R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
J UN M ORITA
E UGENE P LOTKIN
Prescribed Gauss decompositions for Kac- Moody groups over fields
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Prescribed Gauss Decompositions
for Kac-Moody Groups Over Fields.
JUN
MORITA(*) -
EUGENEPLOTKIN (**)
ABSTRACT - We obtain the Gauss
decomposition
withprescribed
torus elementsfor a
Kac-Moody
group over a fieldcontaining sufficiently
many elements.1. Introduction.
Kac-Moody
groups areequipped
with canonicaldecompositions
ofdifferent
types.
Let usnote,
forinstance,
thedecompositions
ofBruhat,
Birkhoff and Gauss. As in the finite dimensional case,
they play
an im-portant
role in calculations with these groups.However,
in the Kac-Moo-dy
case each of thesedecompositions
has its ownspecial
features(see,
for
example, [18]
where J. Tits compares the Bruhat and the Birkhoffdecompositions
andpresents
someapplications).
The aim of this paper is to establish the so-called
prescribed
Gaussdecomposition
forKac-Moody
groups.The
prescribed
Gaussdecomposition appeared
in[16], [9]
for the ge- neral linear group. It wasproved
for allChevalley
and twistedChevalley
groups in
[3], [4], [5].
It turned out to be the main tool forproving
thesubstantial Ore and
Thompson conjectures.
(*) Indirizzo dell’A.: Institute of Mathematics,
University
of Tsukuba, Tsu- kuba, 305-8571,Japan.
(**) Indirizzo dell’A.:
Department
of Mathematics, andComputer Science,
Bar Ilan
University,
52900, Ramat Gan, Israel.Partially
supportedby
grant ISF andby
grant of the Centre of Excellence«Group
theoretic methods in thestudy
ofalgebraic
varieties» of ISF.Namely,
Ore[12] conjectured
that every element of a finitesimple
group is a
single
commutator.The
proof
of this statement for allsimple
groups of Lietype
isgiven
in
[6].
Similar facts are also known for infinitesimple
groups. Let us mention the paper of Ree[14],
whoproved
that every element of a con- nectedsemisimple algebraic
group over analgebraically
closed field is acommutator. A survey on the Ore
problem (not including
the results of Ellers-Gordeev andLev)
can be found in[20].
This paper is the continuation of
[11],
where theprescribed
Gauss de-composition
was established forKac-Moody
groups of rank 2. The gene- ral result becamepossible
due to the recent paper[2]
where theelegant
idea of V. Chernousov gave rise to a uniform
proof
of theprescribed
Gauss
decomposition
for all groups of Lietype.
Wemostly
follow themethod of this paper,
adjusting
it to theKac-Moody
case.We do not consider in this paper Ore and
Thompson type conjectures
for
Kac-Moody
groups. These groups areperfect but, generally
spea-king,
notnecessarily simple,
and their commutator structure can be very delicate.2. -
Kac-Moody
groups.Let A =
(aii)
be an n x ngeneralized
Cartan matrix. Let g be theKac-Moody
Liealgebra
over the field C definedby A according
to thechoice of a realization
17, Il
of A withII = ~ a 1,
... ,a n ~,
the set ofsimple roots,
and11 " _ ~ a i , ... , a n ~,
the set ofsimple coroots,
sati-sfying
= a2~(cf. [7], [10]).
Let d c~*
be the rootsystem
of g withrespect
to the so-called Cartansub algebra I).
Letd +
be theset of
positive (resp. negative)
roots definedby II ,
and L1 re the set of realroots. Set A = L1 Then
(root
spacedecomposition)
and
(triangular decomposition),
where So-
Let C~i be a
Chevalley-Tits
group functor from thecategory
of com-mutative
rings
with 1 to thecategory
of groups,corresponding
to g(cf.
[19]).
For any commutativering
R with 1 and for any aE .LI re,
there exists155
a group
homomorphism xa (~ ) : ~2013~CR),
where R + is the additivegroup of R .
Using
aChevalley basis
for L1re(cf. [19]),
wecan express = exp
(tea )
for t E R . LetG(R)
be thesubgroup
ofQ)(R) generated by
for all aeL1re and ten. We callG(R)
a(standard
orelementary) Kac-Moody
group, whichdepends
on the choice of a root da- tum and here one can choose any root datum(cf. [10], [13], [15], [17], [19]).
SometimesG(R )
is said to be oftype
A . From now on, we assume that R = K is a field and shortenG(K)
to G. We also suppose 1 for all aEL1re and t E K(t ~ 0 ).
Let-.. -- u I , - 1 ’B u
For i =
1,
... , n we definehomomorphisms g
for all a e K and t e 7~ . Let
Ui, Vi
be the rootsubgroups corresponding
to the roots a
i and - a i , respectively.
Thesubgroups Ui and Vi
are defi-ned as follows:
Then
( G , U, T , Y, ~ ~ 1,
... ,~ n ~ )
is atriangular system (see [11]
for thedefinition).
Hence(cf. [11]),
everyKac-Moody
group G over a field has aGauss
decomposition, i.e.,
Set
then
TN ,
and N/T isisomorphic
to theWeyl
group W(cf. [7], [10]).
Wesometimes
identify
W withN/T .
Forw E W,
we writewith w = w T
and w E N,
where w can also be identified with an inverseimage 10
of w under cannonicalepimorphism
Thenand
(cf.[8], [13]). Therefore,
if 1 ~ u E U(resp.
1 ~ v EV),
then there exists w E N such thatfor some i =
1,2, ..., n, where ui
E~/
eUi’ (resp. Vi E Via, ~/
eOtherwise
(resp. -W v -W -’ E Vi’)
for all WE Wand alli
=1,
... , n , whichautomatically
me: , -1(resp.
v E
f l
and u = 1(resp. v
=1).
The aboveproperty
isimportant
for
~7and
we will use it later.For a subset X of
II,
we denoteby L1 x
the subrootsystem
of L1 gene- ratedby X,
andby L1 x
the subset of real rootsgenerated by
X . We defi-11C LliC luiiuwilig
Let ary.
in
W = N/T ,
andidentify a, i With
if necess-PROOF.
(1): Let g
eGx
n U.Then, by
the Bruhatdecomposition
ofGx,
wewrite g
= uhum with u, veTx
and w eW,
whereWx
is the
parabolic subgroup
of Wcorresponding
to X.Using
the Bruhat de-composition
ofG ,
weget w
=1,
whichimplies that 9
can be written asg = uh for some u E
UX
and h E Therefore h = 1 since U n T = 1.Hence, g
= u EUx. Clearly Ux c Gx
n U.Therefore, Gx
n U =Ux.
(2):
Let z EUx
nUi’ .
Then...~ ... ¿,.J,. ¿,.J,. , v v m Bh ~
invariant since
Gx= (Ux, Tx,
LetaieX, ~d X ,
t EE K,
andYEUX.
We write withzi E U2
and s E K.(a)
In the case when s =s
(b)
In the case when s # 0 and0,
where means thevalue of the linear function at
a’( E lj,
where x’
= xp(t)
or x’= xp (t)
for some t’ E K.(c)
In the case when s ~ 0 and~3( a i ) 0,
for s
(4)
can beproved
in the same way as in(3),
and(5)
is ob- vious.Q.E.D.
Let
Z(G)
be the center of the group G. Note thatZ(G)
cT ,
and wecan
explicitly
describe it as follows(cf. [15], [19]):
159
We make a short review here. Let
Using
the Bruhatdecompo-
sition G =
UNU,
one canwrite g = uhwx
E U n
w -1 Vw,
which has aunique expression.
If w #1,
then we choosefl
EL1 +
such that E L1 - . Then =x,~ ( 1 )
g , and we obtain a con-tradiction
by
theuniqueness
ofexpressions.
Inparticular, w
= 1and g
EE UT .
Similarly
we see g E VT .Hence,
we obtainZ( G )
c T = UT n VT.Considering
the action of T on thesubgroups Ui ,
... , IUn , VI ,
... ,Vn ,
itis now easy to see that the center
Z(G)
can beexpressed
as above. Wenotice that we need not assume
«simply
connectedness» here.3. - Theorems.
We shall say that a
Chevalley
orKac-Moody
group G has theprescri-
bed Gauss
decomposition
ifgiven
anarbitrary
elementh * E T,
we have- .. I I I’TTf oL. T’7"’ _1 1
wnere is tne center ot u kL6j, lllJ).
The result on the
prescribed
Gaussdecomposition
forKac-Moody
groups of rank two is as follows.
THEOREM A
([11]).
Let A = 2a
be ageneralized
Cartan ma--b 2
trix with ab ~ 4. Let m =
max {a, 6}.
Let K be a field withK ( >
m + 3.Then every
Kac-Moody
group G over K oftype
A has a Gaussdecomposi-
tion with
prescribed
elements in T.A similar result holds for 0 ~ ab ~ 3 in the rank two case without any restrictions on the
cardinality
of K.Moreover,
in the case ofChevalley
groups, there is a
general
result on the Gaussdecomposition
with pre- scribedsemisimple
elements.THEOREM B
([2]).
Let A be a(finite)
Cartanmatrix,
and let K be a field. Then everyChevalley
group G over K oftype
A has a Gauss decom-position
withprescribed
elements in T.In the
remaining part
of this paper, we will establish thefollowing
main result on
Kac-Moody
groups G. For ageneralized
Cartan matrixA = we set
I l 1
160
MAIN THEOREM.
Suppose K ~ > m + 3.
Then everyKac-Moody
group G over K of
type
A has the Gaussdecomposition
withprescribed
elements in T.
COROLLARY.
Every
noncentral element of aKac-Moody
group G canbe
expressed
as aproduct
of twounipotent
elements in G.4. - An inductive method.
Here we will show the
following proposition.
Denote 1=={1~...~}.
PROPOSITION. Let 7" =
(r, G)
be a group with G normal in r and such that theconjugation
is adiagonal automorphism
of G. Letbe the center of r.
Suppose ] K
> m + 3. Then for every element zg e l, with g e G and rg gZen,
and every elenexists z E G such that
~ E
T,
therefor some v e V and u e U.
This
proposition
can beproved
inexactly
the same way as in[2].
Weproceed by
induction on n . It isalready
known that theProposition
hol-ds for
n = 1, 2 (cf. [2], [111),
which isprecisely
described in Theorem A and Theorem B. Now wesuppose n %
3. To make our inductioncomplete,
we have to assume that the
cardinality
of K isgreater
than m +3,
sincewe
essentially
use the information in case of rank two.Since rU = Ui and G =
UVT U,
we have rG == UW7Y/.
Hence,
for our purpose, we can assume that Tg is of the form zg = ivhu with v eV, h
eT , u
E U.Then,
we fall into one of thefollowing
three cases,
using
the factf l w Uw -1= f 1
and the defini-weW wEW
tion that r induces a
diagonal automorphism
of G.(Case 1)
Let u # 1 or v ~ 1. Then there exists w E Wsatisfying
161
(Case 2)
Let u = v = 1 and the element ig be of the form_ _ ._ 2013T~
.. - ....-- . v - ~
---- f - - - -
(Case 3)
Let u = v = 1 and the element rg be of the formvviv,, "I - ’L G4111A. r - - B,.. 1.
In
(Case 2),
we can find an with t # 0 forsome j
E Isuch that
I., I " , 1 .1 11 1 1
with
1 ~ u~
EUj.
Thus(Case 2)
can be reduced to(Case 1).
Our assum-ption
on the form of igimplies
that we canskip (Case 3).
So let us assumethat we are in
(Case 1)
and fix an i e Iappearing
in the formula for y.Set X = a2, ...,
an_1~,
and Y=la2,
a3, ...,an~.
We writeL L, ~ ~ B ~ l f 1
F° i >
1,
weput
Ty=rh,,, (tl)
andry= (Ty, Gy).
Then--- - . 1 ~ .._1 _ ~ 1 ~ __1 _ ~ 1 ~ 7 --- -.1 --. - --- - - - ----
1. Let us choose K " such that ih *
ha2 (t2 )
is non-central in
Gx ~,
where rx= This isactually possible
since there are
enough
elements in K. Infact,
we can choose ± 1 ifcentralizes both of
U2
andV2,
otherwise wejust put t~’ =1.
In any case, does not centralize
Ga2 = ~ U2, V2),
which
implies
that is not central inApplying
induction to non-central element
....
W 11111 1 ~ we l:au lUlU au eleUleUl1 uy JlAl;il Lll’G6L ;4
with and
.L i ~ i .A-
where Then we 1
with
v" E V, U" e U ; ( t2 ) . Next,
we writeI
Then we
apply
induction toSince y x is noncentral in
T X,
we can find an element ZX EGx
such thatwith
vX E VX, UIE Ux
and rh *= Z x hX . Therefore,
we have
with v"’ E
V,
u"’ EU,
and we are done.If i
= 1,
we take first X and thenY,
andrepeat
the same process as above. Thenby choosing
a n, and a 1 instead of a 2 and a n re-spectively,
wecomplete
theproof
ofProposition.
We can alsoexplain
thematter in the
following
way. The order on II is irrelevant for theproof.
Therefore,
if in our initial order i turns out to be one, we can reorder II in such a way, thattaking
thecorresponding X
and Y we come to the al-ready
considered case i > 1.The main theorem follows from the
proposition
in an obviousway.
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Manoscritto pervenuto in redazione il 2 ottobre 2000.