Discrete groups of affine isometries
isometries of an affine Euclidean space E is discrete (if and) only if there is a. ?–invariant affine subspace F of E such that the restriction homomorphism
Chapitre 4 - Isométries dans un espace affine euclidien
Toute symétrie orthogonale de X (et en particulier toute réflexion) est une isométrie. Démonstration : Soit f une symétrie orthogonale d'axe Y . f est alors
18 Isometries
Every isometry is an affine transformation. Proof. Let F ? Trans(Rn) be an isometry and let y = F(0). Now we may define the.
Sommaire Notations 1. Isométries Affines
L'isométrie affine est la composée de la rotation d'axe ? et de même angle et de la symétrie orthogonale par rapport à ?. C'est un anti-déplacement. 3.5. Pas
On the Linearity of Form Isometries
Form isometries are necessarily affine-linear in the following situations: (1) when H is non- singular andfsurjective (see Ratz [6]); (2) when G is
Chapitre 3 - Isométries affines et vectorielles
un espace affine euclidien. On appelle isométrie affine de X toute application affine f : X ? X qui conserve la distance induite par le produit scalaire.
Definition 7.4.1 Given any two nontrivial Euclidean affine spaces E
An affine isometry is a bijection. Let us now consider affine isometries f:E ? E. If. ?? f is a rotation we call
APPLICATIONS OF AFFINE ROOT SYSTEMS TO THE THEORY OF
(St <r*)~(St'
AFFINE TRANSFORMATIONS IN A RIEMANNIAN MANIFOLD 0
orientable Riemannian manifold an infinitesimal affine transformation is an H(M} the group of all affine transformations M-*M that of all isometries.
Leçon 161 - Isométriqes dun espace affine euclidien de dimension
03-Jun-2017 – Pro : f : E ? E est une isométrie affine ssi c'est une application affine dont la partie linéaire est une application orthogonale. – Cor : ...
JournalofLieTheory
Volume9(1999)321{349
C1999HeldermannVerlag
Discretegroupsofaneisometries
HerbertAbels
CommunicatedbyE.B.Vinberg
testingprocedureisalgorithmic. groupasimage.ISSN0949{5932/$2.50C
HeldermannVerlag
322Abels
wegoalong.IthankE.B.Vinbergforhelpfuldiscussions.
1.TheBieberbachtheorems
recalltheBieberbachtheorems. (v;x)7!x+v,andametricd(x;x+v)=hv;vi12.LetG=Iso(E)bethe
Oof(V;h;i)wedeneanisometry
g=A(x0;t;U)(1.2by) g(x0+v)=x0+t+Uv:(1.3) formula andthedependenceonthebasepointchosen: (1.5)A(x0;t0;U0)=A(x1;t1;U1) haveahomomorphism (1.6):G!O denedby (A(x0;t;U))=U:VwiththegroupoftranslationsofE.
Abels323
duetoBieberbacharebasic. oflinearpartsofisnite. respecttothisbasis( )hasintegerentriesforevery 2. image.Fhenceisdiscrete.
2.Aminimalinvariantsubspace
sucestoanswerourquestionsforoneofthem. twofacts: subspace.324Abels
V 1inV.2.1LemmaandDenition
axisofg. empty. g.V=V1V6=1,then(g)=t1.
andv2V6=1.Then gx=x+(g)+((g)1)v v22V6=1.Then
gx=x+t1+((g)1)v2: nxx n andhencev2=0.Alltheclaimsarenowproved. have(g)2TFandEgintersectsF. greatergeneralitythanpresentlyneeded.Abels325
a)Q1+Q2+Q3=1where1istheidentityofV. c)TheimageofQiiscontainedinUifori=1;2.VontoViareink[T].
P f thepropertyg(S)=0. andQ2=P2Q4.326Abels
xTE1;2=V1\V2.
coecients. b b E2toE1.Wehaveperp(E2;E1)=b1b22W.
Weformalizetheiterationasfollows.
York1965]LA4.1)WehaveMX=`1
n=1XnwhereXnisinductivelydened byX1=XandXn=`1 p+q=nXpXqforn1andthemultiplication M supp(m;n)=supp(m)[supp(n).2XletE
betheaxisof ,andform;ninMSletE(m;n)bethesetEm;n followingproperties: a)E(m;n)Em.Abels327
b)TEm=Vhsupp(m)i. andperp(Em;En)2TFform;ninM. ofE. subspaceofEcontainsaspaceFofthisform. translatesofeachotherbyavectorinV(). followingtwosetsofvectors a)fgxx;g2Sg. b)f(g);g2Sg[fperp(Em;E(g;m));g2Sg: only,by2.9b). (hgh1)=(h)(g)(2.13) by2.1a),and E hgh1=hEg(2.14) bythedenitionoftheaxes.Hence E m=Em(2.15)
328Abels
form2Mand2,wherem7!
mistheuniquemorphismof magmassuchthat7!1for2.So
(2.16)perp(E m;E n)=( )perp(Em;En): theproofof2.10,itremainstoshowthat x02x0+Wfor2andx02Em.
perpendicularfromEntoEmforeveryn2M,since by2.9b).Hencefort=perp(Em;E )wehavex+t2E andthus (2.17) x=x+(g)((g)1)t by2.1e),so x2x+Wfor 2. V bythedenitionofWin2.10.Forx02Emwehave x02x0+Wbfor 2S by2.17. is {invariantforevery2Sandhenceforany
2.Thisimpliesthat
x inaminimal{invariantanesubspaceofE. subgroupsofGsuchthateveryelement ofhasaxedpoint,butthereis nocommonxedpoint.So( )=0forevery2butW6=0.Anexample
Abels329
2 generatesW,bytheBieberbachtheorems. t D 0=f xx; pair( ;b);2S;b2Bj,if(
)b23.Whenisagroupcrystallographic?
crystallographic.Hereisourtest. xx; 2gis discrete. f xx; chaptertwo,inparticular2.9. indexinD(x)andTF=D(x)R=R. F330Abels
isomorphismr1()!r2(). ker(j)=\VandletxbeapointofE. a)D(x)isaZ[()]{modulecontaining. b)D(x)spansTFoverRforeveryx2F. c)Thereisapointx2EwithD(x)1 f. {invariantanesubspaceofE. e)SupposeD(x)Qthen fy;D(y)Qg=x+Q+VF: x=1fP ix, where achosenpointx02E.Foranotherset0iofrepresentativesofmodulowe
have 0i= iiwithi2andthusthepoint y=1fX0ix=1f(X(
i)(ixx)+X ix)2x+1f isinthe1 f{orbitofx.Inparticular,for2thepoint
x=1fP ixisin the 1 f{orbitofx. xandRisanR[()]{module.Itisminimalsincef xx; 2gfor everyx2E. f1{invariant.Then
f1(xx0)2D+D0+VF1:
(x+d)x= xx)+( )dford2Dand21,andsimilarlyforD0.Foranytwopoints
xandx0inEwehavefor 2 x0x0= xx+(( )1)(x0x): xx2D and x0x02D0for F 1in1X h2F1hwf1w2D+D0 whichimpliesthelemmainviewofP h2F1hw2VF1.Abels331
Wearenowreadytoprovetheorem3.1.
isniteandisdiscrete.Hence1 fisadiscretesubgroupofVcontaining xxforevery D(x)1 sincecontainsofniteindex.Assumption.()isnite.
withD(x)discreteisanalogousto2.3: x+DEintersectstheaxisE forevery 2. formx+QofE,e.g.forxasin3.3c) ofgeneratedbytheelement ,letx+Dbe{invariant,x02E and D 0=Z(Dby{invarianceofx+D,thepower
fof2iscontainedinand
f isthetranslationbyf( ).Soxx02D+V( )bylemma3.4.Nowuse TE =V(TE1;2isinQ[F]andperp(E1;E2)isinD.
332Abels
EWeobtainasin2.9:
is{invariant,by3.3c).Foreverypointy2x+1 kwehave yy=( xx)+(( )1)(yx)21 kf; henceD(y)1 ofVgeneratedbyf yy;2g.Thisholdsinparticularforapointy2Em
of3.2followsfrom3.3d). hold (i) 2S. )jW=r( )is 2S. (iii)ThesubgroupofO(W)generatedby( )jW=r(2S,isnite.
Z{submoduleofWgeneratedbyBandf
xx;2Sg.ThenD0isalattice
inW,by(i).Thegroup( )D0iscommensurablewithD0by(ii)forevery2Sandhenceforevery
2.ThenD1=P
2( )D0iscommensurable D1generatedasaZ[()]{modulebyf
xx;2Sgisdiscrete.
Abels333
foragivensubgroupofGtobediscrete. a)thetranslationalpart( )forevery 2 b)perp(Em;En)form;ninM c) xxfor2andx2Em,wherem2MandT(Em\F)=(TF)().
M l((m;n))=l(m)+l(n). thefollowingelements: a)( )forevery 2 b)perp(Em;En)form;n2Mandl(m)N,l(n)N c) xxfor2andx2Em,wherem2M,T(Em\F)=(TF)()and
l(m)N.Ageometricconsequenceisthefollowing
2gis locallynite.ThisholdsinparticularforthesetFr(
)=E \F,2,andthederived
subgroupofGthatthesetfE2gislocallynite.Foranexamplesee
5.3. VH,HasubgroupofthenitegroupF=().
334Abels
2wehave
f)=f(g)and fisatranslation,i.e. f2,so( )21 f.Thenforx02Em0theset
xx,2,iscontainedinthelatticeD(x0),and
D(x0),by3.2.Letw=perp(Em0;E
).Then (x0+w)= x0+( )w and (x0+w)=x0+w+( sincex0+w2E ,hence (3.16) x0x0=( )1)w: so(( )1)wiscontainedinthelattice0=1 f+D(x0).Forevery(
)2Fthereisapolynomialh12Q[t]suchthat h 1(( )1)w=w i=0ti,andwrite )forevery 2.Wemayassumethat0isaZ[F]{module.
1; 2in andx0+wi2E i,wi=perp(Em0;E i),wehaveperp(E 1;E2)=Q3(w2w1)
forQ3asaboveforthetwosubgroupsHi=h( i)i,seetheproofof3.7.Sothe lattice 1M0containsperp(E
1;E2)foreverypair
1;2ofelementsof.
Q3(w2w1)21
M(1+2),x+w1Q1(w1w2)2E(m1;m2)andx+w2+
Q asabove,rstly,x0+1 1Ml(m1)+l(m2)0.Finally,ifx2Em,w=perp(Em;E
),x+w2E ,wehave (x+w)= x+( )w =x+w+( sincex+w2E andhence xx=( )1)w21Ml(m)+10,which
thefootofaperpendiculartoE ,i.e.Em=E(m; )by2.9a)andb).Abels335
4.Whenisagroupdiscrete?
Bofniteindexifisdiscrete.
weshalluseinthealgorithmicprocedure. consistsoftranslationsofFonly.Proof.Let1bethenormalsubgroupofthose
2forwhichr(
11)isthetranslation
)r( 1)for xesTF.ItfollowsthatthenormalsubgroupB=f 2;r( )2TF;( )2L0g ofisabelianandofniteindexin. componentof(()).Corollary4.3.IfisdiscretethenL0:=
(())0isatorusandxesTF. SoB=f 2;r( )2TF;( )2L0gisanabeliannormalsubgroupofof niteindex. ifrhasnitekernel,asfollows.336Abels
expid:tTF!TTF.Then rankA=rankker(r)+rankr(B)+dimT: rankr(B)+dimT: Q. a)r( )isatranslationofFforevery 2S1. b)ThesubgroupofTFgeneratedbyfr(2S1gisalatticeinTF
c)h( )iisconnectedforevery 2S1.Test(i)ThetoriC0((
2S1,commute.
If(i)holds,letTbethesubtorusQ
2S1C0((
))ofO(V)generated bytheC0(( exponentialmap.Pute=expid:tTF!TTF.LetbethesubgroupoftTFgeneratedbyfe1(
2S1g.Test(ii)rankQ=dimF+dimT.
onTFby t=( )t= t1,henceontheproductoTF.Sothefollowing
conditionmakessense.Test(iii)Every
2mapsQtoitself.
2e(Q)for
every2Ssuchthat
(x0)= x0Hereisourlasttest:
Abels337
Test(iv)Thegrouph
12Siisnite.
invariantanesubspaceofE. in4.5passestheTests(i){(iv). relevantobjects(S1;C0(U); and,ofcourse,aproofof4.7.2{dimensionalU{invariantsubspace.
ZandtQbethe
D(X)=f`2t
Qj`(X)2Qg.
4.9.Then
b)FortheLiealgebraLC(U)=LC0(U)wehaveLC(U)=\
`2D(X)ker` c)IfwewriteX=nP i=1 ieiwithrespecttoabasise1;:::;enof,then spannedby1,1;:::;n. thenumbers`(X),`2D(X)\tZ.Henceif`1;:::;`disabasisofthelattice
D(X)\t
ofthenumbers`1(X);:::;`d(X). Z.338Abels
a0+a11++ann.For(a1;:::;an)2QnwehavePn i=1aie i2D(X)i (a0=Pn i=1a1i;a1;:::;an)2Qn1\kerFwheree1;:::;e
nisadualbasisof hencethetoriC(( B index,hence2;Q=Qandthusevery2mapsQtoitself.
m wehave3ande(3)containsalltheelements wehavepicked,onefor every 12Sgwhichxesx0
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