ss="fs6 ls8e ws60">21mn, define
the action


421
:Dih Diff
n
K
by



π
21
41, 0
i
n
ru rue




and



4
1
0,1 ru u
r
.
The quotient 4
K
is the projective plane
22, 2P
containing two cone points of order two. This action ex-
tends to

,
M
bd and we obtain

421
:Dih n


Diff ,
M
bd . The orbifold quotient is
4,


4
Bk W
where

4,,1WIztzt  is the
twisted I-bundle over

22, 2P and
is the homeo-
morphism of induced by
. The orbifold quotient
 
44
,(21),
M
bdb nd
.
5) Quotient type

5,
. Define the following ac-
tions:

521
:Dih Diff
n
K
where



π
21
51,0
i
n
ru rue




and

50,1 ru ru
;


542
:Dih Diff
n
K
where



π
21
5
1
1, 0
i
n
ru ue
r



and


50,1 ru ru
;


542
:Dih Diff
n
K
where



2π
42
51, 0
i
n
ru rue



and



5
1
0,1 ru u
r
;
and


54
:Dih Diff
n
K
where



π
2
51, 0
i
n
ru rue



and



5
1
0,1 ru u
r



. The
orbifold quotient for all these actions is the mirrored disk

22, 2. All these actions extend to

Diff ,
M
bd . If
 

5,,1WIztrzt , where r is a reflection
exchanging a pair of cone points, is the twisted I-bundle
over

22, 2, then the orbifold quotient for these ex-
tended actions is

55
,Bk W

. We obtain
 
55
,21,
M
bdb nd
,
 
35
,21,2
M
bdb nd
,


55
,42,
M
bdb nd

and

55
,4,
M
bd nbd
.
6) Quotient type
6,
. Define actions 66
,:

2
Dih Diff
n
K
as follows:



π
6
1
1, 0
i
n
ru ue
r
and


60,1 ru ru
if
n is even, and



2π
6
1
1, 0
i
n
ru ue
r



and



6
1
0,1 ru u
r
if n is odd. The quotients 5
K
and 5
K
are both a mirrored disk

22 containig
a cone point of order two and two cone points of order
two on the mirror. These actions extend to the prism
manifold
,
M
bd and we obtain
66 2
,:Dih n
Diff ,
M
bd . The orbifold quotient for these actions is
denoted by
66
,Bk W

 where
6,,1WIztrzt
  the twisted I-bundle over
the mirrored disk
22, and r is a reflection leaving
two cone points fixed and exchanging the other two cone
points. The orbifold quotients

5
,Mbd
and
6
,Mbd
are both
6,bnd bnd.
7) Quotient type
7,
. Define
72
:Dih n
Diff
K
and

722
:Dih Diff
n
K
 as fol-
lows:



π
71, 0
i
n
ru rue
,


70,1 ru ru
, and



7
1
1, 0,0ru u
r

,



π
70,1,0
i
n
ru rue



, and
70, 0,1ru ru
. The quotients 7
K
and 7
K
are both a mirrored disk

20 containig four cone
points of order two on the mirror. These actions extend to
the prism manifold
,
M
bd and we obtain


72
:DihDiff ,
n
M
bd
and


722
:DihDiff ,
n
M
bd
 .
If
7,,1WIztrzt
 , where r is a reflection
leaving each cone points fixed, then 7
W is a twisted
I-bundle over the mirrored disk

20. The orbifold
quotient for these extended actions is
6,

6
Bk W
. We obtain
 
77
,2,
M
bd nbd
and
77
,2,2
M
bdnb d
.
3. Prism Manifold Covers of Orbifolds
In this section we give necessary and sufficient condi-
J. E. KALLIONGIS, R. OHASHI
152
tions for when the orbifold

,
i
, 17i, is cov-
ered by a prism manifold. The proofs rely on Section 2.
Proposition 1. For the orbifold

1,
, there exists
a prism manifold cover if and only if either
is odd or
0
(mod 4).
Proof. Suppose
is odd. Then there exists a
-ac-
tion on

1,M
such that
 
1
1, 1,,MM

.
If
is even, write 0
2l
where 0
is odd. If
is odd, then there exists a 0
-action on
2,
l
M
such that
 

01
2, 2,,
ll
MM
.
Suppose now that
and
are both even where 0
(mod 4). Write 0
2
and 0
2m
where 0
and
0
are both odd. Then there exists a 0
2
-action on

0
2,
m
M
such that

0
002100
2,2,2 ,2
mm m
MM
.
For the converse, suppose that
and
are both even
and there is a covering

,,
i
Mbd
. Then ei-
ther

21bn
 and d
, or

21bn
 and
2d
. Since
is even, it follows that 2 divides b. In
the first case, 2 would also divide d, contradicting the
fact that b and d are relatively prime. If 0
(mod 4),
then again 2 divides d giving a contradiction.
Proposition 2. For the orbifold

2,
, there exists
a prism manifold cover if and only if 0
(mod 2).
Proof. Suppose that 0
(mod 2). Write 0
2

.
Then there exists a 0
2
-action on

1,M
such that
 
0
22
1, 1,,MM
.
For the converse, suppose that

2
,,Mbd
. Then
either 2nb
and d
, or 2nb
and 2d
.
Proposition 3. For the orbifold

3,
, there exists
a prism manifold cover if and only if

(mod 2).
Proof. Suppose that

(mod 2) and let 2
d
.
Suppose that 0
 (mod 4), and thus there exists
an integer n such that 4n
 . There exists a 4n
-
action on

1,
M
d such that
 

43
3
1,1,2, 2
,.
n
M
dMd ndnd


If 0
 (mod 4), then write 21
2n


for some
n. There exists a

22 1n
-action on

1,
M
d such that




3
22 1
3
1,1,21,21
,.
n
M
dMdn dn d

 
For the converse, suppose that

3
,,Mbd
.
Then either
21bnd
 and

21bnd
, or
2nb d
and 2nb d
for some n. Subtracting
the two equations in both cases, we obtain 2d
.
Propo sitio n 4. For the orbifold

4,
, there exists
a prism manifold cover if and only if either
is odd or
is odd.
Proof. Since
1
,Mbd
always double covers
4
,Mbd
, using a proof similar to that in Proposition
1 shows that there is a prism manifold covering of
4,
if and only if
or
is odd by [4].
Proposition 5. A prism manifold covering for the orbi-
fold
5,
always exists.
Proof. Suppose
is an odd number. Then
1,M
Dih
-action whose quotient is
5,
.
If β is even, we write 0
2m
where m 1, 0
2n
where 0n, and 0
and 0
are both odd numbers. If
n = 0 or 1n
, then
2,
m
M
and

0
2,
m
M
0
Dih
and
0
2
Dih
-actions respectively, whose
quotient space is
5,
. If n and m are both greater
than 1, or if 1m
and 2n, then

1,M

10
42
Dih m



or a

0
2
Dih
-action respectively,
whose quotient space is
5,
.
Proposition 6. For the orbifold
6,
, there exists
a prism manifold cover if and only if
(mod 2).
Proof. Since
3
,Mbd
double covers
6
,Mbd
and
3
,Mbd
double covers
6
,Mbd
, the re-
sult follows by Proposition 3.
Proposition 7. For the orbifold

7,
, there exists
a prism manifold cover if and only if 0
(mod 2).
Proof. Since
2
,Mbd
double covers
7
,Mbd
and
2
,Mbd
double covers

7
,Mbd
, the re-
sult follows by Proposition 2.
4. Poset of Actions on Prism Manifolds
Recall that two group actions
:DiffGM
and
:DiffGM

are equivalent if there is a homeo-
morphism :hM M
such that
1
:Gh Gh



.
If
:DiffGM
is an action, let :MM

be the orbifold covering map.
Let be the set of equivalence classes of actions on
prisim manifolds which leave a Heegaard Klein bottle
invariant. Now is partially ordered by setting
if there is a covering :
M
M
such that


. Note that the covering :
M
M
is also
a regular covering.
For a pair of positive integers
and
let
1,
denote the equivalence classes of those actions whose
quotient type is
1,
. Note that by Proposition 1 the
set
1,
is nonempty if and only if either
is odd,
or 0
(mod 4). Unless otherwise stated, we assume
from now on that
and
are integers where either
J. E. KALLIONGIS, R. OHASHI
153
is odd or 0
(mod 4).
Let
 
1,, :..,1,
divides,1 (2)and,or2.
bdZZgcd bd
bmoddd
b



 

It follows that

1,
is a partially ordered set under
the ordering

22 11
,,bdbd if 21
bb and 21
dd. Let

1,
be the subset of

1,
consisting of all
ordered pairs
 
1
,,bd
where d
. Note that
 
11
,,
 if
is odd. Moreover, if 0
2
(mod 2) and β is even, then

11
,,2
 
.
Proposition 8. Let

11
,bd and

22
,bd be elements
of the poset

1,
. There exists elements

,bd and

,bd

in

1,
, such that
 
11
,,bdbd and
 
22
,,bdb d, and

11
,,bd bd

and
,bd

22
,bd .
Proof. Let

12
.. ,b gcdbb. Note that since i
d
or 2
for 1,2i, it follows that if
12
min ,ddd,
then i
dd. Thus b divides
and d is
or 2
. If
b
is even, then it follows that 2 divides both 1
bb
and 2

12
.. ,bgcdbb. Thus b
is
odd showing
 
1
,,bd
. Moeover
11
,,bdb d
and
 
22
,,bdb d. Let

12
.. ,blcmbb
and d

12
max ,dd. It follows that b
is odd, and hence
 
1
,,bd

. Furthermore

11
,,bd bd

and
 
22
,,bdb d

.
Corollary 9.

1,
is a lattice where for
11
,bd
and

22
,bd in

1,
the join



112 21212
,,..,,min,bdb dgcdbbdd ,
and the meet



112 21212
,,..,,max,bdb dlcmbbd d .
Furthermore,

1,
is a sublattice of
1,
.
Proposition 10. Let

11
,bd and

22
,bd be elements
of

1,
such that

22 11
,,bdbd. Then there
exists either a standard m
-action 1
on
22
,
M
bd ,
or a standard 2m
-action 1
on

22
,
M
bd , which
we denote by
, and a regular covering
 
22 2211
:, ,,
M
bd MbdMbd

.
Proof. If

22 11
,,bdbd, then 21
bb and 21
dd.
Furthermore 1
d
, or 1
2d
and 2
d
, or 2
2d
.
Now 1
2
bm
b,

11
21bn
, and

21
21bn
 for
some integers m, n1 and n2. Since

1122
21 21bnb n ,
it follows that

21
21 21nmn , and therefore m
must be odd. Since 21
dd
, the only possibilities are
12
dd or 21
2dd. If 12
dd
, then there exists a
m
-action 1
on
22
,
M
bd such that

2222 111
,, ,
M
bd MbdMbd

.
If 21
2dd
, then there exists a 2m
-action 1
on
22
,
M
bd such that

2222 111
,,,
M
bd MbdMbd
.
Proposition 11. Let
,bd be an element of
1,
.
Then there exists either a standard 21n
-action 1
on
,
M
bd , or a standard

22 1n
-action 1
on
,
M
bd
which we denote by
, and a regular covering
1
:, ,,Mbd Mbd

.
Proof. Write 21n
b
. If d
, then there is a
21n
-action 1
such that
11
,, ,Mbd Mbd

.
If 2d
, then there is a

22 1n
-action 1
on
,
M
bd
such that
11
,, ,Mbd Mbd

.
Theorem 12. For each pair of positive integers
and
, the poset
1,
is isomorphic to the poset
1,
.
Proof. Define a function

11
:, ,f
 as
follows: let
1
,,bd
. There exists either a stan-
dard
b
-action if d
, or a standard
2b
-action if
2d
on
,
M
bd , which we denote by
, such that

1
,, ,Mbd Mbd

.
Define
1
,,fbd

.
Suppose

1112 22
,,
f
bd fbd

 . Since 1
and 2
are equivalent, there exists a homeomorphism
1122
:, ,hMbdMbd such that
1
12
Gh Gh


. Since

11
,
M
bd and
22
,
M
bd
are homeomorphic, it follows that 12
bb and 12
dd
,
showing f is one-to-one.
Let
1,

. Then there exist a prism manifold
,
M
bd such that

1
:, ,,Mbd Mbd

.
We may assume that b and d are both positive. By [4], η
is equivalent to one of the standard actions 1
or 1
,
and
1
,21,
M
bdb nd
 or


121,2bn d
respectively, for some positive integer n. Therefore
1
b
(mod 2) and d
or 2d
. If
is either
1
or 1
, then
,fbd
, showing f is onto.
Suppose now that
 
22 11
,,bdbd. Let

111
,fbd
and
22 2
,fbd
where 1
and 2
are the standard
J. E. KALLIONGIS, R. OHASHI
154

11
b

and

22
b

-actions respectively on
11
,
M
bd
and

22
,
M
bd and 1
i
or 2. We have the coverings
 
1111111
:, ,,MbdMbd


and
 
11111 11
:, ,,Mbd Mbd

.
By Proposition 10 there is a standard

12
bb
-action
on

22
,
M
bd where 1
or 2, and a regular covering
covering
 
22 2211
:, ,,
M
bd MbdMbd


.
Since these are standard actions and 2
12 1
b
bb b
it fol-
lows that 21


. This shows that
21
.
Corollary 13.

1,
is a lattice.
We will now consider maximal and minimal elements
in

1,
. Write 0
2n
where 0
is odd. Then
the maximal element in
1,
is

2,
n
if
is
odd, and

2,2
n
if
is even. Note that if
,bd

1,
, then 2n divides b and 2n
b is odd. In de-
scribing the minimal elements let 0
b be the largest odd
divisor of 0
such that

0
.. ,1gcdb
. If
is odd
or if 0n, then the minimal element in
1,
is

0
2,
nb
, otherwise the minimal element is

0
2,2
nb
.
We say an element

11
,bd is directly below
22
,bd
or that

22
,bd is directly above

11
,bd if whenever
 
112 2
,,,bdbdb d, then either

11
,,bd bd
or
 
22
,,bdb d.
Theorem 14. Let

0
011
2,
mb
and

0
022
2,
nc
be the minimal elements in
1
11
,
and
1
22
,
respectively where 1
i
or 2, and let 01
i
km
i
i
bp
and 01
i
sn
i
i
cq
be the prime decompositions. Sup-
pose one of the following holds:
1) 1
and 2
are both odd.
2) 1
and 2
are both even and 00 0mn.
3) 1
and 2
are both even and 00
0mn.
4) 1
even with 00m and 2
odd.
Then

1
11
,
is isomorphic to

1
22
,
if
and only if ks and after reordering ii
mn
for
1, ,ik.
If 1
is odd and 2
is even with 00n, then

1
11
,
is isomorphic to

1
22
,
if and only if
1
ks, after reordering ii
mn
for 1, 2,,1ik,
and 1
k
m.
Proof. We will first assume that 1
and 2
are both
odd. Suppose

11
112 2
:, ,f

 is an iso-
morphism. Now
0
1
2,
m
and

0
2
2,
n
are the maxi-
mal elements of

1
11
,
and

1
22
,
respec-
tively, and

00
12
2, 2,
mn
f
. The elements directly
below
0
1
2,
m
in
1
11
,
are

0
11
2,,,
mp
0
1
2,
m
k
p
and the elements directly below
0
2
2,
n
in
1
22
,
are
 
00
12 2
2,,,2,
nn
s
qq
. Since f
must take the elements directly below

0
1
2,
m
to the
elements directly below
0
2
2,
n
, it follows that ks
.
The elements directly above

01
,b
in
1
11
,
are listed as

00
12
0
11
112 1
12
1
1
2,,2 ,,,
2,
ii
ki
mmm m
mm
ii
ii
mmm
ki
ik
pppp
pp




Similarly the elements directly above

02
,c
are

00
12
0
11
122 2
12
1
2
2,,2 ,,,
2,.
ii
ki
nnnn
nn
ii
ii
nn n
ki
ik
qqq q
qq




By reordering we may assume
 
00
11
12
2,2,
jj
ii
mn
mmnn
ji ji
ij ij
fppq q



for 1jk
. The number of elements in
1
11
,
is

11
k
i
im
, and this equals the number of elements in
1
22
,
which is

11
k
i
in
. Let



0
11
11 111
1
11
,, ,:
2,,.
ji
j
m
mm
ji
ij
b
ppb




Similarly let



0
11
22 222
1
22
,, ,:
2,,.
ji
j
n
nn
ji
ij
c
qqc
 



It follows that

11
112 2
,,
jj
f

. Thus the
number of elements in
1
11
,
j
which is
1
ji
ij
mm
is equal to

1
ji
ij
nn
the num-
ber of elements in
1
22
,
j
. Using the equations
 
1
11
kk
ii
ij i
mn



and

11
jiji
ij ij
mmnn


,
we obtain

11,
11
i
jij j
jij
ij
n
mm
nmn


and this implies that
j
j
mn
for 1jk.
We now suppose that
0
01
2,
mb
and

0
02
2,
nc
are
the minimal elements in

1
11
,
and
1
22
,
respectively, and 01
i
km
i
i
bp
and 01
i
km
i
i
cq
are
the prime decompositions. If
  
1
111
,,b

, then
J. E. KALLIONGIS, R. OHASHI
155
0
1
2i
k
ms
i
i
bp
where 0ii
s
m . Define
  
11
112 2
:, ,f

by


0
12
1
,2 ,
i
k
ns
i
i
fb q
.
It is not hard to check that f is an isomorphism. The proof
in cases (2) - (4) is similar.
We now assume that 1
is odd and 2
is even with
n0 = 0. Since the argument is similar to the previous case,
we will sketch the proof. The elements directly below

0
1
2,
m
in

1
11
,
are
 
00
11 1
2,,,2 ,
mm
k
pp
and the elements directly below
2
1, 2
in
1
22
,
are

122 2
,2,,,2,1,
s
qq

. It follows that k = s +
1. The elements in

1
11
,
directly above

0
01
2,
mb
are
 

00
12
0
11
112 1
12
1
1
2,,2 ,,,
2,,
ii
ki
mmm m
mm
ii
ii
mmm
ki
ik
pppp
pp




and the elements directly above

02
,c
are
 

12
1
11
1222
12
1
12
1
,, ,,,
,,
ii
ki
nn
nn
ii
ii
nn
ki
ik
qqq q
qq






02
,2c
. By relabeling we may assume that

011
12
2, ,
jj
ii
mn
mm n
ji ji
ij ij
fppq q



for 11jk and


01
102
2,,2
ki
mm m
ki
ik
fppc

.
Now
 
1
11
12 1
kk
ii
ii
mn

 

, and for 1jk
Lattice of Finite Group Actions on Prism Manifolds