Jacquet modules of principal series generated by the trivial $K$-type

0.26 MB·English

by  Noriyuki Abe

Jacquet modules of
K
principal series generated by the trivial -type
6
0
0
Noriyuki Abe
2

n
a
J

0 Abstract. WeproposeanewapproachforthestudyoftheJacquetmoduleof
2 aHarish-ChandramoduleofarealsemisimpleLiegroup. Usingthismethod,we
investigatethestructureoftheJacquetmoduleofprincipalseriesrepresentation
]
T generated by the trivial K-type.
R
. §1. Introduction
h
t
a
Let G be a real semisimple Lie group. By Casselman’s subrepresentation theorem, any ir-
m
reducible admissible representation U is realized as a subrepresentation of a certain non-unitary
[
principal series representation. Such an embedding is a powerful tool to study an irreducible ad-

2
missible representation but the subrepresentation theorem dose not tell us how it can be realized.
v
2 Casselman [Cas80] introduced the Jacquet module J(U) of U. This important object retains
6 all information of embeddings given by the subrepresentation theorem. For example, Casselman’s
4
subrepresentation theorem is equivalent to J(U) 6= 0. However the structure of J(U) is very
1
0 intricate and difficult to determine.
6
In this paper we give generators of the Jacquet module of a principal series representation
0
generated by the trivial K-type. Let Z be the ring of integers, g the Lie algebra of G, θ a Cartan
/ 0
h
involution of g , g = k ⊕a ⊕n the Iwasawa decomposition of g , m the centralizer of a in k ,
t 0 0 0 0 0 0 0 0 0
a W the little Weyl group for (g ,a ), e ∈ W the unit element of W, Σ the restricted root system
m 0 0
for (g ,a ), g the root space for α∈ Σ, Σ+ the positive system of Σ such that n = g ,
v: ρ = 0 0 (d0i,αmg /2)α, P = { n α | n ∈ Z}, P+ = { n α| n ∈0Z, n α≥∈Σ0+} a0n,αd
i α∈Σ+ 0,α α∈Σ+ α α α∈Σ+ α α Lα
X U(λ) the principal series representation with an infinitesimal character λ generated by the trivial
P P P
K-type. In this paper we prove the following theorem.
r
a
Theorem 1.1 (Theorem 3.9, Theorem 4.1). Assume that λ is regular. Set W(w) = {w′ ∈
W |wλ−w′λ ∈ 2P+} for w ∈ W. Then there exist generators {v | w ∈ W} of J(U(λ)) such that
w
(H −(ρ+wλ))vw ∈ w′∈W(w)U(g)vw′ for all H ∈ a0,
(Xvw ∈ w′∈W(w)U(gP)vw′ for all X ∈ m0⊕θ(n0).
Graduate School of MathemPatical Sciences, the University of Tokyo, 3–8–1 Komaba, Meguro-ku,
Tokyo 153–8914, Japan.
E-mail address: [email protected]
2000 Mathematics Subject Classification. 22E47.
1
Noriyuki Abe
We enumerate W = {w ,w ,...,w } such that Rew λ ≥ Rew λ ≥ ··· ≥ Rew λ. Set V =
1 2 r 1 2 r i
U(g)v . ThenbyTheorem1.1wehavethesurjectivemapM(w λ) → V /V whereM(w λ)
j≥i wj i i i+1 i
is a generalized Verma module (See Definition 4.4). This map is isomorphic. Namely we can prove
P
the following theorem.
Theorem 1.2 (Theorem 4.5). There exists a filtration J(U(λ)) = V ⊃ V ⊃ ··· ⊃ V = 0
1 2 r+1
of J(U(λ)) such that V /V ≃ M(w λ). Moreover if wλ−λ 6∈ 2P for w ∈ W\{e} then J(U(λ)) ≃
i i+1 i
M(wλ).
w∈W
L This theorem does not need the assumption that λ is regular. In the case of G is split and U(λ)
is irreducible, Collingwood [Col91] proved Theorem 1.2.
For example, we obtain the following in case of g = sl(2,R): Choose a basis {H,E ,E } of
0 + −
g such that RH = a , RE = n , [H,E ] = ±2E and E = θ(E ). Then Σ+ = {2α} where
0 0 + 0 ± ± − +
α(H) = 1. Let λ = rα for r ∈ C. We may assume Rer ≥ 0. By Theorems 1.1 and 1.2, we have the
exact sequence
0 //M(−rα) //J(U(rα)) //M(rα) // 0.
Consider the case λ is integral, i.e., 2r ∈ Z. If r 6∈ Z then this sequence splits by Theorem 1.2. On
the other hand, if r ∈ Z then by the direct calculation using the method introduced in this paper
we can show it does not split. Notice that U(rα) is irreducible if and only if r ∈ Z. Then we have
the following; if λ is integral then J(U(λ)) is isomorphic to the direct sum of generalized Verma
modules if and only if U(λ) is reducible.
Our method is based on the paper of Kashiwara and Oshima [KO77]. In Section 2 we show
fundamental properties of Jacquet modules and introduce a certain extension of the universal
enveloping algebra. An analog of the theory of Kashiwara and Oshima is established in Section 3.
In Section 4 we prove our main theorem in the case of a regular infinitesimal character using the
result of Section 3. We complete the proof in Section 5 using the translation principle.
Acknowledgments
The author is grateful to his advisor Hisayosi Matumoto for his advice and support. He would
also like to thank Professor Toshio Oshima for his comments.
Notations
Throughout this paper we use the following notations. As usual we denote the ring of integers,
the set of non-negative integers, the set of positive integers, the real number field and the complex
number field by Z,Z ,Z ,R and C respectively. Let g be a real semisimple Lie algebra. Fix
≥0 >0 0
a Cartan involution θ of g . Let g = k ⊕ s be the decomposition of g into the +1 and −1
0 0 0 0 0
eigenspaces for θ. Take a maximal abelian subspace a of s and let g = k ⊕ a ⊕ n be the
0 0 0 0 0 0
corresponding Iwasawa decomposition of g . Set m = {X ∈ k | [H,X] = 0 for all H ∈ a }. Then
0 0 0 0
p = m ⊕ a ⊕ n is a minimal parabolic subalgebra of g . Write g for the complexification of
0 0 0 0 0
g and U(g) for the universal enveloping algebra of g. We apply analogous notations to other Lie
0
algebras.
Set a∗ = HomC(a,C) and a∗0 = HomR(a0,R). Let Σ ⊂ a∗ be the restricted root system for
(g,a) and g the root space for α ∈ Σ. Let Σ+ be the positive root system determined by n, i.e.,
α
2
Jacquet modules of principal series generated by the trivial K-type
n = g . Σ+ determines the set of simple roots Π = {α ,α ,...,α }. We define the total
α∈Σ+ α 1 2 l
order on a∗ by the following; for c ,d ∈ R we define c α > d α if and only if there exists
L 0 i i i i i i i i
an integer k such that c = d ,...,c = d and c > d . Let {H ,H ,...,H } be the dual
1 1 k k k+1 k+1 1 2 l
P P
basis of {α }. Write W for the little Weyl group for (g ,a ) and e for the unit element of W. Set
i 0 0
P = { n α | n ∈ Z}, P+ = { n α | n ∈ Z } and P++ = P+ \{0}. Let m be a
α∈Σ+ α α α∈Σ+ α α ≥0
dimension of n. Fix a basis E ,E ,...,E of n such that each E is a restricted root vector. Let
1 2 m i
β be aPrestricted root vector such thaPt E ∈ g . For n = (n ) ∈ Zm we denote En1En2···Enm
i i βi i ≥0 1 2 m
n
by E .
For x= (x ,x ,...,x ) ∈ Zn , we write |x|= x +x +···+x and x! = x !x !···x !.
1 2 n ≥0 1 2 n 1 2 n
For aC-algebraR,letM(r,r′,R)bethespaceofr×r′ matriceswithentriesinRandM(r,R) =
M(r,r,R). Write 1 ∈ M(r,R) for the identity matrix.
r
§2. Jacquet modules and fundamental properties
Definition 2.1 (Jacquet module). Let U be a U(g)-module. Define modules J(U) and J(U)
by
b
J(U) = limU/nkU,
←−
k
Jb(U) = J(U) = {u∈ J(U) | dimU(a)u < ∞}.
a-finite
We call J(U) the Jacquet modulbe of U. b
Set E(n) = lim U(n)/nkU(n).
←−k
Propbosition 2.2. (1) The C-algebra E(n) is right and left Noetherian.
(2) The C-algebra E(n) is flat over U(n).
b
(3) If U is a finitely generated U(n)-module then lim U/nkU = E(n)⊗ U.
b ←−k U(n)
(4) Let S = (Sk) be an element of M(r,nE(n)) and (an) ∈ CbZ≥0. Define ∞n=0anSn =
( k a Sn) . Then ∞ a Sn ∈ M(r,E(n)).
n=0 n k k n=0 n P
b
ProPof. Since Stafford aPnd Wallach [SW82, Theorem 2.1] show that nU(n) ⊂ U(n) satisfies the
b
Artin-Rees property, the usual argument of the proof for commutative rings can be applicable to
prove (1), (2) and (3). (4) is obvious.
Corollary 2.3. Let S be an element of M(r,E(n)) such that S −1 ∈ M(r,nE(n)). Then S is
r
invertible.
b b
Proof. Set T = 1 −S. By Proposition 2.2, R = ∞ Tn ∈ M(r,E(n)). Then SR = RS =
r n=0
1 .
r
P
b
WecanprovethefollowingpropositioninasimilarwaytothatofGoodmanandWallach[GW80,
Lemma 2.2]. For the sake of completeness we give a proof.
3
Noriyuki Abe
Proposition 2.4. Let U be a U(a⊕n)-module which is finitely generated as a U(n)-module.
Assume that every element of U is a-finite. For µ ∈ a∗ set
U = {u ∈ U | For all H ∈ a there exists a positive integer N such that (H −µ(H))Nu= 0}.
µ
Then
J(U) ≃ U .
µ
µ∈a∗
Y
b
Proof. For k ∈ Z put S = {µ ∈ a∗ | U 6= 0, U 6⊂ nkU}. Since U is finitely generated,
>0 k µ µ
dimU/nkU < ∞. Therefore S is a finite set. Define a map ϕ: U → J(U) by
k µ∈a∗ µ
Q
b
ϕ((xµ)µ∈a∗) = xµ (mod nkU) .
 
µX∈Sk k
 
First we show that ϕ is injective. Assume ϕ((xµ)µ∈a∗) = 0. We have µ∈Sk xµ ∈ nkU for all
k ∈ Z . Since nkU is a-stable and S is a finite set, x ∈ nkU for all µ ∈ a∗, thus we have x = 0.
>0 k µ P µ
We have to show that ϕ is surjective. Let x = (x (mod nkU)) be an element of J(U). Since
k k
every element of U is a-finite, we have U = U . Let p : U → U be the projection. The
µ∈a∗ µ µ µ
U(n)-module U is finitely generated and therefore for all µ ∈ a∗ there exists a posibtive integer
L
kµ such that nkµU ∩ Uµ = 0. Notice that if i,i′ > kµ then pµ(xi) = pµ(xi′). Hence we have
ϕ((pµ(xkµ))µ∈a∗) = x.
We define an (a⊕ n)-representation structure of U(n) by (H + X)(u) = Hu− uH +Xu for
H ∈ a, X ∈ n, u∈ U(n). ThenU(n) is aU(a⊕n)-module. By Proposition 2.4 E(n) = U(n) .
µ∈a∗ µ
The following results are corollaries of Proposition 2.4.
Q
b
Corollary 2.5. A linear map
C[[X ,X ,...,X ]] −→ E(n)
1 2 m
n∈Zm anXn 7−→ |n|≤kanEn (mod nkU(n))
≥0 b k
is bijective, where XnP= Xn1Xn2···Xnm for n(cid:16)P= (n ,n ,...,n ) ∈ Zm . (cid:17)
1 2 m 1 2 m ≥0
n
Proof. By the Poincar´e-Birkhoff-Witt theorem {E | n β = µ} is a basis of U(n) . This
i i i µ
implies the corollary since E(n) = U(n) .
µ∈a∗ µ P
n n
We denote the image ofb n∈ZmQanX under the map in Corollary 2.5 by n∈Zm anE .
≥0 ≥0
Corollary 2.6. Let U bePa U(g)-module which is finitely generated as a U(Pn)-module. Assume
that all elements are a-finite. Then J(U) = U.
Proof. This follows from the following equation.
J(U) = J(U) = U = U = U.
a-finite µ µ
 
µ∈a∗ µ∈a∗
Y a-finite M
b  
4
Jacquet modules of principal series generated by the trivial K-type
Put E(g,n) = E(n)⊗ U(g). We can define a C-algebra structure of E(g,n) by
U(n)
(f ⊗1)(1⊗u) = f ⊗u,
b b b
(1⊗u)(1⊗u′) = 1⊗(uu′),
(f ⊗1)(f′⊗1) = (ff′)⊗1,
1 ∂|n|
(1⊗X)(f ⊗1) = f ⊗((ad(E))n)′(X),
n!∂En
n∈Zr
X≥0
where u,u′ ∈ U(g), X ∈ g, f,f′ ∈ E(n), ((ad(E))n)′ = (−ad(E ))nm···(−ad(E ))n1 and
m 1
∂|n| b m!
∂En  amEm= am(m−n)!Em−n.
m∈Zm m∈Zm
X≥0 X≥0
 
Notice that E(g,n)⊗ U ≃ E(n)⊗ U as an E(n)-module for a U(g)-module U. By Propo-
U(g) U(n)
sition 2.2, E(g,n) is flat over U(g). Notice that if b is a subalgebra of g which contains n then
E(n)⊗ Ub(b) is a subalgebrabof E(g,n). Put E(b,bn)= E(n)⊗ U(b).
U(n) U(n)
Let U bbe a U(a⊕n)-module such that U = U . Set
µ∈a∗ µ
b b b b
L
V = (u ) ∈ U there exists an element ν ∈ a∗ such that u = 0 for Reµ < ν .
 µ µ µ 0 µ 
(cid:12)
 µY∈a∗ (cid:12) 
(cid:12)
(cid:12)
Then we can define an a-mo(cid:12)dule homomorphism 
ϕ: E(a⊕n,n)⊗U(a⊕n)U ≃ U(n)µ ⊗U(n) Uµ′ → V
   
µ∈a∗ µ′∈a∗
Y M
b    
by ϕ((fµ)µ∈a∗ ⊗(uµ′)µ′∈a∗) = ( µ+µ′=λfµuµ′)λ∈a∗. Notice that the composition U → U → V is
equal to the inclusion map U ֒→ V.
P
We consider the case U = U(g). Define an (a⊕n)-module structure of U(g) by (H +bX)(u) =
Hu−uH +Xu for H ∈a, X ∈ n, u∈ U(g). We have a map
ϕ: E(g,n) →
b(Pµ)µ∈a∗ ∈ U(g)µ there exists an element ν ∈ a∗0 such that Pµ = 0 for Reµ < ν.
(cid:12)
 µY∈a∗ (cid:12) 
(cid:12)
(cid:12)
Wewriteϕ(P) = (P(µ))µ∈a∗. Pu(cid:12)tP(H,z) = µ(H)=zP(µ) forz ∈ CandH ∈ asuchthatReα(H) >
0 for all α ∈ Σ+.
P
By the definition we have the following proposition.
Proposition 2.7. (1) Assume that U is finitely generated as a U(n)-module. Let ϕ: E(a⊕
n,n)⊗ U → U be an a-module homomorphism defined as above. Then ϕ is coincide
U(n) µ∈a∗ µ
with the map given in Proposition 2.4. In particular ϕ is isomorphic. b
Q
5
Noriyuki Abe
(2) We have (PQ)(λ) = P(µ)Q(µ′) for P,Q ∈ E(g,n) and λ ∈a∗.
µ+µ′=λ
(3) We have P
b
(λ)
n n
anE = anE
 
nX∈Zm≥0 iXniβi=λ
 
for λ ∈ a∗. P
Proposition 2.8. Let U be a U(g)-module which is finitely generated as a U(n)-module. We
take generators v ,v ,...,v of an E(g,n)-module E(g,n) ⊗ U and set V = U(g)v ⊂
1 2 n U(g) i i
E(g,n)⊗ U. Define the surjective map ψ: E(g,n)⊗ V → E(g,n)⊗ U by ψ(f ⊗v) = fv.
U(g) U(g) U(g) P
Assume that there exist weights λ ∈ ab∗ and a positivebinteger N such that (H−λ (H))Nv = 0 for
i i i
abll H ∈ a and 1 ≤ i ≤ n. Let ϕ: E(g,n)⊗ bV → V bebthe map defined as above. Then
U(g) µ∈a∗ µ
there exists a unique map E(g,n)⊗ U → V such that the diagram
U(g) µ∈a∗ µQ
b
Q
b E(g,n)⊗U(g)V ϕ // 77 µ∈a∗Vµ
Q
ψ
b
(cid:15)(cid:15)
E(g,n)⊗ U
U(g)
is commutative. b
Proof. Set U = E(g,n)⊗ U and V = E(g,n)⊗ V. Take f(i) ∈ E(g,n) and v(i) ∈ V such
U(g) U(g)
that ψ( f(i)⊗v(i))= 0. We have to show ϕ( f(i)⊗v(i))= 0. Since V = E(n)⊗ V, we may
i i U(n)
assumePfi ∈ E(n)b. Webcan write f(i) = (fbµ(i))µb∈Pa∗ by the isomorphism E(nb) ≃ µ∈a∗U(n)µ. Since
b b
(i) (i) (i) (i)
V = µ′∈a∗Vµ′,wecanwritevi = µ′∈a∗vµ′ ,vµ′ ∈ Vµ′. Wehavetoshow i Qµ+µ′=λfµ vµ′ = 0
b b
for all λ ∈a∗. Since U is a finitely generated U(n)-module we have U = lim U/nkU = lim U/nkU.
L P ←−Pk P ←−k
It is sufficient to prove f(i)v(i) ∈ nkU for all k ∈Z .
i µ+µ′=λ µ µ′ >0
b b b
Fix λ ∈ a∗ and k ∈ Z . We can choose an element ν ∈ a∗ such that U(n) ⊂ nkU(n).
P>0P 0 Reµ≥ν µ
Then 0 = ϕ( f(i) ⊗v(i)) ≡ f(i)v(ib) (mod nkU). Notice that following two sets are
i i Reµ<ν µ µ′ L
finite.
P P P
b
{µ | Re(µ) < ν and there exists an integer i such that f(i) 6= 0},
µ
{µ′ | there exists an integer i such that v(i) 6=0}.
µ′
This implies f(i)v(i) ∈ nkU.
i µ+µ′=λ µ µ′
The followPingPresult is a corollary of Proposition 2.8.
b
Corollary 2.9. In the setting of Proposition 2.8, we have the following. Let P (1 ≤ i ≤ n) be
i
elements of E(g,n) such that n P v = 0. Then P(λ−λi)v = 0 for all λ ∈ a∗.
i=1 i i i i i
P P
b
6
Jacquet modules of principal series generated by the trivial K-type
§3. Construction of special elements
Let Λ be a subset of P. Put Λ+ = Λ∩ P+ and Λ++ = Λ ∩ P++. We define vector spaces
U(g) ,U(n) ,E(n) and E(g,n) by
Λ Λ Λ Λ
U(g) = {P ∈ U(g)| P(µ) = 0 for all µ 6∈ Λ},
b b Λ
U(n) = {P ∈ U(n) |P(µ) = 0 for all µ 6∈ Λ},
Λ
E(n) = {P ∈ E(n) |P(µ) = 0 for all µ 6∈ Λ},
Λ
E(g,n) = {P ∈ E(g,n) |P(µ) = 0 for all µ 6∈ Λ}.
Λ
b b
Put (nU(n))Λ = nU(n)∩bU(n)Λ and (nE(nb))Λ =nE(n)∩E(n)Λ.
We assume that Λ is a subgroup of a∗. Then U(g) ,U(n) ,E(n) and E(g,n) are C-algebras.
Λ Λ Λ Λ
Let U be a U(g) -module which is finbitely generbated asba U(n) -module. Let u ,u ,...,u be
Λ Λ 1 2 N
generators of U as a U(n) -module. Put u = t(u ,u ,...,u b), U = U/b(nU(n)) U and u = u
Λ 1 2 N Λ
(mod (nU(n)) ). The module U has an a-module structure induced from that of U. By the
Λ
assumption we have dimU < ∞. Let λ ,λ ,...,λ ∈ a∗ (Reλ ≥ Reλ ≥ ··· ≥ Reλ ) be
1 2 r 1 2 r
eigenvalues of a on U with multiplicities. We can choose a basis v ,v ,...,v of U and a linear
1 2 r
map Q: a → M(r,C) such that
Hv =Q(H)v for all H ∈ a,
Q(H)ii = λi(H) for all H ∈ a,

if i > j then Q(H)ij = 0 for all H ∈a,

if λ 6= λ then Q(H) = 0 for all H ∈ a,
i j ij



where v = t(v ,v ,...,v ). Take A ∈ M(N,r,C) and B ∈ M(r,N,C) such that v = Bu and
1 2 r
u = Av.
Set U = E(g,n) ⊗ U.
Λ U(g)Λ
Theboremb 3.1. There exist matrices A ∈M(N,r,E(n)Λ) and B ∈ M(r,N,E(n)Λ) such that the
following conditions hold:
b b
• There exists a linear map Q: a → M(r,U(n) ) such that
Λ
Hv = Q(H)v for all H ∈ a,
Q(H)−Q(H) ∈ M(r,(nU(n))Λ) for all H ∈ a,
if λi−λj 6∈ Λ+ then Q(H)ij = 0 for all H ∈ a,

if λ −λ ∈ Λ+ then [H′,Q(H) ] = (λ −λ )(H′)Q(H) for all H,H′ ∈ a,
i j ij i j ij



where v= Bu.
• We have u =ABu.
• We have A−A∈ M(N,r,(nE(n)) ) and B −B ∈ M(r,N,(nE(n)) ).
Λ Λ
For the proof we need some lemmas. Put w = Bu∈ Ur.
b b
7 b
Noriyuki Abe
Lemma 3.2. For H ∈ athere existsa matrix R ∈ M(r,(nE(n)) ) suchthat Hw = (Q(H)+R)w
Λ
in Ur.
b
Proof. Since w (mod ((nU(n)) U)r) = v, we have Hw − Q(H)w ∈ ((nU(n)) U)r. Hence
b Λ Λ
there exists a matrix R ∈ M(N,r,(nU(n)) ) such that Hw −Q(H)w = R u. Similarly we can
1 Λ 1
choose a matrix S ∈ M(N,(nU(n)) ) which satisfies u= Aw+Su. Put R = R (1−S)−1A. Then
Λ 1
(H −Q(H)−R)w = R u−R (1−S)−1Aw = 0.
1 1
Lemma 3.3. Let λ ∈ C and Q ,R ∈ M(r,C). Assume that Q is an upper triangular matrix.
0 0
Then there exist matrices L,T ∈ M(r,C) such that
λL−[Q ,L] = T +R,
0
(if (Q0)ii−(Q0)jj 6= λ then Tij = 0.
Proof. Since (Q ) = 0 for i > j, we have
0 ij
r
(λL−[Q ,L]) = λL − ((Q ) L −L (Q ) )
0 ij ij 0 ik kj ik 0 kj
k=1
X
r j
= λL − (Q ) L + L (Q )
ij 0 ik kj ik 0 kj
k=i k=1
X X
r j−1
= (λ−((Q ) −(Q ) ))L − (Q ) L + L (Q ) .
0 ii 0 jj ij 0 ik kj ik 0 kj
k=i+1 k=1
X X
Hence we can choose L and T inductively on (j −i).
ij ij
Lemma 3.4. Let H be an element of a such that α(H) > 0 for all α∈ Σ+. Let Q ∈ M(r,C),
0
R ∈ M(r,(nE(n)) ). Assume (Q ) = 0 for i > j. Set L++ = {λ(H) | λ ∈ Λ++}. Then there exist
Λ 0 ij
matrices L ∈ M(r,E(n) ) and T ∈ M(r,(nU(n)) ) such that
Λ Λ
b
Lb≡ 1 (mod (nU(n)) ),
r Λ
(H1r −Q0−T)L = L(H1r −Q0−R),

if (Q0)ii−(Q0)jj 6∈ L++ then Tij = 0,

if (Q ) −(Q ) ∈ L++ then [H,T ]= ((Q ) −(Q ) )T .
0 ii 0 jj ij 0 ii 0 jj ij



Proof. Set L = {λ(H) | λ ∈ Λ} and L+ = {λ(H) | λ ∈ Λ+}. Put f(n) = n β for
i i i
n = (n ) ∈ Zm. Set Λ = {n ∈ Zm | f(n) ∈ Λ}. We define the order on Λ by n < n′ if and only if
i ≥0 P
f(n) < f(n′).
By Corollary 2.5e, we can write R = n∈ΛRnEn where Rn ∈ M(er,C). We have R0 = 0
where 0 = (0) ∈ Λ since R ∈ M(r,(nE(n)) ). We have to show the existence of L and T. Write
i PΛ
n n e
L = 1r + n∈ΛLnE and T = n∈ΛTnE . Then (H1r − Q0 − T)L = L(H1r − Q0 − R) is
equivalent to e b
P P
e f(n)(H)Len −[Q0,Ln]= Tn+Sn−Rn,
8
Jacquet modules of principal series generated by the trivial K-type
where Sn is defined by
n
SnE = T(L−1r)−(L−1r)R.
nX∈Λ
By Proposition 2.7 the above equation is equivalent to
e
n k l k l
SnE = TkLlE E − LkRlE E
f(n)=µ f(k)+f(l)=µ f(k)+f(l)=µ
X X X
forallµ ∈ a∗. NoticethatL0 = T0 = 0. Sn canbedefinedfromthedata{Tk | k< n},{Lk | k< n}
and {Rk |k < n}.
Now we prove the existence of L and T. We choose the Ln and Tn which satisfy
L0 = 0, T0 = 0,
f(n)(H)Ln−[Q0,Ln]= Tn+Sn−Rn,

if (Q0)ii−(Q0)jj 6=f(n)(H) then (Tn)ij = 0.
By Lemma 3.3, we canchoose such Ln and Tn inductively. Put L = 1r + n∈ΛLnEn and
n
T = n∈ΛTnE . Since P
e
P e [H,Tij]= (f(n)(H))(Tn)ijEn = ((Q0)ii−(Q0)jj)Tij,
nX∈Λ
L and T satisfy the conditions of the lemma.
e
Proof of Theorem 3.1. We can choose positive integers C = (C ) ∈Zl such that
i >0
{λ −λ } (λ −λ ∈ Λ++),
{α ∈ Λ++ |α( C H ) = (λ −λ )( C H )} = i j i j
i i i i j i i i (∅ (λi−λj 6∈ Λ++).
P P
The existence of such C is shown by Oshima [Osh84, Lemma 2.3]. Set X = C H . Notice
i i i
that (λ −λ )(X) ∈ {µ(X) | µ ∈ Λ++} if and only if λ −λ ∈ Λ++. By Lemma 3.4, there exist
i j i j
P
T ∈ M(r,(nU(n)) ) and L ∈ M(r,E(n) ) such that
Λ Λ
L ≡ 1 (mod (nE(n)) ),
r b Λ
(X1r −Q(X)−T)L = L(X1r −Q(X)−R),
 b
if λi−λj 6∈ Λ++ then Tij = 0,

if λ −λ ∈ Λ++ then [X,T ] = (λ −λ )(X)T .
i j ij i j ij


Let S ∈ M(N,(nU(n))Λ) such that u = Aw + Su. Put A = (1 − S)−1AL−1, B = LB and
v = (v ,v ,...,v ) = Bu = Lw then ABu = (1−S)−1AL−1LBu = (1−S)−1Aw = u. Moreover,
1 2 r
we have (X1 −Q(X)−T)v = 0. Since [X,T ] =(λ −λ )(X)T , we have (X −λ (X))rv = 0.
r ij i j ij i i
Fix a positive integer k such that 1 ≤ k ≤ l. We can choose a matrix R ∈ M(r,(nE(n)) ) such
k Λ
that H w = (Q(H )+R )w by Lemma 3.2. Set T = H 1 −Q(H )−L(H 1 −Q(H )−R )L−1.
k k k k k r k k r k k
Then we have (H 1 −Q(H )−T )v = 0, i.e., b
k r k k
r r
H v − Q(H ) v − (T ) v = 0
k i k ij j k ij j
j=1 j=1
X X
9
Noriyuki Abe
for each i =1,2,...,r. By Corollary 2.9, we have
r r
H v − Q(H ) v − (T )(X,(λi−λj)(X))v = 0.
k i k ij j k ij j
j=1 j=1
X X
Define T′ ∈ M(r,(nU(n)) ) by (T′) =(T )(X,(λi−λj)(X)). Since (T )(µ) = 0 for µ 6∈ Λ++, we have
k Λ k ij k ij k ij
(T )(λi−λj) (λ −λ ∈ Λ++),
(T )(X,(λi−λj)(X)) = (T )(µ) = k ij i j
k ij µ∈Λ++, µ(XX)=(λi−λj)(X) k ij (0 (λi−λj 6∈ Λ++)
by the condition of C. In particular [H,(T′) ] = (λ −λ )(H) for all H ∈ a. Put Q( x H ) =
k ij i j i i
Q( x H )+ x T′ for (x ,x ,...,x ) ∈ Cl then Q satisfies the condition of the theorem.
i i i i 1 2 l P
PSet ρ = P (dimg /2)α. From the Iwasawa decomposition g = k ⊕ a ⊕ n we have the
α∈Σ+ α
decomposition into the direct sum
P
U(g) = nU(a⊕n)⊕U(a)⊕U(g)k.
Let χ be the projection of U(g) to U(a) with respect to this decomposition and χ the algebra
1 2
automorphism of U(a) defined by χ (H) = H −ρ(H) for H ∈ a. We define χ: U(g)k → U(a) by
2
χ = χ ◦χ where U(g)k = {u ∈ U(g) | Xu = uX for all X ∈ k}. It is known that an image of
2 1
U(g)k under χ is contained in the set of W-invariant elements in U(a).
Fix λ ∈ a∗. We can define the algebra homomorphism U(a) → C by H 7→ λ(H) for H ∈a. We
denote this map by the same letter λ. Put χ = λ◦χ. Set U(λ) = U(g)/(U(g)Kerχ +U(g)k),
λ λ
u = 1 mod (U(g)Kerχ +U(g)k) ∈ U(λ) and U(λ) = U(g) u . Before applying Theorem 3.1
λ λ 0 2P λ
to U(λ) , we give some lemmas.
0
Lemma 3.5. Let u ∈ U(g) where µ ∈ a∗. Then there exists an element x ∈ U(g)k such that
µ
u+x ∈U(a⊕n) .
µ+2P
Proof. Set n = θ(n). We may assume u∈ U(n)µ. Let {Un(n)}n∈Z≥0 be the standard filtration
of U(n) and n the smallest integer such that u ∈ U (n). We will prove the existence of x by the
n
induction on n.
If n = 0thenthelemma is obvious. Assumen > 0. We may assumethatthereexist arestricted
root α ∈ Σ+, an element u ∈ U (n) and a vector E ∈ g such that u = u E . Set
0 n−1 µ+α −α −α 0 −α
E = θ(E ), u = u E , u = E u and u = u −u . Then u+u +u = u+u ∈ U(g)k,
α −α 1 0 α 2 α 0 3 1 2 2 3 1
u ,u ∈ U(g) and u ∈ U (g) . Using the Poincar´e-Birkhoff-Witt theorem and the
1 2 µ+2α 3 n−1 µ+2α
induction hypothesis we can choose an element u ∈ U(a ⊕ n) such that u − u ∈ U(g)k.
5 µ+2P 3 5
Again by the induction hypothesis we can choose an element u ∈ U(a ⊕ n) such that
6 µ+α+2P
u −u ∈ U(g)k. Thenu+u +E u ∈ U(g)k, u ∈ U(a⊕n) and E u ∈ U(a⊕n) .
0 6 5 α 6 5 µ+2P α 6 µ+2α+2P
Lemma 3.6. The following equations hold.
(1) Kerχ ⊂ U(g) .
λ 2P
(2) U(a⊕n)∩(Kerχ +U(g)k) ⊂U(a⊕n) ∩(Kerχ +U(g)k).
λ 2P λ
10