B ERTIN D IARRA
Representative subalgebra of a complete ultrametric Hopf algebra
Annales mathématiques Blaise Pascal, tome 2, n
o1 (1995), p. 99-115
<http://www.numdam.org/item?id=AMBP_1995__2_1_99_0>
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REPRESENTATIVE SUBALGEBRA OF A COMPLETE
ULTRAMETRIC HOPF ALGEBRA
Bertin Diarra
ABSTRACT. Let
(H,
m, c, ~,r)
be a complete ultrametric Hopf algebra over a complete ultrametric valued fieldh’,
e be the unit of H and k the canonical map of K in H. In order words, H is a Banach algebra with multiplication m : :H®H --~ H,
coproduct c : H ->H~H
a continuous algebra homomorphism, inversion or antipode ~ : H --i H a continuous linear map and counit 7 : H ~ K acontinuous algebra homomorphism. The coassociativity and countinary axioms hold, and
We define the representative subalgebra of
H,
i.e. the subalgebra of H generated by thecoefficient "functions" associated with the finite dimensional left H-comodules. Under some conditions on
H, R(H )
is a direct sum of finite dimensional subcoalgebras and is dense in H. But in general,R(H)
is not dense in H. The algebra
T~(~)
is a generalization of the algebra of representativefunctions
on agroup. Notice that when the valuation of K and the norm of H are trivial, one obtains the well known fact that H is equal to its representative subalgebra.
INTRODUCTION.
Let
(H,
m, c, r~,~)
be acomplete
ultrametricHopf algebra
over thecomplete
ultra-metric valued field K. An ultrametric Banach space E over K is said to be a
left
BanachH -comodule if there exists a continuous linear map
~E
: E -+H®E,
calledcoproduct,
such that
(i) (c 01E)
oL~E
=(lH
0 o~E (ii)
A closed linear
subspace
E of E is a(left)
Banach subcomodule of E if cHM.
Let
(E, QE)
and(F, A F)
be two left Banach comodules. A continuous linear mapu : E --~ F is a Banach comodule
morphism
ifOF
o u =(IH
~u)
o~E.
It is associated with any left Banach H -comodule
(E,
the closed linearsubspace
of H
spaned by
the coefficient "functions"(1H
~x’)
oA(x),
x’ EE’,
x EE,
where E’ if the Banach space dual of E.
Furthermore,
letR(H)
be the linearsubspace
ofH
spaned by
all theR(L1 E)
where(E, AE)
is a finite dimensional left H-comodule. ThenR(H)
is a(non
necessaryclosed)sub-Hopf-algebra
ofH ; 1(,(H)
is called therepresentative subalgebra
of H. Ingeneral, ?Z(H)
is not dense in H(cf. ~l~
or[5], ). However,
withadditional conditions on H it will be shown that
R(H)
is dense in H.If E and F are ultrametric Banach spaces over
K,
we denoteby E~F
the com-plete
tensorproduct,
that is thecompletion
ofE ~
F withrespect
to the norm~z~
=Inf
(max
In thesequel
all Banach spaces are ultrametric.)
I - LEFT BANACH COMODULES
I - 1 Tensor
products
of left Banach comodulesLet
(E, DE)
and(F, OF)
be two left Banach comodules. One has the continuous linear mapE®F
--+HEHF
-->HHEF
~HEF
whereand 0
a)
= ~0 ~.Proposition 1 : 0394EF
:E0F
~F0E0F
M &ecoproduct of
aleft
Banach H-comodule structure on
E~F.
Proof :
Put, for
x E Eand y
EF, = 03A3aj~xj
E and0394F(y)
=L. bl~yl
Ej>1 t>1
Therefore,
one has @y)
=03A3 ajbl ® x j ®
yt.j>1 t>1
(i)
It followsimmediately
that(° ~
o @y)
=y~ Y~ °(ajbt)xj
~ yt =j~1 l~1
-
® yt =03A303C3(aj)xj
~03A3 03C3(bl)yl
= x ~ y =1EF(x
®y).
Fromj>1 t>1
what,
one deduces(° ~ IE®F)
o(ii) Also,
one has for x EE,
y E Fa) (c ® 1E)
oDE(x)
=~ C(aj)
~ xj =~ ~
~o9,j
~ x~ =(1H
®DE)
oQE(x)
=j>1
~
=
j>1 j>1 k>1
and
(C
©lF)
° "£ c(bl)
© $lI "£ £ 03B21t,l
©03B22t,l
f§ 91 "(lH
f§0394F)
°4lF(Y)
"t>I I>I
t>1 ~
*
£ bl
©0394F(yl)
"£ £ bt
© pm,t © ym,lI>i t>I m>I
Let
Ez
=E[(z ;, j > i)
U(z
k,; , k >I, j > I)]
be the closed linearsubspace
of Espaned by (z;, j
>1)
u(zk,j,
k >1, j > I),
andF,
=E[(yi, £
>I)
U(ym,i, m > I,I > I)]
be theclosed linear
subspace
of Fspaned by (yl, l
>I, I > I ) .
It is clear that the Banach spacesEz
andF,
are of countabletype. Furthermore,
if z’ eEb
andy’
EF§
onehas
= =
j~1 s~1
=
(1H © I H
©z’)
o(1H
©AE)
o =£ £ z’,
xk,j > aj © yk,;j~1 k~1 and
(I H © I H © y’) O (C §§ I F) O A F(y)
#£ £ y’, yf > Q§ i §§ Q#
°°n>i t>1
~
*
( l H © l H © Y’) ° (l H © 4lF) ° 4lF($l)
"£ £ Y’, Ym,I > bt 6§ pm,I.
I>I m>I
#)
On onehand,
onehas, ( c
@ i O @v)
=£ £ c( ajbl)
@ z; © vi ==
£ £ c(
a ;)c(bi)
l§ z j E9 vi =£ £ (£
E9ai ;) (£ 03B21t,l
l§Qi ) ©z;
© v.s>1
~ ~
t>i
’ ’
On the other
hand,
one has= =
£ ££ I ajbl~
j>1 t>1 k>1 m>I 03B3k,j03C1m,l @ xk,l © ym,l.
Hence,
if z’ eE£
andy’
EF§ ; ; first,
one has~
i I
~"" "j
~I I
~Y"
Yl ~j~1 s~1 l~1 t~1
Q§,i
©03B22t,l
=(lH §§ l H
©z’)
o(c
©IE)
oAE(z) . (1H
~ 1H ©y’)
o(c @ l F)
o ="
(lH
©lH
©Z’)
o(lH
©0394E)
o4lE(") o (1H ~ 1H
f§9’)
o(lH
f§0394F)
o0394F(y).
And, second,
one has"
£ £ Z’, Zk,j
>aj~03B3k,j £ £ y’, ym,l
j>1 k>I I>I m>I
bl ~
03C1m,l =(I H ~ 1H
©z’)
o(1H
©AE)
oAE(z) . (lH ~ 1H
©y’)
o(1H
©AF)
o0394F(y).
Therefore,
for any z’ eEb
and anyy’
eF§,
we have(a)
~ " °y)
SinceEz [resp. Fy]
is of countabletype,
there exist ao >0,
ai > 0 andC
Ex (resp.
C such that for z EEz (resp. ~’
EFy]
one hasz =
03A303BBjej (resp. ~
=03A3 lfl]
with aoSup
a fSup ] (resp,
aoSup| l|
j>1 ~>1 ~>1 ~>1 l>1
~03B6~
~ 03B11Sup | l|] (cf. [4] ).
l>1
Moreover,
one hasExFy
Nco(IN*
xlN*, h’)
andN
co(IN*
x1N*, H®H) (cf. [7]) ;
any Z in(HH)(ExEy)
can be written in theunique
form Z =
03A3 Ajl
® ej ~fl
withAjl
EHH
andao Sup ~Ajl~ ~Z~ a; Sup |Aj,l~.
Let
ea
EEx [resp, f ~
EFy~
be the continuous linear form definedby e~,
ejl >_03B4jj1 [resp. f’l, fl1
>=03B4ll1]. Setting
y)
=Zo
=03A3 A0jl
~ ej ®fl
EHHExFy,
for any 1 and anyl, by (a) ,
one,j,l
has
(1H ® I H
~e’j1
~f y )(Zo)
=03A3 A0jl03B4j1j03B4l1l
=Aoj1 l1
= U. It follows thatZo
=0,
i.e.(c
o0394EF(x ~ y)
=(lH
~ 0394EF)o 0394EF(x
®y).
Fromwhat,
one deduces that(c ®
o =(1H
~ 0394EF) o0394EF.
Corollary
: Let :~1(
resp. N~
be aleft
Banach H.subcomoduleof
E(
resp. F~.
. ThenM®N
is aleft
Banach subcomoduleof E®F,
I - 2 Banach comodule
morphisms
I - 2 - 1
Range
and kernelProposition
2 : : Let u E --~ F be a Banach comodulemorphism.
(i) If
V is a Banach subcomoduleof F,
then is a Banach subcomoduleof
E.(ii)
The closureu(E) of u(E)
is a Banach subcomoduleof
FCorollary
: Let V and W be Banach s’Ubcomoduleof
theleft
Banach H-comodule E ; then V ~ W is a Banach subcomoduleof
E.Proofs : Rather easy, or see
(3~.
Note : One can also see
[3]
for the spaces of comodulemorphisms.
Remark 1 : :
If
M is a Banach subcomoduleof
thele f t
Banach H-comoduleE,
it isinduced on the
quotient
Banach spaceE/M
a structureof
Banachleft
H-comodule such that the canonical map E --~E/M
is a comodulemorphism.
Then, if
u : E -~ F is a Banach comodulemorphism
andif
u isstrict,
the Banachcomodule
E/keru
andu(E)
areisomorphic. Also,
one candefine
the cokernelof u
asbeing F/u(E).
I - 2- 2 Comodule
morphisms
of E into H associatedwith ~ ; R{~)
Put 0 =
!1E
thecoproduct
of the left Banach H-comodule E.Obviously,
H is a leftBanach H-comodule with
respect
to itscoproduct
c.Proposition
3 : : For any x’ EE’,
the linear mapAx’
=(1H
~x’) o 0394
: E ~ H is aBanach comodule
morphism.
Proof : It is easy to see that =
c~x’
=(1H~1H~x’)o(c~1E).
ThereforecoAx~
==(IH®~(1H®x’)o0~)o~=(1H®Az~)o~.
Corollary
1 :{i) ker Ax’
is a closed subcomoduleof
E.(ii) Ax’(E)
is aleft
Banachsubcomodule(
= closedleft coideal) of
H .Corollary
2 : :If
E is a spaceof
countabletype,
one hasker Ax’ ~
Efor
anyProof :
Indeed,
if x’ EE’, x’ ~
0 and 0 a1,
there exists aa-orthogonal
base C E such that
x’, e
1 >= 1 and >=0, j
> 2. Moreover forany j > 1, 0394(ej)
=03A3alj
® et and ej = ; therefore03C3(alj)
=03B4lj
and1>1 ~>1
Ax’(e1)
=(1H
®x’)
o0394(e1)
=a11 ~ 0
since = 1.Corollary
3 : : Assume that H is apseudo-reflexive
Banach space ; i.e. H -+ H" is isometric.Let E be a
simple
Banachleft H -comodule,
i.e. E contains no proper closed subcomo- dule. Then E is a Banach spaceof
countabletype
andAxt
isinjective for
each x’ EE’,
, xi~
0.Proof : If H is
pseudo-reflexive,
it is shown in(3~
that anysimple
Banach left H- comodule is a space of coutable type.Applying Corollary 2,
one sees thatAx~
isinjective for x’ E E’, x’~0
aLet
,Q
E ® E’ -+ K be the continuous linear form defined upon ~x’ )
_x’,
x > .Put po -
(1H
®,Q)
o(0394 ~ 1E’)
o T : :E’®E
~H,
where(x’
®x)
= x ® x’. Then pa is linear and continuous with~03C10394~
Moreover for x’E E’,
x EE,
one has03C10394(x’
®x)
=(lH
®x’)
00394(x).
Put
R(0)
=03C10394(E’E)
the closure of03C10394(E’E)
in H.Obviously, R(Q)
is the closedlinear
subspace
of Hspaned by
the elements 0x’)
oQ(x),
x’E E’,
x EE,
called thecoefficients of the comodule
(E, A).
Proposition
4 : :R(A)
=03C10394(E’E)
is aleft
Banach subcomodule(=
closedleft coideal) of
H.Proof : Since c : H -;
H®H
is linear and is ahomeomorphism
of H ontoc(H) ,
onehas
c(R(0394))
= a closed linearsubspace
of H.It remains to show that if a =
03C10394(x’~x)
= =Ax’(x),
x’ EE’,
x EE;
then
c(a)
EWriting Q(x)
=a j
0 Xj; one hasc(a)
= c oAx’(x)
=(l H
~j~l
Ax’)
oQ(x ) = a j
0 =03A3
a j ~03C10394(x’
®x j )
E 0j>1 j>i
Proposition
5 : :If
thele ft
Banach comodules E andE1
withcoproduct respectively
L1and
03941
areisomorphic,
thenR(A)
=R(03941).
Proof : Let u : : jE 2014~
Ei
be a comoduleisomorphism,
in otherwords, u
islinear,
continuous and
bijective
withAi
o u =(l H
®u)
o A.Moreover,
thereciprocal
map u-l of u satisfies(lH 0
oAi
= A ou’1
and thetranspose
of u,tu
:El
:-~ E’ islinear,
continuous and
bijective
with(tu)"I - tu-1.
Set a =
03C103941(z1)
E03C103941(E’1E1)
and z1 =j>1 y’j~yj, y’j E E’1,
yj E E1, lim ~y’j~ ~yj~
= 0.
There
exist, for j >
1unique x’j
E E’ and x j E E such thatyJ
= =x J
o andyj =
u(x j );
moreoverlim ~x’j~ ~xj~
= o. Therefore a =03C103941(z1) = 03A303C103941(y’j
~yj)
== =
j>1 j>1 j>1
= j>1 0
xl)
o = ~x3}
= pox?
~xj
.Hence
a = Ej>1 j>1
where z =
~ x~
~ xj; that is CLikewise,one
hasj>1
’
03C10394(E’E)
C03C103941(E’1E1).
Therefore = and
R(A)
= 0Assume that E is a
free
Banach space i.e. E ~c0(I, K)
= C K/
limA;
=0).
In other
words,
there exist{e j ) j EI
CE,
ao, al EIR*+
such that any x E E can be writtenin the form x =
03A3 03BBjej, 03BBj ~ K
and aosup al sup|03BBj|.For
any continuouslinear form x’
E’
, one has1 I x’
e j >
I x’ I x’
e j >.
Lete’j
be the element of E’ definedby
>= PutEo
=E[(e’j)j~I],
the closedlinear
subspace of
E’spaned by
Hence each x’ EEo
can be written in theunique
form x’ = E
K, lim
= 0.Moreover,
if v E CE’®E,
one hasv = ljl’l ~ ej, lj ~
K,
lj = 0 anda°
sup~03C5~ ~ «1
sup| lj|.
On the other
hand,
one hasco(I,H)
= CH/lim aj
=0}.
For anyz E
H0E
one has z =03A3aj
~ ej,aj E H withlim ~aj~
= 0 and03B10 sup ~aj~ ~ ~z~
jEI J jEl
ai sup
~aj~. Hence,
if(E, A)
is a left BanachH-comodule,
for 3-E,
one hasA(.r)
=j EI
In
particular 0(et)
= =03A3alj ~ ej
and =jEI jEI jEI
=
03A3alk ~ 03A3akj ~ ej
=jEl kEl kEI jEl
=
03A3 03A3
at.k ® alj ~ e j . Thus one obtainskEI jEI
(1) c(atj)= 03A3alk ~ akj ;
l,j ~ IkEI
Also,
one has(2)
;(3)
I E I.kEI kEI
Proposition
6 : :Ro(A)
=03C10394(E’0E)
is a closedsubcoalgebra of
H. In other wordsc(Ro(ð» C R0(0394)R0(0394)
Proof : Since
(e~
0 is a totalfamily
of and po is linear andcontinuous,
thefamily
0 is total in03C10394(E’0E
=R0(0394)
= the closedlinear
subspace
of Hspaned by
the(1H
0x’)
oA(.r),
x’ EEo,
x E E.To see that C
R0(0394)R0(0394), it
suffices to show that for E I one hase(po(eJ ® ee))
EHowever, by definition, 03C10394(e’j ~ el)
=(1H
= alj E
Ro(A). Then,
one deduces from(1)
that ®el))
=c(alj)
=03A3alj
® akj EkEI
R0(0394)R0(0394).
Note :
If
v =03A3 lje’j
~ el E , one has03C10394(03C5)
=03A3 ljalj
and a ER0(0394) iff
1>,l t,j
there exist vn E
E’®E
such that a = lim03C10394(03C5n).
n-+o
Remark 2 : : Let
(E, 4)
and0~)
be twoisomorphic left
Banach comodules thatare
free
Banach spaces.If
u : E -+E1
is a comoduleisomorphism,
,(ej)j~I
a baseof
Eand
(e j ) j EI
the baseof El defined by
E J =u(e j ); then,
with the above notations, , one hasRa(o)
=R0(03941).
Remark 3 : :
If
dimE = n +~, one hasR(0) - R0(0394)
=pa(E’
®E)
anddimR(0394) n2
II - REPRESENTATIVE SUBALGEBRA
II - 1
Conjugate
comodule of a finite dimensional comoduleLet
(E, L1)
be a( Banach)
left H -comodule of finite dimension and a h’-n
base of E. As
above,
for any x EE,
one has0394(x)
=03A3Aj(x)~ej
andA j (x )
=j=1
n
0394(x)
=03C10394(e’j ~ x); Aj
=(1H~e’j)o0394
E,C(E,H).
Inparticular 03A3
a‘j where.
’ ’
J=l
alj =
Aj(el)
= and we have the relations(1) , (2)
and(3) ,
with I =(l, n].
The relation
(3)
meanshere,
that the matrix A = EMatn(H)
is invertiblewith inverse A-1 =
Fix the base of E and define the linear map C1" : E’ --~ H ~ E’
by setting
n n
0394~ (e J )
~ = ~ei,
1j
n. Hence for x’ =03A3 je’j E E’,
one~‘1 ~~1
n n n
£
~ej
=03A3 A~l(x’)
®e’l.
Lemma 1 :
(E’,
is aleft
H ..comodule.Proof : One verifies that o o r~ = o;
indeed,
if a EH
then ’c(a) _ ~ a~ ~ Hence,
~>i
one has = =
03C3(a)e
and a = = Itc» t>~
follows that = and o o = =
t>i t2:1 t>~
03C3(03C3(a)e)
=u( a).
n n
Since
03C3(alj)
=03B4lj,
one has(o
®lE,)
o0394~(e’j) = 03A3
o o =03A303C3(alj)e’j
=t=i i=i
=
e- 1 j
n. Itfollows, by linearity,
that(cr 01E’)
o Ov =1E’.
Let us remember that where
r(a
0b)
= b ® a.Hence ,
we haven n
co~(alj)
= Therefore = =k=1 l=1
n n n n
-
I I ~(akj)
®~(alj)
~et
- ®(1H ® Ov) ~(akj)
®ek
_~=1 t=l ~=1 :=!
and
Corollary :
:R(6 V)
=~(R(0394)).
Proof :
Identifing
E" withE,
one hasR(6 V)
= Set z= 03A3 03BBljel
®e’j
EE 0 E’;
hence =03A3 03BBlj03C10394~ (el
~ e’j). However ~ e’j) =( 1 H
®et)
o0394~(e’j)
=~(alj)
= ®ee))s
therefore03A3 03BBlj~(03C10394(e’j
®el))
=where zi =
03A3 03BBlje’j
® el EE’
® E. It follows thatR(0394~)
C~(R(0394)).
Thesame formulae show that if a =
03C10394(z1)
ER(0394),
where zi= 03A3 03BBlje’j
~ et EE’
~ E ,one has
~(a)
= 03C10394~(z)
where z =03A3 03BBljel ~ e’j
E E ®E’ ,
hence CR(0394~).
II- 2 Direct sum of Banach comodules
Let
(E9)1~9m
~ ~ be a finitefamily
of left Banach H-comodules with~9
thecoproduct
ofm
n
m
)!
Es.
The direct sum E =equiped
with any normequivalent
to the norm~ -~
=s=1 9;i
m m m
= 1sm max is a Banach space. Put A =
~ G19,
2014 i.e.A~Y~)
~~ ~ =~ L19(x9).
It is- - s=1 s=1 a=l
readily
seen that(E, A)
is a left Banach comodule.Moreover ,
if : E -~ E is theprojection
of E onto then1 H
~ ps is aprojection
ofHE
ontoHEs
and one hasm
HHEE
=~ HEs.
On the other hand pa is a comodulemorphism
i.e.(1H ~ ps)
o A =8= 1
A o ps; furthermore
(1H ~ ps)
o = 0for s ~ t
and .r t EEt.
m
Also,
we have JE~ =~ E~;
theprojections
associated with this direct sum are the9=1
1 ~ m.
m
Proposition
7 : : With the abovenotations,
one has03C10394(E’E)
=03A303C10394s(E’sEs)
~==1
m
and
R(A)
is the closureof ~ R(Da)
in H.a=I
Proof : If
x9
EE~,
xt EEt
ands ~ t,
then = = p.__
m m m m
Set z
= L
E = one has z =£ £
It followsj>1 ~ twl j>1 s=1 t=1
__
m m m m
that
= ~ ~
~x~,j)
o =j>1 J=1 t=1
~
j>1 a=1 t=1
~
m m m m
=
j>1 s=1 t=1 j>1 a=1 a=I j>1
m
= 03A303C10394s(zs). If zJ
EE’sEs
CE’®E,
1 s m, one has =po, (z9). Therefore,
8=1
m
on one
hand,
C03A3
pa,(E’sEs),
and on the otherhand,
po,(E’sEs)
Ca==i
m
03C10394(E’E). Hence,
one has03C10394(EE) = 03A3
po,(E’Es).
One verifiesreadily
thatR(0394)
m
is
equal
to the closure ofV~ R{~a)
in H.s=1 m
Corollary :
:If dimEs
+~, 1 s m, then one hasR(0394) = 03A3 R(0394s)
where~==1
m m
s=1 s=1
Remark 3 : :
If
the comodules(EJ, ~a),
1 s rra , arepairewise isomorphic,
thenfor
m m
the comodule
(E, A)
where E =® Ea,
A =~ 0$,
one hasR()
=R(A~),
1 ~ m.s=1 s=1
II - 3 The
representative subalgebra
of HLet
S(H)
be the set of all elements of the form a =( 1
H ®x’)
o of H where(E, L1 )
is a finite dimensional left H-comodule and x’ E
E’, x
E E. Let usput
dimE = dim0394Lemma 2 : :
S(H)
is amultiplicative, unitary
submonoidof
H.Proof : Set a =
(IH
~x’)
o and b =(IH
®y’)
o03941(y)
ES(H)
where(E, 0394)
and(E, ~1 )
are left H-comodules of finite dimension and x’E E’,
x EE, y’
EEi, y
EEl.
p q P q
One has 0394(x) = 03A3aj®xj, 03941(y) = 03A3bt~yl and 0394E~E1(x~y) = 03A303A3ajbl~yl.
P q
Hence ,
ab =. , a?bt
x’, xj >y’, bl
>=j=i 1=1
(1H ® x’ ® y’)
o ®y)
ES(H).
Since
c(e)
= e ~ e, E = K.e is a left subcomodule of H of dimension1,
one hase =
(IH ® a’)
oc(e)
ES(H).
aLet
R(H)
be the linearsubspace of
Hspaned by S(H).
ThenR(H)
is anunitary
p q
subalgebra
of H.Indeed,
if a =03A303BBjaj
and b =03A3 lbl
are two elements ofR(H),
j=1 ~==1
P q
since
ajbt
ES(H),
one has ab =03A303A303BBj lajbl ~ R(H).
One says thatR(H)
is thej=i l=1
representative subalgebra
of H.Note :
Put,
for the left H-comodule(E, A)
of finitedimension , 5*(A) = {a
=(1H
0x’)
o EH;
x’ EE, x
EE}.
As inProposition 5, depends only
of theisomorphism
class A of(E, A). Furthermore,
one hasS(H)
=U S(A).
0dim0394+~
Also,
it is clear that the K-linear vector spaceR(A)
=03C10394(E’
8E)
isspaned by 5*(A).
Hence one hasR(H)
=U R(A). Moreover,
if(Ei, ðI)
and(E2, A2)
are twodim0394+~
comodules,
thenR(Ai
=R(ð})+R(ð2)
containsR(Ai)
andR(A2)
i.e. thefamily
ordered
by
inclusion is directedupward.
Theorem 1 : : The
representative subalgebra R(H) of
H is such thatc(R(H))
CR(H)
®R(H).
. Moreover(R(H),
m, c, ~,~)
is aHopf algebra.
Proof : It follows from
Proposition
6 and Remark 3 that if A is acoproduct
offinite
dimension,
thenc(R(A))
CR(~) ~ R(A) :
: that isR(A)
is acoalgebra.
SinceR(H)
=U R(0)
is the union ofcoalgebras,
it is acoalgebra.
On the otherhand,
dim0394+~
one deduces from the
Corollary
of Lemma 1 that CR(H).
The Theorem 1 isproved.
II - 4
Simple
comodules of finite dimensionLet be a base of the finite dimensional left H-comodule
( E, 0).
Let usremember that
A j
= is a comodulemorphism.One
sees thatn kerA j
=(0)
1~j~n
and is free in
,C(E, H).
SinceAj(ej)
= ajj, one deduces from(2)
or fromCorollary
2 ofProposition
3 thatej ~ kerAj
andker Aj ~
E.Put
Hj
=Aj(E);
thenHj
is a left subcomodule of H of dimension n.Furthermore,
n n
with
previous notations,
one hasR(0394)
=E) = L Hj
andHj
=03A3K .
alj, alsoj=1 ~-1
J?(A)
is asubcoalgebra
of dimensionn2.
One can havedimR(A) n2 for example,
ifq 9
Eq
=and 0394q
=q ~ 2,
one has =R(A)
anddimR(0394q)
=dimR(0394) ~
t= I t=~
n2 (Qn)2
=(dim(Eq))2.
Definition : : A
left
Banach H ..comodule E is calledsimple
ortopologically
irreducibleif E
is not the null space and does not contain any closed subcomoduledifferent from (0)
and E.
Let
H om.com( E,
be the Banach space of the left Banach comodulemorphisms
of(E, ~)
into andEnd.com(E)
=Hom.com(E, E),
this later is a Banachalgebra.
Remark 4 : Schur’s Lemma.
Let
{E, 0)
and{El, O1 )
be twosimple, finite
dimensionalle f t
H-comodules.(i) If E
andE~
are notisomorphic,
one hasHom.com(E, E~)
=(0)
(ii)
In the alternative case, any non null comodulemorphism of
E intoE~
is an isomor-phism.
Inparticular, End.com(E)
is a(skew) field of finite
dimension(dimE)2. If
Kas
algebraically closed,
thenEnd.com(E)
=K.lE.
Proposition
8 : : Let be asimple
Banachle f t
H-comoduleof finite
dimensionn. Let
(ej)1~j~n
be a baseof
E andAj
=(lH
®fj)
o0394,
1j
n.Then
Hj
=Aj(E) is
asimple left
H ..comoduleof
Hof finite
dimension n. Further-more, there exists J C
~1, n~
such thatR(~) _ ~ Hj (
a direct sumof comodules).
jEJ
Proof : It is the same as in
semi-simple
moduletheory. Indeed,
sinceker Aj ~
E and Eis a
simple
comodule of finite dimension n, the mapAj
E ~Hj
=Aj(E)
is a comoduleisomorphism.
HenceHj
is asimple
comodule of dimension n with base(alj)1~l~n.
If1 ~
j, q
n, one hasHj n Hq
=(0),
orHj
=Hq. Changing
the order if necessary, we my assume that(H1, ... , Hm)
is thefamily
of the distinct comodulesH j ;
m n . Hencem
R(0394) = 03A3Hj, Hj ~ Hq for j ~ q.
j=1
Since
H1
nH2
=(0) ,
one has the direct sum of comodulesHH
(BH2.
Letjo
bethe least
integer >
3 such thatH2) n Hjo
=(0). Hence,
one has the direct sumH1 ~ H2
®Hjo
andfor j jo,
one has(H1
~H2)
nHj ~ (0),
thereforeHj
CH1 ~ H2.
Hence , by induction,
one obtains J ={1, 2, jQ, ... jk
=m~
C[1, m]
and the direct sum of comodules~ Hj
such that forl ~ J, Hl ~ ~ Hj
. It follows thatR(A) = ~ Hj .
jEJ jEJ jEJ
Corollary :
: Let(E, A)
and(Ei, .ðl)
be twosimple left
H-comodule3of finite
dimensionthat are not
isomorphic;
thenAi)
= ,a direct sumof
comodule3.Proof : With
previous notations, put R()
=Hj
andR(Ai) = ~ Hl
. LetjEJ tEL
pj
[resp. p~~
be theprojection
ofR() [resp.
ontoHj [resp. H~~
.Suppose
thatR(A)
n(0);
this finite dimensional comodule must contain at least onesimple
comodule V. There
exists j
E J[resp.
e EL]
such thatpj(V) ~ (0) [resp. p1l(V) ~ (0)] ;
thereforepj(V) = Hj [resp. p1l(V)
=H1l].
Since V issimple,
pj|V[resp. p1l|V]
isan
isomorphism
of V ontoHj [resp. H1l]
. It follows thatHj
andH1l
areisomorphic.
Hence E and
Ei
areisomorphisc ;
a contradiction. ThereforeR(ð)
nR(~1) _ (0)
andR(A
®Ai) =
eR( ð.1).
Remark 5 : : Notations and
hypothesis as
above.If
Kisalgebraically closed,
then theH ~,1 j
n, arepairewise
distinct.Proof :
Indeed,
ifHj
=J?g for j ~ q,
then u =AjoA-1q
is anautomorphism
of the finitedimensional
simple
comoduleHj
.By
Schur’slemma,
one has u = A .1Hj, 03BB E k, 03BB ~
0.Hence
Aj
=03BBAq
and ajj =Aj(ej)
=03BBAq(ej) = 03BBajq.
Therefore03C3(ajj)
= 1 =03BB03C3(ajq) = 03BB03B4jq
=0 ;
a contradiction. ClLet H’ be the Banach space dual of
H ; if we
set fora’,
b’ EH’,
a’ * b’ =( a’ ® b’ ) o c,
thenH’ becomes a
complete
normedalgebra
with unit r. If(E, A)
is a left Banachcomodule, setting
for a’ EH’
, and x EE, a’ .
x =(a’
~lE)
oA(.r),
one induces on E acomplete
normed
right
H’-module structure.Moreover,
if H is apseudo-reflexive Banach
space, thenany closed