R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
J. P. R OSSETTI P. A. T IRAO
Compact flat manifolds with holonomy group Z
2L Z
2(II)
Rendiconti del Seminario Matematico della Università di Padova, tome 101 (1999), p. 99-136
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with Holonomy Group Z2 ~ Z2 (II) (*).
J. P. ROSSETTI - P. A.
TIRAO (**)
ABSTRACT - In this paper we
give
acomplete
classification of compact flat mani- folds withholonomy
groupZ2 EÐ Z2,
with the property that theholonomy
rep- resentationdecomposes
as a direct sum ofindecomposable representations
ofZ-rank
equal
to 1 or 2. We exhibitexplicit
realizations of all the manifolds clas-sified,
computing
the firstintegral homology
groups.Finally,
we compare the results obtained with the known results in low dimensions.Introduction.
It follows from Bieberbach’s work that
classifying compact
flat mani-folds
(cfm’s),
up to affineequivalence,
isequivalent
toclassifying
theirfundamental groups, up to
isomorphism.
In 1965
Charlap (see [3])
gave ageneral approach
to the classifica-tion,
andapplied
it toclassify
allZp-manifolds i.e.,
all cfm’s withcyclic holonomy
group ofprime
order. To thisend,
he used Reiner’s results([10])
on the classification ofintegral representations
of the groupZ~ .
In
[4],
P. Cobb constructed an infinitefamily
of cfm’s withholonomy
and first Betti number zero. This
family
wasenlarged
in[12],
where new infinite families of such manifolds were constructed(*)
Supported partially by
FaMAF and CONICOR.(**) Indirizzo
delgli
AA.: FaMAF, Universidad Nacional de C6rdoba, Ciudad Universitaria, 5000 C6rdoba,Argentina;
e-mail: [email protected];[email protected]
by allowing
certainintegral representations
of nondiagonal type
in the
holonomy representation.
In this paper we shall
give
acomplete
classification ofcompact
flatmanifolds with
holonomy
group with theproperty
that theholonomy representation
is a direct sum ofZ2 ® Z2-indecomposable
rep- resentations of Z-rankequal
to 1 or 2. The cfm’s in this class which have first Betti number zero areexactly
those considered in[12].
In order to obtain this classification we shall
develop Charlap’s scheme, showing
that the basicdifficulty
in this case is to decide when twospecial
classes determine the same flat manifold. This mainstep
iscarried out in Section 5.
Moreover,
we shall exhibitexplicit
realizations of the fundamental groups assubgroups
of We summarize the full classification and the firstintegral homology
groups in the table of Section 7.Finally
wespecialize
our results for lowdimensions, comparing
withthe known classification in
[1]
in the cases of dimension 3 and 4. In par-ticular,
we show that the method constructs allZ2OZ2-compact
flatmanifolds of dimension
3,
and 21 out of theexisting 26,
in dimension 4.For dimensions n = 5 and n =
6,
wegive
the total number ofZ2 EÐZ2-
manifolds constructed in the paper. The
family
westudy
in this paper in- cludes in dimension 5(see
Remark7.4),
the twoisospectral
non homeo-morphic
cfm’s constructed in[5,
p.496].
We note that a full classification of allZ2 ® Z2-manifolds
isyet
unknown for dimensions n ~ 5 .1. - Preliminaries.
Let M be an n-dimensional
compact
flat manifold with fundamental group T. Then M=Rn/T,
F is torsion-freeand, by
Bieberbach first theo- rem, one has a short exact sequencewhere A is free abelian of rank n and 0 is a finite group the
holonomy
group of M. This sequence induces an action of 0 on ll that determines
an
integral representation
of rank n of 0. Thus A becomes aZ(Q)-mod- ule,
which moreover is a free abelian group of finite rank. From now on,by
a0-module,
we will mean aZ( ø )-module which,
as aZ-module,
is freeand of finite rank.
As indicated
by Charlap
in[2],
the classification of allcompact
flat manifolds with
holonomy group (P
can be carried outby
thefollowing steps:
1)
Find all faithful Q-modules A.2)
Find all extensions of 0by A, i.e., compute H2 ( ~ , A).
3)
Determine which of these extensions are torsion-free.4)
Determine which of these extensions areisomorphic
to eachother.
For each
subgroup
K of 0 the inclusion i : induces a restric- tionhomomorphism
resK :H2 ( ~ ; ll ) --~ H2 (K; A).
DEFINITION. A class
A)
isspecial
if for anycyclic
sub-group of
(P, K,
ofprime order,
one has 0.Step 3)
reduces to the determination of thespecial
classesby
virtueof the
following
result.LEMMA 1.1
[3,
p.22].
Let A be a 0-module. The extensionof (P by
Acorresponding
to a EH2 ( ~ ; A)
istorsion-free if
andonly if
a isspecial.
We now state some definitions and a main result in
[3].
DEFINITION. Let A and L1 be Q-modules. A semi-linear map from A to L1 is a
pair ( f , A)
wheref :
is a grouphomomorphism,
A Eand
The Q-modules A and L1 are
semi-equivaLent if f is
a groupisomorphism.
If A = I then A and L1 are
equivalent
viaf.
Let
E(Q)
be thecategory
whoseobjects
are thespecial pointed
W-modules, i.e.,
thepairs ( , a)
where A is a faithful Q-module and a is aspecial
class in~l )
and whosemorphisms
are thepointed
semi-lin-ear maps. That
is, ( f , A)
is a semilinear map from(A, a)
to(A,
suchthat f * (a)
= A *(~8),
where A *(~i)(a~, i) _ Ai)
for any(a, i)
Ei 0 x 0.THEOREM 1.2
[3,
p.20].
There is abijection
between the isomor-phism
classesof
thecategory ~( ~ )
and connectionpreserving diffeo-
morphism
classesof compact flat manifolds
withholonomy
group CP.We consider the
subcategory
of~(~),
whoseobjects
are thepointed
0-modules whichdecompose
as a direct sum of submodules of Z- rank less than orequal
2 and whosemorphisms
are thepointed
semi-lin-ear maps. It is
straightforward
to see that a restricted version of the pre- vious theorem holds for thecategory
COROLLARY 1.3. There is a
bijection
between theisomorphism
classes
of the category
and connectionpreserving diffeomorphism
classes
of compact flat manifolds
withholonomy
such that theholonomy representation decomposes
as a direct sunzof indecompos-
able summands
of
rank less than orequal
two.DEFINITION. Two
integral representations p
andO’
of (P are semi-equivalent
if there exists a unimodular matrix P and anautomorphism
Aof 0 such that
Let L1 be a 0-module and We denote
by A(4 )
the Q- module which has L1 asunderlying
abelian group, with the action of 0 definedby
and L1 are
semi-equivalent
via( f, A), then f : A - A(D)
is a Q-isomorphism.
Inparticular
if(A, a)
and(4 , fl)
areisomorphic special pointed
0-modules in1F( W),
then ~l and L1 aresemi-equivalent.
The asso-ciated
representations p
andQ’
thensatisfy
for some unimodular
P,
and
Q’
aresemi-equivalent.
Therefore,
in order to determine theisomorphism
classes ofIi~((P),
itwill suffice to
classify
theintegral representations involved,
up tosemi-equivalence.
From now on we will
identify integral representations
of 0 with0-modules.
2. - Classification of
$representations.
Every integral representation Q
of a finite group Gdecomposes
as adirect sum of
indecomposable subrepresentations,
but ingeneral,
the in-decomposable
summands are notuniquely
determinedby 9 (see
for in-stance
[11]).
However in the context of this paper, theindecomposable
summands of an
integral representation
will beuniquely
determined up to order andequivalence.
By an representation
we will understand a faithfulintegral
repre- sentationof Z2
thatdecomposes
into a direct sum ofindecomposable representations
of rank less than orequal
2. In this section we willgive
aparametrization
of thesemi-equivalence
classes of~ representations.
It will be convenient to
identify
arepresentation e of Z2
with thethree
integral
matrices inQ(Z2 ?~2 " {0}),
which we will denoteby B1, B2
andWe
begin by recalling
all theindecomposable integral representa-
tions of of rank one and
two,
up toequivalence.
LEMMA 2.1. A
complete
setof representatives of equivalence
class-es
of indecomposable representations of Z2 EÐ Z2 of
rank less thccn orequal
2 isgiven by
-PROOF. It is known that in the
only indecomposable representation
of rank
2 of Z2,
thegenerator
actsby
J(see
for instance[10]).
Given an in-decomposable representation
of of rank 2 we may assume thatone of the
Bi’s
is J. As the otherBi’s
commute with J andsatisfy B 2
= Ithe
only possibilities
for theseBi’s
are ± I and ± J. We notice that J and - J areconjugate by 0 -1 . Thus,
forexample,
therepresentation - I, J, - J
isequivalent to - I, - J,
J andI, J,
J isequivalent
toI, - J,
- J. It is now easy to conclude that
give
a com-plete
set ofrepresentatives
for theequivalence
classes ofindecompos-
able
representations
of rank 2.If Q
isan Grepresentation,
then o isequivalent
towhere the
non-negative integers r,
andl~i
areuniquely
determinedby
o(see
Remark2.4).
Conversely, given non-negative integers r, mi, Li, ki (1 ~
i ~3),
we as-sociate to them the
representation
of rankn = r + m + 2 L + 2 k,
definedby (1),
where m= I mi, 1 = ¿ lie
and k= I ki .
i=l i i=l i i=l i
We now associate to a
given Q
e 1F thetriple
where
mi, li and ki
are as in(1). Conversely,
for eachtriple
as in(2)
and n a nz + 2 1 + 2 k we associate therepresentation
constructed as in(1),
with r=n-m-2L-2k.
A
permutation
of thetriple (2)
is atriple
of the formwhere a e
S3.
PROPOSITION 2.2. There is a
bijective correspondence
between se-mi-equivalence
classesof ’,F’-representations of
rank n andtriples
in theset
~ + 2~ + 2~ ~ ~;
and k > 0 or for at least two indicesi ~
up to
permutation.
PROOF. Given an element in the
set,
we associate theF-representa-
tion constructed in
(1), by taking
r =n - m - 2l - 2k.Choosing only
onetriple
in each class(of
the set up topermutation),
we built thefamily R
ofthese
representations.
Notice that the last condition in the set ensuresthat the
representations of S
are indeed faithful.It is clear that every
Grepresentation
issemi-equivalent
to one inB.
On the other hand if ~o ,
g’ E B
aresemi-equivalent,
we wish to showthat Q
=Q’.
Let P be a unimodular n x n matrix and 0 anautomorphism
of such that
Po(g ) P -1= o ’ ( ~ (g ) ) , VgeZ2EÐZ2. Then,
thereexists a E
,S3
such that for 1 ~ i ~ 3.By
theuniqueness
ofthe
parameters
mentionedabove,
it follows that r = r’ and(mi, Li, kj )
==
(~Q~i~,
for 1 ~ i ; 3. Thusup to
permutation by
Q, and so Q =o ’
as asserted.NOTATION 2.3. We recall
that F
is thefamily
ofrepresentatives
ofsemi-equivalence
classes of~ representations,
constructed in theproof
of
Proposition
2.2.We denote
by Si
1 thesubfamily of 3
ofrepresentations having
thethree elements in the
triple
different. We also denoteby 3 2 (resp. ~3 )
the
subfamily of S
ofrepresentations having
2(resp. 3)
elements in thetriple equal.
Notice
that 9,
is thedisjoint
unionB1
UB2
UR3-
REMARK 2.4. If A is a subset of
(Bi , B2 )
thequotient
is an invariant of the
equivalence class, [ o ],
of o.By using
these quo- tients one can show that theintegers r,
1 ~ i ~3,
in(1)
areuniquely
determinedby [ o ].
Forinstance,
then one hasAlso
Hence Li
can bedetermined,
for alli,
and r is also determined.On the other
hand,
it wasrecently proved (see [8])
that the full Krull- Schmidt theorem holds forintegral representations
of The au-thors wish to thank W.
Plesken,
M. I.Platzech,
L.Levy
and L.Klingler
for useful
correspondence
whichhelped
toclarify
thevalidity
of theKrull-Schmidt
property
in thepresent
case.3. - Some group
cohomology.
In this section we shall
determine,
for eachZ2 E9 Z2-module
~l inIT,
the
cohomology
group~l ), by exhibiting
anexplicit
set ofgenerators.
This
computation
can bepossibly
doneby
othermethods,
like thosein
[13],
but we shall take asimple
mindedapproach
whichonly
uses thebasic definitions.
Since
cohomology
is additive and two modulescorresponding
to se-mi-equivalent representations
haveisomorphic cohomology
groups, it will suffice to consider four cases. From now on we shall writeZ2
=- (B1, B2 ),
andCase
I): H2«B1, B2); A),
where ~l is of rank one and the action is trivial.Case
2): H2 ((B1, B2 ~; ~l ),
where ~l is of rank one andB1
actstrivially
while
B2
actsby - I .
Case
3): H2«B1, B2); A),
where ll is of rank 2 andB1
actstrivially,
while
B2
actsby
J =1 0
.Case
4): H 2((Bl, B2); ~l ),
where ll is of ranktwo, B1
actsby - I
andB2
actsby
J.In all cases we first determine the
cocycles C 2 ( (B1, B2); A), i.e.,
thefunctions
B2 ~ X (Bl , B2 ~ ~ ~l
such that ah = 0. Recall thatBy normalizing
thecocycle h,
we may assume thath(x, I)
=h(I, x)
=0,
for all x E(B1, B2 ~.
Also,
it will be useful to recall that the coboundaries are the functionswhere
f : (B1, B2)
is any function.Thus, evaluating (3)
for x , y ,B2 , B3 ~
we have a linear sys- tem of 27equations
and 91 ~ k ,
l ~3},
in cases1)
and2);
while in cases3)
and4)
we have 54equations
and 18unknowns.
We will make use of the
following
notation. With p, q and r we denote any elementB2 , B3 ~ acting
as I and with iand j
any element in the same setacting
as - I. We will also abbreviateh(Bk , Bl) by ( k , l)
and8h(Bk, Bb Bm) by (k, l, m).
Case
1 ). -
The full set ofequations
is summarized in thefollowing
table
Observations.
Equations
5 follow from 2 and3,
whileequations 2,
3 and 4 areequivalent
toequations 2, 3
and 4’ :(p, q)
=(q, p). Equations
3follow from 2 and 4’.
By
thesymmetry of r and p
in 2 there are threeequations
left. The samehappens
in 4’.Thus the
system
ah =0,
isequivalent
toand so every
cocycle h
has thefollowing general form,
for some a,
y eZ.
Let ha (resp. h~
and be thecocycle
obtainedby letting
a =1, (resp. a = y = 0
andy = 1, a = ~8 = 0).
ThenH2«B1, B2); A)
=( [ ha ], [ h~ ], [ hY ] ). Finally
to determine thecohomology
group, we write down the
coboundary 8f,
whereB2 ~ ~ Z
is any function such thatf (I )
= 0. If welet tk = f (Bk )
we haveIt follows that
ha + 0, h~ -~- 0,
0 and furthermoreha - 0, ha - 0,
0 andfinally 2 ha -~- 2hf3 -- 2 hY ---
0. HenceCase
2). -
As in Case1),
thesystem
of 27equations
can be reduced. It is sufficient to consideronly
thetriples ( i , i , i ), (i, i , j ), ( i , j , i )
and(~ro ,
p,i ),
from which one can conclude that the set oflinearly indepen-
dent
equations
is:Thus,
in this case, everycocycle h
has the formfor some
a, f3 E Z.
Let be as before. It is not difficult to see that
ha --- h~, ha -~-
0and
2 ha --- 0.
ThusIn cases
3)
and4)
wejust give
thegeneral
form of acocycle h,
and therelations satisfied
by
thegenerators
ofC2«(B1, B2); A).
Recall thath:(B1, B2)
xB2 ~ -~ ~ .
We shall write the second coordinate of h below the first one.Case
3).
for some a,
B, y, 6
and E E Z.The
generators satisfy hf3 --- hY --- 0, ha - h, - h6 --,-
0 and2 ha ---
0. Thuswe have
Case
4).
for some a,
f3,
y andIt is
straightforward
toverify
thathY --- hd --- 0,
and thus4. - Restriction functions.
To carry out the third
step
in the classification scheme outlined in Section1,
in thelight
of Lemma1.1,
it is necessary toinvestigate
the re-striction functions to determine the
special
classes in thecohomology
groups.
Since resK:
H2«B1, B2); ~l ) ~ H2 (K; ~l )
and anycyclic subgroup
ofB2 )
isisomorphic
toZ2
we need to determine the groupsH2 (Z2 ; ~l )
for the threeindecomposable Z2-modules, namely,
those modules for which the action of thegenerator
ofZ2
isgiven respectively by (1), ( -1)
and
(i 10).
In the first case we have that H Z2;)
=::Z2,
with
generator given by
whereas in the second and third cases we have
H2 (Z2 ; ~l )
= 0.We next
study
the restriction functions for each of the cases in Sec- tion 3.Summing
upH2 ((Bi ); ~l )
=0,
thus weonly
need to consider the case when K =(Bl ).
We have
Notice that defined in
(4)).
ThusCase
3). -
If i = 2 or i =3, H2 (~Bi ~; ll )
=0,
thus weonly
need to con-sider the case We have
H2«B1); ~l )
=H2«B1); (e1»
E9E9
H2 ( (B2 )i ( e2 ))
=( [ fi ] )
E9( f f2 ~ ) ~ with fi = f By
thisisomorphism
the ele-ment
[ fi ]
+corresponds
to[h]
eH2 ( (Bl ); ~1 )
whereOn the other hand
Notice that
Hence,
it follows thatwhere is the
homomorphism
definedby d ( 1 ) _ - (1, 1).
Case
4). -
SinceH2«B1, B2); A)
=0,
there isnothing
to be done inthis case.
5. -
Isomorphism
classes inIt follows from the discussions in Section 1 that we can view the ob-
jects
in~(Z2 EÐ Z2)
as the set ofspecial
classes inH2 (Z2 ~l )
where~l is a
Z2 ® Z2-module
as above.If a
and f3
are in~(Z2
where a EH2 (Z2 ~l )
and~3
Eif there exists a semi-linear map
( f , A )
suchIn this case, ll and ~l ’ are
semi-equivalent,
thereforeeach
isomorphism
class of is contained inA),
for some module in
S.
Moreover if a andf3
are inA) and a - fl,
then the associatedrepresentation Q
satisfies =o(A(g) ),
and
consequently for Q in F1,
A must be theidentity automorphism.
For each
representation Q
ER,
we consider the matricesB1, B2
andB3 acting
on ~l =( e1,
... ,en ~.
Thus ~1 is the direct sum ofindecomposable submodules,
of the ten classes listed in Lemma 2.1. Let ~l ibe,
fori = 0 , 1,
... ,9,
the submodules that are direct sum ofindecomposable
submodules all of them
equivalent,
in the order considered in Section 2.Thus ll o has rank r and is the sum of trivial
submodules;
A1 has
rank ml ,~12
has rankm,2,...,~19
hasrank2k3-
It follows that
It will be convenient to make use of the
following
notation. Given a EEH2«B1, B2 )); ~l ),
we will write a =(vo,
vl ,... , vs )
with vo E(Z2EÐz2)r,
Vl E
Z21,
... , v6 EZ2 l’
In the coordinates of vo we willidentify ( 1, 0 )
withhi, ( o, 1 )
withh2
and( 1, 1 )
withh3 (Section 4,
Case1)). Finally
if 6 Eie (0, 1 )
we will set3
=( ~ , 0 , ... , 0 )
EZ2
where t could beequal
tom1, m2 , M3,
ll, l2
or19.
LEMMA 5.1. For every a E
H 2((Bl, B2 ); ~l ),
a =(vo,
vl ,... , vs ),
let~
i = 1if 0, ði
i = 0 otherwise. InH2«B1, B2 ); A)
thefollowing equiv-
alences hold:
if
the non zero coordinates in vo ,if
any ,are all
equal
tohj;
if
there are at least twodifferent hi’ s
in vo ,for
1-i-3 .PROOF. To prove the lemma we will construct a linear
isomorphism f :
such0 ,
... ,0, d 1,
... ,d 6)
= a in the first case andin the second.
We will
define f in
each submoduleA j .
We start with~l 2
and~3.
Recall that 1 =
~er+ 1, ~ ~ ~ ~ er+ ml ~·
Ifv1= 0
define 1= I. Ifv, =
(v 1,
... , 0 with11, let jo
be thefirst j
such that~0.
We define
for r + 1 ~ i ~ r + m1 , m1:
v ~ # 0 1.
Notice that(/ ~i)*(0,
9 ... ,0 , 1, 0 ,
9 ... ,0)
= v1, where 1 is inthe jo-th position. By
where is the
transposition interchang- ing ei and ej,
we have that0 ,
... ,0 )
= VI.Finally
we noticethat
1 is
a linearisomorphism.
In a similar way we definefA2 and f A3.
Consider now the submodules
A 4 A 5
andll 6.
If v4 = 0 define == I.
(v 1,
... ,1}, let jo
be thefirst j
such that0. We define
for
r + m + 1 ~ i ~ r + m + 2 ll , ~~0}.
As be-fore,
it follows that( f ~~ 4 )* ( 1, 0, ... , 0)
= v4 .Again f ~~ 4 : A 4 - A 4
is a linearisomorphism.
For A 5
and the definition off
isanalogous.
On the submodules
A 7,
andA 9
we definef
to be theidentity.
We now define This is the most
complicated
situation. If vo = 0o = I. If vo #
0,
wedistinguish
two casesdepending
on whether thenonzero coordinates of vo are all
equal,
or whether there are at least two different nonzero coordinates. In the firstsituation, defining fl 0 in
away similar
1 it
follows that( f ~~ o )* (hj , 0,
... ,0)
= vo . In the sec-ond
situation, if vo = ( v 1, ... , v r ),
wherehl , h2 , let j
be thefirst
index j
such thatv j ~
0 and letj2
be the firstindex j greater than j1
such that
v j ~
0and v j ~
v jl . Thus vo =(0~ ... ~ 0,
+ 1,... , v j2 , v j2 + 1, ... ,
v,),
with vj, =hil’
v j2 =h22 and v j = 0 or v j
=hil j j2.
Thus we defineHence
V o z 2, ~2 ),~ ( hil , h22 , 0 , ... , 0)
= vo . We noticethat/
is alinear
isomorphism.
To finish the
proof
weonly
need to show that( hil , hi2’ 0,
... ,0)
is
equivalent
to( hl , h2 , 0 , ... , 0 ).
By applying
if necessary, we may assume thatil i2 . Thus,
wehave to consider the cases
_ ( h1,
or( hil , h22 ) _ (h2, By taking
in the first and second cases
respectively,
it turns out that a *(hi, h2 , 0 ,
... ,0 )
=(hil’ hi2’ 0 ,
... ,0 ).
Both a’s are linear isomor-phisms. Finally by taking
o =f
o z 1, jl 0 í 2, j2 o Q we obtain therequired f.
It is now convenient to characterize the
special
classes inH 2(~Bl, B2); A).
LEMMA 5.2.
(a)
A classof
theform (o,
10,
6 11... , 3 6)
isspecial
if
andonly
1;i~3. r
( b )
For each1 ~ j ~ 3, (hj, 0 , ... , 0, d 1, ... , d 6)
isspecial if
andonly if6j+ ð j + 3 ~
1.~
( c )
The classes(hl , h2 , 0 , ... , 0, d1,
... ,d6)
arealways special.
PROOF. The lemma follows
directly
from the results in sections 4 and 3. mIn the
sequel
we will determine theequivalences
among thesespecial
classes.
LEMMA 5.3. In
H2«B1, B2); A)
we have thefollowing equiva-
lences :
PROOF.
Let f:
A -A be the linearisomorphism
definedby
It is
straightforward
toverify
thatf
satisfiesf * (vo, 1, T2, d 3, 0, d 5,
==
(vo, 1, 3-2~ T39 19 7~ ?e),
thusproving
the first relation. Theremaining
two follow in a
completely
similar way.LEMMA 5.4. In
H2«B1, B2); A)
thefollowing equivalences
hold:PROOF.
(a)
Letf
be the linearisomorphism
definedby
To understand the effect of
applying f *
we will restrict our attention toWe now recall the
general
form of thecocycles
of Case3)
in Section 3 and the facts that0, ha --- hE -~ 0,
whenB1
acts as I andB2
andB3
actas J. Since on
~er + m + 2 al + ~ ~ er + m + 2l1 + 2 ~ B1
acts as J andB2
as Iwhile B1
and
B2
actby
Jon ~ er + ~rt + 2 h + 2 L2 + ~ ~ er + m + 2 l1 + 2 L~ + 2 )~
we mustinterpret
properly
the table in Case3).
HenceThe
proofs
of( b )
and(c)
areanalogous
to theproof
of(a).
The
proof
of(d)
can be obtained from(a)
and a statement as( b )
withh2
in the secondcoordinate,
inplace
of the first.Thus, by taking
(d)
follows.By using
the above results we can nowgive
an upper bound on the number ofequivalence
classes ofspecial
classes inH2 ( (B1, B2 ); A).
We
begin by considering
the classes with vo = 0. Thusd
+d i + 3 % 1,
for 1 ~ i ~ 3 andby
Lemma 5.3 it suffices to consider the classes with 3 ji + ð i + 3
=1,
for 1 ~ i ~ 3.Therefore,
if vo =0,
there are at most23
= 8equivalence classes, corresponding
to allpossible
choices ofpairs
(54), (ð 2, d5)
and(d3, d6).
In the cases when vo =
0 ,
... ,0)
for some1 ~ j ~ 3,
we willonly
analize the
case j
=1,
since theremaining
two are similar. Thusd 1 + d 4 ~ 1,
and asbefore,
it suffices to consider those classes with 3 1 + 3 4 = 1.By
Lemma 5.4 we may assumed 5 = ~ s = 0.
Hence there areat most 8
equivalence
classes with vo =0 ,
... ,0),
andconsequently,
if we include the cases
with j
= 2or j
=3,
there are at most 3 x 8 = 24 classes of thistype.
Finally
if vo =( hl , h2 , 0, ... , 0 )
we may assumeby
Lemma 5.4 thatd 4 = a 5 = d 6 = 0,
hence there are most 8equivalence
classes of thistype.
If we take into account all three
types,
we see that there are at most 40equivalence
classes ofspecial
classes inH((Bi , B2 ); A).
We see next that
if e
is in the subfamiliesB 2
orB 3 of S (see 2.3),
thereare some new
equivalences
among thespecial
classes describedabove.
Recall that two classes a and
~3
inH2«B1, B2 ~; A)
areequivalent
iff *
a = A* ~3,
with( f, A)
a semi-linear map. At thebeginning
of the sec-tion we
pointed
out that forrepresentations in Si
theautomorphism
Ashould be the
identity.
Forrepresentations
in~2
either A = I or A is theautomorphism
thatpermutes
the two elementshaving
the same associat-ed
triple (see
Section2)
andfixing
the third non trivial element. In thiscase
f
can be chosen tointerchange suitably
the ~l2 s
in order to make( f, A)
a semi-linear map. On the otherhand,
forrepresentations
inB3
the
automorphism
A can bearbitrary
andf
should be chosen so that( f, A)
is semi-linear.EXAMPLE. If A
permutes B1
andB2
then wechoose f mapping
iso-morphically A1 - A2, A4 - A5, A7 - A8, f being
theidentity
onA0, A3,
Notice that in this case,
pairwise, ll 1
andll 2 , ll 4
and~l 5 ,
andA 7
and are of the same rank. Indeed wedefine f ( ei )
= ei + ml for ei EWhen
considering special
classes inH2«B1, B2 ~; A),
with ~l inð2
or83
thepossibility
ofchoosing
A to be a non trivialautomorphism
pro- duces moreequivalences.
These will be listed in Lemmas 5.5 and 5.6.REMARK. Let
B1, B2, B3
be arepresentation
inR2 having
the firstand third associated
triples equal
and the second one different(see Proposition 2.2).
Then and since m2 ~ m3 we have ml == m2 = m3 . Thus the
representation given by B1, B3 , B2
issemi-equivalent
to
B1, B2 , B3
and has the first and second associatedtriples equal.
There-fore we may assume that the
family ð 2
is thedisjoint
union of two sub-families of
representations,
thosehaving
the first and secondtriples equal (~2,1)
and those with the second and thirdtriples equal ( ~2, 2 ).
Thus
ð2 =
1 UU2, 2·
PROOF.
(a), (b)
and(c)
followby choosing ( f, A)
with Apermuting B1
andB2
andf
as in theexample.
(d)
followsby taking ( f , A)
asbefore, except
thatf ( e1 )
= e2 andf(e2)
= el.The
proofs
of( a ’ )-( d ’ )
are similar.LEMMA 5.6. In
H2«B1, B2); A),
with A ER3,
we have:(a)
Two classesof
theform (0,
... ,0, ~ 1, a 2, a 3, 1 - a 1, 1- d 2, 1- d 3) r
having
the same sum 6 1+ 6 2
+ 6 3 areequivalent.
(b) Every
classof
theform (hj, 0 ,
... ,0, d 1, ~, ~ 3, a 4, a 5, a 6) with j
=2,
3 isequivalent
to oneof
theform (j’L1, o , ..., 0, a 1,
3 2,d3, d4, 0, 0).
r
(d)
Two classesof
theform (hl , h2 , 0 , ... , 0, ~, d 2, ~ 3, 0, 0, 0) having r
the same sum 6 1 +
ð 2
+ð 3
areequivalent.
PROOF.
(a)
If 6 1 +ð 2
+ð 3 equals
0 or3,
there isnothing
toprove.
Given two classes a and
~3 with ð 1 + ð 2 + ð 3 = 1,
suppose d i = 1 in aand
3 j
= 1 inB. By taking
A theautomorphism
thatpermutes Bi and Bj
and
f
the linearisomorphism
that mapsll i + 3 H ~l ~ + 3
andone checks
(b) By taking
A to be theautomorphism
thatpermutes B1 and Bj
and
choosing
asuitable f,
we see that thegiven
class isequivalent
to oneof the form
(hl, 0,
... ,0, d 1, ~, ~ 3, ~ 4, d 5, a s).
Thusby
Lemma 5.4(a), ( b )
follows.~
(c)
It followsby choosing ( f, A),
where Apermutes B1
andB2 and f
is as in the
example
before Lemma 5.5.(d) Having
in mind that the vector vo =( h1, h2 , 0 , ... , 0 )
is trans-formed into A * vo
by
any semi-linear map( f, A),
it ispossible
to obtainall the classes with the same sum 3 1 +
d 2 + ð 3,
andhaving
A* vo
in thefirst r coordinates.
Therefore, by
Lemma5.1, (d)
follows.Putting together
the results in Lemmas5.1-5.6,
wegive
inL2
and(~3,
acomplete
set ofrepresentatives
of theequivalence
classes ofspecial
classes
corresponding
torepresentations in F1,
9ð 2
andF3 respectively.
Later we shall prove that in fact two classes in the same set are not
equivalent (except
for theequivalences
stated on theright
of eachline).
,There are 8 classes of each
type,
hence at most 40 classes for a repre- sentation in thefamily F1.
two classes
having
the samesum ,
~1+62)
areequivalent;
two classes
having
the samesum~ 5i+52
areequivalent;
two classes
having
the samesum
ð 1 + ð 2
areequivalent.
Hence there are at most 26 classes for a
representation
in thefamily ð2.
·two classes
having
the samesum, (51+~2+~3
areequivalent;
two classes
having
the samesum , d 2 + ð 3,
areequivalent.
two classes
having
the same sum,ð 1 + ð 2 + ð 3,
areequivalent.
Hence there are at most 14 classes for a
representation
in thefamily F3.
Now we will show that within each
set,
thespecial
classes are in-equivalent.
Forthis,
we notice two facts about aQ-isomorphism f
from Ato A:
(*) This set is built up
considering representations
in~2,
1. There is an anal-ogous set for
representations
in82,2-
·5.7. The action
of Q
on A isQ-diagonalizable. Thus ()
isQ-equivalent
to a direct sum
of Q-characters. Since f is
ag-morphism, if o
acts onA 0 by
a certain
character,
then Q actsby
the same character.5.8.
By considering
Q as anintegral representation (not
as aQ-repre- sentation)
thefollowing
condition can be obtained.LEMMA 5.8.
If f: O-automorphisrrt,
thenfor
any subset Sof Z2 EÐ Z2, f
induces anautomorphism of
abelian groupsThe
proof
of this lemma is not difficult and we shall omit it.In order to show how one can use the conditions
imposed
on aQ-auto- morphism
wegive
an illustrativeexample.
EXAMPLE.
Let Q
be therepresentation
ofZ2 ED Z2
onZJ given by
where J
= (0 1 and let f be
a g-automorphism
of Z3. By
5.7 it follows
that
f (e2 + e3) E e1, e2 + e3>. Moreover, by
Lemma5.8,
E ~ 2 el ,
e2 +e ),
since the class of 61 inKer (B2 - I)/Im (B2
+I)
does notvanish,
while the class of e2 + e3 vanishes in thisquotient.
SinceZei )
has ordertwo, i.e.,
2 a = 0 Va EH2 (x o ; Zel ) (see
Section3),
itfollows
that f * ( 0 , ð)
=( 0 , ~ ’ ).
Theimportant thing
here is that from( 0 , ~ ),
it isimpossible
to obtain( 1, ~ ’ ),
forð,
3 ’ EZe2
Inother
words,
the classes(1, 3)
and(0, d ’ )
are notequivalent.
LEMMA 5.9. Two
special
classes in~1
which areof
twodifferent types
are notequivalent.
PROOF. The
only possible
semi-linear maps in thisfamily
are of theform
( f , I).
SinceI * ( hi ) = hi , Vi,
then it is notpossible
tochange
thetype
of thespecial
class.LEMMA 5.10. The classes in
~1
areinequivalent.
PROOF. The classes
(hl , h2 , 0 , ... , 0, d1, d2, d3, 0, 0, 0)
are notequiv-
alent to each other because of the restrictions in 5.7.
For the classes
(hl , 0 ,
... ,0, d1, d 2, d 3, 1 - d 1, 0, 0),
we can say that two of them are notequivalent
whenthey
have distinctd 2
or6 3 by
5.7.When
they
have the sameð 2
and thesame 6 3
but different 6 1they
areinequivalent by
5.8. Theproofs
for thetypes
withh2
orh3 in place
ofh,
are similar.
Finally,
for the classes(0, ... , 0, d2, d 3,
1-d 1, 1 - d 2, 1 - d 3)
theinequivalence
follows from the fact that two classeshaving
differentd
ifor some i are
inequivalent by
5.8.5.11. In the cases of the sets
62
and~3,
theinequivalence
of all spe-cial classes listed is
proved by entirely
similararguments.
REMARK 5.12. For a
representation Q
withparameters
r ~2,
mi ~ 1and li >
1 for 1 i 3 there existexactly:
40
inequivalent special
classesif Q
isin F1,
26
inequivalent special
classesif Q
is in82
and14
inequivalent special
classesif Q
is in83.
The lowest ranks in which such a Q exists are: 11
for 83,12 for 82
and14 for
6. -
Explicit
realization andintegral homology.
In this section we will
give
anexplicit
realization of the Bieberbach groups rcorresponding
to thespecial
classes classified in Section5,
assubgroups
of isometriesof Rn, i.e., F c--> 0(n) x R’.
To do this webegin by considering
arepresentation in 8
andB1, B2
the matrices associat- ed to Q. We note thatB1
andB2
are in theorthogonal
group0(n).
~If v E Rn ,
letLv
denote the translationby
v. Often we willidentify A
with
LA
in what follows. For each one of thespecial
classes a EE
B2 ); ll )
listed in theprevious section,
we will determinebl
andb2
in Rn in such a way that thesubgroup
ofis an extension of
B2) by A,
with extension class a.We recall that
given
I’ an extension of(Bl , B2 ) by Il,
thecorrespond- ing
extension class is determined as follows. Fix a sectionB2 ) -j 1,
and define the function
f:(B1’ B2)
x(Bl, B2) ~~l by f(X, Y) =
=
s(X) s(Y)
The extension class of r isgiven by [ f ]
EE H2( (B1, B2); A).
Hereafter we
pick
thefollowing
section:Notice
that,
with thisdefinition,
thecorresponding f
satisfiesf(x, I)
==f(I, x)
= 0 for anyB2 ).
Since the
cohomology
isadditive,
it suffices to consider the four inde-composable
cases, as in Section 3. So we will consider the functionf: (B1, B2 ~
xB2 ~ -~ ~l ’,
where ~l ’ isindecomposable
of rank 1 or 2.In each case we will calculate the values
of f
in(x , y )
for x , y EB2 ,
and we will list them in atable,
as in Section 3. It will be easy tofigure
out the class[ f ]
EH 2((Bl, B2); ~l ’ ),
for different choices ofb1
andb2.
We observe that
f (Bl , B2 ) = 0
in all cases, because= s(B1 ) s(B2 ).
Ingeneral f(Bi, Bi)
=s(Bi) s(Bi)
= + The othervalues of
f
are obtainedsimilarly.
Case
1). -
Recall that the action ofB2 ~
is trivial. The valuesof f
are
Notice that
bl
andb2
must be We concludethat,
(Here h1
andh2
are as in Section4.)
Case
2). -
Recall thatB1
=1, B2
=B3
= -1. The valuesof f are
We
get that,
The cases
when B2 = 1, B1 = B3 = -1
andB = 1, Bl = 82 = -1
aresimilar.
Case