On posets of width two with positive Tits form
We give a complete description of the finite posets of width two with the Tits form to be positive. This problem arises in studying the categories of representations of posets of finite type.
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| Cite this: | On posets of width two with positive Tits form / V.M. Bondarenko, M.V. Styopochkina // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 2. — С. 2–35. — Бібліогр.: 21 назв. — англ. |
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irk-123456789-1566192019-06-19T01:29:24Z On posets of width two with positive Tits form Bondarenko, V.M. Styopochkina, M.V. We give a complete description of the finite posets of width two with the Tits form to be positive. This problem arises in studying the categories of representations of posets of finite type. 2005 Article On posets of width two with positive Tits form / V.M. Bondarenko, M.V. Styopochkina // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 2. — С. 2–35. — Бібліогр.: 21 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 15A63, 16G20, 16G60. http://dspace.nbuv.gov.ua/handle/123456789/156619 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We give a complete description of the finite posets
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On posets of width two with positive Tits form |
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On posets of width two with positive Tits form / V.M. Bondarenko, M.V. Styopochkina // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 2. — С. 2–35. — Бібліогр.: 21 назв. — англ. |
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Algebra and Discrete Mathematics |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2005). pp. 20 – 35
c© Journal “Algebra and Discrete Mathematics”
On posets of width two with positive Tits form
Vitalij M. Bondarenko, Marina V. Styopochkina
Communicated by V.V. Kirichenko
Abstract. We give a complete description of the finite posets
of width two with the Tits form to be positive. This problem arises
in studying the categories of representations of posets of finite type.
The quadratic Tits form, introduced by P. Gabriel [1] for quivers, S.
Brenner [2] for quivers with relations and Yu. A. Drozd [3] for posets
plays an important role in representation theory (see also definitions of
the Tits form for wide classes of matrix problems without relations in [4]
and [5]). In particular, there are many results on connections between
representation types of various objects and properties of the Tits forms.
The reader interested in this theme is referred to the monographs [6], [7]
and, e.g., the paper of [8]–[19] (with the bibliographies therein). Above
all one must mention the well known results that a quiver (resp. a poset)
is of finite type if and only if its Tits form is positive (resp. weakly
positive); see [1] and [2], respectively. Our paper is devoted to study the
structure of finite posets with positive Tits form which arise in studying
the categories of representations of posets of finite type.
1. Formulation of the main result
Throughout the paper, all posets (partially ordered sets) are finite. In
considering a poset S = (A, 6) the set A will not be written and therefore
we keep to the following conventions: by a subset of S we mean a subset
of A together with the induced order relation (which is denoted by the
same symbol 6), we write x ∈ S instead of x ∈ A, etc.
2000 Mathematics Subject Classification: 15A63, 16G20, 16G60.
Key words and phrases: minimax (semiminimax) sum, (min, max)-equivalent
posets, positive form, the quadratic Tits form.
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.V. M. Bondarenko, M. V. Styopochkina 21
Let S be a poset. Given nonempty subsets X, Y ∈ S, we write X CY
if x < y for some x ∈ X, y ∈ Y , and X 6 Y if otherwise. We say that
S is a sum of subposets A and B (and write S = A + B) if A ∩ B = ∅
and S = A ∪ B. If any two elements a ∈ A and b ∈ B are incomparable,
this sum is called direct; one denotes such sum by S = A
∐
B. The sum
S = A + B is said to be an one-sided if B 6 A or A 6 B. When the first
(resp. second) case occurs, we say that the sum is right (resp. left); note
that a direct sum is both right and left. Finally, the sum S = A + B is
said to be minimax (resp. semiminimax) if x < y with x and y belonging
to different summands implies that x is minimal and (resp. or) y maximal
in S. 1
For a sum S of posets A and B let R<(A, B) denotes the set of pairs
(x, y) ∈ A × B with x < y. Such a pair (a, b) is said to be short if there
is no other such a pair (a′, b′) satisfying a 6 a′, b′ 6 b. By R<
0 (A, B) we
denote the subset of all short pairs from R<(A, B). We call the order
r0 = r0(A, B) of
R0(A, B) = R<
0 (A, B) ∪ R<
0 (B, A)
the rank of the sum S.
Let S be a poset and ZS∪0 the cartesian product of |S| + 1 copies
of Z (consisting of all vectors z = (zi), i ∈ S ∪ 0), where Z denotes the
integer numbers. The quadratic Tits form of S is by definition the form
qS : ZS∪0 → Z defined by the equality
qS(z) = z2
0 +
∑
i∈S
z2
i +
∑
i<j,
i,j∈S
zizj − z0
∑
i∈S
zi.
This form (as an arbitrary one) is called positive if it takes a positive
value on every nonzero vector z ∈ ZS∪0, and nonpositive if otherwise.
Recall that a linear ordered set is also called a chain. A poset with
the only pair of incomparable elements will be called an almost chain.
The width of a poset S is defined to be the maximum number of pairwise
incomparable elements of S and is denoted by w(S).
If S is a chain, then its Tits form is positive (see, for instance, [20],
or Section 3).
Our aim in this paper is to classifying the posets of width 2 with
positive Tits form.
In the case when the order of a poset is at least 8 we have the following
theorem.
1The notion of a sum have been introduced by the internal way. The external
way is possible but not convenient, however it is natural in some special cases, and in
particular when one has a right (left) minimax sum of two chains.
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.22 On posets of width two with positive Tits form
Theorem 1. Let S be a poset of width 2 and order at least 8. Then
the Tits form of S is positive if and only if one of the following condition
holds:
1) S is a direct sum of two chains;
2) S is an almost chain;
3) S is a one-sided minimax sum of rank 1 of two chains.
From the main result of [18] it follows the above theorem but for a
poset of sufficiently larger order. By Theorem 1 the number of the re-
maining possibilities for orders of posets with the Tits form being positive
becomes explicitly known to us (and is not large). A complete list of the
posets of width 2 and order smaller than 8 with positive Tits form will
be given in Section 4; see Proposition 3 and Theorem 2. Note that this
list is used in the proof of Theorem 1.
We also give a complete list of critical posets (of width 2) with respect
to positivity of the Tits form (see Theorem 3).
2. Subsidiary statements
2.1. Posets with nonpositive Tits form
In this subsection we indicate some posets with the Tits form to be non-
positive. We will see in the next sections that these posets are minimal
ones with nonpositive Tits form; moreover, they form (up to some natural
isomorphisms) the full set of such minimal posets.
When we determine some poset, the corresponding order relation is
given up to transitivity. In the case when the elements of a poset are
denoted by integer numbers, the order relation is denoted by ≺ (to dis-
tinguish between the given relation and the natural ordering of the integer
numbers). An element of a poset S is said to be nodal if it is comparable
to any other element.
Lemma 1. The Tits form is nonpositive for the following posets of order
smaller than 8 and of width 2 :
T1 = {1 ≺ 3, 1 ≺ 4, 2 ≺ 3, 2 ≺ 4},
T2 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6, 2 ≺ 5},
T3 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6, 2 ≺ 5},
T4 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5},
T5 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5},
T6 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5},
T7 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 3 ≺ 7}.
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.V. M. Bondarenko, M. V. Styopochkina 23
(Ti consists of elements 1, 2, . . . , n, where n is the greatest number in its
definition.)
Lemma 2. The Tits form is nonpositive for the following posets of order
8 and of width 2 with nodal elements :
T8 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7},
T9 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 4},
T10 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 2 ≺ 8},
T11 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 4, 2 ≺ 8},
T12 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7, 2 ≺ 8},
T13 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 4, 2 ≺ 5}.
Lemma 3. The Tits form is nonpositive for the following posets of order
8 and of width 2 without nodal elements :
T14 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7},
T15 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 4},
T16 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 2 ≺ 8},
T17 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7},
T18 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7, 2 ≺ 8},
T19 = {1 ≺ 2 ≺ 3 ≺ 4, 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7},
T20 = {1 ≺ 2 ≺ 3 ≺ 4, 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 7, 2 ≺ 8}.
Proof of Lemmas 1–3. The Tits form of T = Ts (s = 1, 2 . . . , 20) is de-
noted by qs(z). The coordinates zi of a vector z ∈ ZT∪0 will be arranged
in the natural way (in increasing order of the integer index i ∈ T ∪ 0).
It is easy to verify that the following numbers are zero:
q1(0, 1, 1,−1,−1), q2(0, 1, 1,−1, 1,−1,−1),
q3(1, 1, 1, 1, 1,−1,−1), q4(1, 2, 1, 1, 1,−1,−1,−1),
q5(0,−2, 1,−1,−1, 1, 1, 1), q6(1,−2, 1, 1,−1, 1, 1, 1),
q7(2, 1, 1, 1, 1, 1, 1,−2), q8(2, 3, 1, 1, 1, 1, 1,−2,−2),
q9(1, 3, 2, 2,−1,−1,−1,−1,−1), q10(3, 2, 2, 1, 1, 1, 1, 1,−3),
q11(1,−2, 2,−1, 1, 1, 1, 1,−1), q12(2, 2, 1, 1, 1, 1, 1,−1,−2),
q13(1, 1, 2, 2, 1,−1,−1,−1,−1), q14(1, 3,−1, 1, 1, 1, 1,−2,−2),
q15(1,−3, 2,−2, 1, 1, 1, 1, 1), q16(2, 2, 2,−1, 1, 1, 1, 1,−3),
q17(0, 3,−1,−1, 1, 1, 1,−2,−2), q18(1, 2, 1,−1, 1, 1, 1,−1,−2),
q19(1,−3, 1, 1, 1,−1,−1, 2, 2), q20(0, 2, 1,−1,−1, 1, 1,−1,−2).
Lemmas 1–3 are immediate from this.
Below we will say “case i)” instead of “the case T = Ti” (i = 1, 2, . . . , 20).
Since the Tits forms of a poset T and the dual poset T op are the
same, the dual lemmas, i.e. those with the dual posets T op
1 –T op
20 instead
of T1–T20, are hold. We always assume that, for a poset T , T op = T as
usual sets (then x < y in T op = S iff x > y in T ).
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.24 On posets of width two with positive Tits form
2.2. Statements on the Tits form
For subsets X and Y of some poset, we write X < Y if x < y for any
x ∈ X, y ∈ Y (clearly, X < Y when X or Y is empty). When S and
S′ are posets, we denote by [S < S′] the disjoint union S ∪ S′ with the
smallest order relation which contains those on S and S′, and such that
S < S′. Obviously, [S < ∅] = S and [∅ < S′] = S′. Singletons {x} are
often identified with the elements x themselves.
Let S be a poset. Recall that an element of S is said to be nodal if
it is comparable to any other element. The set of all nodal elements of S
is denoted by S0; obviously, S0 is a chain. Set S◦ = S \ S0. For a given
element x of S, we denote by NS(x) (or simply N(x) if no confusion is
possible) the subset of all elements y ∈ S that are not comparable to
x; obviously, x is a nodal element of the subset S \ NS(x). We say that
posets S and S′ are 0-isomorphic and write S ∼=0 S′ if S\S0
∼= S′\S′
0 and
|S0| = |S′
0| (the notation T ∼= T ′ means the usual isomorphism). Posets
S and S′ is said to be antiisomorphic (resp. 0-antiisomorphic) if Sop and
S′ is isomorphic (resp. 0-isomorphic).
From the definition of the Tits form it follows immediately the fol-
lowing lemma.
Lemma 4. Let S be a poset, and set S+ = [(S \ S0) < S0] and S− =
[S0 < (S \ S0)]. Then the Tits form of S± is the same as that of S.
The following lemma is a generalization of the previous one (and
follows from the definition of the Tits form too).
Lemma 5. The Tits form of 0-isomorphic or 0-antiisomorphic posets are
isomorphic (i.e. the same up to numbering of their variables).
Now we recall some definitions and assertions from [21].
Let X be a poset and let a be a maximal (resp. minimal) element of
S. Define S↓
a (resp. S↑
a) to be the disjoint union of the subsets {a} and
S \ a with the smallest order relation which contains that on S \ a, and
such that a < N(a) (resp. a > N(a)). Or briefly, S↓
a = (S \ a)∪ a, where
a is a minimal (resp. maximal) element of S↓
a (resp. S↑
a) and a < x in S↓
a
(resp. a > x in S↑
a) iff x ∈ NS(a).
Proposition 1. The Tits forms of the posets S and S↓
a (resp. S↑
a) are
equivalent.
The proposition follows from the following (easily verifiable) equality
for the Tits form q(z) of S and the Tits form q′(z) of S↓
x (resp. S↑
x):
q(z) = q′(z′), where z′0 = z0 − za, z′x = zx for x 6= a and z′a = −za.
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.V. M. Bondarenko, M. V. Styopochkina 25
We call posets S and T (min, max)-equivalent if there are posets
S1, . . . , Sp with p ≥ 0 such that, if one sets S = S0 and T = Sp+1, then
for each i = 0, 1, . . . , p, either Si+1 = (Si)
↓
x (with some maximal x ∈ Si)
or Si+1 = (Si)
↑
y (with some minimal y ∈ Si).
Directly from Proposition 1 it follows the following assertion.
Proposition 2. Let S and T be (min, max)-equivalent posets. Then
their Tits forms are equivalent (and hence simultaneously positive or
not).
Corollary 1. Let T be a poset and L be a chain. Then the Tits forms
of the posets T
∐
L and [T < L] (or[L < T ]) are simultaneously positive
or not.
Indeed, if L = {a < b < . . . < c}, then (T
∐
L)↑↑...↑ab...c = [T < Lop]∼=
[T < L] and (T
∐
L)↓↓...↓cb...a = [Lop < T ] ∼= [L < T ]. And it remains to
apply Proposition 2.
2.3. Statements on one-sided sums
A subset A of a poset S is said to be upper (resp. lower) if x ∈ A whenever
x > y (resp. x < y) and y ∈ A. Recall that the posets T1 and T2 were
defined in Subsection 2.1 (see Lemma 1).
Lemma 6. Let S be a poset of width 2 not containing a subset isomorphic
to the poset T1, and assume that S0 is empty. Then S is a one-sided sum
of two chains. If in addition S does not contain a subset isomorphic to
the poset T2, then the sum is semiminimax.
The assertions of the lemma are obvious if one takes into account that
S has two maximal and two minimal elements and, as a poset of width
2, is a sum of two chains.
For any poset of width 2 we have the following assertion.
Lemma 7. Let S be a poset of width 2 not containing a subset isomorphic
to the poset T1. Then
a) S0 is the union of an upper and a lower subsets;
b) S is a one-sided sum of two chains.
Proof. If a) did not hold, then there would be elements c ∈ S0 and
a, b, d, e ∈ S \ S0 such that a and b (resp. d and e) are incomparable and
a < c, b < c, c < d, c < e; so the subset {a, b, d, e} would be isomorphic to
T1, a contradiction. Further, by the previous lemma S \S0 is a one-sided
sum of two chains, say A and B with B 6 A. Denote by S01 and S02 the
lower and upper “parts” of S0. Then S is a one-sided sum of the subsets
S01 ∪ A = [S01 < A] and B ∪ S02 = [B < S02] which are chains.
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.26 On posets of width two with positive Tits form
Obviously, if b) holds, then S does not contain a subset isomorphic
to T1; further, if the sum is semiminimax, then S also does not contain
a subset isomorphic to T2. So we have the following corollaries from two
previous lemmas.
Corollary 2. A poset S of width 2 is a one-sided sum of two chains if
and only if it contains no subset isomorphic to the poset T1.
Corollary 3. The subset S◦ of a poset S of width 2 is a semiminimax
one-sided sum of two chains if and only if S◦ (or equivalently, S) contains
no subset isomorphic to the poset T1 or T2.
3. Proof of the sufficiency of Theorem 1
Let S be a poset of the form 1), 2) or 3). We prove that the Tits form
of S is positive. By Corollary 1 the first case reduces to the second one
(because, for chains X and Y , the poset [X < Y ] is a chain too, and any
chain is a subset of some almost chain).
If S is an almost chain, then letting c and d to be its only pair of
incomparable elements, one has the (easily verifiable) equality
2qS(z) = z2
0 +
∑
x 6=c,d,
x∈S
z2
x + (zc − zd)
2 + (z0 −
∑
x∈S
zx)2,
and so the form qS(z) is positive.
It remains to prove that the Tits form qS(z) is positive for S to be
the right minimax sum of chains A and B. Let a and b be the minimal
element of A and the maximal element of B, respectively. Then S↓
b is the
direct sum of the chain L = B \ b and the almost chain T = A∪{b} with
b < A\a. Since [L < T ] is an almost chain, the Tits form of S is positive
(via Proposition 1 and Corollary 1).
4. A complete list of the posets of order n < 8n < 8n < 8 with posi-
tive Tits form
In this section we give a complete list of the posets of order smaller than
8 (and width 2) with positive Tits form.
We denote by 〈m〉i the chain {1 + i < 2 + i < · · · < m + i} (for
m, i > 0) and set 〈m〉 = 〈m〉0, 〈m, n〉i = 〈m〉i
∐
〈n〉m+i. Further, for
nonempty chains L and L′, we denote by L ↗ L′ their right minimax
sum of rank 1, and for posets X, Y and Z, we set [X < Y < Z] =
[[X < Y ] < Z]. Finally, we set Dij = 〈i, j〉0, Mij = 〈i〉 ↗ 〈j〉i and
Qij = [〈i〉 < 〈1, 1〉i+1 < 〈j〉i+2].
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.V. M. Bondarenko, M. V. Styopochkina 27
We have the following obvious assertion.
Proposition 3. Let S be a poset of width 2 and order smaller than 8
satisfying one of conditions 1)–3) of Theorem 1. Then S has positive
Tits form (by the previous section) and is isomorphic to a poset Dij with
i, j > 0 and i + j < 8, or Mij with i, j > 0 and 2 < i + j < 8, or Qij with
i, j > 0 and i + j < 6.
The following theorem give (together with Proposition 3) a complete
list of posets of width 2 and order n < 8 with positive Tits form (by
Lemma 5 it is sufficient to consider such posets up to 0-isomorphism and
duality).
Theorem 2. Let S be a poset of width 2 and order n < 8. Then the
Tits form of S is positive if and only if either one of conditions 1)–3)
is satisfied, or S0 is the union of an upper and a lower subsets and S
is 0-isomorphic or 0-antiisomorphic to one of the following poset (which
consists of the elements 1, . . . , n and is a right sum of chains {1 ≺ . . . ≺ i}
and {i + 1 ≺ . . . ≺ n} with i 6 n/2 ) :
A) (of order 5)
R1 = {2 ≺ 3 ≺ 4 ≺ 5, 1 ≺ 4},
R2 = {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5},
R3 = {1 ≺ 2, 3 ≺ 4 ≺ 5, 1 ≺ 4},
R4 = {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5, 1 ≺ 4};
B) (of order 6)
R5 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 5},
R6 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 4},
R7 = {1 ≺ 2 ≺ 6, 3 ≺ 4 ≺ 5 ≺ 6},
R8 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 5},
R9 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 4},
R10 = {1 ≺ 2 ≺ 6, 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 5},
R11 = {1 ≺ 2 ≺ 6, 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 4},
R12 = {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5 ≺ 6, 1 ≺ 4},
R13 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6, 2 ≺ 6},
R14 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6, 1 ≺ 5, 2 ≺ 6},
C) (of order 7)
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.28 On posets of width two with positive Tits form
R15 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6},
R16 = {2 ≺ 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 4},
R17 = {1 ≺ 2 ≺ 7, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7},
R18 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6},
R19 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 4},
R20 = {1 ≺ 2 ≺ 7, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6},
R21 = {1 ≺ 2 ≺ 7, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5},
R22 = {1 ≺ 2 ≺ 7, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 4},
R23 = {1 ≺ 2 ≺ 6, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 4},
R24 = {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 4},
R25 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 2 ≺ 7},
R26 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6},
R27 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6, 2 ≺ 7},
R28 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5, 2 ≺ 7},
R29 = {1 ≺ 2 ≺ 3 ≺ 7, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 6},
R30 = {1 ≺ 2 ≺ 3 ≺ 7, 4 ≺ 5 ≺ 6 ≺ 7, 1 ≺ 5, 2 ≺ 6}.
Proof. The necessity part follows from the Proposition 3 and the following
one.
Proposition 4. Let S be a poset of width 2 and order n < 8 not con-
taining a subset 0-isomorphic to the poset T1, T2 or T3 (of Lemma 1).
Then S is 0-isomorphic or 0-antiisomorphic to one of the following poset:
Dij with j > i > 0 and i + j < 8, Mij with j > i > 0 and 2 < i + j < 8,
Q0j with 0 6 j < 6, Ri with i = 1, 2, . . . , 30, Ti for i = 4, 5, 6, 7 (see
Proposition 3, Theorem 2 and Lemma 1).
Notice that each of the posets Dij , Mij , Qij , Rij and Ti (i 6= 1) is a
right sum of chains {1 ≺ . . . ≺ i} and {i+1 ≺ . . . ≺ n}. And Proposition
4 can be easily proved by exhaustion, up to 0-isomorphism, of all right
sums of two chains, having the order n < 8 (see Lemma 7). Such a
proof can be done more simply if one first describes, up to isomorphism,
the right sums that are semiminimax and without nodal elements (see
Lemma 6), and after this, the others.
Now we prove the sufficiency part (of Theorem 2) showing that the
posets R1–R30 have positive Tits form. Because each poset Rj of order
5 or 6 is a subset of some poset Ri of order 7, it is sufficient to consider
only the posets of order 7, i.e. R15, R16, . . . , R30.
We have the following equalities: (R15)
↑
2
∼= R18, (R18)
↑
3
∼= R26,
(R26)
↑
4
∼= Rop
25 , (R25)
↓
3
∼= R17, (R17)
↑
1
∼= Rop
19 and (R19)
↓
2
∼= R16; (R24)
↑
3
∼=0
R22, (R22)
↑
3
∼=0 R20 and (R20)
↑
3
∼= R27; (R23)
↑
3
∼=0 R21, (R21)
↑
3
∼= R28 and
(R28)
↑
4
∼=0 R29. So each of the sets S1 = {Ri | i = 15, 16, 17, 18, 19, 25, 26},
S2 = {Ri | i = 20, 22, 24, 27} and S3 = {Ri | i = 21, 23, 28, 29} consists of
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.V. M. Bondarenko, M. V. Styopochkina 29
pairwise 0-isomorphic or 0-antiisomorphic posets and hence it is sufficient
to make sure (in view of Proposition 1 and Lemma 5) that R30 and, for
instance, R15, R23 and R24 have positive Tits form. To do this, we use
the well known fact that the Tits form
g(t) = g(t1, t2, . . . , t8) =
8∑
i=1
t2i −
6∑
i=1
titi+1 − t3t8
of the Dynkin graph E8, with the vertices 1, 2, . . . , 8 and the edges
(1, 2), (2, 3), . . . , (6, 7) and (3, 8), is positive (see [1]). For the Tits forms
q15(z), q23(z), q24(z) and q30(z) of the posets R15, R23, R24 and R30,
where z = (z0, z1, . . . , z7), one has the following equalities:
q15(z) = g(t), where t1 = −z6, t2 = −z6 − z7, t3 = z0 − z6 − z7, t4 =
z2 + z3 + z4 + z5, t5 = z3 + z4 + z5, t6 = z4 + z5, t7 = z5, t8 = z1;
q23(z) = g(t), where t1 = z2, t2 = z0 − z1 − z6 − z7, t3 = z0 + z4 +
z5 − z6 − z7, t4 = z0 + z4 − z6 − z7, t5 = z0 − z6 − z7, t6 = −z6 − z7, t7 =
−z6, t8 = z3 + z4 + z5;
q24(z) = g(t), where t1 = z2, t2 = z0 − z1 − z5 − z6 − z7, t3 = z0 + z4 −
z5 − z6 − z7, t4 = z0 − z5 − z6 − z7, t5 = −z5 − z6 − z7, t6 = −z5 − z6, t7 =
−z6, t8 = z3 + z4;
q30(z) = g(t), where t1 = z1 + z2 + z3, t2 = z0 + z1 + z3 − z6 − z7, t3 =
2z0 + z1 − z6 − 2z7, t4 = z0 + z1 − z6 − 2z7, t5 = z0 + z1 − z6 − z7, t6 =
z1 + z4 + z5, t7 = z1 + z5, t8 = z0 − z6 − z7.
Hence each of the Tits forms (of the four fix posets) is equivalent to
the form g(t) and thus is positive, as claimed.
5. Proof of the necessity of Theorem 1 for posets of order 8
Let S be a poset of width 2 and order 8 with positive Tits form. We
prove that then for S one of condition 1)–3) holds. This assertion follows
immediate from Lemmas 1–3 and the next proposition.
Proposition 5. Let S be a poset of width 2 and order 8 not containing
subsets 0-isomorphic or 0-antiisomorphic to the posets T1, T2, . . ., T20.
Then S satisfies condition 1), 2) or 3) (of Theorem 1).
Proof of Proposition. The proof depends on the number s = |S0| (the
smaller s, the more complicated the proof). It is easy to see (taking into
account Lemma 4 and the fact that a poset 0-isomorphic to some almost
chain is an almost chain too) that the subset S0 can be assumed to be
upper. Note that w(S) = 2 implies s 6= 7, 8.
We first prove that if s = 6, 5, 4, 3, 2, then S satisfies condition 2), i.e.
is an almost chain.
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.30 On posets of width two with positive Tits form
By Proposition 4 there are, up to isomorphism and duality, the fol-
lowing possibilities for the poset S◦ of order 8 − s 2:
in the case s = 6, S◦
1 = D11; in the case s = 5, S◦
1 = D12; in the case
s = 4, S◦
1 = D13, S◦
2 = D22, S◦
3 = M22; in the case s = 3, S◦
1 = D14,
S◦
2 = D23, S◦
3 = M23, S◦
4 = R3; in the case s = 2, S◦
1 = D15, S◦
2 = D24,
S◦
3 = D33, S◦
4 = M24, S◦
5 = M33, S◦
6 = R8, S◦
7 = R9, S◦
8 = R13, S◦
9 = R14.
In the case s = 6, S◦ ∼= S◦
1 and therefore S is an almost chain. The
cases s = 5, 4, 3, 2 are impossible since, for each s, every poset Ssi = [S◦
i <
Ls], with S◦
i running through all the posets listed in the corresponding
case and Ls = {8− s + 1 ≺ 8− s + 2 ≺ . . . ≺ 8} being a chain of order s,
contains a, proper or not, subset isomorphic to some Tj : for (s, i) = (4, 2),
(3, 2), (3, 3), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 8), (2, 9), Ssi contains a
subset T ∼= T3; for (s, i) = (4, 1), (3, 1), (3, 4), Ssi contains a subset T ∼=
T4; for (s, i) = (5, 1), (4, 3), (2, 1), Ssi
∼= T9, T13, T8, respectively; for
(s, i) = (2, 7), Ssi contains a subset T ∼=0 T4 (T = Ssi \ {3}).
Now we prove that if s = 1, then S satisfies condition 3) with one of
the chains being one-element.
By Propositions 4 we have in this case the following possibilities for
the subset S◦ of order 7 3: S◦
1 = D16, S◦
2 = D25, S◦
3 = D34, S◦
4 = M25,
S◦
5 = M34, S◦
6 = R18, S◦
7 = R19, S◦
8 = R25, S◦
9 = R26, S◦
10 = R27,
S◦
11 = R28.
In the case S◦ = S◦
1 , the poset S satisfies condition 3) with one of
the chains being one-element. All the other cases are impossible since
every poset Si = [S◦
i < L1], where L1 = {8} is a one-element set and
i = 2, 3, . . . , 14, contains a, proper or not, subset 0-isomorphic to some
P = Tj (j = 1, 2, . . . , 20): for i = 3, 5, 8, Si contains a subset T ∼= T7;
for i = 6, Si contains a subset T ∼= T4; for i = 2, 4, 7, Si
∼= T10, T12, T11,
respectively; for i = 9, 10, 11, Si contains a subset T ∼=0 T3 (T = Si \
{4, 5}, Si \ {3, 6}, Si \ {4, 7}, respectively).
Finally we prove that if s = 0 then S satisfies condition 1) or 3).
We need the following statement which is an analog of Proposition 4
on posets of order 8 without nodal elements.
Proposition 6. Let S be a poset of width 2 and order 8 not containing
a subset isomorphic to the poset T1 or T2 (of Lemma 1), and let S has
no nodal elements.
Then S is isomorphic or antiisomorphic to one of the following poset:
2Since the posets Ti (i = 1, 2, . . . , 20) are considered up to 0-isomorphism or 0-
antiisomorphism (see the formulation of the proposition), S◦ can be described up to
isomorphism and duality. Of course, we do not consider those S◦ that contain a subset
0-isomorphic or 0-antiisomorphic to some Ti.
3See footnote 2.
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.V. M. Bondarenko, M. V. Styopochkina 31
S1 = D17, S2 = D26, S3 = D35, S4 = D44,
S5 = M26, S6 = M35, S7 = M44,
S8 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 6},
S9 = {1 ≺ 2, 3 ≺ 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 5},
S10 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 6},
S11 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 5},
S12 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 6, 2 ≺ 8},
S13 = {1 ≺ 2 ≺ 3, 4 ≺ 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 5, 2 ≺ 8},
S14 = {1 ≺ 2 ≺ 3 ≺ 4, 5 ≺ 6 ≺ 7 ≺ 8, 3 ≺ 8},
S15 = {1 ≺ 2 ≺ 3 ≺ 4, 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 6, 2 ≺ 8},
S16 = {1 ≺ 2 ≺ 3 ≺ 4, 5 ≺ 6 ≺ 7 ≺ 8, 1 ≺ 6, 3 ≺ 8},
S17 = T14, S18 = T15, S19 = T16, S20 = T17,
S21 = T18, S22 = T19, S23 = T20.
The proposition can be easily proved by exhaustion, up to isomor-
phism, of all semiminimax right sums (without nodal elements) of two
chains, having order 8 (see Lemma 6).
Thus, we must consider the cases when S = Si, i = 1, 2, . . . , 16.
The poset S satisfies condition 1) in the cases S = S1, S2, S3, S4 and
condition 3) in the cases S = S5, S6, S7. All the other cases are im-
possible since every poset Si, i = 8, 9, . . . , 16, contains a (proper) sub-
set 0-isomorphic (in fact, isomorphic if i 6= 16) to a poset T = Tp(i)
from Lemma 1. Namely, one has to take p(i) = 4 if i = 8, p(i) = 5 if
i = 9, 10, 12, p(i) = 6 if i = 11, 13, 15, p(i) = 7 if i = 14 and p(i) = 3 if
i = 16 (in the last case T = S \ {4, 5}).
6. Proof of the necessity of Theorem 1 for posets of order
greater than 8
Through this section S is (as in the previous one) a poset of width 2.
We say that a poset S contains (no) subsets of the form T , where T
is a fixed poset (with elements being natural numbers), if there is (no)
subsets of S isomorphic to T .
We need the following assertion.
Proposition 7. A poset S of width 2 (and any order) satisfies one of
condition 1)–3) of Theorem 1 if and only if it does not contain, up to
duality, a subset one of the following forms:
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.32 On posets of width two with positive Tits form
a) {1 ≺ 3, 1 ≺ 4, 2 ≺ 3, 2 ≺ 4};
b) {1 ≺ 4, 2 ≺ 3 ≺ 4 ≺ 5};
c) {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5};
d) {1 ≺ 2, 1 ≺ 4, 3 ≺ 4 ≺ 5};
e) {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5, 1 ≺ 4};
f) {1 ≺ 2 ≺ 5, 3 ≺ 4 ≺ 5, 1 ≺ 3}.
Proof. The necessity part is obvious, and we proceed to a proof of the
sufficiency one. By bop)–fop) we denote the posets dual to b)–f).
Let S contains no subsets of the form a)–f) and bop)–fop), and let S
is not decomposable (otherwise it satisfies condition 1)). Then S is a one-
sided sum of two chains (of nonzero rank), say A = {a1 < · · · < ap} and
B = {b1 < · · · < bq}, where p + q > 2 (p, q > 0) (see Lemma 7). Without
loss of generality one can assume that S is a right sum of them and p 6 q
(taking into account that one can replace S by Sop and renames the
summands). The rank r0 = r0(A, B) of S can not be greater than two,
because otherwise we would have short pairs (x1, y1), (x2, y2), (x3, y3) ∈
A × B with x1 < x2 < x3, y1 < y2 < y3 and this would give the subset
{x2, x3, y1, y2, y3} of the form e).
We consider the cases r0 = 1 and r0 = 2 separately.
The case r0 = 1. Let (ai, bj) be the only short pair. If p = 1, j = 2
(resp. p = 1, j = q), then S satisfies condition 2) (resp. 3)). Otherwise
(in the case p = 1), the poset S contains a subset of the type b), namely,
the subset {a1, b1, b2, bj , bq}, a contradiction.
If either p = 2 and i = 1, j = 1, q, or p = q = 2 and i = j = 2, then S
satisfies condition 3). Otherwise (in the case p = 2), the poset S contains
a subset of the type b), c) or d). Indeed, if i = 1 and j 6= 1, q, then S
contains the subset {a1, a2, b1, bj , bq} of the form d); and if i = 2, q > 2
and j = 2 (resp. j > 2), then S contains the subset {a1, a2, b1, b2, b3}
(resp. {a1, a2, b1, b2, bj}) of the form b) (resp. c)). Again a contradiction.
Finally, if p > 2 and i = 91, j = q, then S satisfies condition 3).
Otherwise (in the case p > 2), the poset S contains a subset of the type
c), cop, d), dop) or bop): if i = p, j = q (resp. i = 1, j = 1), then S contains
the subset {a1, a2, b1, b2, bq} (resp. {a1, a2, a3, b1, b2}) of the form c) (resp.
cop)); if i 6= 1, p and j = 1 (resp. j = q), then S contains the subset
{a1, ai, ap, b1, b2} (resp. {a1, ai, ap, b1, bq}) of the form bop) (resp. dop));
if i 6= p and j 6= 1, q, then S contains the subset {ai, ap, b1, bj , bq} of the
form d); if i = p and j 6= 1, q, then S contains the subset {a1, ap, b1, bj , bq}
of the form b). Again a contradiction.
The case r0 = 2. Let (ai, bj) and (as, bt), where i < s, j < t, be
the corresponding short pairs. Then s − i = 1 and t − j = 1, because
otherwise S would contain a subset of type fop) or f) (that would give
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.V. M. Bondarenko, M. V. Styopochkina 33
a contradiction). If j = 1, s = p, then S satisfies condition 2). If
j 6= 1 (resp. s 6= p), then S contains the subset {ai, as, b1, bj , bt} (resp.
{ai, as, ap, bj , bt}) of the form e) (resp. eop)), a contradiction.
Now it is easy to prove the following proposition.
Proposition 8. Let S be a poset of width 2 and order greater than 8 not
containing subsets 0-isomorphic or 0-antiisomorphic to the posets T1, T2,
. . ., T20. Then S satisfies condition 1), 2) or 3) (of Theorem 1).
Indeed, let S be a poset of order at least 9 and width 2 with no subset
0-isomorphic or 0-antiisomorphic to T1–T20. If none of conditions 1)–3)
were hold, then (by the sufficiency of Proposition 7) the poset S would
contain a subset X of the form a)–f), and, for any subset of S of order
8 containing X4, none of conditions 1)–3) would hold (by necessity of
Proposition 7), a contradiction (to the necessity of Proposition 5).
The necessity of Theorem 1 for posets of order greater than 8 follows
immediately from Proposition 8 and Lemmas 1–3.
7. Critical posets with respect to positivity of the Tits
form
The following theorem give a complete list of critical posets of width 2
with respect to positivity of the Tits form, i. e. minimal posets of width
2 with nonpositive Tits form.
Theorem 3. A poset T of width 2 is critical with respect to positivity
of the Tits form if and only if it is 0-isomorphic or 0-antiisomorphic to
one of the posets Ti (i = 1, 2, . . . , 20).
Indeed, by Lemmas 1–3, Propositions 5 and the sufficiency of Theorem
1, each poset 0-isomorphic or 0-antiisomorphic to Ti, i ∈ {1, 2, . . . , 20},
is critical. The fact that there are no other critical posets follows from
Propositions 4, 5 and 8.
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Contact information
V. M. Bondarenko Institute of Mathematics, National
Academy of Sciences of Ukraine,
Tereshchenkivska 3, 01601 Kyiv, Ukraine
E-Mail: vit-bond@imath.kiev.ua
M. V. Styopochkina Kyiv Taras Shevchenko University,
Volodymyrs’ka 64, Kyiv, 01033, Ukraine
E-Mail: StMar@ukr.net
Received by the editors: 23.06.2005
and final form in 01.07.2005.
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