J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
U
LRICH
H
ALBRITTER
M
ICHAEL
E. P
OHST
On lattice bases with special properties
Journal de Théorie des Nombres de Bordeaux, tome
12, n
o2 (2000),
p. 437-453
<http://www.numdam.org/item?id=JTNB_2000__12_2_437_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On lattice bases
with
special
properties
par ULRICH HALBRITTER et MICHAEL E. POHST
Dedicated to Jacques Martinet on his 60th birthday
RÉSUMÉ. Nous introduisons ici des réseaux
multiplicatifs
de(R>0)r
et déterminons des réunions finies desimplexes
conven-ables comme domaines fondamentaux de sous-réseaux d’indices
finis. Nous définissons pour cela des bases
cycliques
positivesde réseaux arbitraires. Nous utilisons ces bases pour calculer
les cônes de Shintani dans des corps totalement réels de
nom-bres
algébriques.
Nous nous intéressonsplus paxticulierement
auxréseaux en dimensions deux et trois
correspondants
à des corpscubiques
ou quartiques.ABSTRACT. In this paper we introduce
multiplicative
lattices in(R>0)r
and determine finite unions of suitablesimplices
asfunda-mental domains for sublattices of finite index. For this we define
cyclic
non-negative bases inarbitrary
lattices. These bases arethen used to calculate Shintani cones in
totally
realalgebraic
num-ber fields. We
mainly
concentrate our considerations to latticesin two and three dimensions
corresponding
to cubic and quarticfields.
1. Introduction
Since the
guiding
principles
for this article areadopted
fromalgebraic
numbertheory
and since this is thetopic,
where our ideasmainly apply
(see
Section4),
weshortly
introduce some notation foralgebraic
number fields.Throughout
this paper F denotes atotally
realalgebraic
number field ofdegree
n over the rationalnumbers Q .
We assume that it isgenerated
by
a root p of a monic irreduciblepolynomial
Over the real numbers R the
polynomial
f (t)
splits
into aproduct
of linearfactors
We assume p =
p~l~
for theconjugates
p~l~, ...,
pen)
of p as usual. An elementa of F is called
totally
positive
if all itsconjugates
arepositive
real numbers.Arithmetical
problems usually
require computations
withalgebraic
inte-gers contained in
F, i.e.,
those elements of F whose minimalpolynomials
have coefficients in Z.
They
form aring
OF with a Z-basis wi , ... , wn(inte-gral
basis ofF),
the so-called maximal order of F. Then the discriminantdF
of F isgiven
by
Two elements of OF are called
multiplicatively
equivalent
( associated),
iftheir
quotient
is an invertible element(unit)
of oF. The units of oF form afinitely generated
abelian groupUF.
Fortotally
real fields it consists of atorsion
part
and the directproduct
of r = n - 1 infinitecyclic
groups the
generators
ofwhich,
say Cl, ..., Cr, are calledfundamental
units.Equivalence
isusually
tested inlogarithmic
space. For this we consider themapping:
Clearly,
A :=L(UF)
is a lattice in r-dimensional Euclidean space with basisL(sl), ..., L(ér)
and,
consequently,
two elements x, y EOF B
are associated if andonly
iffL(x)
andL(y)
differby
an element of A.In this article we
study
theproblem
ofchoosing
appropriate
bases of lattices A in r-dimensional Euclidean space such that so-called Shintanicones which are of
importance
for ray class zeta functions become moreeasily
accessible. Weproceed
as follows. In Section 2 we define "nice"bases of lattices
(respectively,
sublattices of finiteindex)
and prove theirexistence. In Section 3 we define a
generalized multiplicative equivalence
for vectors with all coordinates
positive
in Euclidean r-space. We show how the bases of Section 2 can be used to obtain fundamental domains for thatmultiplicative equivalence
for r 3 andconjecture
that this is truefor
arbitray
r.Interpreting
them as the intersection of adecomposition
of into Shintani cones with thehyperplane
zn = 1 these can then beused to calculate a Shintani
decomposition
of where r denotes anatural number and
(the
fielddegree) n
is r + 1.We illustrate our considerations
by
anexample
of a fundamental domainfor a
quartic
field. Theapplication
of thatdecomposition
to the calculation of the values of ray class zeta functions atnon-negative
integers
will be done2. Lattice bases with
special
properties
In this section we consider full lattices A in r-dimensional Euclidean space for r > 1. In later sections we shall need the existence of
special
lattice bases for A
(or
at least for a sublattice A of A of finiteindex).
Definition 2.1. A set of r non-zero vectors S
_ ~bl, ..., br I
of A is calledcyclic non-negative,
if the coordinates of the i-th vectorb2 -
(bil, ..., bir)
satisfy
theinequalities
..- > 0(1 ~
i ~
r),
where the secondindices j
are considered modulo r,i.e. j
is to bereplaced by
j - r for j > r.
We will show that every lattice A contains
cyclic
non-negative
sets ofindependent
vectors. In a firststep
we construct acyclic
non-negative
setfor
A,
where allinequalities
between the coordinates of the vectors are strict.Proposition
2.1. A lattice A contains acyclic
non-negative
setof
vectorsbl,
...,br
the coordinatesof
whichsatisfy
bii
>bi,i+l
> ... >bi,i+r-1
>0
(1 ir).
Proof.
Let x1, ..., xr be any basis of A and K an upper bound for the absolute values of the coordinates of the basis vectors.Any
x E R’ has apresentation
x =Aixi
with coefficientsAi
E R. Letm2 E Z
subject
to
I
m2 -Aj
1:!~- 2
( 1 i r)
andput
y := Then we havey = x +
~i 1 (m2 -
and thereforeHence,
anappropriate
choice of xguarantees
> 0. DNext we will show that the
cyclic
non-negative
set obtained in thepropo-sition is
linearly
independent.
Proposition
2.2. Letbl,
...,br
be acyclic
non-negative
setof
vectorsof
a lattice A and denoteby
B := the matrix with rowsb1, ..., br .
Then the determinantof
B isnon-negative.
If
we haveb22
>for
at least r - 1of
the vectorsbl,
...,br
thegiven
cyclic
non-negative
set islinearly
independent.
Proof.
LetDl
:=det(B)
be the determinant of the matrix with rowsbl, ..., br.
We will showDl
> 0by
induction on r. This is trivial for r =1, 2. Hence,
we assume r > 3 in thesequel.
The determinant
Dl
is a linearpolynomial
in each of the entries of thematrix B.
Hence,
it suffices to consider the value ofDl
on the boundariesof those entries. The coefficient of
bll
is the determinant of the matrixinduction
assumption.
As a consequence,Dl
becomes at mostsmaller,
when we
replace
bll
by
b12
in theoriginal
matrix.Again
weproceed
by
induction on the number ofequal
entries in thefirst row of the
given
matrixstarting
from the left. Let C be the matrixwith entries c2~ -
bij
for 2 i or i = 1 whereasClj = for
1 j
~. We conclude from K - 1 to ~.The determinant of C is a linear
polynomial
in cix. We subtract col-umn 1 of C from columns2,..., r.
and obtain for the coefficient of thedeterminant of the matrix E with entries
for 1 i r. We add the last column of E to columns
1, ...,
~ - 1. Then the rows of the new matrix F form acyclic
non-negative
set, and we obtain 0according
to our inductionassumption.
This finishes thestep
from - 1 to ~.Hence,
theoriginal
determinantDl
is bounded from belowby
thede-terminant of the matrix
originating
from Bby replacing
all entries in the first rowby
bln.
Moving
the first row to the last and the first column tothe last and
taking
the common factorbln
of the new last row, we obtainI - - - ,
Again,
the rows of the last matrix form acyclic
non-negative
set.Carrying
out the sameprocedure again,
weget
since
D3
is the determinant of a matrix with 2equal
rows with entries all one. Thus we have also shown the inductionstep
from r - 1 to r.If the additional
premise bi2
>bi,2+1
holds for at least r -1 of the vectorsb1, ..., br
we can rearrange the order of rows and columns such that thiscondition is satisfied for the first r - 1 rows. Then the
proof
aboveclearly
yields
det B > 0. DThe
preceding
propositions
show that we cancompute
cyclic
non-negative
sets in every lattice which
generate
sublattices of finite index. Of course,we are interested in
generating
sublattices of small index. For r = 2 we canprove the existence of
cyclic
non-negative
lattice bases and thatproof
alsoProposition
2.3. For r = 2any lattice A has a
cyclic
non-negative
set{bl, b2}
as a basis.Proof.
Weapply
apurely geometrical
construction which is based onMinkowski’s convex
body
theorem. We note that forcalculating
latticepoints
inpractice
theoccuring
rectangles
will be coveredby ellipses.
Foralgorithmic
details we refer the reader to~5~.
We consider the closed first
quadrant
(JR~O)2.
For the bisectorGl
:=~(x,
y)
E =y~
we letG2, G3
be theparallel
lines toGl
with distance6 > 0. That distance is chosen maximal
subject
toM(A),
whereM(A)
denotes the minimum of thelattice, i.e.,
the norm of a shortest latticevector. This
guarantees
that all latticepoints
betweenG2
andG3
lie in the first and thirdquadrant.
Then we choose the smallest
0-symmetric
closedrectangle
with two sideson
G2
andG3
which contains at least one latticepoint
c,necessarily
on theboundary.
We choose c in the firstquadrant
with maximaldistance,
say e,to
Gl.
We need to discuss two casesdepending
on whether e is zero or not.To
begin
with let us assume e > 0. We letG4,
G5
be theparallel
linesto
Gl
with distance 6’. Then we choose the smallest0-symmetric
closedrectangle
with two sides onG4
andG5
which contains at least one latticepoint
d in the firstquadrant
which is different from c. It is easy to see that c and d lie on different sides ofGl. Hence, putting
them into theright
orderthey
form alinearly
independent cyclic non-negative
set. Also the closedtriangle
with vertices0,
c, d containsexactly
these vertices as latticepoints.
Therefore{c, d}
is a basis of A.In the case e = 0 we consider the
parallel
linesG6, G7
with distanceq =
d(A)/lIcll,
whered(A)
denotes the mesh of the lattice. Let d be a latticepoint
onG6
orG7
which lies in the firstquadrant.
Such apoint
must exist
according
to ourassumptions.
Ordering
c, dappropriately they
form a
cyclic non-negative
set which is a lattice basis because of the volumeof the
parallelogram
spanned
by
them. DWe note that for r = 3 we can show the existence of a
cyclic
non-negative
set
generating
a sublattice of index at most24,
independently
of the lattice A. Theproof
isquite
lengthy
and therefore omitted. Inpractice
wealways
obtained sublattices of much smaller index.
3.
Multiplicative
equivalence
inThe
concept
ofmultiplicative
equivalence
for theintegers
of analgebraic
number field F which we introduced in Section 1 needs to begeneralized
withrespect
tomultiplicative
equivalence
in and Shintani cones.For
multiplicatively
equivalent
totally
positive
elements E OF thererequire
This is
easily
seen to beequivalent
toand therefore also to
This
yields
thefollowing general
definition which isindependent
of number fields.(We
note that r = n -1throughout
thepaper.)
Definition 3.1. Let
.E"
(1
i, j
r)
be fixedpositive
constants.Vec-tors x, y E are called
(multiplicatively)
equivalent,
if there is a vectorm =
(ml,
...,
mr)
E zr suchthat yj
= zjIT~=1
From the definition it is clear that this relation is an
equivalence
relationand we are interested in a "nice" fundamental domain. If we
set Ei :_
(E(l) ...,E~)
(1:S i :S
r)
the vectors x,y areequivalent,
if andonly
ifL(y)
= L (x)
+¿~=1
holds,
where the map L isgiven
by
It is obvious that the vectors
L (Ei )
should belinearly
independent,
we thereforestipulate
this in thesequel.
Hence,
L(E1), ..., L(Er)
are basis vectors of a(logarithmic)
lattice A inR’. It would be
tempting
to take as a fundamental domain for multiplica-tiveequivalence
thepreimage
of the fundamental domain ofA,
i.e. of theset
Even
though
theexponential
functiongenerally
behavesquite
nicely,
such a choice does notyield
anappropriate
fundamental domain.Namely,
itsboundary
is not contained in a finite number ofhyperplanes
which willbe
required
at a laterstage.
Therefore the idea comes up todeform
into a setfl.
Thisprocedure
must be carried out such that theintegral
translations of the new fundamental domain still form a
disjoint
cover ofand that the
boundary
of is contained in a finite number ofhy-perplanes.
To achieve this in astraightforward
manner would mean to connect thepreimages
of the verticesof 11
by hyperplanes,
and to consider thelogarithmic
image
of thatmultiplicative
"fundamental domain " as newfundamental
domain 11
for Aapplying
3.3 . In thesequel
we willstudy
thisapproach
and theoccuring
difficulties in thesimplest
case r = 2. The ideasfor new methods to overcome these difficulties will later form the
guiding
principles
forhigher
dimensions.Let us assume that a =
(a,,
a2)
and b =(bl, b2)
arelinearly independent
vectors in
logarithmic
spacespanning
a lattice A. Inexponential
space theimages
of the vertices of the fundamentaldomain rj
of A are thepoints
We would be
tempted
to connect 1 withEl,
E2
and those two withE12
by
straight
lines. Theimages
of these lines inlogarithmic
space arecongruent
modulo
1,
but this does notgenerally yield
a fundamental domain as isdemonstrated
by
thefollowing example.
Example.
We chooseEl
=(2,4),
E2
=(4, 11).
Then thepreimage
of
rl
has the vertices(1,1), (2, 4), (4,11), (8, 44),
and thestraight
linesconnecting
(1,1), (4, 11),
,respectively
P10
(2,4), (8,44),
intersect in(21,
).
3 We thereforeproceed
as follows. Inlogarithmic
space we assume thatSl
and
Tl
arepiecewise
smooth doublepoint
free curvesconnecting
the latticepoints
0 withL(El),
L(E2),
respectively.
We setS2
:=81
+L(E2),T2
:=Tl
+L(Ei )
andget
thefollowing
lemma.Lemma 3.1. Denote
by
J the setof
points
enclosedby
C :=SlUT2US2
UTl1.
If the
closed curve C is doublepoint
free
then the set(JUC) B (S2
UT2)
is afundamental
domainfor
We note that a fundamental domain for
R /A
yields
a fundamental do-main formultiplicative
equivalence
uponapplying
theexponential
map.Proof.
Theproof
is carried out in severalsteps.
Webegin by simplifying
thedescription.
Forsimplicity’s
sake we assume that the lattice inlogarith-mic space is
spanned
by
the unit vectors(1, 0), (0,1)
and the fundamental domain is therefore[0,1)~.
This can beeasily
achievedby
a lineartrans-formation.
In a first
step
we introduce thepiecewise
smoothboundary
curves. We assume that~2, ~1, 02
arecontinuously
differentiable maps from[0, 1]
into Rsatisfying
0 =§2(0) =
~2(1) _ a/1 (0) =
=and C double
point
free.By
G we denote the(closed)
area surroundedby
C. We will show that the union of the interior of G with
Cl
andC2
without thepoint
(0, 1)
is also a fundamental domain.Since we will
apply
some Fouriertheory
and the theorem of Gauss we introduce severalauxiliary
formulae in the secondstep.
By
the theorem of Gauss anintegral
over G can be evaluatedby
a contourintegral
over theboundary
C of G.Using
theparametrization
of theboundary
of G from above we calculate Fourier coefficients:the same result also holds for
0;
finally
we findIn the third
step
we show that all x E~0, 1)2
are obtained modulo 1. Let us assume that there exists x E~0, 1)2
with(m, n)
+ x not in G for all(m, n)
EZ2 .
Then the set{x
+(m, n)
I
(m, n)
E has apositive
distance to G which is
larger
than apositive
constant e.Hence,
if y E
(0, 1~2
has distance smaller than e from x then the intersection of
{y
+(rrz,
n) I
(m, n)
E7G2~
with G isempty.
Without loss ofgenerality
we can assumethat x E
(0, 1)2
and that the opene/2-neighborhood
of x is contained in(0,1)2.
Consequently,
there exists a function h :~0,1~2
-> R with theproper-ties :
(i) h
isnon-negative
andh(z)
> 0 for all z in the opene/2-neighborhood
of x;
(ii)
h(z)
= 0 for all z E~0,1~2
having
a distance of at least e fromx;
(iii)
h isarbitrarily
often differentiable.The
periodic
continuation of h to allof IEg2
is denoted H. Also H isarbitrar-ily
often differentiable and has therefore anabsolutely
convergent Fourierour
assumption
on x the function H vanishes on G.Hence,
weget
acon-tradiction as follows:
In the fourth and last
step
we show that nopoint
x can occur severaltimes modulo 1. We start to prove this for x in the interior of G. We
assume that also x +
(m, n)
E G for some non-zero(m, n)
EZ~.
Let p
denote the
Lebesgue
measure. We obtainHence,
we must haveequality everywhere
and translations of Gby
differentintegral
vectors cannot have common interiorpoints.
A
slightly
modifiedargument
also shows that aboundary
point
of onetranslate of G cannot be an interior
point
of another one.Eventually,
con-sidering
twoboundary
points
in differenttranslates,
we argueby deforming
the
boundary
curves. DRemark. The last lemma is
proved
in a much moregeneral
version thanneeded.
Namely,
in our case the curves,S’1
andTl
are just images
ofstraight
lines under L. But the latter does not lead to any considerablesimplifica-tion of the
proof
for r = 2.However,
for r > 2 we indeed use the fact that themultiplicative
funda-mental domain in
exponential
space is a2r-gon.
Theapplication
of Gauss’theorem in that situation becomes much easier since the outer normal is
almost
everywhere
well defined on theboundary.
Theproof
for r = 2given
above can therefore
easily
beadopted
to r > 2 and so we omit it.Clearly,
thepremises
of the lemma are not satisfied in theexample
above if we choose thelogarithmic
images
of thestraight
lines as curves. We willsee,
however,
that the lemma becomesapplicable,
ifL(El), L(E2)
form alinearly independent cyclic
non-negative
set inlogarithmic
space.Example
(cont.).
Clearly,
theoriginal
vectorsdo not form a
cyclic
non-negative
set. The reductionprocedure developed
in Section 2 for r = 2
yields
as new(and
cyclic
non-negative)
basisThe
quadrangle
with vertices,, 4
is convex and the last lemma becomes
applicable.
This
phenomenon
isexplained by
the nextproposition.
Proposition
3.1.Any cyclic
non-negative
basisof
vectorsbl, b2
of
a 2-d2mensional lattice A(in
logarithmic
space)
yields
a convexfundamental
domain
f or multiplicative equivalence.
Proof.
Let x2
=exp(bil),
yi =
exp(bi2)
for i =l, 2. Hence,
we have 1yl zi ,
1
x2 y2. We first show that theslope
of thestraight
linethrough
(1,1)
and(ziz2 ,
YlY2)
islarger
than thatthrough
(1,1)
and(zi ,
yl)
and smaller than thatthrough
(1,1)
and(X2, Y2).
From ourassumptions
we obtainsuccessively
as well as
hence,
in case x2 > 1(i
=1, 2),
Finally,
we need to show that thepoint
(xl x2,
is not in the interiorof the
triangle
with vertices{1,1),
(X2, y2).
This isagain
obvious from the size conditions.0
Although
inprinciple
the same ideas can be used in dimensions r > 2 thedescription
of the fundamental domain inexponential
space becomes muchmore
complicated.
The reason for this is that for vertices of a maximalbounding
r - 1-dimensionalparallelotope
of the fundamental domain ofA the
preimages
of these vertices inexponential
space will notnecessarily
lie in a
hyperplane
anymore.Hence,
we need todecompose
each of these surfaces in anappropriate
way. This will beexplained
in thesequel.
The
general procedure
isroughly
as follows. Theoriginal
fundamentaldomain in
logarithmic
space isdecomposed
into a finite numberof simplices.
For each such
simplex
we consider theimages
of its vertices inexponential
space.
They
form the vertices of asimplex
there. What we then need toshow is that the
logarithmic
image
of the union of theexponential simplices
becomes a fundamental domain for the lattice inlogarithmic
space.We
decompose
the fundamental domain of the lattice A with basisbi , ... , br
iFor o, E S,
we setr
and
consequently
obtain(we
put xu(O)
:= 0 Via ESr)
The accent at the last union indicates that sets
occuring
several times are consideredonly
once. Then all sets of the fourfold union aredisjoint
and and allcomponents
are opensimplices
in theircorresponding
dimension.(Without
the accent we had a union over 2rr! sets. With the accent this number decreases from 48 to 26 for r =3.)
The vertices of every such
simplex
aremapped
intoexponential
spaceand the
images
become the vertices of acorresponding simplex
there. Thesimplex
definedby
P( a)
in this way inexponential
space will be denotedby
Pe(a).
We will show that for a, T ESr
the closures intersectin the closed
simplex
which isspanned
by
thepreimages
of the vertices ofP( a) n P(T)
thusexactly mirroring
the situation inlogarithmic
space.Because of the definition of
Pe ( a)
it suffices to show that r the setsQ
andPe (T)
areseparated by
ahyperplane
inexponential
space which contains thepreimage
ofP(a)np(r).
Therefore we canglue
theexponential
simplices
Pe (a)
in the same manner as the fundamental domain of A isglued
as indicatedby
thedecomposition
offl
into thesimplices
P(~).
Thus weget
a2’’-gon
(which
is notnecessarily
convex)
whoseimage
inlogarithmic
space hasboundary
sets which arecongruent
modulo 1.Again
we canapply
Gauss’ theorem to prove that we indeed have a fundamental domain.To avoid
complications
at first it should be assumed that theexponential
simplices
Pe(oa)
contain interiorpoints.
The other cases are then obtainedby
performing
limits. We omit details since allimportant
steps
werealready
explained
in theproof
of Lemma 3.1 and thesubsequent
remark.Of course, the
practical problem
remains to show that the union ofex-ponential
simplices
constructed as above indeed satisfies therequirements
of the
preceding paragraph.
Weconjecture
that this is ingeneral
the casewhen we choose a
cyclic
non-negative
basis for A in thebeginning.
AFor
proving
that different openexponential
simplices
do not havecommon
points
it suffices to show that their vertices areseparated by
ahyperplane.
For a betterunderstanding
we enumerate the vertices of theexponential
simplices
from 0 to 7 inbinary
representation,
e.g., 6 =(o, l,1)
denotes the vertex whose coordinates are the
exponential
values of thecoordinates of the vector
b2
+b3.
In the
sequel
weonly
consider the six closed three-dimensionalsimplices.
We order them as follows:For
example, concerning
the intersection ofsimplices
(1)
and(2)
we showthat the
point
1 isseparated
frompoints 2,6 by
theplane
determinedby
the threepoints 0,3,7 .
We do thisby considering
thesign
of the determinant whose first two rows are the vectorsspanning
theseparating plane
and thelast row is the vector from 0 to the
point
underconsideration, i.e.,
1 or2,6.
In this way we need to discuss a total of 15 intersections. Because ofoccuring symmetries
it suffices to consider 6 different determinants.We recall that
bi, b2, b3
is acyclic
non-negative
basis. The coordinates of these vectors are denotedby
bi
=(b21,
bi2,
for 1 i 3. Inexponential
space we write
Bij
:=(1 ~
i, j
3)
for abbreviation. Then theplane
through
0,3,7
isspanned by
the vectorsB12 B22 B32
-1, B13 B23 B33 -
1 ) (BIIB21 -
1, B12 B22 - 1 B13 B23 -
1)
and for thepoint
1 we need to determine the
sign
of the determinantWe do this
similarly
as in Section 2. We write the determinant as apolyno-mial
812, 813, B21, B22 , B23 , B31, B32 ,
B33 ) ~
Thispolynomial
is lin-ear in each of its 9 variables.Hence,
weonly
need to discuss itssign
on theboundary
of each variable. Instead ofjust
considering
29
cases we reducethe number of variables
by
3, i.e.,
we discusspolynomials
in 6 variablesWe therefore start to consider the
sign
of the coefficientPi
ofBll
in P: P =Pl Bll
+P2. Repeating
thisprocedure
we obtainsuccessively:
At this
stage
the coefhcientpolynomial
Plll
is apolynomial
inonly
6variables
Bij (1 j
3 ,
i =2, 3).
Writing
it aspolynomial
inB22,
thenthe coehicients as
polynomials
inB23
thesign
ofPlil
(and
similarly
for any suchpolynomial
in the same 6variables)
can becomputed
fairly easily.
Next we set
Bl3
= 1 inPll
and determine thesign
of thatpolynomial.
These twosigns
take care of the coefficientPI1
ofB12
(in
the coefficientPl
ofBll
inP),
i.e.,
they
determine thesign
ofP,
forB12
large.
This is followedby
a discussion of the caseB12
=813.
We need to consider apolynomial
P,
=PllBl3
+Pl2
in the new main variableBl3
which can betreated
similarly.
’We note that
considering
a coefficient of one variable orsetting
onevari-able to 1
yields
apolynomial
with one variableless,
but thedegrees
in the other variablesstay
invariant.However,
if we need toreplace
one variableby
another one, say Aby
B,
thedegree
of of B in theresulting polynomial
can increase.
Fortunately, only quadratic polynomials
occur. For those we need to consider the coefhcient of thequadratic
term,
the value at apotential
extremalpoint
(derivative vanishing),
and the value obtainedby
replacing
the variableby
the next smaller one. For allquadratic
polyno-mials it turned out that their extremal values were outside the considered
interval of the main variable.
For
discussing
the different cases we wrote a smallMaple
program[2]
which made these
investigations
a lot easier. As weconjectured
it turnedout that the
signs
for any of theseparametrized
determinants were the same for allpotential
values of theparameters
and did indeedseparate
theconsidered
simplices.
We thereforeproved
the next theorem. Theorem 3.1. Let r 3and fl
=Uo-ESr
P(a)
be the standarddecompo-sition 3.7
of
afundamental
domaingiven
by
acyclic
non-negative
basis inlogarithmic
space. Let0,
v2 (a~),
...,Vr+l(a)
be the r + 1 verticesof
P(a)
and1,
E(vj(a))
(2 j
r +1)
be theirpreimages
inexponential
space. LetThen ,~’ is a
fundamental
domainfor multiplicative
equivalence as
intro-duced inDefinition
3.1.Remark.
(i)
If thecyclic
non-negative
basis in the theorem is denotedbl, ..., br,
then we haveEi
=exp(bi) (1 i
r)
in Definition3.1,
wherethe
exponential
map isapplied
coordinatewise.(ii)
The amount of work forproving
a similar theorem inhigher
dimensions growsexponentially,
but the cases r =4, 5, 6
areeasily
within thescope of
these methods. Also we
emphasize
that forexplicitly
given
lattices it is noproblem
to checknumerically
whether thecorresponding
simplices
can be
separated
as demonstrated for r = 3.4. Shintani
decomposition
Let f be a module of the
totally
realalgebraic
number fieldF,
i.e. f is anintegral
ideal of F combined with a subset of the infiniteplaces.
In thesequel,
weonly
need the idealpart,
which we also denoteby
f. Let b be a(fractional)
idealcoprime
tof,
hencerepresenting
a narrow ray class bmodulo f. The
corresponding
ray class zeta function is definedby
where the summation is over all
integral
ideals of b modulo f. We recall Shintani’s ideas[7]
forevaluating
that zeta function atnon-positive
integers.
Besides the full unit group
UF
of F we need thesubgroup
oftotally
positive
unitsut,
respectively
thesubgroup
ut (f)
oftotally
positive
ray class units. Bothsubgroups
are of finite index inUF.
A full set ofgenerators
for them iseasily
obtained from fundamental units ofUF.
Example.
Fordetermining
UF
wejust
need to findrepresentatives
of those classes ofUF/UF,
which containonly totally
positive
units. This isstraightforward
from thesigns
of theconjugates
of the fundamental unitsof
UF
andrequires
only binary
arithmetic.The group
ut
(respectively
acts on as described in theprevious
section. Shintani showed that there exists a finitenumber,
say s,of open
simplicial
conesCj
(1 j s)
such thatThe union on the
right-hand
side isdisjoint.
The opensimplicial
cones areFollowing
Shintani we set for S Cand
note,
thatR(j,8)
isfinite,
when S is a subset of a fractional ideal. Inthe
sequel
we use the notation x E S + 1 for x -1 E S. For m e N Shintani obtains:where the
Aj
are matrices the rows of which span one of the ssimpli-cial cones and the
Bm
aregeneralized
Bernoullipolynomials.
Once thesimplicial
cones areknown,
allremaining
calculations arequite
straightfor-ward. This will be demonstrated in a
forthcoming
paperby
the authors.In the remainder of this section we show how to
apply
thegeneral theory
ofmultiplicative
fundamental domainsdeveloped
in Section 3 in thisspecial
situation.
It is of
importance
that we needmultiplicative
(in)dependence
only
for non-zerointegers
of F of the same norm. This is used to eliminate onedimension. Whereas Shintani
[7]
and Okazaki[4]
do thisby
requiring
thetrace of the
integers
to be one, we rather follow Reidemeister’sprecedent
[6]
and normalize the last coordinate to one.In the
sequel
we assume that ~1, ..., are a basis of the unit group underconsideration
(consisting
only
oftotally
positive
units).
Then the lattice Ain
logarithmic
space has the basisFrom these vectors we determine a
cyclic
non-negative
basisbl, ...,
br
whichgenerates
a lattice A of finite index t in A. Wejust
have tokeep
in mind that the elements which need to be determined in thesimplicial
conescoming
fromA
will occur withmultiplicity
t. As outlined in Section 3 the fundamental domain of A withrespect
tobi, ..., br
isdecomposed
inthe standard way into r!
simplices
and thecorresponding simplices
for themultiplicative
fundamental domain are calculated.Example.
Let F =Q( p)
for a root p of thepolynomial x4+x3-4x2-4x+1.
The discriminant of F is 1125.(Up
toisomorphy
we could alsoput F
=15
+6V5),
but in that case theequation
order would not bemaximal.)
The unit groupUF
of F isgenerated by -1
and the three fundamental unitsel =
3 p -
p3,
e2 = 1 +
p, e3 = 3 -
p2.
None of the fundamental units istotally
positive,
only
theproduct
of all 3.Hence,
the index ofUF
inUF
is 4.Choosing
the order of theconjugates
of pappropriately,
we obtain acyclic non-negative
basis of A from thetotally
positive
unitsThis kind of calculations is very easy with a software
package
like KANT~1~.
References
[1] M. DABERKOW ET AL., KANT V4. J. Symb. Comp. 24 (1997), 267-283.
[2] D. REDFERN, The Maple handbook: Maple V release 3. Springer Verlag, New York (1994).
[3] R. OKAZAKI, On an effective determination of a Shintani’s decomposition of the cone Rn+.
J. Math. Kyoto Univ. (1993), 1057-1070.
[4] R. OKAZAKI, An elementary proof for a theorem of Thomas and Vasquez. J. Number Th. 55 (1995), 197-208.
[5] M. POHST, H. ZASSENHAUS, Algorithmic Algebraic Number Theory. Cambridge University
Press 1989.
[6] K. REIDEMEISTER, Über die Relativklassenzahl gewisser relativ-quadratischer Zahlkörper.
Abh. Math. Sem. Univ. Hamburg 1 (1922), 27-48.
[7] T. SHINTANI, On evaluation of zeta functions of totally real algebraic number fields at
non-positive integers. J. Fac. Sci. Univ. Tokyo IA 23 (1976),393-417.
Ulrich HALBRITTER Mathematisches Institut Universitat zu K61n Weyertal 86-90 50931 K61n Germany E-mail : uhalb(Qmath.uni-koeln.de Michael E. POHST Fachbereich 3 Mathematik, MA 8-1
Technische Universitat Berlin
Straiie des 17. Juni 136 10623 Berlin
Germany