Decomposition Theory of a Certain Class of O N ( en ) Yuan-Ching Huang (黃苑青) Department of International Business, Yu Da College of Business No. 168, Hsueh-Fu Rd, Chaochiao Township, Miaoli County, 361, Taiwan, R.O.C.. Tel: 037-651188-6025 Fax: 037-651201 Email: [email protected] Abstract If O N (e n ) has a certain period, then O N (e n ) has a nice decomposition theory. Furthermore, there exists an interesting correspondence between the irreducible factors over Z[x] of x k 1 , where k is the number of periodic points of O N (e n ) and the orthogonal invariant subspaces of O N (e n ) . Key words: Cuntz algebras, permutative representations, irreducible subspaces, decomposition, periods 1. Introduction Recall that the Cuntz algebra of order N , denoted by O N , is the unital C * -algebra generated by N isometiries {si }iN1 with the following relation: si* s j ij 1 ; i1 si si* 1 N to B(H ) (bounded operators on a Hilbert Space H ). Let H be a separable Hilbert space with an orthonormal basis {en }n . A representation of O N on H with respect to is a permutative if there exist i : , i 0,..., N 1 such that i () j () , for all i j such that Eq. (1) N 1 S *j (em ) 0 . j1 , , j k 1 are When defined, let j k be the unique j Z N such that S *jk S *j1 (em ) 0 . A function system is called multiplicity-free if ([Cun]) A representation on O N is a *-homomorphism from O N m , (m) ( j1 , j2 ,, jk ,) Z N , where j1 is the unique element j in Z N {0, , N 1} the coding map is injective. map σ We say that the coding is partially injective if n and i1 ,, ik Z N and (n) ( i1 ik (n)) , imply n i1 ik (n) . In this case, the function system is then called to be regular. For a H , denote ON (a) the closure of the linear {SI S (a) Si1 Si2 * J J ( j1 , span * j Sik S S , j ), i1 , , ik , j1 , of S (a) | I (i1 , * j 1 * j1 , ik ), , j Z N and k , } . i () Eq. (2) Define i is injective Eq. (3) R(m) m1 , where m1 is the unique element in i 1 R : as follow: Let and such that j (m1 ) m for some j Z N . S j (en ) e j ( n) , for all j 0, , N 1, where ON (n) { I R (n) i1 i2 S j (s j ) k Eq. (4) The collection of such maps 0 ,, N 1 satisfying Eq. (1) to Eq. (3) is called a (branch) function system of order N . The coding map : Z N Z N is denoted as follows: i 1 Let , im ), m, k and i1 , i RR m m , Denote R(n) | I (i1 , , im Z N } . Jorgensen and Bratteli [BJ] have shown that if the function system { j } jZ N is regular then O N (e n ) , for all n is an irreducible invariant subspace of the corresponding permutative representation on O N . That is to say, if the function system is regular, then the decomposition theory of is completely known. they showed that when Furthermore, is regarded as a i Qm . For this m , find p(m) Qm . Then there exist I p (m ) and J p (m) such that S I p ( m) S J* p ( m) (e1 ) e p(m) representation of UHF N , then without assuming SI p ( m ) S J*p ( m ) (c1 ) m cm a linear combination of regularity, UHF N (en ) the closure of the linear {c j } j , where m C span {S I S J* (en ) | I (i1 , | Q1 || Qm | i1 , , ik , j1 , , ik ), J ( j1 , , jk ), k , , jk Z N } But, we also have ( UHF N is the uniform hyperfinite algebra of order N ) is invariant and irreducible for all n . In [Hua], it is shown that if K ON (en ) is a S J p ( m) S I*p ( m) (e p(m) ) e1 S J p ( m ) SI*p ( m ) (cm ) m c1 a linear combination of where m C proper invariant subspace of O N with a reduced {c j } j , basis (Definition 6, [Hua]) then ON (n) possesses a | Qm || Q1 | period. Eqs. (5) cm 1 develop (Theorem 9, [Hua]). the In this paper, we will decomposition between invariant subspaces of an O N ( en ) with a certain period and an interesting Eq. (5) connection between m Eq. (6) and (6) imply | Q1 || Qm | ; hence S I p ( m ) S J* p ( m ) (c1 ) . So the proof is complete. Q.E.D. invariant subspaces of O N and the irreducible factors p (x) over Z [x] of x k 1 0 , where k is the number of periodic points. (Definition 6, [Hua]). Our next goal is to find the decomposition theory for such a reducible O N (e n ) . Before going to the theoretical proof, let us look at an example. 2. Main Results Proposition 1 Suppose O N ( en ) is reducible with Suppose there exist a1 , a 2 , a3 , a 4 ON (n) such that a proper subspace K and there is a j0 such that some j0 Z N . S j0 (ai ) ai 1 , i 1, 2,3 and S j0 (a4 ) a1 We claim that and (a k ) a1 , K1 ON (ea1 ea2 ea3 ea4 ) then K ON (1en1 en ) for some , K 2 ON (ea1 ea2 ea3 ea4 ) i C, i {1, , } , and {n1 , , n } {a1 , , ak } . K3 ON (ea1 ea3 ), j0 (ai ) ai 1 , 1 i k 1 for and K 4 ON (ea1 ea3 ) Proof: Let {ci }i be a reduced basis for K , where ci (c1i , c 2i , , c mi , ) . be an i0 such that i c a0 1 Since there must 0 , without loss of generality, we assume i0 1 . Let c1 where Q1 , and c 0, j Q1 . 1 j c1j e j , jQ1 We claim are proper invariant subspaces of O N (e n ) . In order to see K 2 is proper it suffices to show that H 2 span{S mj0 (ea1 ea2 ea3 ea4 ) | m 0,1,2,3} is a proper subspace of K span{ea1 , ea2 , ea3 , ea4 } C 4 . ( j ) (i), i, j Q1 . If not, then there exists a But it is easy to see that H 2 is the 1-dimensional finite string L such that S L* (ei ) 0 S L* (e j ) . space Observe that S L S L* (C1 ) C1 , which leads to a contradiction. Since Q1 {a1 , , a k } . cm cim ei iQm Let , where a1 Q1 , we have m 2 be given, put Qm and cim 0 , of K generated by ea1 ea2 ea3 ea4 . Similarly, dim H 3 2 dim H 4 2 , and the vector dim H 1 1 , , where H i span{S (bi ) | m 0,1, 2,3}, i 1,3, 4 , b1 ea1 m j0 ea2 ea3 ea4 , b3 ea1 ea3 , and b4 ea1 ea3 . Suppose dim K 0 1 , then we Furthermore, it is not hard to see that K 4 is the K 0 is irreducible. orthogonal complement of K 3 , K1 K 2 K 4 and have K 1 is actually the orthogonal complement of K 2 in S j0 ( xi eai ) ( xi eai ) the subspace Hence K K4 K3 K4 . k k i 1 i 1 K1 K2 K3 with K 1 , K 2 , and K 3 pairwisely x j x j 1 , j 2, , k and x1 xk orthogonal. x2 x1 , x3 2 x1 , , xk k 1 x1 , x1 k x1 Moreover, there is an interesting correspondence between K 1 , K 2 , K 3 and the k 1 (Since x1 0) so we conclude that if is a k th root of unity, irreducible factors of x 4 1 in Z[x] . x 1 ( x 1)( x 1)( x 1) 4 2 K ON (i 1 i 1eai ) k then We make the following correspondence is an irreducible K1 x 1 p1 ( x) subspace. K 2 x 1 p 2 ( x) K K ' {0} if and ' are disjoint k th K 3 x 1 p3 ( x) Now it 2 Later on, we will show that in general there is such correspondence between the irreducible factors easy to see 0 . 1 0 ' ( ' ) root of unity since with dim( K i ) deg( pi ( x)) . is 1 dim( K ) k dim H 0 k that Because , we conclude that k 1 K ON (en ) . Q.E.D. of x k 1 in Z[x] and some special invariant subspaces of O N (e n ) . The next result carries out our earlier promise. It is obvious that K 1 and K 2 are irreducible, but is K 3 irreducible? The answer is no, and we introduce the following theorem that decomposes O N (e n ) into irreducibles. Corollary 3 {a1 , a 2 , , ak } i 1, p1 ( x) Theorem 2 Suppose O N (e n ) is reducible with period {a1 , a 2 , , a k } , and j0 (ai ) ai 1 , i 1, , k 1 , j0 (ak ) a1 . Then, O N (e n ) Let O N (e n ) be reducible with period , k 1 , pr ( x) and j0 (ai ) ai 1 j0 (ak ) a1 . If xk 1 where pi (x) is irreducible over Z[ x] , then each pi (x) corresponds to a proper invariant subspace K i ON ( x ij ea j ) jQi Qi {1,2,, k} , x ij 1 , j Qi . decomposes into k disjoint irreducible subspaces as where follow: Furthermore, Ki K j , i j, and ON (en ) Ki . r O N (e n ) k 1 O N (e a1 e a2 k 1e ai1 ) . Proof: , Let H 0 span{ea1 , , eak } C k . i 1 In k order to show that K 0 O N ( xi eai ) , xi C is a i 1 proper subspace of H 0 , it suffices to show that K 0 {S mj0 ( k xi ea ) | m 0,1,, k 1} i 1 i is a proper Proof: For i {1, 2, pi (x) be {i ,1 , i ,2 , ni k j 1 1 , r} , let the complex roots of , i ,ni } . Define K i O N ( ( i , j ) 1 ea ) O N ( x ij ea j ) jQi where Qi {1,2,, k} , x ij 0 , j Qi . dim( K i ) ni deg( pi ) and x 1 i j coefficients of pi ( x) are either 1 or -1. Then since the To show The idea of the proof is to find all that K i K j , i j , it suffices to show that if the K 0 such that dim K 0 1 which implies that x k 1 p ( x) q ( x ) , where p( x), q( x) Z[ x] , then subspace of H 0 . the corresponding subspaces orthogonal. Kp k O ( a p ( a ) 0 Kp and are Kq Now N 1 1 ea ) , K q x 1 ON (a1 a2 a3 a4 a5 a6 ) k O ( q( )0 N j 1 j 1 ea j ) . ( a 1 1 where is a complex root of x3 1 . k ea , j 1 ea j ) x 2 x 1 ON (a1 1a2 12 a3 j 1 1 ( ) ( ) 2 ( ) k 1 is a k -th root of unity, Eq. (7) is 0 unless . By the complex conjugate root theorem, we know that is also a root of q( x) 0 ; hence cannot be a root or p( x) 0 ; so . k i 1 deg( pi ), r decompose. Furthermore, i 1, the Ki 's, Hence the proof is complete. The following example , r, Q.E.D. shows the ON (a1 a2 a4 a5 ) where 1 and i 3 i is a complex number, i 1, 2 . The case k 7 is similar to the case k 3 . In fact, if k a ( x 1)( x k 1 x k 2 is prime, then x7 1 1) is the factorization over Z[ x] ; hence they are all similar. k 8 , x8 1 ( x 1)( x 1)( x 2 1)( x 4 1) x 1 ON (a1 a2 a8 ) correspondence between irreducible factors of x 1 x 1 ON (a1 a2 a3 a4 in Z[ x] and orthogonal invariant subspaces of some x 2 1 ON (a1 a3 a5 a7 ) special cases. x 4 1 ON (a1 a5 ) . k 15 a6 ) ON (a1 2 a2 22 a3 25 a6 ) Eq.(7) It suffices to show that Eq. (7) is 0, for all , such that p( ) 0 and q( ) 0 . Since since x 2 x 1 ON (a1 a2 2 a3 a4 a5 2 a6 ) ON (a1 a2 a4 a5 ) Consider k x 1 ON (a1 a2 a3 a4 a5 a6 ) a7 a8 ) k 9 , x9 1 ( x 1)( x 2 x 1)( x 6 x3 1) Example 4 x 1 ON (a1 a2 k 2 , x 2 1 ( x 1)( x 1) x 2 x 1 ON (a1 a2 a3 a4 a7 a8 ) x 1 ON (a1 a2 ) x 6 x 3 1 ON (a1 a4 ) . x 1 ON (a1 a2 ) . k 10 , x10 1 ( x 1)( x 4 x3 x 2 x 1) k 3 , x 1 ( x 1)( x x 1) 3 a9 ) 2 ( x 1)( x 4 x3 x 2 x 1) x 1 ON (a1 a2 a3 ) x 2 x 1 ON (a1 a2 ) ON (a2 a3 ) ON (a3 a1 ) and ON (a1 a2 ) ON (a1 a2 a3 ) x 1 ON (a1 a2 a10 ) x 1 ON (a1 a2 a3 a4 a9 a10 ) x 4 x3 x 2 x 1 ON (a1 a2 a6 a7 ) 2 ON (a1 a2 a3 ) , where is a complex root of x 4 x3 x 2 x 1 ON (a1 a2 a6 a7 ) . 2 x3 1 0 . A Remark on UHFN k 4 , x 1 ( x 1)( x 1)( x 1) 4 2 If is a permutative representative of ON x 1 ON (a1 a2 a3 a4 ) on H , then there is a natural way to view as a x 1 ON (a1 a2 a3 a4 ) representation of UHFN . relation on by x 2 1 ON (a1 a3 ) . k 5 , x 1 ( x 1)( x x x x 1) 5 4 3 2 n m S I S J* (en ) em , for some finite string I , J x 1 ON (a1 a2 a3 a4 ) with | I || J | . x x x x 1 ON (a1 a2 ) . given n , 4 3 Define an equivalence 2 k 6 , x 6 1 ( x 1)( x 1)( x 2 x 1)( x 2 x 1) Theorem 2.7 in [BJ2] shows that ON (en ) UHFN (em ) References m is the decomposition of into ON (en ) [BJ] O. Bratteli and P. E. T. Jorgensen, “Iterated UHFN - irreducible, where m runs through all the Function equivalent class in ON (en ) . Representations of the Cuntz Algebra”, Mem. The sum is finite if and only if there is an a ON (en ) such that I (a) a for some finite I . Suppose ON (en ) is ON - reducible with a period {a1 , , ak } such that S j0 (ai ) ai 1 , S j0 (ak ) a1 , for some j0 Z N . By Theorem 2, ON (en ) (a1 a2 k 1ak ) . k 1 and Permutative Amer. Math. Soc. 139 (1999) no. 663. [Cun] J. Cuntz, “ C * -Algebra Gemerated by Isometries”, Comm. Math. Phys., 57 (1977), 173-185. [DaPi] K. Davidson and D. R. Pitts, “Invariant Subspaces and Hyper-reflexivity for Free Semigroup Algebras”, Proc. London Math. Soc. We claim that UHFN (a1 a2 78 (1999), 401-430. k 1ak ) ON (a1 a2 UHFN (a1 a2 [Hua] k 1ak ) for all k - th root of unity . I (i1 , ”On Permutative O N ”, Yu-Da Academic [KaRi] R. Kadison and J.R. Ringrose, “Fundamentals k 1ak ) . of the Theory of Operator Algebras”, Amer. Math. To show the others direction, let be given, where Huang, Journal, Vol. 9, (2005) 283-290. k 1ak ) b S I S J* (a1 a2 Yuan-Ching Representation of It is trivial that ON (a1 a2 Soc., (1997). k 1ak ) J ( j1 , , ip ) , , jq ) . [Tak] M. Takesaki, “Theory of Operator Algebras I”, Springer-Verlag, New York-Heidelberg-Berlin, Firstly, note that (S j0 )m (a1 a2 k 1ak ) m (a1 a2 (1979). k 1ak ) . Suppose q p , then let I 0 ( j0 , j0 , b S I S J* (a1 a2 , j0 ) k 1ak ) S I S J* ( q p S I0 (a1 a2 q p S I S J*0 (a1 a2 k 1ak )) k 1ak ) where J 0 is some finite string with J 0 p I . Similarly, if q p , then we can find I 0 such that I 0 J with b SI0 S J* (a1 a2 for some Systems k 1ak ) Z . This shows that UHFN (a1 a2 k 1ak ) ON (a1 a2 k 1ak ) . Hence ON (en ) UHFN (a1 a2 k 1 k 1ak ) is the decomposition into UHFN - irreducibles.
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