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Journal of Lie Theory
Volume ?? (??) ??–??
c
?? Heldermann Verlag
Version of November 7, 2018
Moment map and Gelfand transform for the enveloping
algebra
Didier Arnal and Mohamed Selmi
Abstract. Describing the Gelfand construction for the analytic states on an
universal enveloping algebra, we characterize pure states and re-find the main result
of a preceding work with L. Abdelmoula and J. Ludwig on the separation of unitary
irreducible representations of a connected Lie group by their generalized moment
sets.
Mathematics Subject Classification 2010: 22D10, 22D20, 52A05.
Key Words and Phrases: Universal enveloping algebra; Analytic states; Moment
set of a representation.
1. Introduction
There is a direct relationship between representations of the C?-algebra of a locally
compact group Gand unitary representations of G. The Gelfand construction is
the main tool to describe the cyclic and irreducible representations of a C?-algebra
A, starting with a state or a pure state on A(see [D1] for instance). In this paper,
we consider a real connected and simply connected Lie group G, and the complex
enveloping associative algebra A(g) of its Lie algebra g. We then define the notion
of analytic state ϕon A(g), and perform the Gelfand construction, starting with ϕ.
This construction was described by K. H. Neeb in [Ne] for the large class of BCH-Lie
algebras. We present it here in the really simpler setting of finite dimensional Lie
algebras. The result is a unitary representations of Gitself. This representation is
irreducible if and only if ϕis a pure state.
Conversely, associated to any unitary representation πof G, there is a notion
of generalized moment map and set (see [W, BLS, ABLS, AALS]): each smooth
vector fin the space Hπof the representation πdefines naturally a state Φπ(f)
(see Section 2 for definition). It is proved in [AALS] that, if πis irreducible, the
convex hull Jπ,Cof this set of states characterizes the representation π.
In the present paper, we re-find this result, by using the Gelfand construction
when fis an analytic vector, and proving that, if πis irreducible, the set of Φπ(f)
(fanalytic) is the subset of analytic pure states in Jπ,Cand the disjointness of the
sets {Φπ(f) : fanalytic}for two non-equivalent irreducible representations of G.
The starting point of this work is a personal remark of Jean Ludwig. We
thank him for his ideas and suggestions.
2Arnal and Selmi
2. The moment map
In this section, we recall definitions and main results for the moment map of
a unitary representation.
Let Gbe a connected and simply connected real Lie group, and πa unitary
representation of Gon an Hilbert space Hπ. Let us denote by gthe (real) Lie algebra
of Gand A(g) the complex universal enveloping algebra of g. The associative algebra
A(g) acts naturally on the space H∞
πof C∞vectors in Hπ, we denote by dπ the
corresponding representation.
Definition 2.1. Let fbe a non vanishing C∞vector in Hπ. The moment of π
in fis by definition the element Ψπ(f) of the real linear dual of A(g) defined by:
Ψπ(f)(A) = Re1
ihdπ(A)f, f i
hf, f i.
The map Ψπ:H∞
π\ {0} → (A(g))?
Ris the moment map of π.
The range Iπof the map Ψπis the moment set of π, we denote by Jπthe
convex hull of Iπ.
Since each complex linear form ϕon A(g) can be written as ϕ(A) = ψ(iA) +
iψ(A), where ψis the real linear map ψ(A) = Im(ϕ(A)), we put also:
Definition 2.2. Let fbe a non vanishing C∞vector in Hπ. The complex
moment of πin fis by definition the element Φπ(f) of the linear dual (A(g))?
defined by:
Φπ(f)(A) = hdπ(A)f, f i
hf, f i.
The map Φπ:H∞
π\ {0} → (A(g))?is the complex moment map of the
representation π.
The range Iπ,Cof the map Φπis the complex moment set of π, we denote by
Jπ,Cthe convex hull of Iπ,C.
These objects were studied in [BLS, ABLS, AALS, ABB]. The main result is:
Theorem 2.3. ([AALS]) Let πand ρbe two irreducible unitary representations
of G, then πand ρare equivalent if and only if Jπ=Jρ, if and only if Jπ,C=Jρ,C.
Especially, if Gis solvable, for each regular, integral coadjoint orbit Ois
associated a family of irreducible unitary representations πO,χ (see [AK, B]). In
[ABB], the authors present a direct construction, allowing to re-find Oand χfrom
IπO,χ .
Since g⊂A(g), we can restrict the real forms Ψπ(f) to g. Then [f]7→
Ψπ(f)|gis the moment map for the strongly Hamiltonian action of Gon the natural
symplectic manifold P(H∞) (see [W, ASZ]). The denomination ‘moment map’ for
the map Ψπis coming from this observation.
Arnal and Selmi 3
3. Gelfand construction for A(g)
Let (X1, . . . , Xd) be a basis of the Lie algebra g. Recall that each unitary represen-
tation πof Ghas a dense subspace Hω
πof analytic vectors (see [N]), that is vectors
ffor which there is r > 0, C > 0 such that for each n, each 1 ≤i1, . . . , in≤d,
kdπ(Xi1. . . Xin)fk ≤ rC nn! (1)
From now one, we restrict ourselves to the space Hω
π,1of analytic unit vectors
f:f∈ Hω,kfk= 1. We thus put
Φω
π= Φπ|Hω
π,1, Iω
π,C= Φω
π(Hω
π,1),and Jω
π,C= Conv(Iω
π,C).
Observe that if π0is a representation unitarily equivalent to π, there is a
unitary operator U:Hπ→ Hπ0such that π0(x) = U◦π(x)◦U−1, and for each
f∈ Hω
π,1,U f is in Hω
π0,1and
Φω
π0(Uf )(A) = hdπ0(A)U f, U fi=hU dπ(A)f, U fi= Φω
π(f)(A).
Therefore Φω
π0(Uf )=Φω
π(f) and Iω
π0,C=Iω
π,C.
Now A(g) is an involutive algebra, when we equip it with the involution (called
principal anti-isomorphism in [D2]):
(αY1. . . Yn)?= (−1)nαYn. . . Y1,
if Yi∈g,α∈C.
Therefore we shall now repeat the usual Gelfand construction for the involu-
tive algebra (A(g), ?) as for C?-algebras (see for instance [D1]).
Definition 3.1.
1. A linear form ϕ:A(g)→Cis positive if ϕ(A?A)≥0, for each A∈A(g) .
2. A state is a positive linear form ϕon A(g) , such that ϕ(1) = 1. We denote
by S(A(g)) the set of states of A(g) .
3. A state ϕis analytic if there is a basis (X1, . . . , Xd) of g, and positive constants
r,Csuch that, for any n, any 1 ≤i1, . . . , in≤d,
|ϕ(Xi1. . . Xin)| ≤ rCnn! (2)
Fix C, the set of states ϕfor which there is r=rϕ>0 such that the relation
(2) holds for any n,i1, . . . , inis denoted by Sω
C. The set of analytic states of
A(g) is denoted by Sω(A(g)): Sω(A(g)) = ∪C>0Sω
C. Clearly, the set Sω(A(g))
does not depend of the choice of the basis.
4. A state ϕis pure if the only state ψsuch that there exists a > 0 such that
aψ(A?A)≤ϕ(A?A) for each A∈A(g) is ϕitself. Denote by P(A(g)) (resp.
Pω(A(g))) the set of pure states (resp. of pure analytic states).
4Arnal and Selmi
Relation with the Neeb theory
Recall that in [Ne], K. H. Neeb studies systematically the integrability problem
for functional ϕon A(g), where gis a BCH-Lie algebra. To solve this problem, he
defines analytic functional as a linear form ϕsuch that Pn
ϕ(Xn)
n!converges for each
Xin a 0-neighborhood in g. If gis finite dimensional, a functional ϕis analytic if
and only if the relation (2) holds.
An important result of Neeb is the following: let ϕbe a functional on A(g),
pa sub-multiplicative semi-norm on g(p([X, Y ]) ≤p(X)p(Y)). Put:
kϕs
nkp= sup n|1
n!X
σ∈Sn
ϕ(Xσ(1) . . . Xσ(n))|:p(Xi)≤1 (1 ≤i≤n)o.
Now Theorem 6.10 in [Ne] says that if Gis a BCH-Lie group with Lie algebra g, if ϕ
is an analytic functional which is positive and such that there is a sub-multiplicative
semi-norm pon gfor which Pn
kϕs
nkptn
n!converges for some t > 0, then there is a
unique unitary representation (πϕ,Hϕ) of Gand a vector fin Hϕsuch that
ϕ(A) = hdπϕ(A)f, f i.
It is not difficult to prove that if Gis a finite dimensional Lie group, the above
condition holds for any analytic state ϕin the sense of Definition 3.1, therefore the
next theorem is a direct consequence of Theorem 6.10 in [Ne]. However, to be
complete, we present here a direct proof of this theorem.
Theorem 3.2. If πis a unitary representation of Gand fa analytic unit vector
for π, then: Φω
π(f) : A7→ hdπ(A)f, f iis an analytic state for A(g).
Conversely, let ϕbe an analytic state on A(g), then there is a unitary repre-
sentation τof Gand an analytic vector f∈ Hω
τ,1such that:
ϕ= Φω
τ(f).
Proof. Let πbe a unitary representation of Gand f∈ Hω
π,1. By definition,
Φω
π(f) is a positive form: Φω
π(f)(A?A) = hdπ(A?A)f, f i=kdπ(A)fk2≥0, and a
state since Φω
π(f)(1) = hf, f i= 1. Now since fis analytic, Formula (1) says that
there is r > 0, C > 0 such that:
|Φω
π(f)(Xi1. . . Xin)|=|hdπ(Xi1. . . Xin)f, f i| ≤ kdπ(Xi1. . . Xin)fk ≤ rCnn!
Thus Φω
π(f)∈Sω(A(g)).
Conversely, if ϕ∈Sω(A(g)), put Annϕ={A∈A(g) : ϕ(A?A) = 0}. Since
ϕis positive, the relation b(A, B) = ϕ(B?A) defines a positive form b, linear in the
argument A, anti-linear in the argument B. By the Cauchy-Schwarz inequality:
|b(A, B)| ≤ (b(A, A)) 1
2(b(B, B)) 1
2.
Therefore, Annϕis the kernel of b: Annϕ={A∈A(g) : ϕ(B?A) = 0 (B∈A(g))},
it is a left ideal. Put V=A(g)/Annϕ, and let dτ be the natural A(g)-action on V:
dτ(A)[B]=[AB].
Arnal and Selmi 5
The form bdefines a scalar product ([A]|[B]) = b(A, B) = ϕ(B?A) on V. Com-
pleting V, we get an Hilbert space H=V. Let A=Xj1. . . Xja. For any n, any
1≤i1. . . in≤d, the square of the norm of the vector dτ (Xi1. . . Xin)[A] is:
k[Xi1. . . XinA]k2=|ϕ(A?Xin. . . Xi1Xi1. . . XinA)| ≤ rC2n+2a(2n+ 2a)!
Using the Stirling formula: n!∼√2πn n
en, if C0> C , we get:
C2n+2a(2n+ 2a)!
((C0)nn!)2∼1
√π
1
√n
C2a
e2a2n+ 2a
2n2n
(2n+ 2a)2aC
C02n
22n.
Since 2n+2a
2n2n→e2aand (2n+ 2a)2aC
C02n→0, there is rA>0 such that:
k[Xi1. . . XinA]k2≤rA(2C0)nn!2.(3)
This implies that the vectors in Vare analytic for the representation of g, thus
Theorem 1 in [FSSS] proves there is a unique unitary representation τof Gon H
such that V⊂ H∞, and dτ(X)[A] = [X A]: on V,dτ is the representation of A(g)
associated to τby differentiation.
Finally [1] is an analytic vector for τ(take A= 1 in the preceding computa-
tion), and by definition:
Φω
τ([1])(A) = hdτ(A)[1],[1]i= ([A]|[1]) = ϕ(A),
for each A∈A(g) , this finishes the proof.
The relation between pure states and irreducible representations of Gis:
Proposition 3.3.
1. Let πbe an irreducible unitary representation of G, and f∈ Hω
π,1an analytic
unit vector for π, then the state Φω
π(f)is pure.
2. Conversely, if ϕis a pure state, then the representation τdefined in Theorem
3.2 is irreducible.
Proof.
1. Suppose a > 0 and ϕis a state such that aϕ(A?A)≤Φω
π(f). Since dπ(Xn)f∈
dπ(A(g))f, for any n, if gis a vector in (dπ(A(g))f)⊥, then g∈(π(G)f)⊥,
thus gis orthogonal to the closure Kof the vector space generated by π(G)f.
But Kis a closed vector subspace of Hπ, invariant under the action of π(G),
and containing f6= 0 , therefore K=Hπ,g= 0, and W=dπ(A(g))fis a
dense subspace in Hπ.
Now since kdπ(A)fk2= Φω
π(f)(A?A)≥aϕ(A?A),
Annf={A∈A(g) : dπ(A)f= 0}= AnnΦω
π(f)⊂Annϕ.
6Arnal and Selmi
Thus the surjective map A7→ dπ(A)finduces a bijective map [A]7→ dπ(A)f
between the spaces V=A(g)/Annfand W. Moreover the Hermitian form
b([A],[B]) = ϕ(B?A) is well defined on Vand:
a2|b(A, B)|2≤a2b(A, A)b(B, B)≤ kdπ(A)fk2kdπ(B)fk2.
By passing to the completion of these spaces, there is a positive self-adjoint
operator Ton Hbounded by 1, such that, for any Aand Bin A(g) ,
ϕ(B?A) = hT dπ(A)f, dπ(B)fi.
But Tcommutes with dπ : indeed for each A,Band Cin A(g),
hT dπ(AB)f, dπ(C)fi=ϕ(C?AB) = ϕ((A?C)?B)
=hT dπ(B)f, dπ(A?C)fi
=hdπ(A)T dπ(B)f, dπ(C)fi.
thus by the density of W,T dπ(A) = dπ(A)Ton W, by T-continuity, for each
xin G,T π(x) = π(x)T, on W. This implies Tcommutes with π(x) on Hπ
for each x. By the Schur Lemma T=λId, but 1 = ϕ(1) = hT f, f i=λ,
ϕ= Φω
π(f), Φω
π(f) is a pure state.
2. Suppose now ϕis a pure analytic state, perform the Gelfand construction for
ϕ, as in Theorem 3.2. With the notation of the Theorem, the vector f= [1]
is analytic in Hτ, and ϕ= Φω
τ(f).
Let Kbe an invariant closed subspace of Hτ, let us decompose fon the direct
sum K ⊕ K⊥into f=f1+f2. If f1= 0, then Kbeing the closure of the
projection of dτ(A(g))fis vanishing. Suppose now f16= 0. Since Kand K⊥
are invariant, for any n, any 1 ≤i1, . . . , in≤d,
kdτ(Xi1. . . Xin)fk2=kdτ (Xi1. . . Xin)f1k2+kdτ (Xi1. . . Xin)f2k2≤rCnn!.
Thus f1is an analytic vector, and
Φω
τ(f1)(A?A) = hdτ(A?A)f1, f1i=kdτ (A)f1k2≤ kdτ (A)fk2=ϕ(A?A).
Since ϕis pure, this implies Φω
τ(f1) = ϕ, and choosing A= 1, f2= 0 . Thus
the only closed invariant subspaces of Hτare {0}and Hτ:τis irreducible.
Proposition 3.3 means that there is a canonical map F:Pω(A(g)) →b
G,
defined by F(ϕ) = [τ] where τis defined in Theorem 3.2. Since ϕ= Φτ([1]), this
map is onto. Now let [π] be in b
G, by definition, F−1([π]) = {Φπ(f) : f∈ Hω
π,1}:
F−1([π]) = Iω
π,C.
Corollary 3.4. Let πand ρtwo unitary irreducible representations of G. If π
and ρare equivalent, then:
Iω
π,C=Iω
ρ,Cand Iω
π=Iω
ρ.
If πand ρare not equivalent, then:
Iω
π,C∩Iω
ρ,C=∅and Iω
π∩Iω
ρ=∅.
Arnal and Selmi 7
4. Convex hulls
Let Xbe a convex subset in a real vector space. Recall that the extremal points in
Xare points vsuch that v=av1+ (1 −a)v2, with 0 ≤a≤1 and vi∈X,v16=v2,
implies a= 0 or a= 1. We denote by Ext(X) the set of extremal points in X.
Proposition 4.1. The subsets Sω
C,Sω(A(g)) (C > 0) of (A(g))?are convex,
and:
Ext(Sω(A(g))) = Pω(A(g)).
Proof. Since Sω
C⊂Sω
C0if C≤C0, and Sω(A(g)) = ∪C>0Sω
C, the convexity of
Sω(A(g)) is a consequence of the convexity of each Sω
C. Let now 0 ≤a≤1 , ϕ,ψ
in Sω
C, then θ=aϕ + (1 −a)ψis clearly a state and
|θ(Xi1. . . Xin)| ≤ a|ϕ(Xi1. . . Xin)|+ (1 −a)|ψ(Xi1. . . Xin)| ≤ (arϕ+ (1 −a)rψ)Cnn!
Therefore Sω
Cis convex.
Let θbe an analytic state. Suppose first θis not extremal: there is 0 <a<1
and ϕ6=ψin Sω(A(g)) such that θ=aϕ + (1 −a)ψ. Thus aϕ(A?A)≤θ(A?A), for
any A. This means that θis not a pure state.
Conversely, suppose that θis a non pure analytic state, more precisely,
suppose θ∈Sω
C. Let πbe the unitary representation of Gin Hπ=A(g)/Annθ
and f∈ Hω
πas in Theorem 3.2, such that θ= Φω
π(f). Since θis not pure, πis
not irreducible (Proposition 3.3): there is a non trivial invariant closed subspace K
in Hπ. Decompose fon Hπ=K ⊕ K⊥into f=f1+f2. Then fi6= 0. Indeed if
f2= 0, then dπ(A(g))f=dπ(A(g))f1⊂ K, taking the closure, we get: Hπ=K,
a contradiction. Let pbe the orthogonal projection onto K, since Kand K⊥are
invariant, pcommutes with π(x) for each x. Therefore p(Hω
π)⊂ Kω
π|K. Thus f1,f2
are analytic vectors in Hπ. Put ϕ= Φω
π(f1
kf1k), ψ= Φω
π(f2
kf2k), and a=kf1k2. Then:
θ= Φπ(kf1kf1
kf1k+kf2kf2
kf2k) = aϕ + (1 −a)ψ.
This finishes the proof of the proposition.
Let us now come back to the complex analytic moment set Iω
π,Cof a unitary
irreducible representation πand its convex hull, Jω
π,C. We have:
Proposition 4.2. The following holds:
Jω
π,C∩Pω(A(g)) = Jω
π,C∩Ext(Sω(A(g))) = Iω
π,C.
Proof. Let θbe in Jω
π,C∩Pω(A(g)), thus there are k,ϕi∈Iω
π,C,ai>0
(1 ≤i≤k), such that Piai= 1 and:
θ=X
i
aiϕi.
8Arnal and Selmi
Especially, ϕ1is a state such that a1ϕ1(A?A)≤θ(A?A). Since θis pure, this implies
ϕ1=θ,θ∈Iω
π,C.
Conversely, since πis irreducible, each state Φπ(f) in Iω
π,Cis pure and in
Jω
π,C. This proves the proposition.
Observe that Theorem 2.3 (see [AALS]) is now a direct consequence of Propo-
sition 4.2. Recall the notation Jπ= Conv(Iπ), Jπ,C= Conv(Iπ,C) .
Corollary 4.3. Let πand ρbe two unitary irreducible representations of a con-
nected Lie group G, then πand ρare equivalent if and only if Jπ=Jρ.
Proof. Let e
Gbe the universal covering group of Gand eπ,eρthe representations
of e
Gcorresponding to πand ρ. Then dπ =deπ,Iπ,C=Ieπ,C, and Jπ,C=Jeπ,C.
Let f∈ Hω
eπ=Hω
π, then
Φeπ(f)∈Jω
eπ,C∩Pω(A(g)) = Iω
eπ,C.
On the other hand, if Jπ=Jρ, then this analytic state is in Jeρ,C, thus we also have:
Φeπ(f)∈Jω
eρ,C∩Pω(A(g)) = Iω
eρ,C.
Therefore eπ∼eρ, and π∼ρ.
The converse being evident, this proves the corollary.
5. Complete convex sums
In [AALS], the complete moment set e
Jπ,Cof a unitary representation πis defined
by:
e
Jπ,C=n∞
X
n=0
Φπ(fn) : fn∈ H∞
π,X
nkfnk2= 1,X
nkdπ(A)fnk2<∞for all A∈A(g)o.
Similarly, let us define now the complete states ψ:
Definition 5.1. An analytic state ψis the sum of a convex series of states if there
exists a sequence (ϕn) of states and a sequence (an) of positive real numbers such
that:
ψ=
∞
X
n=0
anϕn.
Lemma 5.2. If ψ=Pnanϕnis the sum of a convex series of states, then for
each n,an6= 0 implies that ϕnis an analytic state.
Proof. Fix a basis {X1, . . . , Xd}of g, since ψis analytic, there are r > 0 and
Arnal and Selmi 9
C > 0 such that, for any k, and any 1 ≤i1, . . . , ik≤d,
|ψ(Xik. . . Xi1Xi1. . . Xik)|= (−1)kψ(Xik. . . Xi1Xi1. . . Xik)
=X
n
an(−1)kϕn(Xik. . . Xi1Xi1. . . Xik)
≤rC2k(2k)!.
Thus if an>0 ,
|ϕn(Xik. . . Xi1Xi1. . . Xik)| ≤ r
an
C2k(2k)!.
Now since the form (A, B )7→ ϕn(B?A) on A(g) is positive, the Cauchy-Schwarz
inequality |ϕn(B?A)|2≤ϕn(B?B)ϕn(A?A) holds and:
|ϕn(Xi1. . . Xik)| ≤ rr
an
Ckp(2k)!.
Now, the Stirling formula implies:
(2k)! ∼√4πk 2k
e2k
∼2√πk 22kk!
√2πk 2
=22k
√πk (k!)2,
or p(2k)! ∼(πk)−1/42kk!, this proves that ϕnis an analytic state.
From now on, we suppose that if an= 0 , then ϕn(Xi1. . . Xik) = 0 for any
k > 0 (ϕnis the map called augmentation in [D2]).
Observe that, if τnis the unitary representation such that ϕn= Φω
τn(fn)
defined in Theorem 3.2, then putting f0
n=√anfn, and denoting ⊕τnthe direct
sum of the τn, we get:
k⊕nf0
nk2=X
nkf0
nk2=X
n
an=ψ(1) = 1.
Moreover, for any Ain A(g),
k⊕ndτn(A)f0
nk2=X
nkdτn(A)f0
nk2=X
n
ankdτn(A)fnk2
=X
n
anΦω
τn(fn)(A?A) = ψ(A?A)<∞
Therefore ψ= Φω
⊕τn(⊕f0
n) belongs to Iω
⊕τn,C.
Proposition 5.3. Let ψ=Pnanϕnbe the sum of a convex series of analytic
states and π(resp. τn) the representation associated to ψ(resp. to ϕn) by Theorem
3.2. Then πis unitarily equivalent to a subrepresentation of ⊕nτn.
Proof. Recall that Hπis the completion of the pre-Hilbert space V=A(g)/Annψ,
equipped with the bilinear form:
([A]|[B]) = ψ(B?A).
10 Arnal and Selmi
As above, denote ϕn= Φτn(fn), and f0
n=√anfn. Define thus the map U:V→
⊕nHτnby
U([A]) = X
n
dτn(A)f0
n.
Observe that Uis well defined because, if Abelongs to Annψ,
0 = ψ(A?A) = X
n
anhdτn(A?A)fn, fni=X
n
ankdτn(A)fnk2.
Thus U([A]) = 0. Moreover Uis linear and an isometry, since
kU([A])k2=X
n
ankdτn(A)fnk2=ψ(A?A) = k[A]k2.
We can thus extend Uto the space Hπ.
We saw that if Cis such that ψ∈Sω
C, then each ϕnis in Sω
2C. Fix the basis
{X1, . . . , Xd}of gand define the norm on gby kPixiXik= sup |xi|. Let Xbe in
gsuch that kXk<1
4Cd , then we claim that the following holds:
U(π(exp X)[A]) = X
k
1
k!U(dπ(Xk)[A]) = X
kX
n
1
k!dτn(XkA)f0
n
=X
nX
k
1
k!dτn(XkA)f0
n
=X
n
τn(exp X)dτn(A)f0
n= (⊕nτn)(exp X)U([A]).
Indeed, thanks to the inequality (3) for any C0such that 2C > C0> C , we saw
there is rA>0 such that:
kdτn(XkA)f0
nk ≤ X
i1,...ik|xi1. . . xik|kdτn(Xi1. . . XikA)f0
nk
≤rA(2C0)kk!dk
(4Cd)k≤rAk!C0
2Ck
.
Therefore by Lebesgue’s theorem of dominated convergence, we can exchange the
sums over kand nin the above computation, this proves our claim.
Since Gis connected and simply connected, this implies that for each xin
G, each f∈ Hπ,
U(π(x)f)=(⊕nτn)(x)(Uf ).
This proves the proposition.
As in [AALS], if πis a unitary representation of G, we denote by eπthe
representation ℵ0πsum of a countably many representations unitarily equivalent to
π. Moreover the proof of Proposition 3.2 in [AALS] says that e
Jπ,C=Ieπ,C. Therefore
Theorem 1.2 of [AALS] is a direct corollary of Proposition 5.3. Recall that two
unitary representations πand ρare called quasi equivalent if the representations eπ
and eρare unitarily equivalent.
Arnal and Selmi 11
Corollary 5.4. Two unitary representations πand ρof a connected Lie group G
are quasi equivalent if and only if e
Jπ,C=e
Jρ,C, if and only if their complete moment
sets Ieπand Ieρcoincide.
Proof. By considering the extensions of the representations πand ρto the
universal covering group of G, we can suppose that Gis simply connected.
If πand ρare quasi-equivalent, then eπand eρare equivalent and Ieπ=Ieρ.
Conversely, if Ieπ=Ieρ, then Iω
eπ,C=Iω
eρ,C. Pick f∈ Hω
π,1, then Φω
π(f) is in
Iω
π,C⊂Iω
eπ,C=Iω
eρ,C. This means there is a sequence (fn) of vectors in Hω
ρ,1, a sequence
(an) of positive numbers, such that:
ψ= Φω
π(f) = X
n
anΦω
ρ(fn) = Xanϕn.
For each ndefine Hτnas the closure of the subspace dρ(A(g))fnof Hρ, and τn
as the restriction of ρto Hτn. Thus τnis the representation coming from ϕnby
Theorem 3.2. Now Proposition 5.3 implies that πis a subrepresentation of ⊕nτn,
thus a subrepresentation of eρ.
Therefore eπis a subrepresentation of e
eρ=eρ. Interchanging the roles of πand
ρ, we obtain that eρis a subrepresentation of eπ, an usual argument proves that eπ
and eρare equivalent (see [D1] or [AALS]).
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Didier Arnal
Institut de Math´ematiques de Bour-
gogne UMR CNRS 5584
Universit´e de Bourgogne Franche
Comt´e
B.P. 47870
F-21078 Dijon Cedex
France
Didier.Arnal@u-bourgogne.fr
Mohamed Selmi
Laborattoire Physique Math´ematique,
Fonctions sp´eciales et Applications
LR 11 ES 35
Universit´e de Sousse
Ecole Sup´erieure des Sciences et de
Technologie de Hammam Sousse
Rue Lamine Abassi 4011 H.Sousse
Tunisie
Mohamed.Selmi@fss.rnu.tn
