Ver\"anderungen \"uber einen Satz von Timmesfeld - I. Quadratic Actions

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by  Adrien Deloro

Veränderungen
3 über einen Satz von Timmesfeld
1
0
I. Quadratic Actions
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u Adrien Deloro
A
December 12, 2013

5

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R
Ich hätteglücklich geendet, aberdiese Nro. 30, das Thema, rißmich
G unaufhaltsam fort. Die quartblätter dehnten sich plötzlich aus zu
. einem Riesenfolio, wo tausend Imitationen und Ausführungen jenes
h
Themas geschrieben standen, die ich abspielen mußte.
t
a
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[

Abstract
2
v We classify quadratic SL2(K)- and sl2(K)-modules by crude compu-
2 tation, generalizing in the first case a Theorem proved independently by
8 F.G. Timmesfeld and S. Smith. The paper is the first of a series dealing
1 with linearization results for abstract modules of algebraic groups and
0
associated Lie rings.
.
1
0
3
General Foreword
1
:
v
Myhopeisthattheresultsgatheredhereafterwillsuggesttothereadera
i
X simpleidea: awholechapterofrepresentation theorycouldbewrittenat
the basic level of computations, without the help of algebraic geometry.
r
a For I wish in this article and in others which may follow to study how
much of geometric information on modules is already prescribed by the
inner constraints of algebraic groups seen as abstract groups. In a sense
theproblemisakintotheonesolvedbyBorelandTitsintheircelebrated
work on abstract homomorphisms of algebraic groups. But here instead
of morphisms between abstract groups we deal with abstract modules.
The central question is thefollowing.
LetKbeafieldandGbetheabstractgroupofK-pointsofan
algebraic group. Is every G-modulea KG-module?
Keywords: quadraticaction,G-module,Lieringsl(2,K),Lieringrepresentation
MSC2010: 20G05, 20G15,17B10, 17B45
1
It would be obscene to hope for a positive answer. The question is
not asked literally; one should at least require the algebraic group to be
reductive, if not simple. Moreover it may be necessary to bound the
complexity of modules in some sense yet tobe explained.
Since there is no K-structure a priori and therefore no notion of a
dimension over K, one may focus on actions of finite nilpotence length,
that iswhere unipotentsubgroupsact unipotently. Thissettingseems to
me more natural than that of M modules (where centralizer chains are
C
stationary); to support this impression one may bear in mind that the
class of M modules is not stable by going to a quotient, an operation
C
which is likely to be relevant here. One could also make various model-
theoretic assumptions, hoping that they would force configurations into
the world of algebraic geometry; this did not seem natural either, since
the first computations one can make are much too explicit for logic to
play a deep role here. The future might bring contrary evidence; as for
now, purenilpotence seems more relevant.
So let us makeour question more precise.
Let K be a field and G be an algebraic group. Let G = GK.
Understand therelationships between:
• KG-modulesoffinitelength• G-modules of finite length.
One may be tempted to tackle the question by reducing it to actions
of theassociated Lie algebra. Two difficulties appear.
• TheLiealgebra“remembers”thebasefield,inasensewhichweshall
notexplicithere;inanycaseonereadilyseesthatactionsoftheLie
algebracantakeplaceonlyinnaturalcharacteristic(thatofthebase
field). YetthegroupGdoesnotnecessarilyrememberitsbasefield,
sincetherearevariousisomorphisms theextremecasesofwhichare
over finite fields, such as SL (F ) ≃ PSL (F ). These pathologies
3 2 2 7
canbeeliminatedbyreasonableassumptionsonthealgebraicgroup
and on thefield, and we shall deal only with decent cases.
• Above all the Lie algebra g which is a K-vector space, can appear
only if V is already equipped with a K-linear structure; since it is
not a priori, all one can hope for instead is the Lie ring, which is
the(neitherassociativenorunitary)ringunderlyingtheLiealgebra
when one forgets its vector space structure. The representations of
theLiealgebraareexactlytheKg-modules,wheregisseenasaLie
ring. Similarly, the universal object in this context will not be the
enveloping algebra, but theenveloping ring.
And anyway nothing guarantees that reducing a group action to an
action of its Lie ring (if possible) is any simpler than directly linearizing
themodule. ItthuslooksliketheintroductionoftheLiealgebrawill not
solve any question but bring new ones. Our central problem extends as
follows:
LetKbeafieldandGbeanalgebraicgroup. LetG=GK and
g=(LieG)K beitsLiealgebra,seenasaLiering. Understand
therelationships between:
2
• KG-modulesoffinitelength• Kg-modules of finitelength
• G-modules of finitelength • g-modulesof finitelength.
Iwanttospeakfortheideathatthereareindeedgoodcorrespondences
between these categories. Here again this should not be taken literally:
onemayrequirethefieldtohavesufficientlylargecharacteristicandmany
roots of unity.
Thearchetypeofaneffectivelinearizationisthefollowingresult,proved
independentlybyStephenSmithandFranzGeorgTimmesfeld (thelatter
mathematician actually did not require simplicity).
AsimpleSL (K)-moduleonwhichtheunipotentsubgroupacts
2
quadratically is a KSL (K)-module.
2
It is not known whether the same holds over a skew-field. We shall not
enter the topic, as all our results rely on heavy use of the Steinberg rela-
tions. Actuallyanalternativetitlemighthavebeen: “G-modulesandthe
Steinberg relations”.
ThepresentworkisconstructedasaseriesofvariationsontheSmith-
Timmesfeld theme, showing theunexpectedrobustness of theunderlying
computation. Encountered difficulties and provable results will provide
equally important information: one must determine the limits of this
computation in order to understandits deep meaning.
• These variations will not be to the taste of geometers: from their
point of view, I shall state only partial trivialities in an inadequate
language. But to defend these pages, and with a clear sense of pro-
portions,IwillappealtotheBorel-Titsfamousresult. Theideanow
is to understand to what extent inner constraints of abstract struc-
tures determine their representation theory. There is no rational
structurehere; everythingis done elementarily.
• These variations could perhaps amuse group theorists, who must
sometimesdealwithaustereobjectswithnocategoricalinformation.
Experts in finite group theory will nonetheless be upset by thelack
of depth of my results, and by the efforts they cost: but the fields
here may be infinite,and there is no character theory.
• Thevariationsmayatleastbeusefultologicians. Thosewithanin-
terest inmodel-theoretic algebra often encounterabstract permuta-
tiongroups;thesesometimesturnouttobegroupsactingonabelian
groups, and one needs results from more or less pure group theory
to complete thediscussion.
I confess that the present work takes place in a general context, far
from model theory: I got carried away by the subject. To conclude this
general foreword, I would love as much as the reader to suggest a con-
jecturedescribinginprecisetermsaphenomenonof“linearityofabstract
modules of structures of Lie type”;I would loveto but I cannot, because
it is too early.
My heart-felt thanks to Alexandre Borovik, for he believes in crazy
ideas.
3
1 The Setting
In this article we study quadratic actions of SL (K) and sl (K) on an
2 2
Expanded. abelian group.
Thearticlescitedinthenextfewparagraphsarebynomeansrequired
in order to understand the hopefully self-contained present work. Only
the reader with some knowledge of the topic will find interest in this in-
troduction;theotherreadermayfreelyskipit. Suchaliminarydigression
is merely meant to provide some historical background on the notion of
quadraticity which lies at the center of our first article. The results we
shall quote are not used anywhere and they bear no relationship to the
rest of theseries nor toits general spirit.
To the readerversed in finitegroup theory theword quadraticity will
certainly evoke a line of thought initiated by J. Thompson: the classifi-
cation of quadratic pairs, consisting of a finite group and a module with
certain properties which we need not make precise. J. Thompson’s semi-
nal yet unpublished work [6] was quite systematically pursued by Ho [3]
among others, and more recentlycompleted byA.Chermak [1]using the
classification of the finite simple groups. This strain of results aims at
pushing the group involved in a quadratic pair towards having Lie type.
Its purposemay therefore be called group identification.
As A. Premet and I. Suprunenko [4] put it, in [6] and [3] “groups
generated by quadratic elements are classified as abstract finite groups
and corresponding modules are not indicated explicitly.” The article [4]
by A. Premet and I. Suprunenko we just quoted attempts at remedying
the lack of information on themodule by listing finite groups of Lie type
and representations thereof such that the pair they form is quadratic in
J. Thompson’s sense. This orthogonal line could conveniently be named
representation zoology. Yet one then deals with a representation instead
of a general module and this is much more accurate data.
Asamatteroffact weshalladoptneitherthegroup identification nor
the representation zoology approach but a third one which qualifies as
module linearization: given amore-or-less concretegroupofLietypeand
an abstract module, can one retrieve a linear structure compatible with
the action? Such a trend can be traced to a result of G. Glauberman
[2,Theorem4.1]whichhavingamongitsassumptionsbothfinitenessand
quadraticity turns an abelian p-group into a sum of copies of the natu-
ral SL2(Fpn)-module. Following S. Smith [5, Introduction], it was F.G.
Timmesfeld who first asked whether similar results identifying the nat-
ural SL (K)-module among abstract quadratic modules would hold over
2
possibly infinite fields. As one sees this involves reconstructing a linear
structure without the arsenal of finite group theory. Answers were given
by F.G. Timmesfeld [7, Proposition 2.7] and S. Smith [5].
Ofcoursemattersarealittlemoresubtlethanthisroughhistoricalac-
countasonemaybeinterestedinsimultaneous identification: actuallyG.
Glauberman[2,Theorem4.1]alsoidentifiedthegroup,andthiscombined
direction has been explored extremely far by F.G. Timmesfeld [8].
We shall follow the line of pure module linearization. Our group or
Lie ring is explicitly known to be SL (K) or sl (K); given a quadratic
2 2
module, we wish to retrieve a compatible linear geometry. Parts of the
4
present article, namely the Theme and Variations n◦1–n◦3, are no orig-
inal work but are adapted from F.G. Timmesfeld’s book [9]. What we
add tothe existing literature is thereplacement of an assumption on the
unipotentsubgroupbyanassumptiononasingleunipotentelement,and
the treatment of theLie ring sl (K).
2
The liminary digression endshere. Ourmain result is thefollowing.
Variationsn◦7,n◦3,andn◦12. LetKbeafieldofcharacteristic6=2,3,
G = SL (K) or sl (K), and V be a G-module. Suppose that there is a
2 2
unipotent element u(resp., nilpotent element x) of G acting quadratically
Fixed on V, meaning that (u−1)2 or x2 is zero in EndV. If G=sl (K) where
2
K has characteristic 0, suppose inaddition that V is 3-torsion-free. Then
V is the direct sum of a G-trivial submodule and of copies of the natural
representation G.
The result for the Lie ring sl (K) (Variation n◦12) seems to be new.
2
The result for thegroup SL (K) is a non-trivialstrengthening (Variation
2
n◦7) of F.G. Timmesfeld’s work (Variation n◦3), as the assumption is
now only about one unipotent element, not about a unipotent subgroup;
Revised however the argument works only in characteristic 6= 2,3. It could be
expectedfromJ.Thompson’sworkincharacteristic≥5andHo’sdelicate
extension to characteristic 3 (see the introductory digression above) that
the case p=3 would be quiteharderif not different.
In the case of the Lie ring G = sl (K), one can produce counter-
2
Revised examples in characteristic 3 but this requires the “opposite” nilpotent
element y to behave non-quadratically. In the case of the group G =
SL (K),I donot know.
2
ThereadermayalsofindofinterestVariationn◦8,whoselengthyproof
indicatesthatreducinganSL (K)-moduletoansl (K)-moduleisnotany
2 2
simpler than directly linearizing theformer.
Thecurrentsection§1isdevotedtonotationsandbasicobservations.
In §2 the core of the Smith-Timmesfeld argument for quadratic SL (K)-
2
modules is reproduced; it will be generalized in following papers whence
ourpresentrecastingit. StillonquadraticSL (K)-modules,§3.1bearsno
2
noveltybut§3.2may. In§4,theLieringsl (K)anditsquadraticmodules
2
are studied.
K, G Notation. Let K be a field and G be the K-points of SL or sl .
2 2
G will thusdenoteeither a group G or a Lie ring g.
1.1 The Group
G Notation. Let G be the group SL (K).
2
uλ, tλ Notation. For λ∈K (resp. K×), let:
1 λ λ 0
u = and t =
λ 0 1 λ 0 λ−1
(cid:18) (cid:19) (cid:18) (cid:19)
u, i One simply writes u=u and i=t ∈Z(G).
1 −1
If the characteristic is 2, one has i=1.
5
U,T Notation. Let:
U = u :λ∈K+ ≃K+ and T = t :λ∈K× ≃K×
λ λ
B Let B=U⋊(cid:8)T =N (U)(cid:9). (cid:8) (cid:9)
G
B is a Borel subgroup of G and U is its unipotent radical, which is a
maximal unipotent subgroup;T is a maximal algebraic torus.
Relations.
• u u =u ;
λ µ λ+µ
• t t =t ;
λ µ λµ
• tµuλtµ−1 =uλµ2.
Note that in characteristic 6= 2, every element is a difference of two
squares: consequently hT,ui=T ⋉U.
0 1
w Notation. Let w= .
−1 0
(cid:18) (cid:19)
Relations. One has w2 =i and wtλw−1 =tλ−1 =t−λ1.
Relations. uλwuλ−1wuλw=tλ, and in particular (uw)3 =1.
The natural(left-) module NatSL (K) corresponds tothenaturalac-
2
tion of Gon K2.
1.2 The Lie Ring
g Notation. Let g be the Lie ring sl (K).
2
hλ, xλ, yλ Notation. For λ∈K, let:
λ 0 0 λ 0 0
h = , x = , y =
λ 0 −λ λ 0 0 λ λ 0
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
h,x,y One simply writes h=h , x=x , y=y .
1 1 1
u,t Notation. Let:
u= x :λ∈K+ ≃K+ and t= h :λ∈K+ ≃K+
λ λ
b Let b=u⊕(cid:8)t=Ng(u). (cid:9) (cid:8) (cid:9)
b is a Borel subring of g and u is its nilpotent radical; t is a Cartan
subring.
Relations.
• [h ,x ]=2x ;
λ µ λµ
• [h ,y ]=−2y ;
λ ν λν
• [x ,y ]=h .
µ ν µν
Thenatural(left-)moduleNatsl (K)correspondstothenaturalaction
2
of g on K2.
6
1.3 The Module
V Notation. Let V be a G-module, that is a G- or g-module.
ThenamesoftheelementsofGwillstilldenotetheirimagesinEndV.
∂λ, ∂ Notation. When G=G, one lets for λ∈K: ∂λ =uλ−1∈EndV. One
simply writes ∂ =∂ .
1
Relations.
• ∂ ◦∂ =∂ ◦∂ ;
λ µ µ λ
• tλ∂µ =∂λ2µtλ;
• ∂ =∂ +∂ +∂ ◦∂ .
λ+µ λ µ λ µ
Proof of Claim. The first claim is by abelianity of U; the second comes
from the action of T on U. Finally, denoting by u the corresponding
λ
elementinthegroupring(ormorepreciselyitsimageinEndV),onehas:
∂ = u −1=u u −1=(u u −u )+(u −1)
λ+µ λ+µ λ µ λ µ µ µ
= ∂ u +∂ =∂ (∂ +1)+∂ =∂ ∂ +∂ +∂
λ µ µ λ µ µ λ µ λ µ
as desired. ♦
Ei(V) Notation. WhenG=g,oneletsfori∈Z: Ei(V)={a∈V :h·a=iv}.
When there is no ambiguity on the module, one simply writes E .
i
Eachh (resp. x ,resp. y )mapsE intoE (resp. E ,resp. E ).
λ µ ν i i i+2 i−2
Added One should however be careful that if the module contains torsion, the
various E ’s need not be in direct sum.
i
ℓU(V),ℓu(V) Notation. The length of V is the smallest integer, if there is one:
• when G=G, such that [U,...,U,V]=0 (U-length);
• when G=g, such that u...u·V =0 (u-length).
A length 2 module is called quadratic.
Clearly,ifV issimple(i.e. withoutaproper,non-trivialG-submodule),
thenV eitherhasprimeexponent,oristorsion-freeanddivisible. Weshall
not always assume this.
The group G is said to act trivially on V if it centralizes it, that is if
the image of G in EndV is {Id}; the Lie ring g is said to act trivially on
V if it annihilates it, that is if the image of g in EndV is {0}. We then
say that V is G- (respectively g-) trivial. The following observations will
be used with no reference.
Observation. Suppose that G=g=sl (K). Let V be a g-module.
2
1. IfKhas characteristic pand V isp-torsion-free, then V isg-trivial.
2. If K has characteristic 0 and V is torsion, then V is g-trivial.
Proof of Claim. Fix a∈V \{0} and any element z of g.
1. If K has characteristic p, then g has exponent p. Supposethat V is
p-torsion-free;thenpz·a=0impliesthatz·a=0: gannihilatesV.
7
2. If K has characteristic 0, then g is divisible. Suppose that V is
torsion; let n be the order of a. Then n 1z ·a = 0 = z ·a: g
n
annihilates V. ♦
(cid:0) (cid:1)
The case of the group is hardly less trivial.
Observation. Suppose that G = G = SL (K). Let V be a G-module.
2
Suppose that ∂ is nilpotent in EndV.
1. IfKhascharacteristic pandV isp-torsion-free, thenV isG-trivial.
2. If K has characteristic 0 and V is torsion, then V is G-trivial.
Proof of Claim.
1. We show that u centralizes V. Otherwise, from the assumptions,
there is a ∈ ker∂2\ker∂. Let a = ∂(a ) ∈ ker∂\{0}. Since K
2 1 2
has characteristic p and V is p-torsion-free, up·a =a =a +pa
2 2 2 1
impliesa =0: acontradiction. HenceucentralizesV. SoC (V)is
1 G
anormalsubgroupofGcontaininganelement oforderp: it follows
that C (V)=G(this still holds of K=F or F ).
G 2 3
2. If K has characteristic 0, the previous argument is no longer valid.
Since V splits as the direct sum of its p-torsion components, one
may assume that V is a p-group. We further assume that V has
exponent p.
For any a ∈ V and any integer k, one has uk ·a = k ∂i(a);
i≤k i
since∂hasfiniteorderℓ,fork≥ℓoneevenhasuk·a= ℓ k ∂i(a).
P i=0(cid:0) (cid:1)i
But since V has exponent p, for k big enough (independently of a)
one findsuk·a=a. P (cid:0) (cid:1)
Hence uk centralizes V. Here again the normal closure of uk is
G, which must centralize V. We finish the argument. Let Vpn be
the G-submodule of V of exponent pn. Then G centralizes every
Vpn/Vpn−1. But G = SL2(K) is perfect; it therefore centralizes V.
♦
2 The Natural Module
Theme. Let K be a field, G = SL (K), and V be a simple G-module of
2
U-length 2. Then there exists a K-vector space structure on V making it
isomorphic to NatSL (K).
2
This theorem was proved by F.G. Timmesfeld in a more general con-
text (Theorem 3.4 of chapter I in his book [9]) and independently by S.
Smith [5]. Let usadapt theproof to ournotations.
Proof. The assumption means that [U,U,V] = 0. Let Z = C (U), so
1 V
thatU centralizesV/Z . Recallthatonelets∂ =u −1∈EndV. These
1 λ λ
functions map V to Z and annihilate Z .
1 1
Observe that by simplicity, C (G)=0.
V
8
2.1 Finding a Decomposition
Step 1. Z ∩w·Z =0.
1 1
Proof of Claim. Z ∩w·Z =C (U,wUw−1)=C (G)=0. ♦
1 1 V V
Recall that i denotes the central element of G (i=1 in characteristic
2).
Step 2. For all a ∈Z , ∂ (w·a )=it ·a .
1 1 λ 1 λ 1
Proof of Claim. Letb1 =∂λ(w·a1)andc1 =∂λ−1(w·b1);byassumption,
b and c lie in Z . Then:
1 1 1
uλ−1wuλw·a1 = uλ−1w·(w·a1+b1)
= i·a +w·b +c
1 1 1
=(u w)−1t ·a = w−1u t ·a
λ λ 1 −λ λ 1
= iwt ·a
λ 1
Soi·a +c =w·(it ·a −b )∈Z ∩w·Z =0andtheclaimfollows. ♦
1 1 λ 1 1 1 1
2.2 Linear Structure
Notation 3. For λ∈K and a ∈Z , let:
1 1
λ·a = t ·a
1 λ 1
λ·(w·a ) = w·(λ·a )
1 1
(cid:26)
Step 4. This defines an action of K on Z ⊕w·Z .
1 1
Proof of Claim. It is clearly well-defined. The action is obviously multi-
plicative on Z ⊕w·Z , because each term is T-invariant. Moreover one
1 1
has:
• on Z :
1
(λ+µ)·a = t ·a =i·∂ (w·a )
1 λ+µ 1 λ+µ 1
= i·(∂ (w·a )+∂ (w·a )+∂ ∂ (w·a ))
λ 1 µ 1 λ µ 1
= i·∂ (w·a )+i·∂ (w·a )=t ·a +t ·a
λ 1 µ 1 λ 1 µ 1
= λ·a +µ·a
1 1
• on w·Z :
1
(λ+µ)·(w·a ) = w·((λ+µ)·a )=w·(λ·a +µ·a )
1 1 1 1
= w·(λ·a )+w·(µ·a )
1 1
= λ·(w·a )+µ·(w·a )
1 1
and everythingis proved. ♦
Step 5. G is linear on Z ⊕w·Z .
1 1
Proof of Claim. Clearly hT,wi acts linearly. Moreover ∂ is trivially lin-
λ
ear on Z . Finally ∂ (µ·(w·a )) = ∂ (w·(µ·a )) = it ·(µ·a ) =
1 λ 1 λ 1 λ 1
µ·(it ·a )=µ·∂ (w·a ) so ∂ is linear, and u is therefore too. ♦
λ 1 λ 1 λ λ
9
V being simple is additively generated by the G-orbit of any a ∈
1
Z \{0}, and one then sees that V ≃K2 as the natural G-module. This
1
finishes theproof.
Remark. Note that although there are a priori several K-vector spaces
structures such that G acts linearly (twist the action by any field auto-
morphism), our construction is uniquely defined. It is functorial: if V
1
andV aretwosimpleSL (K)-modulesandϕ:V →V isamorphism of
2 2 1 2
SL (K)-modules, thenfor our construction ϕis K-linear.
2
3 First Variations
3.1 Centralizers
ThestatementsofthissubsectioncanbefoundinF.G.Timmesfeld’sbook
[9].
Variation n◦1. Let K be a field of characteristic 6= 2 with more than
three elements, G= SL (K), and V be a G-module. Suppose that V has
2
U-length at most 2. Then G centralizes C (i).
V
Proof. We assume that the central involution i centralizes V and show
thatGdoestoo. Byourassumptions, Gcentralizes the2-torsion compo-
nent V of V. Recall that one writes ∂ =∂ .
2 1
Let a∈C (U), b=∂(w·a), and c=∂(w·b). Onethen has:
V
uwuw·a = uw·(w·a+b)
= a+w·b+c
=w−1u·a = w·a
Hencew·(a−b)=a+c. ButbyassumptionontheU-length,b,c∈C (U),
V
so b−a∈C (U,wUw−1)=C (G). Let us resume:
V V
uwuw·a = uw·(w·a+(b−a)+a)
= a+(b−a)+(w·a+(b−a)+a)
=w−1u·a = w·a
Therefore2b=0,thatisb∈V ≤C (G),whencea=(a−b)+b∈C (G).
2 V V
As a conclusion G centralizes C (U). But by assumption on the U-
V
length, U centralizes V/C (U), so by the same argument G centralizes
V
V/C (U)aswell. NowGbeingperfect bytheassumptionsonK,Gdoes
V
centralize V.
Variation n◦2 ([9, Lemma 3.1 of chapter I]). Let K be a field of char-
acteristic 6= 2 having more that three elements, G = SL (K), and V be
2
a G-module of U-length 2 satisfying C (G) = 0. Then for any λ ∈ K×,
V
[u ,V] = [U,V] = C (U) = C (u ). In particular C (u ) does not
λ V V λ V λ
depend on λ.
10