奈良教育大学学術リポジトリNEAR
Non‑splitting Extensions of Hypergroups of Order Two
著者 KAWAKAMI Satoshi, SAKAO Masafumi, TAKEUCHI Tatsuya
journal or
publication title
奈良教育大学紀要. 自然科学
volume 56
number 2
page range 7‑13
year 2007‑10‑31
URL http://hdl.handle.net/10105/653
1. Introduction
Let H and L be finite commutative hypergroups.
A finite commutative hypergroup K is called an exten- sion of L by H if the sequence
1 H K L 1
is exact, i.e. if the quotient hypergroup K/H is isomor- phic to L. Here, the notions of subhypergroup, quo- tient hypergroup and isomorphism between hyper- groups are taken from papers
(1), (14), a source from which all the elementary knowledge needed in the sequel will be taken.
There exist several methods to construct exten- sions of hypergroups from given ones. These methods lead to an insight into the structure of hypergroups.
One of the methods is based on the notion of hyper- group join as introduced by Jewett
(8)and further devel- oped by Dunkl-Ramirez
(2), Fournier-Ross
(3), Voit
(11), Vrem
(13), and Zeuner
(16). The method of substitution introduced by Voit
(11)is a generalization of the join which provides extensions of hypergroups.
In the previous papers
(5), (7)we constructed splitting extensions by applying the notion of a field of commu- tative hypergroups. At the same time Voit
(12)reported to classify all hypergroup structures on two disjoint tori and two disjoint real lines, which was interpreted for us
to determine all extensions of hypergroups of order two by locally compact abelian groups and respectively.
Stimulated by his work, we calculated all extensions of hypergroups of order two by some finite abelian groups.
It is an important problem to determine all extensions of hypergroups in order to understand their full struc- ture. In this paper we do a promising task to find all extensions of some finite commutative hypergroups which was a remained problem in the previous paper
(7). This work has been done by developing some result in bachelor s thesis
(10)by Takeuchi in 2007. We appreciate referee s valuable comments on this paper.
In this paper we determine all extensions K of L(p) of order two by finite abelian groups H as described in the following theorem.
Theorem Let K be a commutative hypergroup extension of a hypergroup L(p) (0 p ＜ 1) of order two by a finite abelian group H, which means that there exists a hypergroup homomorphism from K onto L(p) such that Ker = H. Let H( r
1) denote the stability group of H at s
0∈ S : =
−１( r
1) and ω ( r
1) be the normalized Haar measure of H( r
1) and Q( r
1) := H/H( r
1). Then we have S = {hs
0: h∈H} Q( r
1), s
＊0= hs
0, s
＊0s
0= qω( r
1) + pcs
0for h ∈Q( r
1) and c ∈M
１(Q( r
1)) such that hc
＊= c.
All extensions K of L (p) by H are characterized in this way. Each equivalence class of such extensions K is Bull. Nara Univ. Educ., Vol. 56, No.2 (Nat. ) , 2007
Non-splitting Extensions of Hypergroups of Order Two
Satoshi KAWAKAMI, Masafumi SAKAO * and Tatsuya TAKEUCHI **
(Department of Mathematics, Nara University of Education, Nara 630-8528, Japan) (Received May 7, 2007)
Abstract
The purpose of this paper is to investigate extension problem for the category of finite com- mutative hypergroups. In fact, we determine all extensions of hypergroups of order two by finite abelian groups. (AMS Subject Classification : 43A62, 20N20.)
Key Words : hypergroup, extension
*The University of the Air, **Sigaraki Junior High School
determined by a subgroup H( r
1) of H, h ￣∈ Q( r
1)/Q
2( r
1), and ￣ c = ｛ bc : b
2= w ( r
1) ｝ such that hc
＊= c where Q
2( r
1) = {b
2∈ Q ( r
1): b∈Q ( r
1)}. Moreover, the extension K is splitting if and only if h = b
2and c = b ω( r
1) for some b∈
Q( r
1).
2. Preliminaries
We recall some notions and facts on finite commu- tative hypergroups from (1) and (14). K:= (K, A) is called a finite commutative hypergroup if the following con- ditions (1)〜(6) are satisfied.
(1) A is a ＊ -algebra over with the unit c
0. (2) K={ c
0, c
1, ..., c
n} is a linear basis of A.
(3) K
＊= K.
(4) c
ic
j= n
kijc
k, where n
kijis a non-negative real number such that
c
＊i= c
jn
ij0＞0 and c
＊i≠ c
jn
0ij= 0.
(5) n
kij= 1 for any i, j.
(6) c
ic
j= c
jc
ifor any i, j.
The weight of an element c
i∈ K is defined by w(c
i) := (n
0i j)
−1where c
j= c
＊i, and the total weight of K is given by w(K) := Σ
ni=0w(c
i).
Let M
1(K) denote the set of probability measures on K, i.e.
M
1(K) := {c = a
kc
k: a
k0 (k = 0, 1, ..., n), a
k= 1}.
For c = Σ
nk=0a
kc
k, support of c is defined by supp(c) := {c
k: a
k≠0. k = 0, 1, … , n}.
Let ω(K) denote the normalized Haar measure of K which is given by
ω(K) = c
k.
Let L be a hypergroup L(p) = { r
0, r
1} of order two, which is determined by r
21= q r
0+ p r
1, q = 1 − p, 0 p ＜ 1 and H = {h
0, h
1, ..., h
n} be a finite commutative hypergroup. Then, a hypergroup join H∨L of H by L is defined by
H∨L = {h
0, h
1, ..., h
n, s
0}, h
ks
0= s
0(k = 0, 1, ..., n), s
20= q ω (H) + ps
0. Let K be an extension of L by a finite abelian group H and let denote a quotient mapping from K onto L.
For s ∈ S: =
−1( r
1), H( r
1) denotes the stability group of H at s, i.e.
H( r
1) = {h ∈ H: hs = s}.
Here we note that H( r
1) does not depend on s but only on S. If we can take a mapping : L K such that
6
= id, ( r
0) = h
0and ( r
1)
2= ω(H( r
1)) ( r
21), we called K a splitting extension of L by H in the previous paper
(5). For a fieldη: L ∋r H( r ) ⊂ H based on L, we constructed a splitting extension K (H,η , L) of L by H in our papers
(5), (7). In our situation that H is an abelian group and L = { r
0, r
1}, we note that an extension K of L by H is splitting if and only if K=K(H,η, L) for some subgroup H( r
1) of H, which is also given as the substi- tution S(Q ( r
1)×L,Q ( r
1) H) of Q ( r
1) by H in Q( r
1) × L(p) in the sense of Voit
(11)where Q ( r
1) = H/H( r
1) .
3. Extensions of hypergroups of order two
Let L = { r
0, r
1} be a hypergroup of oreder two where r
0is the unit of L. Since the hypergroup struc- ture of L is determined by
r
21= q r
0+ p r
1, q = 1 − p, 0 p ＜ 1.
We denote this hypergroup L of order two by L(p) (0 p ＜ 1). We note that L(0)
2. Let H = {h
0, h
1, ..., h
n} be a finite abelian group where h
0is the unit of H.
We calculate all extensions K of L = L(p) by some concrete groups H. Let be a canonical projection of K onto L. Then K is written as the disjoint union of H and S where H = Ker and S=
−1( r
1). Hence K is writ- ten as K = { h
0, h
1, ..., h
n, s
0, s
1, ..., s
m} where H = {h
0, h
1, ..., h
n} and S = {s
0, s
1, ..., s
m}. Let H( r
1) denote the sta- bility group of H at s
0∈ S, i.e.
H( r
1) = {h ∈ H: hs
0= s
0}.
Proposition 1. Each element s
k∈ S is given by s
k= h
n(k)s
0(k= 0, 1, ..., m) where h
0, h
n(1), ..., h
n(m)are repre- sentatives of the coset H/H( r
1).
Proof. If s
k∈supp(hs
0) for h∈H, then supp (h
＊s
k)
⊂supp (h
＊hs
0). Since H is a group, h
＊h = h
0so that supp (h
＊hs
0) = supp(h
0s
0) = supp (s
0) = {s
0}. Hence we see that h
＊s
k= s
0, namely s
k= hs
0. By the fact that
S = Hs
0= supp(hs
0), we get the desired conclusion.
［Q. E. D.］
Let ω ( r
1) denote the normalized Haar measure of H( r
1).
Proposition 2. For the above s
0∈S, s
＊0s
0= qω( r
1) + pcs
0where c is a probability measure on H such that hc
＊= c for h ∈ H with s
＊0= hs
0, and h
kc = c for h
k∈ H( r
1) so that ω( r
1)c = c. In this case, we have s
02= qh
＊ω( r
1) + ph
＊cs
0.
Proof. It is easy to see that s
＊0s
0is written as
∪
h∈Hw(c
k) w(K)
n
Σ
k=0n
Σ
k=0 nΣ
k=0 nΣ
k=0 nΣ
k=0S a t o s h i K a w a k a m i
・M a s a f u m i S a k a o
・T a t s u y a T a k e u c h i
8
s
＊0s
0= qc
0+ pcs
0for some c
0, c ∈ M
1(H). By the formula (s
＊0s
0)
＊= s
＊0s
0, we see that c
＊0= c
0and hc
＊ω( r
1) = cω( r
1) for h∈H such that s
＊0= hs
0.
For h
k∈ H ( r
1), the fact h
ks
0= s
0implies that h
kc
0= c
0and h
kc = c. Then we have ω ( r
1) c
0= c
0and ω ( r
1)c = c. For h
k∈ / H ( r
1), the fact h
ks
0≠s
0implies that h
0∈ / supp ((h
ks
0)
＊s
0). Hence we see that h
0∈ / supp (h
＊kc
0), namely h
k∈ / supp (c
0). Therefore supp(c
0) must be H( r
1) which implies that c
0= ω ( r
1).
［Q. E. D.］
Let Q( r
1) denote the quotient group of H by H ( r
1) and sometimes we identify
M
1(Q( r
1)) = {c ∈M
1(H) :ω( r
1)c = c}.
Then we have seen that an extension K = H ∪S of L(p) by H is determined by a subgroup H ( r
1) of H, s
0∈S Q( r
1), h∈Q( r
1), and c ∈M
1(Q ( r
1)) such that s
＊0= hs
0and hc
＊= c. Therefore we denote such an extension K by K(H( r
1), s
0, h, c). Let K
1= H ∪ S
1and K
2= H ∪ S
2be two extensions of L(p) by H and
1[resp.
2] be a canonical quotient mapping from K
1[resp. K
2] onto L(p). Then K
1is called to be equivalent to K
2as extensions if there exists a hypergroup isomorphism from K
1onto K
2such that (h) = h for all h∈H and
26=
1. When we take r
0∈ S, k∈Q ( r
1), and d∈M
1(Q ( r
1)) such that r
＊0= kr
0and kd
＊= d, we have another extension K(H ( r
1), r
0, k, d) of L(p) by H.
Proposition 3 Two extensions K (H( r
1), s
0, h, c) and K(H
1( r
1), r
0, k, d) are mutually equivalent as exten- sions of L(p) by H if and only if H
1( r
1) = H( r
1) and there exists b ∈Q( r
1) such that r
0= b
＊s
0, k = b
2h, and d = bc.
Proof. Suppose that K=K(H ( r
1), s
0, h, c) and K
1= K(H
1( r
1), r
0, k, d) are mutually equivalent as extensions.
Then it is clear that H
1( r
1) =H( r
1) so that we may assume K
1=K=H∪S where S=Hs
0= {hs
0: h∈H} Q ( r
1). Hence for r
0∈ S, there exists b ∈ Q ( r
1) such that r
0= b
＊s
0. By the relations that s
＊0= hs
0, r
＊0= kr
0and
r
＊0r
0= q ω ( r
1) + pdr
0= q ω ( r
1) + pcs
0= s
＊0s
0, we obtain that k = b
2h and d = bc.
Conversely if H
1( r
1) = H ( r
1), r
0= b
＊s
0, k = b
2h, and d = bc for some b∈Q( r
1), it is easy to see that the correspon- dence : s
0r
0induces a hypergroup isomorphism from K = K(H ( r
1), s
0, h, c) onto K
1= K(H
1( r
1), r
0, k, d).
［Q. E. D.］
Theorem Let K be a commutative hypergroup extension of a hypergroup L(p) (0 ≦ p＜1) of order two
by a finite abelian group H, which means that there exists a hypergroup homomorphism from K onto L(p) such that Ker = H. Let H ( r
1) denote the stability group of H at s
0∈S =
−1( r
1) and ω( r
1) be the normal- ized Haar measure of H( r
1) and Q ( r
1) := H/H( r
1). Then we have S = {hs
0: h ∈ H} Q ( r
1), s
＊0= hs
0, s
＊0s
0= q ω ( r
1) + pcs
0for h∈Q ( r
1) and c∈M
1(Q ( r
1)) such that hc
＊= c.
All extensions K of L(p) by H are characterized in this way. Each equivalence class of such extensions K is determined by a subgroup H ( r
1) of H, h ￣∈ Q ( r
1)/Q
2( r
1), and ￣ c = ｛bc : b
2= w ( r
1)｝ such that hc
＊= c where Q
2( r
1) = {b
2∈Q ( r
1) : b∈Q ( r
1)}. Moreover, the extension K is splitting if and only if h = b
2and c = b ω ( r
1) for some b ∈ Q ( r
1).
Proof. These statements follow immediately from Proposition 1, 2 and 3, so that we omitt the details.
［Q. E. D.］
Remark When H =
n(n is an odd number ) we see that | Q ( r
1)/Q
2( r
1) | = 1 for any subgroup H( r
1) of H.
When H=
n(n is a prime number) there are no non- trivial subgroups H ( r
1) of H so that splitting extensions K of L(p) by H are only K = H×L(p) and K = H ∨L (p).
4. Applications and Examples
Under these preparations we calculate all extensions K of L(p) by concrete abelian groups H =
2,
3,
4,
5,
6
, and
2×
2. We denote the order of K by |K|.
Model 1. H =
2and L = L(p).
H = {h
0, h
1}, h
21= h
0.
(1) Case of |K | = 4, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, s
0, s
1}, s
1= h
1s
0.
(1−a) K = K
a(t) (0 t 1) which is characterized by s
＊0= s
0, s
20= qh
0+ pts
0+ p (1 − t) s
1.
(1 −b) K = K
bwhich is characterized by s
＊0= s
1, s
20= qh
1+ s
0+ s
1. (2) Case of |K | = 3, i.e. H( r
1) = H.
K = H∨L(p) = {h
0, h
1, s
0} which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ ps
0. Remark 1. K
a( t) K
a(1−t) .
1. K is a splitting extension of L(p) by H if and only if K = K
a(1) = H×L(p) or H∨L(p).
2. When p = 0, we see that L =
2and K
b=
4which is known as a non-splitting extension of
2by
2in the category of finite abelian groups.
q 2 q 2
p
2
p
2
Model 2. H =
3and L = L(p).
H = {h
0, h
1, h
2}, h
13= h
0, h
21= h
2, h
22= h
1, h
＊1= h
2, h
＊2= h
1. (1) Case of |K|= 6, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, h
2, s
0, s
1, s
2}, s
k= h
ks
0( k = 0, 1, 2).
K = K( t) (0 t 1) which is characterized by s
＊0= s
0, s
20= qh
0+ pts
0+ (1 − t) s
1+ (1 − t )s
2. (1−b) K = K
b(t) (0 t 1) which is characterized by
s
＊0= s
1, s
20= qh
2+ pts
1+ (1−t )s
0+ (1−t) s
2. (1 −c) K = K
c(t) (0 t 1) which is characterized by s
＊0= s
2, s
20= qh
1+ pts
2+ (1 − t) s
1+ (1 − t )s
0. (2) Case of |K|= 3, i.e. H( r
1) = H.
K=H∨L(p) which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ h
2+ ps
0. Remark 2.
1. We remark that K
b(t) K
c(t) K(t) as extensions of L(p) by H.
2. K is a splitting extension of L(p) by H if and only if K = K(1) = H × L(p) or H ∨ L(p).
Model 3. H =
4and L = L(p).
H = {h
0, h
1, h
2, h
3}, h
41= h
0, h
k1= h
k(k = 1, 2, 3), h
＊1= h
3, h
＊2= h
2.
(1) Case of |K| = 8, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, h
2, h
3, s
0, s
1, s
2, s
3}, s
k= h
ks
0(k = 0, 1, 2, 3).
(1 −a) K = K
1a(t, u) (0 t 1, 0 u 1, 0 t+u 1)
which is characterized by
s
＊0= s
0, s
20= qh
0+ pts
0+ (1 −t −u) s
1+ pus
2+ (1 −t −u) s
3(1 −b) K = K
1b(t) (0 t 1) which is characterized by s
＊0= s
1, s
20= qh
3+ ts
0+ (1−t)s
1+ (1−t)s
2+ ts
3. (2) Case of |K| = 6, i.e. H( r
1) = {h
0, h
2}.
K = {h
0, h
1, h
2, h
3, s
0, s
1}, s
1= h
1s
0.
(2 −a) K = K
2a(t) (0 t 1) which is characterized by s
＊0= s
0, s
20= h
0+ h
2+ pts
0+p(1−t) s
1. (2 −b) K = K
2bwhich is characterized by
s
＊0= s
1, s
20= h
1+ h
3+ s
0+ s
1. (3) Case of |K| = 5, i.e. H( r
1) = H.
K = {h
0, h
1, h
2, h
3, s
0}.
K = H∨L(p) which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ h
2+ h
3+ ps
0. Remark 3. K
1a(t, u) K
1a(u, t), K
1b(t) K
1b(1 −t),
K
2a(t) K
2b(1−t).
K is a splitting extension of L(p) by H if and only if K = K
1a(1, 0) = H × L(p), K
2a(1) = S(Q( r
1) × L(p), Q( r
1) H), or H∨L(p) where Q( r
1) = H/H( r
1)
2and S(Q( r
1)×
L(p), Q( r
1) H) is substitution of Q( r
1) by H in Q( r
1)×
L(p) in the sense of Voit
(11).
Model 4. H =
5and L = L(p).
H = {h
0, h
1, h
2, h
3, h
4}, h
51= h
0, h
k1= h
k(k = 1, 2, 3, 4), h
＊1= h
4, h
＊2= h
3.
(1) Case of |K|= 10, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, h
2, h
3, h
4, s
0, s
1, s
2, s
3, s
4}, s
k= h
ks
0(k = 0, 1, 2, 3, 4).
K = K(t, u) (0 t 1, 0 u 1) which is charac- terized by
s
＊0= s
0, s
20= qh
0+ pts
0+ (1 −t )us
1+ (1−t)(1−u)s
2+ (1−t )(1−u) s
3+ (1 −t)us
4. (2) Case of |K|= 5, i.e. H( r
1) = H.
K = {h
0, h
1, h
2, h
3, h
4, s
0}.
K = H ∨ L(p) which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ h
2+ h
3+ h
4+ ps
0. Remark 4.
K is a splitting extension of L(p) by H if and only if K = K(1, 0) = H×L(p) or H∨L(p).
Model 5. H =
6and L = L(p).
H = {h
0, h
1, h
2, h
3, h
4, h
5},
h
61= h
0, h
k1= h
k(k = 1, 2, 3, 4, 5), h
＊1= h
5, h
＊2= h
4, h
＊3= h
3. (1) Case of | K |= 12, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, h
2, h
3, h
4, h
5, s
0, s
1, s
2, s
3, s
4, s
5}, s
k= h
ks
0(k = 0, 1, 2, 3, 4, 5).
(1 −a ) K = K
1a(t, u, w) (0 t 1, 0 u 1, 0 w 1, 0 t + u 1) which is characterized by
s
＊0= s
0, s
02= qh
0+ pts
0+ (1−t−u)ws
1+ (1−t−u) (1−w)s
2+ us
3+ (1−t −u)(1−w)s
4+ (1−t−u)ws
5.
(1−b) K = K
1b(t, u) (0 t 1, 0 u 1, 0 t + u 1)
which is characterized by
s
＊0= s
1, s
20= qh
5+ ts
0+ us
1+ (1 − t − u)s
2+ (1 − t − u)s
3+ us
4+ ts
5. (2) Case of |K|= 9, i.e. H( r
1) = {h
0, h
3}.
K = {h
0, h
1, h
2, h
3, h
4, h
5, s
0, s
1, s
2}, s
1= h
1s
0, s
2= h
2s
0. K = K
2(t) (0 t 1) which is characterized by s
＊0= s
0, s
20= h
0+ h
3+ pts
0+ (1 − t)s
1+ (1 − t)s
2. (3) Case of |K|= 8, i.e. H( r
1) = {h
0, h
2, h
4}.
K = {h
0, h
1, h
2, h
3, h
4, h
5, s
0, s
1}, s
1= h
1s
0. (3−a) K = K
3a(t) (0 t 1) which is characterized by
s
＊0= s
0, s
20= h
0+ h
2+ h
4+ pts
0+ p(1−t )s
1. (3−b) K = K
3bwhich is characterized by
s
＊0= s
1, s
20= h
1+ h
3+ h
5+ s
0+ s
1. (4) Case of |K|= 7, i.e. H( r
1) = H.
K
4= {h
0, h
1, h
2, h
3, h
4, h
5, s
0}.
p 3 p 3 q 3 q 3 q 3
q 3 q 3 q 3
p 2 p
2 q
2 q 2
p 2 p 2 p
2
p 2 p 2 p 2
p 2 p
2 p 2
p 2 p
2
q 5 q 5 q 5 q 5 q 5
p 2 p
2 p
2
p 2
p 4 q 4 p 4 q 4
p 2 p 2 p 2 q 2
p 2 q 2
p 2 p
2 p
2 p 2
p 2 p
2 q 3 q 3 q 3
p 2 p
2
p 2 p
2
p 2 p
2
S a t o s h i K a w a k a m i
・M a s a f u m i S a k a o
・T a t s u y a T a k e u c h i
10
K
4= H ∨ L(p) which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ h
2+ h
3+ h
4+ h
5+ ps
0Remark 5. K
1a(t, u, w) K
1a(u, t, w), K
1b(t, u) K
1b(1 −t−u, u), K
3a(t) K
3a(1−t).
K is a splitting extension of L(p) by H if and only if K = K
1a(1, 0, 0) =H × L(p), K
2(1) = S(
3× L(p),
3H), K
3a(1) = S(
2×L(p),
2H) or H∨L(p) where S(
k×L(p),
k
H) (k = 2, 3) is the substitution of
kby H in
k× L(p) in the sense of Voit
(11).
Model 6. H =
2×
2and L = L(p).
H = {h
0, h
1, h
2, h
3},
h
2k= h
0, (k = 1, 2, 3), h
1h
2= h
3, h
2h
3= h
1, h
1h
3= h
2, h
＊k= h
k(k = 1, 2, 3).
(1) Case of |K|= 8, i.e. H( r
1) = {h
0}.
K = {h
0, h
1, h
2, h
3, s
0, s
1, s
2, s
3}, s
k= h
ks
0(k = 0, 1, 2, 3).
(1−a) K = K
1a(t, u, w) (0 t 1, 0 u 1, 0 w 1, 0
t + u + w 1) which is characterized by
s
＊0= s
0, s
20= qh
0+ pts
0+ pus
1+ pws
2+ (1 − t − u − w)s
3. (1−b) K = K
1b(t) (0 t 1) which is characterized by
s
＊0= s
1, s
02= qh
1+ ts
0+ ts
1+ (1−t )s
2+ (1−t)s
3. (1 −c) K = K
1c(t) (0 t 1) which is characterized by
s
＊0= s
2, s
02= qh
2+ ts
0+ (1 − t)s
1+ ts
2+ (1 − t)s
3. (1−d) K = K
d1(t) (0 t 1) which is characterized by
s
＊0= s
3, s
02= qh
3+ ts
0+ (1−t)s
1+ (1−t)s
2+ ts
3. (2) Case of |K|= 6, i.e. H( r
1) = {h
0, h
k} (k = 1, 2, 3).
K = {h
0, h
1, h
2, h
3, s
0, s
1}, s
1= h
k's
0(k'= 1, 2, 3, k'≠k).
(2 −a−k) K = K
a(2,k)(t)(k = 1, 2, 3) (0 t 1) which is characterized by
s
＊0= s
0, s
20= h
0+ h
k+ pts
0+ p(1−t) s
1. (2 −b−1) K = K
b(2,1)
which is characterized by s
＊0= s
1, s
20= h
2+ h
3+ s
0+ s
1. (2 −b− 2) K = K
b(2,2)
which is characterized by s
＊0= s
1, s
20= h
1+ h
3+ s
0+ s
1. (2 −b−3) K = K
b(2, 3)
which is characterized by s
＊0= s
1, s
20= h
1+ h
2+ s
0+ s
1. (3) Case of |K|= 5, i.e. H( r
1) = H.
K = {h
0, h
1, h
2, h
3, s
0}.
K = H∨L(p) which is the join of H by L(p) and characterized by
s
＊0= s
0, s
20= h
0+ h
1+ h
2+ h
3+ ps
0. Remark 6. K
1a(t, u, w) K
1a(u, t, 1−t−u−w)
K
1a(w, 1 −t−u−w, t) K
1a(1 − t − u − w, w, u), K
＊1(t) K
＊1(1 −t) (＊＝a, b, c), K
a(2,k)(t) K
a(2,k)(1 −t).
K is a splitting extension of L(p) by H if and only if K = K
1a(1, 0, 0) = H × L(p), K
a(2,1)(1) = S(
2× L(p),
2H), or H∨L(p).
Calculations of each model are done by applying Proposition 1, 2, 3, and Theorem. For the case H =
6, we will show how to calculate extensions and to obtain the above results.
Calculation of Model 5. Case of H =
6. A probability measure c on H = {h
0, h
1, h
2, h
3, h
4, h
5} is written as
c = a
0h
0+ a
1h
1+ a
2h
2+ a
3h
3+ a
4h
4+ a
5h
5, a
k0 (k = 0, 1, ..., 5), a
k= 1.
(1) Case of H( r
0) = {h
0}.
We see that
c
＊= a
0h
＊0+ a
1h
＊1+ a
2h
＊2+ a
3h
＊3+ a
4h
＊4+ a
5h
＊5= a
0h
0+ a
1h
5+ a
2h
4+ a
3h
3+ a
4h
2+ a
5h
1= a
0h
0+ a
5h
1+ a
4h
2+ a
3h
3+ a
2h
4+ a
1h
5. (1−a) Case of s
＊0= s
0.
By the condition that c
＊= c , we get a
1= a
5and a
2= a
4. Set
a
0= t, a
3= u, a
1= a
5= (1−t −u)w, a
2= a
4= (1 −t−u)(1−w), where
0 u 1, 0 t 1, 0 w 1, 0 t + u 1.
Hence we obtain by Proposition 2, s
02= qh
0+ pcs
0= qh
0+ pts
0+ p (1 −t−u) ws
1+ p(1 −t −u) (1− w)s
2+ p u s
3+ p (1−t −u)(1− w) s
4+
p(1 −t−u)ws
5. (1 −b) Case of s
＊0= s
1= h
1s
0.
In this case, by Proposition 2 we see that, s
02= qh
＊1+ ph
＊1cs
0, h
1c
＊= c.
By the formula
h
1c
＊= a
1h
0+ a
0h
1+ a
5h
2+ a
4h
3+ a
3h
4+ a
2h
5, we obtain
a
0= a
1, a
2= a
5, a
3= a
4. Set parameters t and u by
a
0= a
1= t, a
2= a
5= u, a
3= a
4= (1 −t −u), where
0 t 1, 0 u 1, 0 t + u 1.
We note that
1 2
1 2 1 2 1
2
1 2 1
2
1 2 1
2 1 2
1 2
5
