain
(
m∑
i=1
xnr+i −
m−1∑
i=1
xnr+1−i − znr+1−m+1
)
+(xnr+m+1 − znr+m+2) = xnr+1
,
102 On the dimension of Kirichenko space
whence
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i = znr+1−m+2 + znr+m+1,
and lemma is proved.
Lemma 10. For arbitrary p, 1 ≤ p ≤ lr −1, equalities, which are second
line of (18) are corollaries of element relations and Gorenstein ones.
Proof. Consider equality anr+1,k + ak,nr+1 = anr+1,nr+1 = xnr+1
for k =
ns + p, p > 0.
For p = 1 obtain anr+1,k, using (11) for k = nr + 1 and m = lr.
More over, according to (6), ak,nr+1 = xk − zk+1, k 6= ns, whence obtain(
−
lr−1∑
i=1
xnr+i +
lr−2∑
i=0
xns+1−i + zns+1−lr+2
)
+(xns+1−zns+2) = xnr+1
, that
is
lr−2∑
i=0
xns+1−i + zns+1−lr+2 + xns+1 − zns+2 =
nr+1∑
i=n1+1
xi
For 2 ≤ p ≤ lr−1 obtain anr+1,k using (11), for k = ns+l, substituting
l = p; m = lr, whence
(
−
lr−1∑
i=1
xnr+i +
p−1∑
i=1
xns+p−i +
lr−p−1∑
i=0
xns+1−i + zns+1−lr+p+1
)
+ (xns+p − zns+p+1) = xnr+1
, that is
p−1∑
i=0
xns+p−i +
lr−p−1∑
i=0
xns+1−i + zns+1−lr+p+1 − zns+p+1 =
nr+1∑
i=nr+1
xi.
For p = ns+1 − ns the formula (6) gives ans+1,nr+1 = xns+1
− zns+1
whence obtain
lr−1∑
i=0
xns+1−i + zns+1−lr+1 − zns+1 =
nr+1∑
i=nr+1
xi,
which gives the first line of (18).
Lemma 11. For every p, lr ≤ p ≤ ls − 1 equalities, which are second
line of (18) are corollaries of element relations and Gorenstein relations.
Proof. Consider equality anr+1,k + ak,nr+1 = anr+1,nr+1. Let k = ns + p.
M. Plakhotnyk 103
For lr − 1 < p < ls from the formula (11) for k, ns + m ≤ k ≤
ns+1, and m = nr+1 − nr obtain anr+1,k = −
lr−1∑
i=1
xnr+i +
lr−1∑
i=1
xk−i +
+ zk−lr+1, and as ns + k < ns+1, then ak,nr+1 = xk − zk+1. Thus obtain(
−
lr−1∑
i=1
xnr+i +
lr−1∑
i=1
xns+p−i + zns+p−lr+1
)
+ (xns+p − zns+p+1) = xnr+1
,
that is
lr−1∑
i=0
xns+p−i + zns+p−lr+1 − zns+p+1 =
nr+1∑
i=nr+1
xi,
which is necessary.
3. Gorenstein relations are corollaries of element relations
and count ones
For an arbitrary i, 1 ≤ i ≤ n the equality ai,k + ak,σ(i) = ai,σ(i) for k = i
is valid. Later for every i we will consider equalities ai,k +ak,σ(i) = ai,σ(i)
and ak,i + ai,σ(k) = ak,σ(k) for k > i, which coincides with the whole set
of Gorenstein relations.
Lemma 12. Equalities ai,k + ak,σ(i) = ai,σ(i) and ak,i + ai,σ(k) = ak,σ(k)
for 0 < i < n1 and i < k ≤ n1 are corollaries of element relations and
count ones.
Proof. For i = 1 Consider equalities ak,i + ai,σ(k) = ak,σ(k). As the first
line of matrix is zero, these equalities go to ak,1 = ak,σ(k) which is corollary
of (1).
For i = 2 consider equalities ak,i + ai,σ(k) = ak,σ(k), which are equiv-
alent to equalities ak,2 + a2,σ(k) = ak,σ(k), which take place in order
to (1), (2) and denotations.
For 3 ≤ m ≤ n1 consider equality ak,m + am,σ(k) = ak,σ(k), which is
ak,m +am,σ(k) = xk. If k < n1, then according to (3) and (4) this equality
is equivalent to
(
m−1∑
i=2
xi −
m−2∑
i=1
xk−i
)
+
(
m−1∑
i=1
xk+1−i −
m−1∑
i=2
xi
)
= xk,
which is identity.
For i = 1 consider equalities ai,k + ak,σ(i) = ai,σ(i). they take place in
order to (2) and because the first line of matrix is zero.
For i = 2 consider equalities ai,k + ak,σ(i) = ai,σ(i).
Consider case, when n1 > 2. In this case σ(2) = 3, and for finding
out the formula for ak,σ(i) one may use (4), whence ak,3 = x2 −xk−1, and
according to (3) we obtain a2,k = xk−1, whence these equalities are valid.
For n1 = 2 one obtain equalities a2,k + ak,1 = a2,1. According to (3)
for k 6= ns + 1 they are xns+1
+ xns
= x2, which coincides with (12)
104 On the dimension of Kirichenko space
For 3 ≤ m ≤ n1 the equality am,k +ak,σ(m) = am,σ(m) according to (1)
is equivalent to am,k + ak,m+1 = xm.
As k ≤ n1, then according to (4) and (5) it is equivalent to identity
(
m−1∑
i=1
xk−i −
m−1∑
i=2
xi
)
+
(m+1)−1∑
i=2
xi −
(m+1)−2∑
i=1
xk−i
= xm.
For k = n1 consider equality ak,2 + a2,σ(k) = ak,σ(k), that is an1,2 +
a2,1 = xn1
, whence according to denotations x2 = xn1
, and it is corollary
of (13).
For k = n1 and m, 3 ≤ m ≤ n1−1 consider equality ak,m +am,σ(k) =
ak,σ(k) for m, 3 ≤ m ≤ n1, that is an1,m + xm = xn1
. And according
to (4) one may rewrite it in the form
(
m−1∑
i=2
xi −
m−2∑
i=1
xn1−i
)
+ xm = xn1
,
that is
m∑
i=2
xi =
m−2∑
i=0
xn1−i, which is the same as
m−1∑
i=1
(x1+i −xn1−i+1) = 0,
which is corollary of (13).
Lemma 13. For every i, 1 ≤ i ≤ n1 and s, 1 ≤ s ≤ q − 1 equalities
ai,k + ak,σ(i) = ai,σ(i) and ak,i + ai,σ(k) = ak,σ(k) for k, ns < k ≤ ns+1 are
corollaries of element relations and count ones.
Proof. For i = 1 consider equalities ai,k + ak,σ(i) = ai,σ(i) and ak,i +
ai,σ(k) = ak,σ(k). They are equivalent to equalities a1,k + ak,2 = a1,2 and
ak,1 + a1,σ(k) = ak,σ(k). The former of these equalities are valid in order
to (2) and because the first line of matrix is zero. Last is valid in order
to definitions, (1), and because the first line of matrix is zero.
Let n1 > 2. Then σ(2) = 3. For i = 2 consider equalities ai,k +
ak,σ(i) = ai,σ(i) and ak,i + ai,σ(k) = ak,σ(k). They are equal to equalities
a2,k +ak,3 = a2,3 and ak,2 +a2,σ(k) = ak,σ(k). According to (4), one obtain
ak,3 =
{
x2 − xns+1
, if k = ns + 1
x2 − xk−1, if k 6= ns + 1
, and according to (3)
a2,k =
{
xns+1
, if k = ns + 1
xk−1, if k 6= ns + 1
, whence the first equality takes place.
Second equality takes place in order to (1), (2) and denotations.
In the case n1 = 2, equality ai,k+ak,σ(i) = ai,σ(i) for i = 2 is equivalent
to a2,k+ak,1 = x2 one. From (3) and denotations, for k = ns+1 we obtain
xns+1
+ xk = x2 , which is corollary of (12). If ns + 2 ≤ k ≤ ns+1, then
the former equality is equivalent to xk−1 + xk = x2, which is corollary
of (12).
For every m, 3 ≤ m ≤ n1 − 1 equality am,k + ak,σ(m) = am,σ(m)
according to (1) is equivalent to am,k + ak,m+1 = xm one. Fix arbitrary
0 < s < q − 1. For using (4) and (5) consider some different cases for k.
M. Plakhotnyk 105
For k = ns + 1 obtain the identity
(
m−2∑
i=0
xns+1−i −
m−1∑
i=2
xi
)
+
(
m∑
i=2
xi −
m−2∑
i=0
xns+1−i
)
= xm.
For k = ns + l, 2 ≤ l ≤ m − 1 equality am,k + ak,m+1 = xm is equivalent
to
(
l−1∑
i=1
xk−i +
m−l−1∑
i=0
xns+1−i −
m−1∑
i=2
xi
)
+
(
m∑
i=2
xi −
l−1∑
i=1
xk−i−
−
m−1−l∑
i=0
xns+1−i
)
= xm one, which is identity.
For k, ns+m ≤ k ≤ ns+1 equality am,k+ak,m+1 = xm is equivalent to(
m−1∑
i=1
xk−i −
m−1∑
i=2
xi
)
+
(
m∑
i=2
xi −
m−1∑
i=1
xk−i
)
= xm one, which is identity
also.
For m = n1 the equality am,k + ak,σ(m) = am,σ(m) according to (1) is
equivalent to an1,k + xk = xn1
. Consider cases.
Let k = ns + 1. Then according to (5) this equality is equivalent to(
n1−2∑
i=0
xns+1−i −
n1−1∑
i=2
xi
)
+xns+1 = xn1
one, that is
n1−2∑
i=0
xns+1−i+xns+1 =
n1∑
i=2
xi, and coincides with (14).
For k = ns+l and arbitrary l, 2 ≤ l ≤ n1−1 the equality an1,k +xk =
xn1
is equivalent to
(
l−1∑
i=1
xns+l−i +
n1−l−1∑
i=0
xns+1−i −
n1−1∑
i=2
xi
)
+ xns+l =
xn1
, that is
l∑
i=1
xns+i+
n1−l−1∑
i=0
xns+1−i =
n1∑
i=2
xi, whence xns+1+
l∑
i=2
xns+i+
n1−2∑
i=0
xns+1−i−
n1−2∑
i=n1−l
xns+1−i =
n1∑
i=2
xi. Subtracting (14) from this equality
one may obtain equivalent equality
l∑
i=2
xns+i −
n1−2∑
i=n1−l
xns+1−i = 0, that
is
l∑
i=2
xns+i −
l∑
i=2
xns+1−n1+i = 0, whence
l∑
i=2
(
xns+i − xns+1−n1+i
)
= 0,
which is corollary of (15).
For ns+n1 ≤ k ≤ ns+1 the equality an1,k+xk = xn1
according to (5) is
equivalent to
(
n1−1∑
i=1
xk−i −
n1−1∑
i=2
xi
)
+ xk = xn1
one, that is
n1−1∑
i=0
xk−i =
n1∑
i=2
xi. From the second line of equality (15) it follows that equalities
xns+1 = xns+n1+1, xns+2 = xns+n1+2 and so on up to xns+1−n1
= xns+1
take place, whence the sequence is periodic with period n1. It means that
n1−1∑
i=0
xk−i−1 is the sum of n1 elements of this sequence, and this sum equals
106 On the dimension of Kirichenko space
the sum of n1 last elements, whence the equality
n1−1∑
i=0
xk−i =
n1−1∑
i=1
x1+i
according to (14) one may transform to
n1−1∑
i=0
xns+1−i =
n1−2∑
i=0
xns+1−i +
xns+1, whence xns+1−n1+1 = xns+1, which is (15) for p = 1. Now for
every m, 3 ≤ m ≤ n1 − 1 consider equality ak,m + am,σ(k) = ak,σ(k).
For k = ns + 1, using (4) and (5) one obtains(
m−1∑
i=2
xi−
m−3∑
i=0
xns+1−i
)
+
(
2−1∑
i=1
x(k+1)−i +
m−2−1∑
i=0
xns+1−i −
m−1∑
i=2
xi
)
=
xns+1, which is identity.
For k = ns + l, and every l, 2 ≤ l ≤ m − 1 according to (4)
and (5), the equality ak,m + am,k+1 = ak,k+1 may be transformed to(
m−1∑
i=2
xi −
l−1∑
i=1
xk−i −
m−l−2∑
i=0
xns+1−i
)
+
(
l∑
i=1
xk+1−i +
m−l−2∑
i=0
xns+1−i−
−
m−1∑
i=2
xi
)
= xns+l one, which is identity.
For ns + m ≤ k ≤ ns+1 equality ak,m + am,k+1 = ak,k+1 may be
transformed to
(
m−1∑
i=2
xi −
m−2∑
i=1
xk−i
)
+
(
m−1∑
i=1
xk+1−i −
m−1∑
i=2
xi
)
= xk
one, which is identity.
Consider equality ak,n1
+ an1,σ(k) = xk, ns + 1 ≤ k ≤ ns+1.
For k = ns + 1 obtain ans+1,n1
+ an1,ns+2 = xns+1 and using (4)
and (5), this is equivalent to(
n1−1∑
i=2
xi −
n1−3∑
i=0
xns+1−i
)
+
(
1∑
i=1
xns+2−i +
n1−3∑
i=0
xns+1−i −
n1−1∑
i=2
xi
)
=
xns+1, which is identity.
For k = ns + p, 1 < p < n1 − 1 one may obtain ans+p,n1
=
n1−1∑
i=2
xi −
p−1∑
i=1
xns+p−i −
n1−p−2∑
i=0
xns+1−i, and an1,ns+p+1 =
p∑
i=1
xns+p−i+1 +
+
n1−p−2∑
i=0
xns+1−i −
n1−1∑
i=2
xi from the relations (4) and (5). Substituting
these values to the equality ans+p,n1
+ an1,ns+p+1 = xns+p, obtain equiv-
alent equality
(
n1−1∑
i=2
xi −
p−1∑
i=1
xns+p−i −
n1−p−2∑
i=0
xns+1−i
)
+
(
p∑
i=1
xns+p−i+1 +
n1−p−2∑
i=0
xns+1−i −
n1−1∑
i=2
xi
)
= xns+p, which is identity
also.
For k = ns + p, n1 ≤ p one may obtain
(
n1−1∑
i=2
xi −
n1−2∑
i=1
xns+p−i
)
+
+
(
n1−1∑
i=1
xns+p+1−i −
n1−1∑
i=2
xi
)
= xns+p, which is identity, in order to the
M. Plakhotnyk 107
equality ak,n1
+ an1,σ(k) = xk according to (4) and (5).
Lemma 14. For arbitrary r, 0 < r < q − 1, equalities anr+m,k +
ak,σ(nr+m) = anr+m,σ(nr+m) and ak,nr+m + anr+m,σ(k) = ak,σ(k) for
m, 1 ≤ m ≤ lr and k, nr + 1 < k ≤ nr+1
are corollaries of element relations and count ones.
Proof. Consider equality anr+1,k + ak,nr+2 = xnr+1. Using denotations
and (8) one may obtain, that it is equivalent to zk +(xnr+1−zk) = xnr+1,
which is identity.
Consider equality anr+2,k + ak,σ(nr+2) = xnr+2.
If lr > 2, then using (9) and (10) one may obtain that it is equivalent
to (xk−1 −xnr+1 + zk−1)+ (xnr+1 +xnr+2 −xk−1 − zk−1) = xnr+2, which
is identity.
If lr = 2, then equality anr+2,k + ak,σ(nr+2) = xnr+2 is equivalent to
anr+2,k + ak,nr+1 = xnr+2. As nr + 1 < k ≤ nr+1 then according to
conditions of the lemma this equivalence is possible to consider only in
the case when k = nr+1 = nr + 1, but in this case it is trivial.
For arbitrary m, 3 ≤ m ≤ lr − 1 consider equality
anr+m,k + ak,σ(nr+m) = xnr+m.
As nr + m < k < nr+1, this equality is anr+m,k + ak,nr+m+1 = xnr+m
and according to (10) and (11) is equivalent to
(
−
m−1∑
i=1
xnr+i +
m−1∑
i=1
xk−i+
+zk−m+1 ) +
(
m∑
i=1
xnr+i −
m−1∑
i=1
xk−i − zk−m+1
)
= xnr+m one, which is
identity.
Consider equality ak,nr+1 + anr+1,σ(k) = xk for k 6= nr+1. Using (6)
and denotations one may obtain that it is equivalent to xk−zk+1+zσ(k) =
xk, which is identity.
Consider equality ak,nr+2 + anr+2,σ(k) = xk for k 6= nr+1. Using (8)
and (9) one may obtain that it is equivalent to (xnr+1 − zk) + (xk −
xnr+1 + zk) = xk one, which is identity.
For arbitrary m, 3 ≤ m ≤ lr consider equality ak,nr+m +anr+m,σ(k) =
ak,σ(k) for nr +m < k < nr+1. According to (11) and (10) it is equivalent
to
(
m−1∑
i=1
xnr+i −
m−2∑
i=1
xk−i − zk−m+2
)
+
(
−
m−1∑
i=1
xnr+i +
m−1∑
i=1
xk+1−i+
+zk−m+2) = xk one, which is identity.
For k = nr+1 consider equalities ak,nr+m + anr+m,σ(k) = ak,σ(k), that
is anr+1,nr+m + anr+m,nr+1 = xnr+1
for 1 ≤ m < lr.
For m = 1 one obtains anr+1,nr+1 + anr+1,nr+1 = anr+1,nr+1, which is
valid.
108 On the dimension of Kirichenko space
For m = 2 the equality anr+1,nr+m + anr+m,nr+1 = xnr+1
is
anr+1,nr+2 + anr+2,nr+1 = xnr+1
and according to (6) and (8) it is equiv-
alent to (xnr+1 − znr+1
) + (xnr+2 − znr+3) = xnr+1
one, which follows
from (17) for m = 2.
For 3 ≤ m ≤ lr−1 consider equality anr+1,nr+m+anr+m,nr+1 = xnr+1
,
and, according to (10) and (6) this equality is equivalent to
(
m−1∑
i=1
xnr+i−
−
m−2∑
i=1
xnr+1−i − znr+1−m+2
)
+ (xnr+m − znr+m+1) = xnr+1
, whence
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i = znr+1−m+2 + znr+m+1, which follows from (17)
for 3 ≤ m ≤ lr − 1. For m = lr equality anr+1,nr+1
+ anr+1,nr+1 = xnr+1
is corollary of (7).
Lemma 15. For an arbitrary r, 0 < r < q − 1, equalities anr+m,k +
ak,σ(nr+m) = anr+m,σ(nr+m) and ak,nr+m + anr+m,σ(k) = ak,σ(k) for every
s, r < s < q; m, 1 ≤ m ≤ lr and k, ns + 1 < k ≤ ns+1 are corollaries of
element relations and count ones.
Consider equality anr+1,k+ak,nr+2 = xnr+1. According to denotations
and (8) it is equivalent to zk + (xnr+1 − zk) = xnr+1, which is identity.
Consider equality ak,nr+1 + anr+1,σ(k) = xk. According to (6) and
denotations it is equivalent to
{
xk − zns+1, if k = ns+1
xk − zk+1, if k 6= ns
+ zσ(k) = xk,
which is identity.
Consider equality anr+2,k + ak,σ(nr+2) = xnr+2.
If lr > 2, then σ(nr + 2) = nr + 3, and according to (9) and (10) it is
equivalent to
{
xns+1
− xnr+1 + zns+1
, if k = ns + 1
xk−1 − xnr+1 + zk−1, if k 6= ns + 1
+
+
{
xnr+1 + xnr+2 − xns+1
− zns+1
, if k = ns + 1.
xnr+1 + xnr+2 − xk−1 − zk−1, if k 6= ns + 1
= xnr+2, which is
identity.
If lr = 2, then equality anr+2,k + ak,σ(nr+2) = xnr+2 is equivalent to
anr+2,k + ak,nr+1 = xnr+2. Let ns+1 − ns > 2. Then according to (6)
and (9) for k = ns+1 it is the same as (xk − zns+1) + (xk−1 − xnr+1 +
zk−1) = xnr+2. For k = ns + 1 it is the same as (xns+1
− xnr+1 +
+ zns+1) + (xk − zk+1) = xnr+2. And for ns + 1 < k < ns+1 the equality
anr+2,k + ak,nr+1 = xnr+2 is equivalent to (xk−1 − xnr+1 + zk−1) + (xk −
zk+1) = xnr+2. Note, that all of these equalities are corollaries of (18).
Consider equality ak,nr+2 + anr+2,σ(k) = xk. According to (9) it is
equivalent to anr+2,σ(k) =
{
xns
− xnr+1 + zns
, if k = ns
xk − xnr+1 + zk, if k 6= ns
, whence, ac-
M. Plakhotnyk 109
cording to (8) it is equivalent to
(xnr+1 − zk) +
{
xns
− xnr+1 + zns
, if k = ns
x(k+1)−1 − xnr+1 + zk, if k 6= ns
= xk which is
identity.
For arbitrary 3 ≤ m ≤ lr consider equality ak,nr+m + anr+m,σ(k) =
ak,σ(k). Consider cases for k.
For k = ns+1 according to (1) it is ans+1,nr+m+anr+m,ns+2 = xns+1,
whence, according to (10) and (11) it is equivalent to
(
m−1∑
i=1
xnr+i −
−
m−3∑
i=0
xns+1−i − zns+1−m+3
)
+
(
−
m−1∑
i=1
xnr+i +
2−1∑
i=1
x(ns+2)−i+
m−2−1∑
i=0
xns+1−i + zns+1−m+2+1
)
= xns+1, which is identity.
For k = ns + l, and l, 2 ≤ l ≤ m − 1, according to (10) and (11) the
equality ans+l,nr+m+anr+m,ns+l+1 = xns+l is equivalent to
(
m−1∑
i=1
xnr+i −
−
m−l−2∑
i=0
xns+1−i −
l−1∑
i=1
xk−i − zns+1−m+l+2
)
+
(
l∑
i=1
x(k+1)−i −
m−1∑
i=1
xnr+i+
m−l−2∑
i=0
xns+1−i + zns+1−m+l+2
)
= xns+l, which is identity.
For every l, ns +m ≤ l ≤ ns+1−1 equality ak,nr+m +anr+m,k+1 = xk
according to (10) and (11) is equivalent to(
m−1∑
i=1
xnr+i −
m−2∑
i=1
xk−i − zk−m+2
)
+
(
−
m−1∑
i=1
xnr+i +
m−1∑
i=1
x(k+1)−i+
+zk−m+2) = xk, which is identity.
Consider equality anr+m,k+ak,σ(nr+m) = xnr+m for m = lr and obtain
anr+1,k + ak,nr+1 = xnr+1
.
For k = ns + 1 obtain anr+1,ns+1 + ans+1,nr+1 = xnr+1
, whence, ac-
cording to (11) and (6), it is equivalent to
(
−
lr−1∑
i=1
xnr+i +
lr−2∑
i=0
xns+1−i+
zns+1−lr+2
)
+(xns+1 − zns+2) = xnr+1
, whence
lr−2∑
i=0
xns+1−i+zns+1−lr+2+
xns+1 − zns+2 =
lr∑
i=1
xnr+i which is (18) for p = 1.
For k = ns+l and l, 2 ≤ l ≤ lr−1, the equivalence anr+1,k+ak,nr+1 =
xnr+1
is equivalent to
(
−
lr−1∑
i=1
xnr+i +
l−1∑
i=1
xk−i +
lr−l−1∑
i=0
xns+1−i+
+zns+1−lr+l+1
)
+ (xns+l − zns+l+1) = xnr+1
, whence
l−1∑
i=0
xns+l−i+
110 On the dimension of Kirichenko space
+
lr−l−1∑
i=0
xns+1−i + zns+1−lr+l+1 − zns+l+1 =
lr∑
i=1
xnr+i =
nr+1∑
i=nr+1
xi, which
is (18) for 2 ≤ p ≤ lr − 1.
For k = ns+l and l, lr ≤ l ≤ ls−1, the equivalence anr+1,k+ak,nr+1 =
xnr+1
according to (11) and (6) is equivalent to
(
−
lr−1∑
i=1
xnr+i+
+
lr−1∑
i=1
xns+l−i + zns+l−lr+1
)
+ (xns+l − zns+l+1) = xnr+1
, whence
lr−1∑
i=0
xns+l−i + zns+l−lr+1 − zns+l+1 =
nr+1∑
i=nr+1
xi, which is (18) for lr ≤ p ≤
ls − 1.
For k = ns+1 the equality anr+1,k + ak,nr+1 = xnr+1
according to (11)
and (6) is equivalent to
(
−
lr−1∑
i=1
xnr+i +
lr−1∑
i=1
xns+1−i + zns+1−lr+1
)
+
+
(
xns+1
− zns+1
)
= xnr+1
, whence
lr−1∑
i=0
xns+1−i + zns+1−lr+1 − zns+1 =
nr+1∑
i=nr+1
xi, which coincides with the first line of (18).
4. Lemma about permutation
Let’s prove lemma for counting the number of cycles the permutation,
whose bottom line of standard representation is (l + 1, . . . , n, 1, . . . , l) is
decomposed to. This lemma will be used for counting the dimension of
Kirichenko space.
Lemma 16. The permutation
π =
(
1
l + 1
· · ·
n − l n − l + 1
n 1
· · ·
n
l
)
is decomposed to (n, l) cycles, where (n, l) is greatest common divisor of
numbers n and l. In this case each cycle consists of numbers which give
the same remainder in division by (n, l).
Proof. Remark, that if n is divisible by l, then lemma is obvious. Let n
be indivisible by l.
Denote by ξ(π) = ξ(n, l) the quantity of cycles in the decomposition of
the permutation of the type, specified above. Remark, that this quantity
is equal to one for permutation
π−1 =
(
1
n − l + 1
· · ·
l l + 1
n 1
· · ·
n
n − l
)
,
M. Plakhotnyk 111
whence obtain relation ξ(n, l) = ξ(n, n − l).
Consider the cases n > 2l and n < 2l. Let n > 2l.
For n > 3l we will show that ξ(n, l) = ξ(n − l, l) which will give us a
reason to reduce this case to 2l < n < 3l one. Write out the permutation
π, decomposing it into blocks, as
π =
(
1
l + 1
· · ·
l
2l
l + 1
2l + 1
· · ·
2l
3l
2l + 1
3l + 1
· · ·
n − l
n
n − l + 1
1
· · ·
n
l
)
.
Pay attention to that cycles of π which contain elements of the
first block, which is interval [1, l]. Consider some element xi ∈ [1, l] of
some cyclic trajectory, and it’s obvious that for the previous element
xi−1 of this trajectory we have enclosure xi−1 ∈ [n − l + 1, n], which
means, that xi−1 belongs to the last block of π. It is also obvious that
for the next element of trajectory xi+1 we have inclosure xi+1 ∈ [l+1, 2l]
which means, that is belongs to the second block of π. More over, as
lengthes of the first, second and the last cycles are equal to each other,
then for arbitrary element xi−1 ∈ [n − l + 1], which belongs to some
cyclic trajectory, an element xi+1 of the same trajectory will belong to
interval [l + 1, 2l].
That is why the quantity of cycles, which the permutation π is de-
composed to, is equal to one for permutation π1, where
π1 =
(
l + 1
2l + 1
· · ·
2l
3l
2l + 1
3l + 1
· · ·
n − l
n
n − l + 1
l + 1
· · ·
n
2l
)
.
Reducing all numbers, which figure in the record of the permutation
π1 on l we will not change the permutation itself and so we will not change
the quantity of cycles it is decomposed to, and obtain
π2 =
(
1
l + 1
· · ·
l
2l
l + 1
2l + 1
· · ·
n − 2l
n − l
n − 2l + 1
1
· · ·
n − l
l
)
.
So, ξ(π) = ξ(π2), whence ξ(n, l) = ξ(n − l, l). Repeating some times,
if necessary, these reasonings, obtain ξ(n, l) = ξ(n − pl, l), where 2l <
n − pl < 3l. Then ξ(π2) = ξ(π̃2), where
π̃2 =
(
1
l + 1
· · ·
l
2l
l + 1
2l + 1
· · ·
n1 − l
n1
n1 − l + 1
1
· · ·
n1
l
)
,
and n1 = n − pl. Let us consider the permutation π3 = π̃−1
2 , and denote
k = n1 − l. As 2l < n1 < 3l, then 3k < 2n1 < 4k. So the permutation π3
112 On the dimension of Kirichenko space
looks like
π3 =
(
1
k + 1
· · ·
n1 − k
n1
n1 − k + 1
1
· · ·
k
2k − n1
k + 1
2k − n1 + 1
· · ·
n1
k
)
.
Notice that number k does not belong to the first line of the first block,
i.e. k > n1 − k (because 4k > 2n1), which gives possibility to decompose
the permutation π3 to three blocks, taking k as a border between second
and third block.
We may note also, that number 2n1 − 3k belongs to the first block,
because 2n1 − 3k < n1 − k (as 2n1 < 4k), and the number 2n1 − 2k
belongs to third block, because k +1 < 2n1 − 2k +1 (as 2n1 > 3k). That
is why, the permutation π3 may be represented as
π3 =
(
1
k + 1
· · ·
2n1 − 3k
2n1 − 2k
2n1 − 3k + 1
2n1 − 2k + 1
· · ·
n1 − k
n1
n1 − k + 1
1
· · ·
k
2k − n1
k + 1
2k − n1 + 1
· · ·
2n1 − 2k
n1 − k
2n1 − 2k + 1
n1 − k + 1
· · ·
n1
k
)
,
that is both the first and third blocks are decomposed to two blocks.
Pay attention on elements of the forth block. Like the way we have
considered the permutation π, it is easy to see that each cycle which
contains some element xi of the forth block contains some element xi−1
of the first block, more over, each cyclic trajectory which contains some
element of the first block contains some element of the forth block on the
next place.
Pay attention on elements of the fifth block. Each cycle, which
contains some element xi ∈ [2n1 − 2k + 1, n1], contains also xi−1 ∈
[2n1 − 3k + 1, n1 − k] (from the second block) and xi+1 ∈ [1, 2k − 1]
(from the third block), that is why, like in considering the permutation
π, it is possible to regard that elements of interval [2n1 − 3k + 1, n1 − k]
(second block) go accordingly to elements of interval [1, 2k − n1] (third
block) which gives us possibility to state that the number of cycles, the
permutation π3 is decomposed to, is equal to one for permutation
π4 =
(
1
2k − n1 + 1
· · ·
2n1 − 3k
n1 − k
2n1 − 3k + 1
1
· · ·
n1 − k
2k − n1
)
.
M. Plakhotnyk 113
So one may see that
ξ(n1, n1 − k) = ξ(n1 − k, 2k − n1) = ξ(l, n1 − 2l) = ξ(l, n1 mod l) =
= ξ(l, n mod l). Let n < 2l. then the permutation π looks like
π =
(
1
l + 1
· · ·
n − l
n
n − l + 1
1
· · ·
l
2l − n
l + 1
2l − n + 1
· · ·
n
l
)
.
Reasoning like in consideration of the permutation in the case n > 3l
it is possible to state that permutation π is decomposed to the same
quantity of cycles as permutation π5
π5 =
(
1
2n − l + 1
· · ·
n − l
l
n − l + 1
1
· · ·
l
2l − n
)
,
whence ξ(n, l) = ξ(l, 2l−n) = ξ(l, l−(2l−n)) = ξ(l, n−l) = ξ(l, n mod l).
So we have obtained that for every n and l the equality ξ(n, l) =
ξ(l, n mod l) takes place, which is the equality used in the algorithm of
finding the greatest common divisor which, as known, may be obtained
by this step, after using it some number of times.
Show, that all the numbers which belong to the same cycle give the
same remainders in division by (n, l). Let x be an element of some cycle.
If x+ l ≤ n, then the next element of this cycle is x+ l, which, obviously,
gives the same remainder in dividing by l as x does and so the same one
in dividing by (n, l).
If x+ l > n then the next element of the cycle is x+ l−n, which also
gives the same remainder in dividing by (n, l) as x does.
5. The dimension of Kirichenko space.
The purpose of this section is to count the defect of the matrix of count
relations which is the dimension of Kirichenko space.
Lemma 17. Consider the relation
l1
ls∑
i=1
xns+i = ls
l1∑
i=2
xi. (14a)
Systems of relations (14; 15) and (14a; 15) are equivalent.
Proof. Consider the system of relations
n1−2∑
i=0
xns+1−i + xns+1 =
n1∑
i=2
xi, (14)
114 On the dimension of Kirichenko space
{
xns+p = xns+1−(n1−p) 1 ≤ p ≤ n1
xns+p = xns+p−n1
n1 < p ≤ ns+1 − ns
(15)
According to the first line of the last relation, the relation (14) is
equivalent to
n1∑
i=1
xns+i =
n1∑
i=2
xi (14b)
one.
From second line of equalities (15) we may obtain xns+n1+1 = xns+1,
xns+n1+2 = xns+2 and so on up to xns+1
= xns+1−n1
, that is the set of
parameters {xns+1, . . . , xns+1
} is periodic with period n1. Here consider
the periodicity in the sense, that if i − j is divisible by n1, then xi = xj .
Without bounding of generality one may consider this set as consisted of
elements {xns+1, . . . , xns+n1
}.
It goes from the first line of equalities (15) that the last n1 elements
of a set {xns+1, . . . , xns+1
} coincide with {xns+1, xns+2, . . . , xns+n1
} cor-
respondingly.
Taking into account the fact that the set {xns+1, . . . , xns+1
}, is pe-
riodic, it is easy to see that the problem of calculating the quantity of
different parameters of the set {xns+1, . . . , xns+n1
} is equivalent to one
of finding the quantity of different cycles, the permutation π,
π =
(
1
l + 1
· · ·
n1 − l n1 − l + 1
n1 1
· · ·
n1 − 1 n1
l − 1 l
)
,
which the first line corresponds to the last n1 elements of a set
{xns+1, . . . , xns+1
} (which are equal to former ones), and second line cor-
responds to the last elements of a set {xns+1, . . . , xns+1
} is decomposed
to.
According to lemma 16 (about permutation), the set of parameters
{xns+1, . . . , xns+1
} is decomposed to (ls, l1) sets of pairwise equal ones,
and indices of elements of each cycle give the same remainders in division
by (ls, l1).
Consider ones more the equality (14b) which is
n1∑
i=1
xns+i =
n1∑
i=2
xi.
Taking into account facts proved above, sum of first n1 parameters is
equal to sum of all (ls, l1) pairwise unequal ones taken
l1
(ls, l1)
times.
So, it is obvious, that
(ls, l1)
l1
l1∑
i=1
xns+i is sum of one gang of pairwise
M. Plakhotnyk 115
unequaled parameters, whence
l1∑
i=1
xns+i =
l1
ls
ls∑
i=1
xns+i, which is
l1
ls∑
i=1
xns+i = ls
l1∑
i=2
xi. (14a)
Using the same steps (with revers order) one may prove that the
system of equations (14; 15) is a corollary of (14a; 15).
Lemma 18. If n1 = 2, then the system of equations
{
xk−1 + xk = x2 ns + 2 ≤ k ≤ ns+1,
xns+1
+ xk = x2 k = ns + 1
(12)
is a partial case of relations system (14a; 15).
Proof. Write out the matrix of relation (12) and obtain
1 1 0 0 · · · 0 x2
0 1 1 0 · · · 0 x2
...
. . .
. . .
. . .
. . .
...
...
0 · · · 0 1 1 0 x2
0 0 · · · 0 1 1 x2
1 0 · · · · · · 0 1 x2
If one subtract from each line (except the first) the previous one, and
add all the lines to the last one then the matrix which determine the
system of equations it will appear. This matrix is equivalent to former
one, where (just for convenience) it is possible to add the last line of a
former matrix to penultimate place, and obtain matrix
1 0 −1 0 · · · 0 0
0 1 0 −1 · · · 0 0
...
. . .
. . .
. . .
. . .
...
...
0 · · · 0 1 0 −1 0
−1 0 · · · 0 1 0 0
0 –1 · · · 0 0 1 0
2 2 · · · · · · 2 2 lsx2
where bold type is used for added line. It’s obvious, that this system
coincides with the system of relations (14a, 15) for n1 = 2.
116 On the dimension of Kirichenko space
Thus, the problem of calculating the dimension of Kirichenko space
has reduced to finding the defect of matrix K whose columns correspond
to parameters {x2, . . . xn, zn1+2,1, zn1+3,1 . . . zn,1, zn2+2,2, . . . zn,2 . . .
. . . , znq−1+2,q−1, . . . zn,q−1} and lines correspond to relations (13; 14a; 15-
18) ordered in a natural way. It’s obvious, that this matrix looks like
K =
(
X 0
X ′ Z
)
=
X 0 0 · · · 0
X1 Z1 0 · · · 0
X2 0 Z2 · · · 0
...
...
. . .
. . .
...
Xq−1 0 · · · 0 Zq−1
,
where matrices X, X1, . . . , Xq−1, Z1, . . . Zq−1 are block matrices, matrix
X corresponds to relations (13), and each block
(
Xr 0 Zr 0
)
cor-
responds to count relations (14a; 15-18) for every r, 1 ≤ r ≤ q − 1.
Blocks which correspond to different values of r have similar form and
in some sense are differ only by dimension. Consider their structure in
more details.
Like in proving relations, we may fix arbitrary r, 1 ≤ r ≤ n − 1 and
so, for admissible k instead of zk,r we will write zk.
Consider the system of relations
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i = znr+m+1 + znr+1−m+2, 2 ≤ m ≤ lr − 1. (17)
Fix arbitrary m0, 2 ≤ m0 ≤ lr − 1. It’s obvious that parameter
znr+m0+1 will be met in the equality for m = m0 and in the equality for
such m = m1, that nr+1 −m1 + 2 = nr + m0 + 1, because znr+m0+1 may
be met not only as first, but also as second item in relations (17). Let’s
find indices for z, which will appear in relations for m = m1. Sure, one
of them will be nr + m0 + 1. As nr+1 − m1 + 2 = nr + m0 + 1, then
m1 = nr+1 − nr − m0 + 1, whence the index, we need, is nr + m1 + 1 =
nr+1−m0+2, which coincides with index of second element in the relation
for m = m0. It means, for any relation in (17), there is one more, which
has the same right side. Show, that these two relations coincide, that
is they are one relation, written out two times. This statement may be
formulated as such lemma.
Lemma 19. For every r, 1 ≤ r ≤ q − 1 and every m, 2 ≤ m ≤ lr − 1
denote
f(m) =
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i.
Then f(m) = f(nr+1 − nr − m + 1).
M. Plakhotnyk 117
Proof. Let’s prove, that f(m) − f(nr+1 − nr − m + 1) = 0.
f(m) − f(nr+1 − nr − m + 1) =
(
m∑
i=1
xnr+i −
m−2∑
i=0
xnr+1−i
)
−
−
(
nr+1−nr−m+1∑
i=1
xnr+i −−
nr+1−nr−m−1∑
i=0
xnr+1−i
)
=
(
nr+m∑
i=nr+1
xi−
−
nr+1∑
i=nr+1−m+2
xi
−
(
nr+1−m+1∑
i=nr+1
xi −
nr+1∑
i=nr+m+1
xi
)
= 0
Thus, lines of matrix, which correspond to fixed r and relation (17)
looks like
Xr,2 0 1 0 0 · · · 0 0 1 0
Xr,3 0 0 1 0 · · · 0 1 0 0
Xr,4 0 0 0 1 · · · 1 0 0 0
...
... 0 0 0
. . . 0 0 0 0
Xr,lr−3 0 0 0 1
. . . 1 0 0 0
Xr,lr−2 0 0 1 0 · · · 0 1 0 0
Xr,lr−1 0 1 0 0 · · · 0 0 1 0
where by the bold font we type block-lines, and the first half of lines of
this part of K are pairwise different, but others are copies of some first
lines.
The problem of calculating the defect of this part of K is reduced to
calculating the quantity of independent parameters in the set (17). This
set is decomposed into independent pairs. Each such pair is relation,
which looks like sum zi + zj equals to some expression, depends only on
the set x2, . . . , xn. It is obvious, that if nr+m0+1 6= nr+1+m0+2 for any
m0, then one of parameters znr+m0+1, znr+1+m0+2 we may consider as
independent, and the other as expressed through independent ones, and
so in this case, the relation (17) for m = m0 gives us one independent
parameter. If for some m0 we have equality nr +m0 +1 = nr+1 +m0 +2,
then znr+m0+1 is expressed through parameters of the set x2, . . . , xn and
is dependent. That’s why the quantity of independent parameters equals
lr − 2
2
, if lr is even, and equals
lr − 3
2
, if lr is odd, which is
[
lr − 2
2
]
in
general case and is equal to the defect of matrix, determined by (17).
Consider the part of matrix K which corresponds to the system of
relations (18), and looks like
(
X(r) 0 Z(r) 0
)
(indices are bracketed
118 On the dimension of Kirichenko space
to make difference with denotations for blocks, which appeared in some
previous propositions) which may be written in the matrix form as
Z(r) =
1 0 · · · 0 −1 0 · · · 0 f0
0 1 · · · 0 0 −1 · · · 0 f1
...
...
. . .
. . .
...
...
0 0 1 0 0 · · · −1 fls−lr−1
−1 0 · · · 0 1 0 · · · 0 fls−lr
0 −1 · · · 0 0 1 · · · 0 fls−lr+1
...
...
. . .
. . .
...
...
0 0 · · · −1 0 0 · · · 1 fls−1
.
Let’s show, that by means of interchanging lines and columns, this matrix
may be reduced to block-diagonal form. The possibility of such reduc-
ing means that the set of parameters, which correspond to columns of
this matrix may by decomposed into such disjunct join, that there is no
relation, which connects parameters from different sets. Write out the
connections between parameters zns+1, . . . , zns+ls in the form of permu-
tation π, where equality π(i) = j will mean that there is an equation
zns+j − zns+i = fj+1 in the system (18). From the fact that for each
i, 1,≤ i ≤ ls, parameter zns+i is met only two times i.e. once with
positive sign and once with negative one, we obtain that the quantity of
blocks the left part of Z(r) may be decomposed to is the quantity of cycles
the permutation π is decomposed to.
The first two lines of system (18) give that π will be like
π =
(
· · · ls − lr + 1 ls − lr + 2 · · · ls
· · · 1 2 · · · lr
)
.
And in the same time, the last line of (18) gives, that π looks like
(
1 · · · ls − lr · · ·
lr + 1 · · · ls · · ·
)
,
and in general π is
π =
(
1 2 · · · ls − lr ls − lr + 1 · · · ls
lr + 1 lr + 2 · · · ls 1 · · · lr
)
.
According to lemma 16 it is decomposed into composition of (ls, lr) =
t cycles, and each cycle in this decomposition contains exactly w =
ls
t
elements. Note, that all elements from each cycle in this decomposition
give the same remainders in division by t.
M. Plakhotnyk 119
Make the simultaneous interchanging of lines and columns of matrix
Zr determined by permutation τ which looks as(
1 π(1) π2(1) . . . πw−1(1) 2 π(2) · · · πw−1(2) · · ·
1 2 3 . . . w w + 1 w + 2 · · · 2w · · ·
· · · t · · · πw−1(t)
· · · ls − w + 1 · · · ls
)
.
After this matrix Z(r) will be transformed into
Zr,1 · · · 0 Fr,1
...
. . .
...
...
0 · · · Zr,t Fr,t
where Fr,i is column of functions f̃(i−1)t+1, f̃(i−1)t+2, . . . f̃(i−1)t+w, ob-
tained during transformations, and matrices Zr,i are equal and look like
Zr,i =
1 −1 0 · · · 0
0 1 −1 · · · 0
...
...
. . .
. . .
...
0 0 · · · 1 −1
−1 0 · · · 0 1
.
Let’s do transformations with lines of each block of obtained matrix.
In each block of obtained matrix add all previous lines to each one except
the first line (because it has no previous), and obtain matrix
Z̃r,i =
1 −1 0 · · · 0 f̃(i−1)t+1
1 0 −1 · · · 0 f̃(i−1)t+1 + f̃(i−1)t+2
...
...
. . .
. . .
...
...
1 0 · · · 0 −1
w−1∑
j=1
f̃(i−1)t+j
0 0 · · · 0 0
w∑
j=1
f̃(i−1)t+j
.
Now, my means of addition of all columns of matrix Z̃r,i except the
first one to other columns of matrix K with necessary coefficients we
may make zero all elements of Z̃r,i except ones, belongs to it’s first sub
diagonal, which will stay be equal to −1.
Show that the sum
w∑
j=1
f̃(i−1)t+j doesn’t depend on i. The last propo-
sition is a corollary of next lemma.
120 On the dimension of Kirichenko space
Lemma 20. For every common divisor t of numbers lr and ls and for
every b, 1 ≤ b ≤ t the equality
(ls/t)−1∑
i=0
f̃b−1+it =
lr
t
ls∑
i=1
xns+i−
ls
t
lr∑
i=1
xnr+i
takes place.
Proof. As functions f̃0, . . . f̃ls−1 are obtained from the set of ones
{f0, . . . fls−1} after interchanging of lines and columns of matrix, then
for every p, 0 ≤ p ≤ lq − 1 the equality f̃p = τ∗(fτ(p+1)−1), where τ∗ is
automorphism of the set {xns+1, . . . xns+1
} such that τ∗(xns+i) = xns+τ(i)
for every i, 1 ≤ i ≤ ls takes place. As for every b, 1 ≤ b ≤ t the
set {b − 1 + it, 1 ≤ b ≤ t} is invariant for τ , then the proposition
of lemma is equivalent to
(ls/t)−1∑
i=0
fb−1+it =
lr
t
ls∑
i=1
xns+i −
ls
t
lr∑
i=1
xnr+i,
which we will prove. Denote gk = fk −
nr+1∑
i=nr+1
xi. It is obvious that it is
necessary and enough for proving lemma to prove that
(ls/t)−1∑
i=0
gb−1+it =
lr
t
ls∑
i1
xns+i, because items −
nr+1∑
i=nr+1
xi are met exactly once in each fk,
and that is why alter taking the sum of fk we will obtain that quantity of
appearing of −
nr+1∑
i=nr+1
xi is
ls
t
. Prove, that
(ls/t)−1∑
i=0
gb−1+it =
lr
t
ls∑
i1
xns+i.
Let us write out the matrix G, which columns correspond to parameters
xns+1, . . . , xns+1
and ls lines correspond to relations g0, . . . gls−1.
When k runs through the interval [0, lr − 1], then lines of matrix G
which correspond to relations gk look like
G∗∗ =
0 0 0 · · · · · · 1 1 1 · · · 1 1 1
1 0 0 · · · · · · 0 1 1 · · · 1 1 1
1 1
. . .
. . . · · · 0 0
. . .
. . . 1 1 1
1 1
. . . 0 · · · 1 0 0
. . .
. . . 1 1
1 1 . . . 1 · · · 1
. . . 0
. . . 0
. . . 1
1 1 . . . 1 1 1 · · · 1 0 · · · 0 1
, or
G∗∗ =
0 0 0 · · · · · · 0 1 1 1 · · · 1 1 1
1 0 0 · · · · · · 0 0 1 1 · · · 1 1 1
1 1 0 · · · · · · 0 0 0 1 · · · 1 1 1
1 1
. . .
. . . · · · 0 0 0
. . .
. . . 1 1 1
1 1 . . . 1 0 0 0 0 0 · · · 0 1 1
1 1 . . . 1 1 0 0 0 0 · · · 0 0 1
M. Plakhotnyk 121
which depends on whether 2lr ≥ ls, or not, where the first line of G∗∗
contains lr identities and ls − lr zeros. This matrix is decomposed in
the natural way to blocks Gl and Gr, the former of which corresponds
to parameters xns+1, . . . , xns+1−lr and the last - to xns+1−lr+1, . . . , ns+1
ones.
When k runs through interval [lr, ls − 1], then the part of matrix G,
whose lines correspond to relations gk for these k looks like
G∗ =
1 1 1 · · · · · · 1 1 0 0 0 · · · 0
0 1 1 · · · · · · 1 1 1 0 0 · · · 0
0 0 1 · · · · · · 1 1 1 1 0 · · · 0
. . .
. . . · · ·
. . .
...
0 0 · · · 0 1 1 1 · · · · · · 1 1 0
where the former lr elements of the first line are equal to identity, and
the last ones are zero.
The statement of this lemma in terms of matrix G is the following.
If we write out the lines of g with the step t, and arbitrary beginning
b, 1 ≤ b ≤ t then the sum of ones in each column will be the same and
will be equal to
lr
t
. Note that as the quantity of lines of matrix G∗∗ is
equal to lr which is divisible by t, then such writing out the lines means
that we obtain a matrix Gb, which consists of writing out one above
another matrices G∗
b and G∗∗
b , each of them is obtained by writing out
the lines of matrices G∗∗ and G∗ with the step t starting from b.
Denote by Glb and Grb those parts of matrix G∗∗
b which are obtained
during writing out the lines with step t. Let’s show that matrix Gb
will be column-block matrix, i.e. it will be consisted of such vertical
blocks with
ls
t
lines that each of them will have the same element in
each line. We will say about such matrix, that it satisfies column-block
condition. Let’s show, that for G∗∗
b the correspond widthes of blocks will
be b− 1, t, t, . . . , t, t− b + 1, b− 1, t, t, . . . , t, t− b + 1, and the first
blocks of matrices Glb and Grb have width b − 1.
For proving this proposition, consider matrices Glb, Grb and G∗
b sep-
arately.
Consider matrix Glb. The first line of this matrix have b− 1 ones and
zeros at the other places, and satisfies column-blocks condition. Each
next line of this matrix will have t ones more at the beginning and t zeros
less at the end and so, satisfies the column-block condition also.
Consider matrix Grb. The first line of this matrix have b − 1 zeros
and ones at the other places, and satisfies the column-block condition.
Each next line of this matrix will have t zeros more at the beginning and
t ones less at the end and so, satisfies the column-block condition also.
122 On the dimension of Kirichenko space
Consider matrix G∗
b . The first line of this matrix have b − 1 zeros at
the beginning, then lr ones, and at last (ls − lr − t) + (t− b + 1) zeros at
the end, and satisfies the column-block condition. Each next line of this
matrix has t zeros more at the beginning, and t zeros less at the end, and
so, satisfies the column-block condition also.
Let us count the quantity of ones in the first column of matrix Gb that
is equal to this quantity for the first block-column. If b = 1 then each of
lr
t
lines except the first one of matrix Glb will have some number of ones
at its beginning, and the first one will have zero there, so the quantity of
ones in the first block of matrix Glb is
lr
t
− 1. At the same time, the first
line of the matrix G∗
b has one at its beginning and other ones have zero
at their beginnings, so the quantity of ones of the first block of matrix
Gb is
lr
t
.
If b > 1, then the first element of each line of matrix Glb will be equal
to one, but the first element of each line of matrix G∗
b will be equal to
zero, whence the quantity of ones at first places of all lines of Gb is equal
to
lr
t
.
It’s easy to see that every next block of Glb has exactly 1 one less,
but every next block of G∗
b has 1 one more, so general quantity of ones
is not changed and is equal to
lr
t
for all columns from the first up to one
number ls − lr.
Count the quantity of ones in the last column of matrix Gb. The last
element of each of
lr
t
lines of matrix Grb is equal to one, and in the came
time the last element of each line of G∗
b is equal to zero. That is why the
quantity of ones in the last column of Gb is equal to
lr
t
.
It’s easy to see that moving left from last block of Grb each block we
meet, will have 1 one less then previous in matrix Grb and 1 one more
in the matrix G∗
b , saving the general quantity of ones in columns with
numbers from ls − lr +1 up to ls. So the quantity of ones in each column
of Gb is equal to
lr
t
.
Corollary 1 of lemma: The part
(
X(k) 0 Z(k) 0
)
of matrix K
which corresponds to count relations (18) for fixed r by transforming lines
and columns may be reduced to being like
Ns =
lr
ls∑
i1
xns+i − ls
lr∑
i1
xnr+i 0 0 0
0 0 E 0
,
M. Plakhotnyk 123
where E is identity matrix of order ls − (ls, lr).
Corollary 2 of lemma: Using transformations of lines and columns,
matrix K =
(
X 0
X ′ Z
)
may be reduced to being like
X 0 0
X ′′ 0 0
0 E 0
,
where E in such identity matrix that
(
E 0
)
has defect
q−1∑
s=2
q∑
r=s+1
(ls, lr) +
q−2∑
s=2
[
ls − 2
2
]
, and matrix X ′′ looks like
X ′′ =
0 L3 −L2 0
0 L4 0 −L2
...
. . .
0 Lq−1 0 0 · · · −L2 0
0 Lq 0 0 · · · 0 −L2
0 0 Ls+1 −Ls 0
0 0 Ls+2 0 −Ls
...
. . .
0 0 Lq 0 0 −Ls
...
0 0 Lq−1 −Lq−2 0
0 0 Lq 0 −Lq−2
0 0 0 Lq −Lq−1
where all elements at full places are zero, and by Li we denote a matrix
which has one line all whose elements are equal to li. In this case the
quantity of blocks-columns of matrix is equal to q, the width of first of
them is equal to l1 − 1, and the widthes of others are equal to li, i > 1.
Let’s go to describing a matrix X.
According to relations (13; 14a; 15), it is like X =
(
A
B
)
, where
A =
1 0 · · · 0 0 −1
0 1 · · · 0 −1 0
. . .
0 −1 · · · 0 1 0
−1 0 · · · 0 0 1
X3
X4
. . .
Xq
124 On the dimension of Kirichenko space
where the first block has l1 − 1 columns, and each next one has li, i > 1,
columns, and Xi is like
Xi =
1 0 · · · 0 −1 0 · · · 0
0 1 · · · 0 0 −1 · · · 0
...
...
. . .
. . .
...
0 0 1 0 0 · · · −1
−1 0 · · · 0 1 0 · · · 0
0 −1 · · · 0 0 1 · · · 0
...
...
. . .
. . .
...
0 0 · · · −1 0 0 · · · 1
,
where the sub diagonal, whose elements are equal to −1 has number n1,
i.e. the element of first line of Xi, which is equal to −1, is in the column
number n1 + 1.
Matrix B looks like
B =
L2 −L1 0 · · · 0 0
L3 0 −L1 · · · 0 0
...
. . .
Lq−1 0 0 · · · −L1 0
Lq 0 0 · · · 0 −L1
where Li means the same as in matrix X ′′.
Consider block matrix
(
X
X ′′
)
.
Note that widthes of blocks of matrices A, B, and X ′′ are equal, and
its easy to see, that if we will subtract the first column of each block of
the width l1 − 1, l2, . . . lq, from all other it’s columns, then each block
of matrices B and X ′′ will have number li at the first place and zeros
at the others. Denote them by L̃i, and matrices which appear after
transformation of B and X ′′ by B̃ and X̃ ′′.
Let’s find the rang of matrix
(
B̃
X̃ ′′
)
. It’s obvious that it is not less
then q − 1, because its first q − 1 lines are linear independent.
Let’s consider the first q − 1 lines and arbitrary other line, which
contains elements li and lj , obtaining
det
l2 −l1 0 · · · 0 0
l3 0 −l1 · · · 0 0
...
. . .
lq−1 0 0 · · · −l1 0
lq 0 0 · · · 0 −l1
0 li · · · −lj · · · 0
=
M. Plakhotnyk 125
= (−l1)
q−3 · det
lj −l1 0
li 0 −l1
0 li −lj
= (−l1)
q−3 · (l1lilj − l1lilj) = 0. And,
the rang of matrix
(
B̃
X̃ ′′
)
is equal to q − 1.
That is why the rang of matrix
(
X
X ′′
)
is equal to one of matrix
A minus q − 1, And we may generalize our calculations of dimension of
Kirichenko space.
Thus, if q > 2, then the dimension of Kirichenko space may be written
out as
[
l1
2
]
+
q∑
s=2
(ls, l1)+
q∑
r=2
[
lr − 2
2
]
+
q∑
r=2
q∑
s=r+1
(ls, lr)−q+1 =
q∑
r=1
[
lr
2
]
+
q∑
r=1
q∑
s=r+1
(ls, lr) − 2q + 2.
If q = 2, then matrix X ′′ will be absent, whence the dimension of
Kirichenko space will be
[
l1
2
]
+
[
l2 − 2
2
]
+ (l2, l1) − 1 =
[
l1
2
]
+
[
l2
2
]
+
(l2, l1) − 2
If q = 1, then all count relations will be represented by equality (13),
and the dimension of Kirichenko space will be equal to
[n1
2
]
.
Thus, we have proved the main theorem
Main Theorem. Let σ be a permutation, which is composition of
cycles of lengthes l1, . . . , lq. Then the dimension of Kirichenko space is
equal to
2−2q+
q∑
r=1
[
lr
2
]
+
1
2
∑
r 6=s
(ls, lr), where (a, b) denotes the greatest common
divisor of numbers a and b.
References
[1] M. A. Dokuchaev, V. V. Kirichenko, A. V. Zelensky, V. N. Zhuravlev Gorenstein
Matrices, Algebra and discrete mathematics N1, 2005 pp. 8-29.
[2] M. Hazewinkel, N. Gubareni and V. V. Kirichenko, Algebras, Rings and Modules,
V.I, Mathematics and Its Applications, V.575, Kluwer Academic Publishers, 2004.
[3] H. Fujita, Full matrix algebras with structure systems, Colloq. Math. 2003. 98(2)
pp. 249-258.
[4] H. Fujita, and Y. Sakai, Frobenius full matrix algebras and Gorenstein tiled orders,
Communications in algebra, 2006, 34: pp. 1181-1203.
[5] V. V. Kirichenko, Quasi-Frobenius rings and Gorenstein orders Trudy Mat. Steklov
Inst. 148, (1978) 168-174 (in Russian), English translation in Proceedings of the
Steklov Institute of Mathematics (1980), issue 4, pp. 171-177.
126 On the dimension of Kirichenko space
[6] V. V. Kirichenko, A. V. Zelensky, V. N. Zhuravlev, Exponent matrices and tiled
orders over discrete valuation rings, Intern. Journ. of Algebra and Computation,
Vol 15, N. 5 and 6, 2005, pp. 997-1012.
[7] M. Plakhotnyk, On the dimension of the space of Gorenstein matrices for
some types of correspond permutations, 5th Iinternational Algebraic Conference
in Ukraine, 2005, p. 157.
[8] K. W. Roggenkamp, V. V. Kirichenko, M. A. Khibina, V. N. Zhuravlev, Gorenstein
tiled orders, Comm. Algebra, 29(9), 2001, pp. 4231-4247.
Contact information
M. Plakhotnyk Department of Mechanics and Mathemat-
ics, Kyiv National Taras Shevchenko Univ.,
Volodymyrska str., 64, 01033 Kyiv, Ukraine
E-Mail: Makar_Plakhotnyk@ukr.net
Received by the editors: 31.10.2005
and in final form 06.10.2006.
|