Download The natural partial order on variants of linear

The 22nd Annual Meeting in Mathematics (AMM 2017)
Department of Mathematics, Faculty of Science
Chiang Mai University, Chiang Mai, Thailand
The natural partial order on variants of linear
transformation semigroups
Pongsan Prakitsria†‡ and Pattanachai Rawiwanb
a
Faculty of Science at Sriracha
Kasetsart University, Sriracha Campus, Chonburi 20230, Thailand
b
The Institute for the Promotion of Teaching Science and Technology
Sukhumvit Rd., Bangkok 10110, Thailand
Abstract
For a semigroup S and a ∈ S, define an operation ∗ by x ∗ y = xay for all x, y ∈ S. Then
(S, ∗) is a semigroup called a variant of S with respect to a. In this paper, we characterize
the natural partial order on variants of the linear transformation semigroups L(V ) where
V is a vector space. Furthermore, we also provide necessary and sufficient conditions for
elements in variants of L(V ) to be minimal, maximal, lower and upper cover elements.
Keywords: linear transformation semigroup, natural partial order, minimal (maximal) element, lower (upper) cover, variant.
2010 MSC: Primary 20M20; Secondary 06A06, 20M10.
1
Introduction
Variants of a semigroup have long been studied. In 1967, Magill [5] considered the semigroup
of functions from a nonempty set X to a nonempty set Y with an operation defined by f ∗g = f θg
where θ : Y → X is fixed. Hickey [3] introduced, in 1983, a variant of an arbitrary semigroup,
that is, for a semigroup S and a ∈ S, define a binary operation ∗ by x ∗ y = xay for all x, y ∈ S.
Then (S, ∗) is a semigroup and this is called a variant of S with respect to a. Denote by (S, a)
the semigroup (S, ∗). It can be seen that if S has an identity 1, then (S, 1) is the semigroup S.
In 1952, Wagner [10] defined the natural partial order on inverse semigroups. Later,
Hartwig [2] and Nambooripad [8], independently introduced the natural partial orders on regular semigroups. Finally, in 1986, Mitsch [7] defined the natural partial order ≤ on any semigroup
S by
a ≤ b if and only if a = xb = by and a = xa for some x, y ∈ S 1
where S 1 is the semigroup S adjoined 1 ∈
/ S as its identity if S has no identity, otherwise S 1
is S. Furthermore, he provided many equivalent forms of the natural partial order on S as
follows: for any a, b ∈ S, (i) a ≤ b;
(ii) a = xb = by and a = ay for some x, y ∈ S 1 ;
(iii) a = xb = by and a = xa = ay for some x, y ∈ S 1 .
†
Corresponding author.
Speaker.
E-mail address: [email protected] (P. Prakitsri), ball
‡
Proceedings of AMM 2017
[email protected] (P. Rawiwan).
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In this paper, we write a < b to mean a ≤ b and a 6= b. For any distinct a, b ∈ S, a is said to
be a lower cover of b if a < b and there is no c ∈ S such that a < c < b. From this definition, b
is called an upper cover of a.
The problem of determining necessary and sufficient conditions for elements in semigroups
to be related is of interest. Many researchers characterize the natural partial order on various
transformation semigroups, see also [1, 4, 6]. In 2005, Sullivan [9] studied the natural partial
order on linear transformation semigroups. In this paper, we pay our attention on the natural
partial order on variants of linear transformation semigroups.
Let V be a vector space and L(V ) the set of all linear transformations on V . It is obvious
that L(V ) is a semigroup under composition. The semigroup L(V ) is called the full linear
transformation semigroups on V . Denote dim V by the dimension of V . For each α ∈ L(V ),
let ker α and im α respectively represent the kernel of α and the image of α. We use rank α to
stand for the rank of α. The zero map on V and the identity map on V are denoted by 0V and
1V , respectively. A set {xi ∈ V | i ∈ I} will be written in short by {xi }i∈I where I is an index
set. Throughout this paper, all functions act on the right-hand side of the argument. For any
α, β ∈ L(V ), let
E(α, β) = {v ∈ V | vα = vβ}.
From [9], this set is a subspace of V and it is easy to see that E(α, β) ⊆ V αβ −1 for all
α, β ∈ L(V ). Moreover, if V αβ −1 = E(α, β), then ker β ⊆ ker α.
Sullivan characterized the natural partial order on L(V ) and determined minimal and maximal elements in L(V ) as follows.
Theorem 1.1. [9] Let α, β ∈ L(V ). Then α ≤ β on L(V ) if and only if im α ⊆ im β and
V αβ −1 = E(α, β).
Theorem 1.2. [9] Let α ∈ L(V ). Then
(i) α is a minimal nonzero element in L(V ) if and only if rank α = 1.
(ii) α is a maximal element in L(V ) if and only if α is a monomorphism or an epimorphism.
We give a notation that will be used to represent some linear transformations on V . Note
that any linear maps can be defined on a basis of V . Let {xi }i∈I ∪ {yj }j∈J be a basis of V where
I, J are index sets. If α ∈ L(V ) is defined by xi α = u and yj α = vj for all i ∈ I, j ∈ J, we write
α=
{xi }i∈I
u
yj
vj
.
j∈J
The following are useful quoted results for our work.
Proposition 1.3. [1] Let α ∈ L(V ) and let B1 be a basis of ker α and B a basis of V extended
from B1 . Then the following statements hold.
(i) For any v1 , v2 ∈ B \ B1 , v1 = v2 if and only if v1 α = v2 α.
(ii) (B \ B1 )α is a basis of im α.
Proposition 1.4. [1] Let α, β ∈ L(V ) be such that ker β ⊆ ker α and let B1 , B2 , B3 be disjoint
linearly independent sets such that B1 , B1 ∪B2 and B1 ∪B2 ∪B3 are bases of ker β, ker α and V ,
respectively. If vα = vβ for all v ∈ B3 , then V αβ −1 = E(α, β).
Proposition 1.5. [1] Let α, β ∈ L(V ) be such that im α ⊆ im β and V αβ −1 = E(α, β). Then
α=
{xi }i∈I ∪ {yj }j∈J
0
zk
uk
and β =
k∈K
{xi }i∈I
0
yj
vj
zk
uk
j∈J,k∈K
where {xi }i∈I , {xi }i∈I ∪ {yj }j∈J , {uk }k∈K , {vj }j∈J ∪ {uk }k∈K , {xi }i∈I ∪ {yj }j∈J ∪ {zk }k∈K are
bases of ker β, ker α, im α, im β and V , respectively.
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2
Main results
Our purpose is to characterize the natural partial order on variants of L(V ), that is, the
semigroup (L(V ), θ) where θ ∈ L(V ). The following proposition is helpful for our main theorem.
Proposition 2.1. Let S be a semigroup and a, b, u ∈ S. If a ≤ b on (S, u), then a ≤ b on S.
Proof. Suppose that a ≤ b on (S, u). Then a = xub = buy and a = auy for some x, y ∈ S 1 .
Hence a ≤ b on S.
One may expect that ≤ is a partial order on (L(V ), θ) where θ ∈ L(V ). However, ≤ fails to
be a partial order on (L(V ), θ) for some θ ∈ L(V ), see the following example.
Example 2.2. Let V be an infinite dimensional vector space and B a basis of V . Then there
is a partition {B1 , B2 } of B such that |B| = |B1 | = |B2 |. Define α, θ ∈ L(V ) by
B1 v
B2 v
α=
and θ =
.
0 v v∈B
0 v v∈B
2
1
Observe that αθ = 0V . Then αθµ = 0V 6= α for all µ ∈ L(V ). Hence α α on (L(V ), θ).
Therefore, ≤ is not a partial order on (L(V ), θ).
It is interesting to determine when the natural partial order ≤ is a partial order on (L(V ), θ).
Proposition 2.3. ≤ is a partial order on (L(V ), θ) if and only if θ is an isomorphism.
Proof. Suppose that ≤ is a partial order on (L(V ), θ). Then 1V ≤ 1V on (L(V ), θ) and hence
there are λ, µ ∈ L(V ) such that 1V = λθ1V = 1V θµ. That is 1V = λθ = θµ. This implies that
θ is an isomorphism.
For the converse, assume that θ is an isomorphism. Let α, β, γ ∈ L(V ). Then α = θ−1 θα =
αθθ−1 and so ≤ is reflexive. For the antisymmetric, suppose that α ≤ β and β ≤ α on (L(V ), θ).
Thus α ≤ β and β ≤ α on L(V ) by Proposition 2.1. Hence α = β. Next, assume that α ≤ β
and β ≤ γ on (L(V ), θ). Then there are λ1 , µ1 , λ2 , µ2 ∈ L(V ) such that
α = λ1 θβ = βθµ1 and α = αθµ1 = λ1 θα,
β = λ2 θγ = γθµ2 and β = βθµ2 .
This implies that α = λ1 θλ2 θγ = γθµ2 θµ1 . Now consider α = βθµ1 = βθµ2 θµ1 . Thus
α = λ1 θα = λ1 θ(βθµ2 θµ1 ) = (λ1 θβ)θµ2 θµ1 = αθµ2 θµ1 . Hence α ≤ γ on (L(V ), θ). Therefore
≤ is a partial order on (L(V ), θ).
We now provide necessary and sufficient conditions for elements in (L(V ), θ) to be related
under the natural partial order where θ is an isomorphism.
Theorem 2.4. Let α, β, θ ∈ L(V ) be such that θ is an isomorphism. Then α ≤ β on (L(V ), θ)
if and only if im α ⊆ im β and V αβ −1 = E(α, β).
Proof. Suppose that α ≤ β on (L(V ), θ). By Proposition 2.1, α ≤ β on L(V ). Hence, by
Theorem 1.1, im α ⊆ im β and V αβ −1 = E(α, β).
For the sufficiency, suppose the conditions hold. By Proposition 1.5, we write α and β as
{xi }i∈I ∪ {yj }j∈J zk
{xi }i∈I yj zk
and β =
α=
0
uk k∈K
0
vj uk j∈J,k∈K
where {xi }i∈I , {xi }i∈I ∪ {yj }j∈J , {uk }k∈K , {vj }j∈J ∪ {uk }k∈K , {xi }i∈I ∪ {yj }j∈J ∪ {zk }k∈K are
bases of ker β, ker α, im α, im β and V , respectively. Now define λ ∈ L(V ) by
{xi }i∈I ∪ {yj }j∈J
zk
λ=
.
0
zk θ−1 k∈K
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Then α = λθβ. Since {uk }k∈K ∪ {vj }j∈J is linearly independent, extend it to a basis {wl }l∈L ∪
{uk }k∈K ∪ {vj }j∈J of V . Since θ is an isomorphism, we obtain {wl θ}l∈L ∪ {uk θ}k∈K ∪ {vj θ}j∈J
is a basis of V . Next, define µ ∈ L(V ) such that
{wl θ}l∈L ∪ {vj θ}j∈J uk θ
µ=
.
0
uk
k∈K
Hence α = βθµ and α = αθµ. Therefore α ≤ β on (L(V ), θ).
From Theorem 2.4, the following corollary holds by taking θ = 1V .
Corollary 2.5. Let α, β ∈ L(V ). Then α ≤ β on L(V ) if and only if im α ⊆ im β and
V αβ −1 = E(α, β).
The below corollary is obtained from Theorems 1.1 and 2.4
Corollary 2.6. Let α, β, θ ∈ L(V ) be such that θ is an isomorphism. Then α ≤ β on L(V ) if
and only if α ≤ β on (L(V ), θ).
By the above results, the characterizations for minimal nonzero elements and maximal
elements in (L(V ), θ) can be proved similar to Theorem 1.2, see [9] for more details, and we
omit the proof of the following theorem.
Theorem 2.7. Let α, θ ∈ L(V ) be such that θ is an isomorphism. Then
(i) α is a minimal nonzero element in (L(V ), θ) if and only if rank α = 1.
(ii) α is a maximal element in (L(V ), θ) if and only if α is a monomorphism or an epimorphism.
Finally, we give a characterization for an element in (L(V ), θ) to be a lower cover of an
element in (L(V ), θ).
Theorem 2.8. Let α, β, θ ∈ L(V ) be such that θ is an isomorphism and α < β on (L(V ), θ).
Then α is a lower cover of β in (L(V ), θ) if and only if dim(ker α/ ker β) = 1. In other words,
β is an upper cover of α in (L(V ), θ) if and only if dim(ker α/ ker β) = 1.
Proof. Assume that dim(ker α/ ker β) = 1. Suppose that α < γ ≤ β on (L(V ), θ) for some
γ ∈ L(V ). Then V αγ −1 = E(α, γ) and V γβ −1 = E(γ, β) by Theorem 2.4. Thus ker β ⊆ ker γ (
ker α. Let B1 be a basis of ker β and B2 a basis of ker α containing B1 . As dim(ker α/ ker β) = 1,
we obtain |B2 \ B1 | = 1. Let u ∈ B2 \ B1 . Then B2 = B1 ∪ {u}. Observe that if u ∈ ker γ,
then B1 ∪ {u} is a basis of ker γ which is impossible
since ker γ ( ker α. Hence u ∈
/ ker γ. Let
P
w ∈ ker γ. Thus w ∈ ker α. We write w =
ai xi + bu where xi ∈ B1 , ai , b are scalars, i ∈ I
i
and IPis a finite index set. Then 0 = wγ = 0 + buγ. Since uγ 6= 0, we get b = 0. Hence
ai xi ∈ ker β, and so ker γ = ker β. As γ ≤ β on (L(V ), θ), we obtain im γ ⊆ im β. To
w=
i
see γ = β, let v ∈ V . Then vγ ∈ im γ. Since im γ ⊆ im β, we have vγ = u0 β for some u0 ∈ V .
It follows that u0 ∈ V γβ −1 = E(γ, β) and hence u0 γ = u0 β = vγ. Thus v − u0 ∈ ker γ = ker β,
so vβ = u0 β = vγ. This implies that γ = β. Hence α is a lower cover of β in (L(V ), θ).
To see the necessity, assume that dim(ker α/ ker β) 6= 1. By Theorem 2.4, im α ⊆ im β and
V αβ −1 = E(α, β). Since α 6= β, we have dim(ker α/ ker β) > 1. By Proposition 1.5, write α and
β as
{xi }i∈I ∪ {yj }j∈J zk
{xi }i∈I yj zk
α=
and β =
0
uk k∈K
0
vj uk j∈J,k∈K
where {xi }i∈I , {xi }i∈I ∪ {yj }j∈J , {uk }k∈K , {vj }j∈J ∪ {uk }k∈K , {xi }i∈I ∪ {yj }j∈J ∪ {zk }k∈K are
bases of ker β, ker α, im α, im β and V , respectively. Since |{yj }j∈J | = dim(ker α/ ker β) > 1,
we let j0 ∈ J. Define γ ∈ L(V ) by
{xi }i∈I ∪ {yj }j∈J\{j0 } yj0 zk
γ=
.
0
vj0 uk k∈K
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It follows that im α ( im γ ( im β. Furthermore, V αγ −1 = E(α, γ) and V γβ −1 = E(γ, β) hold
as the sets {xi }i∈I ∪ {yj }j∈J\{j0 } , {yj0 } and {zk }k∈K , and the sets {xi }i∈I , {yj }j∈J\{j0 } and
{yj0 } ∪ {zk }k∈K fulfill Proposition 1.4, respectively. Then α < γ < β on L(V ) by Theorem 2.4.
Hence α is not a lower cover of β in (L(V ), θ).
Acknowledgment. The authors are grateful to the referee(s) for his/her careful reading of the
manuscript and useful comments.
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