Contents lists available atScienceDirect
Journal
of
Symbolic
Computation
www.elsevier.com/locate/jsc
Integrality
and
arithmeticity
of
solvable
linear
groups
✩
A.S. Detinko
a,
D.L. Flannery
a,
W.A. de
Graaf
baDepartmentofMathematics,NationalUniversityofIreland,Galway,Ireland bDepartmentofMathematics,UniversityofTrento,Italy
a
r
t
i
c
l
e
i
n
f
o
a
b
s
t
r
a
c
t
Articlehistory:
Received25October2013 Accepted10March2014 Availableonline11August2014
Keywords:
Arithmeticgroup Algorithm Algebraicgroup Lattice
We develop a practical algorithm to decide whether a finitely generated subgroupofasolvable algebraicgroupG isarithmetic. This incorporates a procedure to compute a generating set of an arithmetic subgroup of G. We also provide a simple new algorithmfor integralitytesting offinitelygenerated solvable-by-finite lineargroups over the rational field. The algorithms have beenimplementedin Magma.
©2014ElsevierLtd.All rights reserved.
1. Introduction
For K
≤
GL(
n,
C)
anda subringR ofC
,denote K∩
GL(
n,
R)
by KR.LetG be analgebraic groupdefinedover
Q
.Thispaperisconcernedwiththequestion:Given a finitely generated subgroup H of GQ, is H an arithmetic subgroup of G? (
∗
) RecallthatH
≤
GQisanarithmeticsubgroupofG if
itiscommensurablewithG
Z,i.e.,H
∩
GZ=
HZhas finiteindexinboth H and GZ. (More generally,amatrix group issaid tobe arithmetic ifitis an arithmetic subgroupofsome algebraic
Q
-group (Segal,1983, p. 119).)The significanceof (∗
) is evidenced by,e.g., Sarnak (2012). Decidability of(∗
) hasnot beensettled. However, itseems to be undecidable,evenforG
=
SL(
n,
C)
(Miller,2013).Weprovethat(
∗
)isdecidablewhen G is solvable,
by givingapracticalalgorithmtoanswerthisquestion.✩ TheauthorsweresupportedbyScienceFoundationIrelandgrant11/RFP.1/MTH3212.
E-mailaddresses:alla.detinko@nuigalway.ie(A.S. Detinko),
dane.flannery@nuigalway.ie
(D.L. Flannery),degraaf@science.unitn.it(W.A. deGraaf).
http://dx.doi.org/10.1016/j.jsc.2014.08.011
Asfurthermotivationforarithmeticitytestingweobservethatapositiveanswerenablestheuse ofarithmetic group theoryto investigatethegroup athand.Forexample,theautomorphism group ofafinitelygeneratednilpotentgroupisisomorphictoanarithmeticgroup(Segal,1983,Corollary 9, p. 122). There are polycyclic groupsthat are not isomorphic to any arithmetic group (Segal,1983, Proposition3,p. 259).Althoughwecantestwhetherapolycyclicsubgroupof
G
Qisarithmetic,testing arithmeticityofpolycyclicgroupsinthisbroadercontextisstillopen.Our approach draws on recent progress (de Graaf and Pavan, 2009; Faccin et al., 2013) in the constructionofgeneratingsetsof
G
ZwhenG is
unipotentoratorus.Wecombinethosetoconstruct afiniteindexsubgroupofG
ZforasolvablealgebraicgroupG—a
resultofinterestinitsownright.WeadaptmethodsforcomputingwithSF (solvable-by-finite)lineargroups(Detinko et al.,2011, 2013a, 2013b).TodecidewhetherH is commensurablewith
G
Z,we usean algorithmfromDetinko
et al. (2013b) to compute the rankof an SF linear group. We alsodesign a simple newalgorithm totestwhetherafinitelygeneratedSF subgroupofGL(
n,
Q)
isconjugatetoa subgroupofGL(
n,
Z)
. Sinceintegralityisanimportantlineargroupproperty(Dixon,1971,Theorem 3.5,pp. 54–55),thisis anotherusefulresultofthepaper.Integralitytestingappearsasacomponentofearlieralgorithmsto testfiniteness(Babaiet al.,1993) andpolycyclicity(AssmannandEick,2007).Thepaperisorganized asfollows.Section 2givesaprocedure toconstructageneratingset ofa finiteindexsubgroupof
G
Z,whereG is
solvablealgebraic.InSection3weconsidertestingfiniteness of|
K:
KZ|
forK
≤
GL(
n,
Q)
.Ifthisindexisfinite,weexplainhowtofind KZand g∈
GL(
n,
Q)
such that Kg≤
GL(
n,
Z)
.Section 4 thendescribesa newalgorithmto test integralityofSF subgroups of GL(
n,
Q)
.OurmainalgorithmispresentedinSection5.Adiscussion ofexperimentalresultsderived froma Magma (Bosmaet al.,1997) implementationofthealgorithmsconcludesthepaper.2. Computinganarithmeticsubgroupofasolvablealgebraicgroup
Let
G be
asolvablealgebraicQ
-group.Inthissectionwecombineresultsfromde Graaf
andPavan (2009)andFaccin
et al. (2013)toconstructageneratingsetofafiniteindexsubgroupofG
Z.Thefirstresultis
Vinberg
et al. (2000,Proposition 7.2(3)).Proposition2.1.
Suppose that G
=
A N, where both A and N are algebraic subgroups of G, defined overQ
. LetΓA
,ΓN
be arithmetic subgroups of A and N respectively. ThenΓA
Γ
Nis an arithmetic subgroup of G.From
Chevalley (1955,
Ch. 5,§3,No. 5,Propositions 20and 21)wegetthefollowing.Proposition2.2.
Let G be connected and solvable, with Lie algebra
g
⊂ gl(
n,
C)
. Letn
be the ideal ofg
con-sisting of all nilpotent elements ofg
. Then there is a subalgebrad
ofg
, consisting of commuting semisimple elements, such thatg
= d
⊕ n
. Moreover, bothn
,d
are algebraic, and if we let N, D be the corresponding connected subgroups of G, then G=
DN.Let G,
g
be asin Proposition 2.2. We suppose thatg
is givenby a basis consisting ofmatrices havingentriesinQ
.Inde Graaf (2009)
,an algorithmthat computesbasesofsubalgebrasd
,n
with thepropertiesofProposition 2.2
is given.Here we donotgo intothedetails butonly brieflyrecall thebasicsteps.1. ComputeaCartansubalgebra
h
ofg
.2. Let
a
1,
. . . ,
ar beabasisofh
,andleta
i=
si+
ni betheJordandecompositionofa
i.3. Let
d
be the space spanned by s1,
. . . ,
sr and letn
be the space spanned by n1,
. . . ,
nralong withthe Fitting1-component
g
1(h)
.(The latteris the spacei≥1
[h
i,
g]
, where[h
i,
g]
=
[h,
[h,
. . . ,
[h,
g
]
. . .
]]
(i factorsh
).)Starting fromthegiven basisof
g
,we compute generators ofan arithmetic group in GZ by the following.GeneratingArithmetic
(
G)
Input:abasisoftheLiealgebra
g
ofthesolvablealgebraicgroupG.
Output:ageneratingsetforafiniteindexsubgroupof
G
Z.1. Computebasesof
d
andn
asabove.DenotebyD, N the
connectedsubgroupsofG with
respective Liealgebrasd
,n
.2. Computegenerators of
N
Z usingthealgorithmfromde Graaf
andPavan (2009),andgenerators of DZusingthe algorithmfromFaccinet al. (2013).(These algorithmstake asinputbasesofd
andn
respectively.)3. Returntheunionofthetwogeneratingsetsobtainedinthepreviousstep.
We notethat
GeneratingArithmetic
iscorrect.Indeed,let H be thegroup generatedbyits output.ThenH
≤
GZ.Ontheother hand,H is anarithmetic subgroupofG by Proposition 2.1
.So Hhasfiniteindexin
G
Z.GeneratingArithmetic
(
G)
formsavitalpartofourmainalgorithm.NoticethatG need
not be connected.Indeed, let G◦ be theconnected component ofthe identity of G. Since|
GZ:
G◦Z|
is finite,GeneratingArithmetic
(
G◦)
returnsageneratingsetofanarithmeticsubgroupofG.
3. IntegralityandGL
(
n,
ZZZ)
-interceptsThissectionandthenextdependonsomeideasfrom
Babai
et al. (1993).An element or subgroup ofGL
(
n,
Q)
that has a conjugatein GL(
n,
Z)
issaid to beintegral.
LetH
≤
GL(
n,
Q)
.Lemma3.1.
(See
Babai et al., 1993, Proposition 2.3.) H is integral if and only if there exists a positive integer d such that dh∈
Mat(
n,
Z)
for all h∈
H .Proof. Suppose that Hg
≤
GL(
n,
Z)
for some g∈
GL(
n,
Q)
. Thenm
2H⊆
Mat(
n,
Z)
wherem is any commonmultipleofthedenominatorsoftheentriesin g and g−1.Supposethat
dH
⊆
Mat(
n,
Z)
.Letg be
anymatrixwhosecolumnscompriseaZ
-basis ofthelattice generatedbydH
Z
n.Then g∈
GL(
n,
Q)
andHg≤
GL(
n,
Z)
.Call d
=
d(
H)
asin Lemma 3.1 a common denominator for H . As the above proof demonstrates, knowingd
(
H)
isequivalenttoknowing g∈
GL(
n,
Q)
suchthatH
g≤
GL(
n,
Z)
.Amethodtoconstruct d may beextractedfromBabai
et al. (1993,Section 2);wegiveanalgorithmthatisamodificationof thisforourpurposesinSection4.IfH is finitelygeneratedthenwecalculate g from d by meansof thefollowing(cf.Babai
et al.,1993,Section 3).BasisLattice
(
S,
d)
Input:afiniteset S
⊆
GL(
n,
Q)
andd
=
d(
H)
,H=
S. Output:abasisforthelatticegeneratedbydH
Z
ninZ
n.1.
L
:=
dZ
n.2. While
∃
h∈
S∪
S−1 suchthath
L
L
doL
:=
the lattice generated byL
∪
hL
.
3. ReturnabasisofL
.Wewrite L
≤
f H to indicatethatthesubgroupL has
finiteindexinH .
Proof. Suppose that H1 is integral. By Lemma 3.1, d1H1
⊆
Mat(
n,
Z)
for some d1∈ Z
. Choose a transversal{
h1,
. . . ,
hr}
for the cosets of H1 in H , and let d2 be a positive integer such thatd2hi
∈
Mat(
n,
Z)
foralli.
Sinced
1d2H⊆
Mat(
n,
Z)
,H is
integralbyLemma 3.1
again.Lemma3.3.
Suppose that d is a common denominator for H , and let
L
be the lattice generated by dHZ
n.(i)
L
⊆ Z
n⊆
1d
L
.(ii) H acts
by left multiplication on the (finite) set of lattices lying between
L
and 1d
L
.(iii) HZis the stabilizer of
Z
nunder the action in (ii).Proof. (i)Clear:
dH
⊆
Mat(
n,
Z)
andd
Z
n⊆
dHZ
n⊆
L
.(ii)Wehave
H
L
=
L
.ThusH acts
onthesetoflatticesL
suchthatL
⊆
L
⊆
1dL
.(iii)Let
h
∈
H . Ifh∈
HZ thenZ
n=
h(
h−1Z
n)
⊆
hZ
n, so hZ
n= Z
n. Conversely, ifhZ
n= Z
n theneveryentryin
h is
aninteger.Proposition3.4.
H is integral if and only if H
Z≤
f H .Proof. Thisisaconsequenceof
Lemmas 3.2 and
3.3.Thefollowingprocedureconstructs HZ ifH is finitelygeneratedand
d
(
H)
isknown.IntegralIntercept
(
S,
d)
Input:afiniteset
S
⊆
GL(
n,
Q)
andd
=
d(
H)
,H
=
S. Output:ageneratingsetofHZ.1.
L
:=
the lattice generated byBasisLattice
(
S,
d)
. 2.Λ
:=
the set of all latticesL
suchthatL
⊆
L
⊆
1dL
.3. Returnageneratingsetforthestabilizerof
Z
n undertheactionofH on
Λ
.Remark3.5. Step(3)maybecarriedout usingstandardalgorithms forfinitepermutation groupsto obtainatransversalfor HZ in H , thenwritingdownSchreiergenerators.More practicalapproaches are possiblein specialsituations; say forpolycyclic H (e.g., arithmetic H
≤
GQ andsolvable G). In thatcaseanalgorithmcanbebasedonanorbit-stabilizer approach.Thisissimilartothealgorithm describedinLaue
et al. (1984).Such analgorithmenumeratestheorbitofZ
n underactionbyH ,
in theprocessobtaininggeneratorsofthestabilizer.Theefficiencyofthisapproachdependsheavilyon orbitsize.4. Integralityofsolvable-by-finitesubgroupsofGL
(
n,
Q
Q
Q)
We next describe how to test integrality of a finitely generated SF subgroup of GL
(
n,
Q)
, and computeacommondenominatorifthegroupisintegral.Let H
=
h1,
. . . ,
hr≤
GL(
n,
Q)
.We have H≤
GL(
n,
R)
where R=
1bZ
= {
a/
bi|
a∈ Z,
i
≥
0}
forsome positive integer
b determined
by theentries inthe hi andh−i1.Forany prime p∈ Z
notdi-viding
b,
reductionmodulo p of matrixentriesdefinesacongruencehomomorphismϕ
p:
GL(
n,
R)
→
GL
(
n,
p
)
. If H is SF and p=
2 then Hp=
kerϕ
p∩
H is torsion-free andunipotent-by-abelian (see Dixon, 1985, Lemma 9, or Detinko et al., 2011, Section 2.2.1). Assume that p has been so chosen wheneverH is
SF.Denotetheunipotent radicalof K
≤
GL(
n,
Q)
byU
(
K)
.Replace K if necessarybya conjugatein block triangularformwithcompletelyreduciblediagonal blocks.Ifπ
denotes projection of K ontoLemma4.1.
The following are equivalent.
(i) H is
integral.
(ii)
π
(
H)
is integral.(iii)
π
(
Hp)
is integral.Proof. By
Babai
et al. (1993,Theorem 2.4),afinitelygeneratedsubgroup K of GL(
n,
Q)
isintegral if andonlyif{
tr(
x)
|
x∈
K}
⊆ Z
.Thus(i)⇔
(ii).Since|
π
(
H)
:
π
(
Hp)|
= |
H:
HpU(
H)
|
<
∞
,Lemma 3.2
gives(ii)
⇔
(iii).Denotethecharacteristicpolynomialof
h
∈
GL(
n,
C)
byχ
h(
X)
.Proposition4.2.
(Cf. Lemma 8 and Theorem 9 of
Assmann and Eick, 2007.) Suppose that each element of H is equal to hm11
· · ·
hmr
r for some mi
∈ Z
. Then H is integral if and only if each hiis integral, i.e., χhi(
X)
∈ Z[
X]
and det(
hi)= ±
1 for1≤
i≤
r.Proof. If
χ
hi(
X)
∈ Z[
X]
has constant term±
1 for all i then hi⊆ {
n−1j=0ajhij
|
aj∈ Z}
. So thereexist positive integers di, 1
≤
i≤
r, such that dihi⊆
Mat(
n,
Z)
. Hence d=
d1· · ·
dr is a commondenominatorfor
H .
If H is polycyclicwithpolycyclicsequence
(
h1,
. . . ,
hr),then H satisfies thehypothesisofProposi-tion 4.2.Anygeneratingsetof H is similarlysuitablewhen
H is
abelian.Lemma4.3.
Suppose that H
≤
K≤
GL(
n,
Q)
and the normal closure HKis finitely generated abelian. Then HKis integral if and only if each hiis integral.
Proof. If the hi are integral then Proposition 4.2 guaranteesthat H is integral. Since HK isfinitely
generated, there are x1
,
. . . ,
xt∈
H and y1,
. . . ,
y
t∈
K such that HK=
xiyi:
1≤
i≤
t. ByProposi-tion 4.2again,
H
K isintegral.Lemma4.4.
Suppose that H is SF and Y is a normal generating
set for Hp(p=
2),i.e., H
p=
YH. Then H is integral if and only if each element of Y is integral.Proof. If each element of
Y is
integral then thesameholds forπ
(
Y)
. Hence thefinitelygenerated abeliangroupπ
(
Hp)=
π
(
Y)
π(H)isintegralbyLemma 4.3
.TheclaimnowfollowsfromLemma 4.1
. We computeY from
apresentationP
ofϕ
p(H)
≤
GL(
n,
p
)
onϕ
p(h1),
. . . ,
ϕ
p(hr)usingthe func-tionNormalGenerators
asinDetinko
et al. (2011, Section 3.2): thisevaluates therelatorsofP
, replacingϕ
p(hi) everywhere by hi, 1≤
i≤
r. Proposition 4.2and Lemma 4.4consequently provideourstraightforwardproceduretotestintegralityof
H .
IsIntegralSF
(
S)
Input:afinitesubset
S of
GL(
n,
Q)
suchthatH
=
SisSF. Output:true
ifH is integral;false
otherwise.1. Y
:= NormalGenerators(
S)
.2. Ifeveryelementof
Y is
integralthenreturntrue
; elsereturnfalse
.Remark4.5.When H is finitewe have Hp
=
1,soIsIntegralSF
always returnstrue
forsuchA major class of SF groups is PF (polycyclic-by-finite) groups. According to Assmann and Eick (2007, Corollary 10), one may test integrality of a PF subgroup H of GL
(
n,
Q)
after computing a polycyclicsequence andtransversalfora finiteindexpolycyclic normalsubgroupof H . Bycontrast,IsIntegralSF
doesnotrequireH to bePFandjusttestsintegralityofseveralmatricesinHp.ToconjugateanintegralSFgroup
H into
GL(
n,
Z)
,ortocomputegeneratorsofitsfiniteindex sub-group HZviaIntegralIntercept
,wemustknowd
(
H)
.Themethodbelowtofindthiscommon denominatorisasimplificationofBabai
et al. (1993,p. 120)forSFinput.Firstdetermineablockuppertriangularconjugateof
H with
completelyreduciblediagonalblocks; this maybe done as in Detinko et al. (2013b,Section 4.2). Let{
a1,
. . . ,
am}
⊆
π
(
H)
be a basis ofthe envelopingalgebra
π
(
H)
Q. Ifc is a common multiple ofthe denominators of entries intheai then d1
:=
c det(
[
tr(
aiaj)]
1≤i,j≤n)=
d(
π
(
H))
. Define ui=
hi−
π
(
hi) and vi=
h−i1−
π
(
h−i1)
. Let d2=
en−1 wheree is
acommonmultipleofdenominatorsofentriesintheu
i and vi.Eachelementof H is asumofterms x
=
π
(
g1)
w1· · ·
π
(
gk)wkwhere gj∈
H and wj=
1 orsomeu
i or vi.Sinceπ
(
h)
uj andπ
(
h)
vj arenilpotent,ifthereareatleastn occurrences
ofu
jsand vjs inx then x=
0.Thus
dH
⊆
Mat(
n,
Z)
whered
=
dn1d2.Notethatforcompletelyreducible H (e.g., H is finite),d
(
H)
=
d(
π
(
H))
=
d1.5. Arithmeticitytestinginsolvablealgebraicgroups
Inthissectionweapplyresultsof
Detinko
et al. (2013b)andtheprevioussectionstotestwhether afinitelygeneratedsubgroupH
≤
GQofasolvablealgebraicgroupG is
arithmetic.DenotetheHirschnumberofa group K with finitetorsion-freerankby h
(
K)
.Finitelygenerated SFsubgroupsofGL(
n,
Q)
havefinitetorsion-freerank(Detinkoet al.,2013b,Proposition 2.6). Lemma5.1.Let L be a finitely generated SF subgroup of
GL(
n,
Q)
, and let K≤
L. Then K≤
f L if and only ifh
(
K)
=
h(
L)
.Proof. ThisfollowsfromProposition 2.3andCorollary 3.3of
Detinko
et al. (2013b).Corollary5.2.
H is an arithmetic subgroup of G if and only if H is integral and
h(
H)
=
h(
GZ)
. In particular, H≤
GZis arithmetic if and only if h(
H)
=
h(
GZ)
.Proof. This is immediate from Proposition 3.4 and Lemma 5.1, using that h
(
H)
=
h(
HZ)
whenHZ
≤
f H .Remark5.3. If G is a unipotent
Q
-group then we have a more general statement (Detinko et al., 2013b, Lemma 3.7): H≤
GQ is arithmetic in G if andonly if h(
H)
=
h(
GQ)
, i.e., h(
H)
equals the dimensionofG.
Fornon-unipotentsolvable(evenabelian)G this
isnottrue;G
Qneednotevenhave finiterank.Wenowstateourarithmeticitytestingalgorithm.Thisusestheprocedure
HirschNumber
fromDetinko et al. (2013b, Section 4.4), which returns h
(
K)
fora finitely generated SF subgroup K ofGL
(
n,
Q)
.IsArithmeticSolvable
(
S,
G)
Input:afinitesubset S of GQ,
G a
solvablealgebraicgroup. Output:true
ifH=
Sisarithmetic;false
otherwise. 1. IfIsIntegralSF
(
S)
= false
thenreturnfalse
. 2. T:= GeneratingArithmetic(
G)
.3. If
HirschNumber
(
S)
= HirschNumber(
T)
thenreturnfalse
; elsereturntrue
.Table 1
Runningtimes(inseconds)ofthestepsinIsArithmeticSolvable.
n dimg(n) h(G(n)Z) IsInt(S) T:= GA(G(n)) h(S) h(T)
2 6 5 0.12 0.05 0.51 0.68
3 15 14 0.29 0.19 2.10 2.28
4 28 27 0.86 1.12 11.28 12.77
Remark5.4.Steps(1)and(3)arejustifiedby
Corollary 5.2
.IfG is
unipotentthen H is integral(Segal, 1983,Lemma 2,p. 111),andstep(1)canbeomitted.6. Practicalperformance
We have implemented our algorithms in Magma. The implementation relies on the package
‘Infinite’ (Detinko et al.,2013c) andproceduresavailableat
de Graaf (2013)
.Althoughthemaingoal hasbeentoestablishthatarithmeticityisdecidable,inthissectionweshowthatouralgorithmscan beappliedinpracticetonontrivialexamples.One ofthemainbottlenecks of
GeneratingArithmetic
liesinthecomputation ofthetoruspart.Let
g
= d
⊕ n
andD be
asinProposition 2.2
.ThefirststepofthealgorithmgiveninFaccin
et al. (2013)forcomputinggeneratorsofD
ZconstructstheassociativealgebraA with
unitygeneratedbyd
. Subsequently A is writtenasadirectsumofnumberfieldsQ(
α
)
.Foreachsuch field,generators of the unit group ofZ[
α
]
arecomputed. Butthe algorithm forthistask (asimplemented in Magma) becomes extremelydifficultto applyin practiceifthedegreeofQ(
α
)
is toolarge,say30 or more. Ontheother hand,ifall fieldsthat occurare equaltoQ
thenthecomputation ofgenerators of DZistrivial.Forthesereasonsweconstructedtestexampleswherethefieldextensionshavedegree
≤
2 (andsomeextensionsofdegree2 dooccur).Thisconstructionworksasfollows.FirstwedefineaLie algebrag(
n)
⊂ gl(
2n,
C)
,n
≥
2.Forthiswedividethematricesofgl(
2n,
C)
into2×
2 blocks.OurLie algebrag(
n)
isthedirectsumg(
n)
= t
⊕ n
oftwosubalgebras.Thesubalgebrat
hasdimensionn,
and theith
basiselementhasonitsith
diagonalblockthematrix0 1 2i
−
1 0andzeroselsewhere.
Let
e
i,jbetheelementarymatrixwith1 inposition(
i,
j
)
andzeroselsewhere.Thenn
isspannedbythe
e
i,jwhere(
i,
j
)
appearsinablockabovethediagonal.Sodimn
=
4 n2
.
The Lie algebra
g(
n)
is algebraic,and welet G(
n)
<
GL(
2n,
C)
be the connectedalgebraic group withLiealgebrag(
n)
.Nowlet
m be
a2n×
2n matrixproducedbythefollowingrandomizedconstruction.Theentriesofm are randomlyanduniformlychosenfrom
(
0,
0,
1)
(sowemakeittwiceaslikelythata0 ischosen). Wecontinuetoproducematriceslikethisuntilthedeterminantisnot0 or±
1.Set
G(
n)
=
mG(
n)
m−1.Thisisanalgebraic groupwithLiealgebram
g(
n)
m−1.The Liealgebra has basis Bn consisting of all mum−1, for u in the above constructed basis ofg(
n)
. Let g1,
. . . ,
gr begeneratorsofanarithmeticsubgroupof
G
(
n)
Z.For1≤
i≤
r let kibechosenrandomlyanduniformlyfrom
{
1,
2}
.ThenS
= {(
mgim−1)
ki|
1≤
i≤
r}
generatesasubgroupofG(
n)
Q(andnotethatitalwaysis arithmetic).The inputon whichwe testedour implementationof
IsArithmeticSolvable
istheset
S,
togetherwiththealgebraicgroupG(
n)
givenbyitsLiealgebra,inturngivenbyitsbasis Bn.In Table 1 we report on the running times of our algorithm with the input as above for n
=
2,
3,
4. Allexperiments were performed on a 3.16 GHzmachine running Magma V2.19-9. The first three columns ofTable 1list n, dimg(
n)
,andthe Hirschnumber ofG(
n)
Z. The other columnslist running times ofIsIntegralSF
(
S)
, T:= GeneratingArithmetic(
G(
n))
,HirschNumber
(
S)
and
HirschNumber
(
T)
.From
Table 1
we seethat ouralgorithm isefficientenough tohandlequite nontrivialexamples. Moreover,thebulkoftherunningtimegoesintocomputingHirschnumbers.Ofcourse,thisdepends onourparticulartestexample.Ifwetookexampleswhereitisdifficulttocomputegeneratorsof DZ,thenamuchlargerproportionofthetimewouldgointocomputinganarithmeticsubgroup.However, thecurrent implementationis rathersensitive torandomness ofthe input. Occasionallyit happens that generators in T have coefficientswith many digits (up to 10, forexample). This then causes problemswhencomputingtheHirschnumber.So wehaveaveragedtimesover 50 runsinorderto dampentheeffectsofrandomnessoftheinput.
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