Table Of ContentINDECOMPOSABLE MODULES OF SOLVABLE LIE ALGEBRAS
PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
7
1
0 Abstract. We classify all uniserial modules of the solvable Lie algebra g = hxi⋉V,
2
where V is an abelian Liealgebra over an algebraically closed field ofcharacteristic 0
b andxisanarbitraryautomorphismofV.
e
F
8
1. Introduction
]
T Let F be analgebraicallyclosedfieldofcharacteristic0. Allvectorspaces, including
R allLiealgebrasandtheirmodules,areassumedtobefinitedimensionaloverF.
. Recallthatamoduleissaidtobeindecomposableifitcannotbedecomposedasthedi-
h
rectsumoftwonon-trivialsubmodules. Naturally,knowingallindecomposablemodules
t
a of a given Lie algebra would provide a complete description of all its modules. Unfor-
m
tunately, the problem of classifying all indecomposable modules of a given Lie algebra
[ -thatisnotsemisimpleorone-dimensional-isvirtuallyunsolvable,eveninthecaseofthe
two-dimensional abelian Lie algebra, as observed in a celebrated paper by Gelfand and
2
v Ponomarev[GP].
4 In spite of this fact, many types of indecomposable modules of non-semisimple Lie
8 algebras have been recently classified, see for example [CGS, CMS, CS, CS1, D, DdG,
0
DP,DR,P]
0
Inallthese papersthecentralidea isto considerparticularclasses ofindecomposable
0
. modulesforwhichacompleteclassificationcanbeachieved.Besidestheirreduciblemod-
2
ules,thesimplesttypeofindecomposablemoduleis,inacertainsense,theuniserialone.
0
Thisisa modulehavingauniquecompositionseries, i.e. anon-zeromodulewhosesub-
7
1 modulesformachain. Alternatively,suchmodulescanbedefinedasfollows.
: LetgbeagivenLiealgebraandletU beanon-zerog–module.Thesocleseries
v
i 0=soc (U)⊂soc (U)⊂···⊂soc (U)=U
X 0 1 k
r ofU isinductivelydefinedbydeclaringsoci(U)/soci−1(U)tobethesocleofU/soci−1(U),
a
thatis, thesum ofallirreduciblesubmodulesofU/soc (U), for1 ≤ i ≤ k. ThenU is
i−1
uniserialifandonlyifthesocleseriesofU hasirreduciblefactors.
Inthelastyears,theclassificationoftheuniserialmodulesofimportantclassesofsolv-
ableandperfectLiealgebrashasbeenachievedinvariousresearchpapers[CGS,CS,CS1,
Pi, C]. In particular, [Pi] and [C] classify a wider class of modules, called cyclic in [Pi]
andperfectcyclicin[C],overtheperfectLiealgebrassl(2)⋉F2 andsl(n+1)⋉Fn+1,for
F =C,respectively.
Theaimofthispaperistoproceedfurtherinthestudyofuniserialmodules. Weshall,
indeed, classify the uniserial modules of a distinguished class of solvable Lie algebras,
namely those of the form g = hxi ⋉ V, where V is an abelian Lie algebra and x is an
arbitraryautomorphismofV.
2010MathematicsSubjectClassification. 17B10,17B30.
Keywordsandphrases. uniserialmodule;indecomposablemodule;Clebsch-Gordanformula.
1
2 PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
A properidealaof g isofthe forma = W, whereW isan x-invariantsubspaceof V.
Thus, either W = V and g/a (cid:27) hxi is one-dimensional, or else g/a (cid:27) hxi ⋉ V, where
V =V/W ,(0)andxistheautomorphismthatxinducesonV.
Weknowfrom[CS]alluniserialmodulesoveranabelianLiealgebraaswellasalluni-
serialg-moduleswhenxisdiagonalizable.Thus,itsufficestoclassifyallfaithfuluniserial
g-moduleswhenxisnotdiagonalizable.Inthisregard,ourmainresultsareasfollows.
In §2 we construct a family of non-isomorphic faithful uniserial representations of g
whenxactsonV viaasingleJordanblockofsizen>1. Thisfamilyconsistsofallmatrix
representations
(1.1) R →gl(n+1), R →gl(n+1), R →gl(n+1),
α,k,X α,n α,1
where
α∈ F, 1<k <n, X ∈ M ,
k−1,n−k
aswellasthematrixrepresentations
(1.2) R →gl(n+2),
α,a
whichexistonlyforoddn,andwhere
α∈F, a=(a ,...,a ), a =1, a =0foralleveni.
1 n 1 i
In §3, we show that if x acts on V via a single Jordan block of size n > 1, then ev-
ery faithful uniserial g-module is isomorphic to one and only one of the representations
appearingin(1.1)and(1.2).
In §4 we deal with the general case. By our results from §3, we may assume that x
has e Jordan blocks, where e > 1. Moreover, as indicated above, we may also assume
that x isnotdiagonalizable. Underthese assumptions. Theorem4.1givesnecessaryand
sufficientconditionsforgtohaveafaithfuluniserialmoduleandclassifiesallsuchmodules
whenevertheseconditionsaresatisfied.
Indeed,let
(1.3) V =V ⊕···⊕V
1 e
beadecompositionofV intoindecomposableF[x]-submodules(thismeansthatxactson
eachV viaaJordanblock)ofdimensions
i
n=n ≥···≥n ,
1 e
wheren > 1because xisnotdiagonalizable. Foreach1 ≤ i ≤ e,considerthesubalgebra
g =hxi⋉V ofg,wherex = x| .
i i i i Vi
SupposethatghasafaithfuluniserialrepresentationR:g→gl(d).Thentherestriction
R :g →gl(d)isalreadyuniserial.Inparticular,
1 1
d =n+1ord =n+2.
In the first case, R is isomorphic to a unique R , the automorphism x has a single
1 α,k,X
eigenvalueλ,andtheJordandecompositionofxis
(1.4) Jn(λ)⊕Jn2(λ)⊕···⊕Jne(λ),
where
(1.5) n ≤n−2, n ≤n−4, n ≤n−6,..., n ≤n−2(e−1),
2 3 4 e
and
(1.6) e≤min{k,n+1−k}.
INDECOMPOSABLEMODULESOFSOLVABLELIEALGEBRAS 3
In the second case, R is isomorphicto a uniqueR , n is odd, the automorphism x has
1 α,a
twoeigenvaluesλand2λ,andtheJordandecompositionofxis
(1.7) Jn(λ)⊕J1(2λ),
sothate=2andn =1.
2
Conversely,if x hasa single eigenvalue,1 < k < n, and(1.5)-(1.6) are satisfied, then
R can be extendedto a faithfuluniserial representationof g. In fact, let M be
α,k,X k,n+1−k
subspaceofgl(n+1)consistingofallmatrices
0 N
, N ∈ M .
0 0 ! k×n+1−k
LetvbeageneratoroftheF[x]-moduleV andset
1
Jk(α) 0
A=R (x)= ∈gl(n+1), E =R (v)∈M .
α,k,X 0 Jn+1−k(α−λ) ! α,k,X k,n+1−k
Then the extensions of R to a faithful uniserial representation of g are given by all
α,k,X
possibleF[t]-monomorphismsV →M suchthatv→E,wheretactsviaad x−λ1
k,n+1−k g g
on V and via ad A − λ1 on M . Moreover, all such extensions produce
gl(n+1) gl(n+1) k,n+1−k
non-isomorphicrepresentationsofg. Illustrativeexamplesareprovidedin§5.
Likewise, if x hasJordandecomposition(1.7), then R can be extendedto a faithful
α,a
uniserialrepresentationofg.Wedetermineallsuchextensionsandprovethattheyproduce
non-isomorphic representations of g (this case is much simpler than the above and no
examplesarerequired).
Finally, a necessary and sufficient condition for g to have a faithful uniserial repre-
sentation is that x has Jordan decomposition (1.4) and (1.5) holds, or that x has Jordan
decomposition(1.7)andnisodd.
Perhapssurprisingly, the representationtheory of sl(2), and in particularthe Clebsch-
Gordanformula,playsadecisiveroleinourstudyandclassificationofuniserialg-modules.
2. Constructionofuniserialrepresentations
Given p ≥ 1 and α ∈ F, we write J (α) (resp. Jp(α)) for the lower (resp. upper)
p
triangularJordanblockofsize p andeigenvalueα. We also let Ei,j ∈ gl(p)standforthe
matrixwithentry(i, j)equalto1andallotherentriesequalto0.
Wesupposethroughoutthissectionthatg=hxi⋉V,whereV isanabelianLiealgebra
andx∈gl(V)actsonV viaasingle,lowertriangular,Jordanblock,say J (λ),relativetoa
n
basisv ,...,v ofV. Thecaseλ = 0isallowed. Themultiplicationtableforgrelative
0 n−1
toitsbasisx,v ,...,v is:
0 n−1
(2.1) [x,v ]=λv +v ,[x,v ]=λv +v ,...,[x,v ]=λv .
0 0 1 1 1 2 n−1 n−1
Wemaytranslate(2.1)into
(2.2) (ad x−λ1 )kv =v , 0≤k≤n−1,
g g 0 k
and
(2.3) (ad x−λ1 )nv =0.
g g 0
Proposition2.1. Givenpositiveintegers p,q,letM besubspaceofgl(p+q)consisting
p,q
ofallmatrices
0 N
N = , N ∈ M .
0 0 ! p×q
b
4 PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
Givenα,λ∈F,set
A= Jp(α)⊕Jq(α−λ).
Letθstandfortheendomorphismad A−λ1 ofgl(p+q)restrictedtoitsinvariant
gl(p+q) gl(p+q)
subspaceM . Thenθisnilpotentwithelementarydivisors
p,q
tp+q−1,t(p+q−1)−2,t(p+q−1)−4,...,t(p+q−1)−2z,
where
z=min{p−1,q−1}.
Moreover,givenanyN ∈ M ,theminimalpolynomialofN withrespecttoθistp+q−1 if
p×q
andonlyifN ,0.
p,1
b
Proof. For a ≥ 0, let V(a) stand for the irreduciblesl(2)-modulewith highest weight a.
TheClebsch-Gordanformulastatesthat
(2.4) V(a)⊗V(b)(cid:27)V(a+b)⊕V(a+b−2)⊕V(a+b−4)⊕···⊕V(a+b−2r),
where
r =min{a,b}.
Let
0 1 1 0 0 0
e= ,h= , f =
0 0 ! 0 −1 ! 1 0 !
standforthecanonicalbasisofsl(2). Itiswell-knownthateactsnilpotentlyonV(a)with
asingleelementarydivisor,namelyta+1. ItfollowsfromtheClebsch-Gordanformulathat
eactsnilpotentlyonV(a)⊗V(b)withelementarydivisors
ta+b+1,t(a+b+1)−2,t(a+b+1)−4,...,t(a+b+1)−2r.
Fora≥0,letR :sl(2)→gl(a+1)bethematrixrepresentationaffordedbyV(a)givenby
a
(2.5) R (h)=diag(a,a−2,...,−a+2,−a),
a
(2.6) R (e)= Ja+1(0),
a
(2.7) R (f)=diag(0,a,2(a−1),3(a−2),...,3(a−2),2(a−1),a)J (0).
a a+1
Now
V(a)⊗V(b)(cid:27)V(a)∗⊗V(b)(cid:27)Hom(V(a),V(b)),
where
(2.8) (y·φ)(v)=y·φ(v)−φ(y·v), y∈sl(2),φ∈Hom(V(a),V(b)),v∈V.
Itfollowsthat
(2.9) V(a)⊗V(b)(cid:27)M ,
a+1,b+1
where
0 N R (y) 0 0 N
(2.10) y· = a , , y∈sl(2).
0 0 ! " 0 Rb(y) ! 0 0 !#
Ontheotherhand,lettingB= Jp(0)⊕Jq(0),wereadilyverifythat
(ad B)N =(ad A−λ1 )N, N ∈ M ,
gl(p+q) gl(p+q) gl(p+q) p×q
whichmeansthatθistherestbrictionofadgl(p+q)BtoMp,qb. Settinga= p−1andb=q−1
andusing(2.6)aswellas(2.10),wededucethatθisnothingbuttheactionofeonM .
a+1,b+1
ThestatedelementarydivisorsforθnowfollowfromthoseoftheactionofeonV(a)⊗V(b).
INDECOMPOSABLEMODULESOFSOLVABLELIEALGEBRAS 5
Using (2.5) and (2.10) we find that, for 0 ≤ i ≤ r, the h-eigenspace of M with
p,q
eigenvalue−(a+b)+2i,sayS(i),consistsofallQsuchthattheentriesofQoutsideofits
ithlowerdiagonalareequalto0. Herethe0thlowerdiagonalconsistsofposition(p,1),
b
the 1st lower diagonalof positions (p,2),(p−1,1), the 2nd lower diagonalof positions
(p,3),(p−1,2),(p−2,1),andsoon.
EachlowestweightvectorofM generatesanirreduciblesl(2)-submodule.Inviewof
p,q
themultiplicity-freedecomposition(2.4)andtheisomorphism(2.9),weseethatforeach
0≤i≤r,thereisoneandonlyone0, E(i)∈S(i),uptoscaling,suchthat
(2.11) b fd·E(i)=0.
LettingW(i)bethesl(2)-submodulegeneradtedbyE(i),wehave
(2.12) Mp,q =W(0)⊕···d⊕W(r).
GivenanarbitraryN ∈ M ,letuswriteN intermsof(2.12). Wehave
p×q
N =w(0)+w(1)··b·+w(r), w(i)∈W(i),
where, b
w(0)=α E(0)+α e·E(0)+···+α ea+b·E(0), α ∈ F.
0 1 a+b i
FromthefirstpartoftheTheorem,weknowthat, relativetotheactionofe, theminimal
d d d
polynomialof E(0) is ta+b+1, while ta+b−1 annihilates all w(i), i > 0. It follows that the
minimal polynomialof N is ta+b+1 if and only if α , 0. On the other hand, given that
d 0
E(i)∈S(i),itfollowsthateveryE(i),i>1,hasentry(p,1)equalto0,whereasentry(p,1)
b
ofE(0)isnot0.Moreover,using(2.6)and(2.10)wefindthatifP=e·Q,thenentry(p,1)
d
ofPisequalto0foranyQ. Thus,N ,0ifandonlyifα ,0,asrequired. (cid:3)
p,1 0
b b
Proposition2.2. Givenα ∈ F, positiveintegers p,q,and N ∈ M suchthat N , 0,
p×q p,1
considerthelinearmapR=R :g→gl(p+q)givenby
α,p,q,N
Jp(α) 0
x7→A=
0 Jq(α−λ) !
and
0 N
v 7→(ad A−λ1 )k , 0≤k ≤n−1.
k gl(p+q) gl(p+q) 0 0 !
ThenRisarepresentationofgifandonlyif p+q−1 ≤ n,inwhichcaseRisuniserial.
Moreover,Risafaithfulrepresentationifandonlyif p+q−1=n.
Proof. Byconstruction,Rpreservesthefollowingrelationsofg:
[v,v ]=0, (ad x−λ1 )kv =v , 0≤k≤n−1.
i j g g 0 k
On the other hand, due to Proposition 2.1, (ad A − λ1 )nN = 0 if and only if
gl(p+q) gl(p+q)
p+q−1≤n,whichmeansthatRpreservesthelastdefiningrelationofg,namely
b
(ad x−λ1 )nv =0,
g g 0
if and only if p+q−1 ≤ n. Thus, condition p+q−1 ≤ n alone determines whether
R is a representationor not. Suppose that indeed p+q−1 ≤ n. It is obviousthat R is
uniserial. Moreover,R(v ),...,R(v )arelinearlyindependentifandonlyiftheminimal
0 n−1
polynomialofNwithrespecttoad A−λ1 hasdegreen. ByProposition2.1,this
gl(p+q) gl(p+q)
happensifandonlyif p+q−1=n. (cid:3)
b
6 PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
Givenα∈F,1≤k≤n,andX ∈ M ,weset p=k,q=n+1−kand
k−1,n−k
0 X
N = ∈ M .
1 0 ! k,n+1−k
ThenProposition2.2ensuresthat
R =R :g→gl(n+1)
α,k,X α,p,q,N
is a faithfuluniserial representationof g. In the extreme cases k = n and k = 1 there is
noX,andN isrespectivelyequalto
0
.
01.. ∈ Mn×1and(1,0,...,0)∈ M1×n.
ThecorrespondingrepresentationswillberespectivelydenotedbyR andR
α,n α,1
Givenanyα∈ Fanda=(a ,...,a )∈Fnsuchthata =1,weconsiderthelinearmap
1 n 1
R :g→gl(n+2)definedasfollows:
α,a
α 0 0
x7→A= 0 Jn(α−λ) 0 ,
0 0 α−2λ
0 a 0
vk 7→(adgl(n+2)A−λ1gl(n+2))k 0 0 en , 0≤k ≤n−1,
where{e ,...,e }isthecanonicalbasisofthe0col0umn0spaceFn.
1 n
Lemma2.3. R isarepresentationofgifandonlyifnisoddanda =0foralleveni,in
α,a i
whichcaseR isuniserial.
α,a
Proof. Bydefinition,R preservesrelations(2.2)and(2.3).WenextdeterminewhenR
α,a α,a
preservesrelations[v,v ]=0. LettingN = Jn(0),wehave
i j
0 a 0 0 (−1)kaNk 0
(adgl(n+2)A−λ1gl(n+2))k 0 0 en = 0 0 Nken , 0≤k≤n−1.
Thus[R(v ),R(v )]=0iff(−01)ka0Nk+0je=(−10)jaNk+j0e iffa 0 =0fork+ jodd.Since
k j n n n−k−j
a =1,thelastconditionisequivalenttonoddanda =0,foranys. Thisprovesthefirst
1 2s
assertion.Asuniserialityisclear,theproofiscomplete. (cid:3)
Finally, in the extreme case n = 1, given any α ∈ F and ℓ ≥ 2 we have the faithful
uniserialrepresentationT :g→gl(ℓ)givenby
α,ℓ
x7→diag(α,α−λ,...,α−(ℓ−1)λ), v 7→ Jℓ(0).
0
Definition2.4. Givenpositiveintegersℓ,d,d ,...,d suchthatd +···+d =dandℓ>1,
1 ℓ 1 ℓ
andamatrixA ∈ M ,weconsiderAaspartitionedintoℓ2 blocksA(i, j) ∈ M . Wesay
d di×dj
thatAisblockuppertriangularifA(i, j)=0foralli> j,andstrictlyblockuppertriangular
ifA(i, j)=0foralli≥ j. If0≤i≤ℓ−1,bytheithblocksuperdiagonalofAwemeanthe
ℓ−iblocksA(1,1+i),A(2,2+i),...,A(ℓ−i,ℓ).
INDECOMPOSABLEMODULESOFSOLVABLELIEALGEBRAS 7
Lemma2.5. Givenpositiveintegersℓ,d1,...,dℓ,withℓ>1,setJi = Jdi(0),1≤i≤ℓ,and
letGbethesubgroupofGL(d)consistingofallX ⊕···⊕X suchthatX ∈U(F[J]).Let
1 ℓ i i
E ∈ M bestrictlyblockuppertriangular,withdiagonalblocksofsizesd ×d ,...,d ×d .
d 1 1 ℓ ℓ
Let E ∈ M ,...,E ∈ M be the blocksin the first block superdiagonalof E,
1 d1×d2 ℓ−1 dℓ−1×dℓ
andsupposethatthebottomleftcornerentryofeachE isnon-zero.
i
Then E isG-conjugateto a matrix H suchthateachofthe blocks H ,...,H in the
1 ℓ−1
first block superdiagonal of H has first column equal to the last canonical vector, and
H haslastrowequaltothefirstcanonicalvector. Likewise, E isalsoG-conjugatetoa
ℓ−1
matrix H such thateach ofthe blocks H ,...,H in the first block superdiagonalof H
1 ℓ−1
haslastrowequaltothefirstcanonicalvector,andthefirstcolumnof H isequaltothe
1
lastcanonicalvector.
Proof. RecallthatA ∈ F[Jp(0)]ifandonlyifAisuppertriangularandallitssuperdiago-
nalshaveconstantvalue.Forinstance,atypicalelementofA∈F[J4(0)]hastheform
α β γ δ
0 α β γ
(2.13) A= .
It is clear that the unit groupof F[Jp(000)] ac00ts trα0ansαβitivelyfrom the left (right) on the set
ofcolumn(row)vectorsofFp thathavenon-zerolast(first)entry. Moreover,itisequally
clearthatifU ∈ U(F[Jp(0)])and B ∈ M (resp. B ∈ M )thenthefirstcolumn(resp.
q×p p×q
lastrow)of BU (resp. UB)isthatof Bscaledbya non-zeroconstant. Itfollowsatonce
fromtheseconsiderationsthatwecanfindX ,...,X (resp. X ,...,X )sothatforanyX
1 ℓ−1 2 ℓ ℓ
(resp. X )theresultingX ∈GwillconjugateE intoamatrixH suchthatthefirstcolumn
1
(resp. last row) of every H is equal to a non-zeroscalar multiple, say by α, of the last
i i
(resp. first)canonicalvector. MakingasecondselectionofscalarmatricesY ,...,Y and
1 ℓ
conjugatingHbytheresultingY =Y ⊕···⊕Y ,wecanmakeallα =1above.Finally,by
1 ℓ i
suitablychoosingX (resp. X )withwith1’sonthediagonalandtakingallotherX =1 ,
ℓ 1 i di
we can makethe last row (resp. firstcolumn)of H (resp. H ) equalto the first (resp.
ℓ−1 1
last)canonicalvector. (cid:3)
Proposition2.6. Supposeλ,0andn>1. ThentherepresentationsR ,R ,R and
α,k,X α,n α,1
R arenon-isomorphictoeachother.
α,a
Proof. Considering the eigenvalues of the image of x as well as their multiplicities, the
only possible isomorphisms are easily seen to be between R and R , or R and
α,k,X α,k,Y α,a
R .
α,b
SupposefirstT ∈GL(n+1)satisfies
TR (y)T−1 =R (y), y∈g.
α,k,X α,k,Y
ThenT commuteswithR (x)= Jk(α)⊕Jn+1−k(α−λ),andthereforeT =T ⊕T ,where
α,k,X 1 2
T (resp. T )isapolynomialinJk(0)(resp. Jn+1−k(0))withnon-zeroconstantterm. Thus
1 2
TR (v )T−1 =R (v )
α,k,X 0 α,k,Y 0
translatesinto
0 X 0 Y
(2.14) T = T
1 1 0 ! 1 0 ! 2
Explicitly writing T and T , as in (2.13), we infer from (2.14) that T = α1 = T ,
1 2 1 n+1 2
whenceX =Y.
8 PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
SupposenextS ∈GL(n+2)satisfies
SR (y)S−1 =R (y), y∈g.
α,a α,b
Asabove,S = J1(β)⊕S ⊕J1(γ),whereS isapolynomialinJn(0)withnon-zeroconstant
2 2
termandβ,γarenon-zero,butthen
SR (v )=R (v )S
α,a 0 α,b 0
forcesS tobeanon-zeroscalarmatrix,whencea=b.
(cid:3)
3. Classificationofuniserialrepresentations
Lemma3.1. LetT :h→kbeahomomorphismofLiealgebras.Then
T((ad y−µ1 )kz)=(adT(y)−µ1)kT(z), y,z∈h,µ∈ F,k≥0.
h h k k
Proof. Thisfollowseasilybyinduction. (cid:3)
Theorem3.2. ConsidertheLiealgebrag=hxi⋉V,whereVisanabelianLiealgebraand
x ∈ GL(V)actsonV viaasingleJordanblock J (λ),λ , 0. LetRbeafaithfuluniserial
n
representationofg. Then
(a)Ifn>1thenRisisomorphictooneandonlyoneoftherepresentationsR ,R ,
α,k,X α,n
R ,R .
α,1 α,a
(b)Ifn=1thenRisisomorphictooneandonlyoneoftherepresentationsT .
α,ℓ
Proof. LetU beafaithfuluniserialg-module,sayofdimensiond. Lie’stheoremensures
theexistenceofabasisB={u ,...,u }ofU suchthatthecorrespondingmatrixrepresen-
1 d
tationR:g→gl(d)consistsofuppertriangularmatrices.
Since x ∈GL(V),wehave[g,g]=V,whenceR(v)isstrictlyuppertriangularforevery
v∈V. Set
A=R(x)andE =R(v ), 0≤k ≤n−1.
k k
Inviewof[CS,Lemma2.2],wemayassumethatAsatisfies:
(3.1) A =0wheneverA , A .
ij ii jj
Moreover,from[CS,Lemma2.1],weknowthatforevery1 ≤ i < d thereissomey ∈ g
i
suchthat
(3.2) R(y) ,0.
i i,i+1
Step1. IfA , A then(E ) ,0andA −A =λ.
i,i i+1,i+1 0 i,i+1 i,i i+1,i+1
Indeed,(2.2),(2.3)andLemma3.1imply
(3.3) (ad A−λ1 )kE =E , 0≤k ≤n−1,
gl(d) gl(d) 0 k
and
(3.4) (ad A−λ1 )nE =0.
gl(d) gl(d) 0
SinceAisuppertriangularandE isstrictlyuppertriangular,(3.3)and(3.4)give
0
(3.5) (A −A −λ)k(E ) =(E ) , 0≤k≤n−1,1≤i<d.
i,i i+1,i+1 0 i,i+1 k i,i+1
and
(3.6) (A −A −λ)n(E ) =0, 1≤i<d.
i,i i+1,i+1 0 i,i+1
INDECOMPOSABLEMODULESOFSOLVABLELIEALGEBRAS 9
Fixisuchthat1≤i<dandA ,A . By(3.1),wehave
i,i i+1,i+1
(3.7) A =0.
i,i+1
Combining(3.2),(3.5)and(3.7)weobtain
(3.8) (E ) ,0.
0 i,i+1
From(3.6)and(3.8)wededuce
A −A =λ.
i,i i+1,i+1
Step2. Wehave
(3.9) A= A ⊕···⊕A , A ∈gl(d),
1 ℓ i i
whereeachA hasscalardiagonal,sayofscalarα,and,settingα=α ,wehave
i i 1
α =α−(i−1)λ.
i
Thisfollowsatoncefrom(3.1)andStep1.
Step 3. Let us write each E in block form compatiblewith (3.9), that is, with diagonal
k
blocksofsizesd ×d ,...,d ×d . ThenalldiagonalblocksofeveryE areequalto0.
1 1 ℓ ℓ k
Indeed,supposei≤ jand
A =···=A .
i,i j,j
SettingUr =span{u ,...,u },weseethatthesection
1 r
Ui,j =Uj/Ui−1
of U is a (uniserial) g-module of dimension e = j − i + 1. Let T : g → gl(e) be the
corresponding matrix representation relative to the basis u + Ui−1,...,u + Ui−1 of U.
i j
ThenT(x)isuppertriangularwithscalardiagonal,soad T(x)isnilpotent.Ontheother
gl(e)
hand,sinceT isaLiehomomorphism,Lemma3.1gives
(ad T(x)−λ1 )nT(v )=0, 0≤k≤n−1.
gl(e) gl(e) k
ItfollowsthateveryT(v )isageneralizedeigenvectorofad T(x)forthedistincteigen-
k gl(e)
valuesλand0. WeinferthateveryT(v ) = 0. Itfollowsthatalldiagonalblocksofevery
k
E areequalto0.
k
Step4. ReferringtotheblockdecompositionofE usedinStep3,ifi < jand j , i+1,
k
thenblock(i, j)ofE is0forall0≤k ≤n−1.
k
Indeed,recallingDefinition2.4,welet
S(1),S(2),...,S(ℓ−1)
bethesubspacesofgl(d)correspondingtotheblocksuperdiagonals1,2,...,ℓ−1,andset
S =S(1)⊕···⊕S(ℓ−1).
ThenS(i)isthe generalizedeigenspaceofad AactingonS forthe eigenvalueiλ, for
gl(d)
all1 ≤ i ≤ ℓ−1. Ontheotherhand,everyE ∈ S byStep3,while(3.3)and(3.4)imply
k
thatevery E belongsto the generalizedeigenspaceof ad A forthe eigenvalueλ. We
k gl(d)
concludethateveryE isinS(1).
k
10 PAOLOCASATI,ANDREAPREVITALI,ANDFERNANDOSZECHTMAN
Step5. WemayassumewithoutlossofgeneralitythatAisinJordanform
(3.10) Jd1(α)⊕Jd2(α−λ)⊕···⊕Jdℓ(α−(ℓ−1)λ).
Indeed,by(3.2)andStep3,thefirstsuperdiagonalofeveryA appearingin(3.9)con-
i
sists entirely of non-zero entries. Thus, for each 1 ≤ i ≤ ℓ there is X ∈ GL(d) such
i i
that
XiAiXi−1 = Jdi(α−(i−1)λ).
Set
X =X ⊕···⊕X ∈GL(d).
1 ℓ
Then XAX−1 is equal to (3.10) and XE X−1 is strictly block upper triangular with each
k
block(i, j), j,i+1,equalto0.
Step6. Ahasatleast2Jordanblocks.
Ifnot,V isannihilatedbyR,bySteps3and4,contradictingthefactthatRisfaithful.
Step7. d +d ≤n+1forall1≤i<ℓ.
i i+1
ApplyProposition2.2tosuitablesectionsofU.
Step8.Withoutlossofgeneralitywemayassumethatthefirstcolumnofeachblockalong
the first block superdiagonalof E is equal to the last canonicalvector, and that the last
0
rowofthelastoftheseblocksisequaltothefirstcanonicalvector.
ThisfollowsfromLemma2.5.
Final Stepwhen n = 1. Supposen = 1. Thenalld = 1byStep 7,so Steps5, 6and8
i
yield that R is isomorphic to a representation T . As these representations are clearly
α,ℓ
non-isomorphictoeachother,theTheoremisproveninthiscase.
Weassumefortheremainderoftheproofthatn>1.
Step9. AhasatleastoneJordanblockofsize>1.
Ifnot,d +d < n+1forallibyStep7. Sincethe x-invariantsubspacesofV forma
i i+1
chain,itfollowsfromProposition2.2thatR(v )=0,contradictingthefaithfulnessofR.
n−1
Step10.LetJdi(α−(i−1)λ)beaJordanblockofAofsize>1ofA,asensuredbyStep9.
Theni≥ℓ−1andi≤2.
Letusfirstseethati≥ℓ−1. Suppose,ifpossible,thatAhasconsecutiveJordanblocks
Ja(β),Jb(β−λ),Jc(β−2λ)witha>1.
Case1. b = 1. ConcentratingonasuitablesectionofU,asintheproofofStep3,wesee
thatghasamatrixrepresentationP:g→gl(4)suchthat
β 1 0 0 0 0 0 0
0 β 0 0 0 0 1 0
P(x)= , P(v )= .
HeretheshapeofP(E )00ise00nsuβre−0dλbySβte−0p2sλ3,4and80. Now00 00 00 10
0
P(v )=[P(x),P(v )]−λP(v )=E1,3,
1 0 0
