On Theta Ranks of Irreducible Characters of a Finite Classical Group
aa r X i v : . [ m a t h . R T ] F e b ON THETA RANKS OF IRREDUCIBLE CHARACTERS OF AFINITE CLASSICAL GROUP
SHU-YEN PAN
Abstract.
In this paper, we interpret the Theta rank of an irreducible char-acter of a finite classical group in terms of the data under the Lusztig corre-spondence. Then we show that the parabolic asymptotic rank and the Thetarank agree for any irreducible character of a classical group.
Contents
1. Introduction 12. Theta Correspondence and Lusztig Correspondence 43. Θ-ranks of Irreducible Characters 114. Θ-ranks and Other Ranks 23References 291.
Introduction G = G n be a finite symplectic group Sp n ( q ), a finite orthogonal groupO +2 n ( q ), O − n +2 ( q ), O n +1 ( q ), or a finite unitary group U n ( q ), U n +1 ( q ) defined overa finite field f q where q is a power of an odd prime. For an irreducible character ρ of G n , several notions of “rank” attached to ρ are defined in [GH20], namely the U -rank (denoted by U -rk( ρ )), the asymptotic rank (denoted by A -rk( ρ )), and the tensor rank (denoted by ⊗ -rk( ρ )). It is known that U -rk( ρ ) ≤ n . An irreduciblecharacter ρ of G n is said to be of low U -rank if U -rk( ρ ) < n (resp. U -rk( ρ ) < n − G n is a symplectic group or a unitary group (resp. an orthogonal group).Because the tensor rank can be interpreted as the first occurrence of ρ in theTheta correspondence for some dual pair involving G n , in this article the tensorrank will be called the Θ -rank and denoted by Θ-rk( ρ ). A similar notion of theΘ-rank which is called level and denoted by l ( ρ ) is also defined in [GLT20] and Date : February 19, 2021.2010
Mathematics Subject Classification.
Primary: 20C33; Secondary: 22E50.
Key words and phrases. theta rank, theta correspondence, eta correspondence, reductive dualpair. [GLT19]. Due to different normalizations, we can see that l ( ρ ) = Θ-rk( ρ ) , if G is a symplectic group or a unitary group; Θ-rk( ρ ) , if G is an orthogonal group.Note that Θ-rk( ρ ) is always even if G is an orthogonal group. Moreover, we willmodify the definition of the asymptotic ranks and define the parabolic asymptoticrank (denoted by A -rk( ρ )).Of course, it is expected these ranks are closely related. More precisely, Gurevich-Howe make the following conjecture in [GH20]: Conjecture 1.1 (Gurevich-Howe) . Let ρ be an irreducible character of a finiteclassical group G . Then (1) A -rk( ρ ) = ⊗ -rk( ρ ) . (2) If ρ is of low U -rank, then U -rk( ρ ) = ⊗ -rk( ρ ) . As indicated in [GH20], if the conjecture is true, the Theta correspondence pro-vides a method to construct an irreducible character of a given U -rank or A -rank.When G is a general linear group, (1) is already proved in [GH20] theorem 11.4.One of the purposes of the article is prove the similar result for other classicalgroups.1.2. Let E ( G ) denote the set of irreducible characters of G . It is known that theset E ( G ) partitions into subsets E ( G ) s , called Lusztig series , indexed by the set ofconjugacy classes of semisimple elements s in the connected component of the dualgroup G ∗ of G . Elements in E ( G ) are called unipotent . It is known by Lusztigthat unipotent characters ρ = ρ Λ ∈ E ( G ) are parametrized by the similar classesof symbols Λ = (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) ∈ S G satisfying certain conditions where a i , b j are non-negative integers such that a >a > · · · > a m and b > b > · · · > b m . Our first result ( cf . Proposition 3.4) is tointerpret the Θ-rank of a unipotent character ρ Λ in terms of symbols:(1.2) Θ-rk( ρ Λ ) = min { r (cid:0) a − (cid:1) ) , r (cid:0) − b (cid:1) ) } , if G is Sp or O;min { rk U (Λ r (cid:0) a − (cid:1) ) , rk U (Λ r (cid:0) − b (cid:1) ) } if G is Uwhere rk(Λ) = m X i =1 a i + m X j =1 b j − $(cid:18) m + m − (cid:19) % , rk U (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) = m X i =1 a i + m X j =1 b j + | m − m | − ( m + m )( m + m − . Note that by replacing Λ by another symbol in the similar class if necessary, wecan always assume that both rows { a i } , { b j } of Λ are non-empty. HETA RANKS 3 G is a symplectic group or an orthogonal group. For s ∈ ( G ∗ ) , wecan define G (0) , G ( − ) , G (+) where G (0) is a product of unitary groups or generallinear groups, and G ( − ) , G (+) are given as follows: • if G = O ǫ n , then G ( − ) = O ǫ ′ n ( − ) , and G (+) = O ǫ ′ ǫ n (+) ; • if G = Sp n , then G ( − ) = O ǫ ′ n ( − ) , and G (+) = Sp n (+) ; • if G = O n +1 , then G ( − ) = Sp n ( − ) , and G (+) = Sp n (+) ,for some ǫ ′ = ± , and n ( − ) , n (+) depending on s . Then we have a (modified) Lusztigcorrespondence Ξ s : E ( G ) s → E ( G (0) ( s ) × G ( − ) ( s ) × G (+) ( s )) . The mapping Ξ s is two-to-one, i.e., Ξ s ( ρ ) = Ξ s ( ρ · sgn) when G is an odd-orthogonalgroup; and Ξ s is a bijection otherwise. Then we can writeΞ s ( ρ ) = ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) for ρ ∈ E ( G ) s where Λ ( − ) , Λ (+) are certain symbols.Next suppose G is a unitary group. As described in Subsection 3.4, for theLusztig correspondence L s : E ( G ) → E ( C G ∗ ( s )) we can write(1.3) L s ( ρ ) = r O i =1 t i O j =1 ρ ( ij ) . Note that t = q +1 and ρ (1 j ) is a unipotent character of U n j ( q ) for j = 1 , . . . , q +1.Now we have the following formula ( cf . Proposition 3.12 and Proposition 3.17)for the Θ-rank of a general irreducible character in terms of data from Lusztigcorrespondences: Theorem 1.4.
Let ρ ∈ E ( G ) . (1) If G = Sp n , then Θ-rk( ρ ) = min { n − n (+) + Θ-rk( ρ Λ (+) ) , n − n ( − ) + Θ-rk( ρ Λ ( − ) ) + 1 } . (2) If G = O ǫ n or O n +1 , then Θ-rk( ρ ) = min { n − n (+) + Θ-rk( ρ Λ (+) ) , n − n ( − ) + Θ-rk( ρ Λ ( − ) ) } . (3) If G = U n , then Θ-rk( ρ ) = min { n − n j + Θ-rk( ρ (1 j ) ) | j = 1 , . . . , q + 1 } . η -correspondence for a reductive dual pair instable range by Gurevich-Howe in [GH20]: Theorem 1.5.
Let ρ ∈ E ( G n ) where G n is Sp n , O +2 n , O − n +2 , O n +1 , U n , or U n +1 . If Θ-rk( ρ ) ≤ n , then U -rk( ρ ) = A -rk( ρ ) = A -rk( ρ ) = Θ-rk( ρ ) . SHU-YEN PAN
Finally, by using the same arguments in [GH20] section 11, we can show theagreement of the parabolic asymptotic ranks and the Theta ranks:
Theorem 1.6.
Let ρ ∈ E ( G ) where G is a symplectic group, an orthogonal groupor a unitary group. Then A -rk( ρ ) = Θ-rk( ρ ) . Theta Correspondence and Lusztig Correspondence
Basic notations.
Let f q be a finite field of q where q is a power of an oddprime p , and let f q denote a fixed algebraic closure of f q .Let G be a symplectic group, an orthogonal group or a unitary group definedover f q , and let G denote the finite classical group of rational points. The characterof an irreducible representation of G is called an irreducible character of G , andthe set of all irreducible characters of G is denoted by E ( G ). The character of aone-dimensional representation is also called a linear character of G . It is provedby Lusztig that the set E ( G ) admits a partition E ( G ) = G ( s ) ⊂ ( G ∗ ) E ( G ) s indexed by conjugacy classes ( s ) of semisimple simple elements s in the connectedcomponent ( G ∗ ) of the dual group G ∗ of G . Each E ( G ) s is called a Lusztig series ,and the element in E ( G ) are called unipotent .The trivial character of G is denoted by .2.2. Unipotent characters of a symplectic group or an orthogonal group.
The set of partitions of n is denoted by P ( n ). If µ ∈ P ( n ), we also write | µ | = n .An ordered pair of two partitions µ, ν , denoted by (cid:2) µν (cid:3) , is called a bi-partition of n if (cid:12)(cid:12)(cid:2) µν (cid:3)(cid:12)(cid:12) := | µ | + | ν | = n . The set of all bi-partitions of n is denoted by P ( n ).Now we recall the notations of symbols from [Lus77]. Note that for our con-venience, the definition here is slightly different from the original one. A β -set A = { a , a , . . . , a m } is a finite subset (possibly empty) of non-negative integerswritten in (strictly) decreasing order, i.e., a > a > · · · > a m . A symbol Λ = (cid:18) AB (cid:19) = (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) HETA RANKS 5 is an ordered pair of two β -sets A, B . On the set of symbols we define an equivalencerelation “ ∼ ” generated by (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) ∼ (cid:18) a + 1 , a + 1 . . . , a m + 1 , b + 1 , b + 1 , . . . , b m + 1 , (cid:19) . The set of similar classes of symbols is denoted by S .For a symbol Λ = (cid:0) a ,...,a m b ,...,b m (cid:1) , we define its defect and rank bydef(Λ) = m − m , rk(Λ) = m X i =1 a i + m X j =1 b j − $(cid:18) m + m − (cid:19) % . It is not difficult to see that two equivalent symbols have the same defect and thesame rank. For a symbol Λ = (cid:0) AB (cid:1) we define its transpose Λ t = (cid:0) BA (cid:1) . It is obviousthat rk(Λ t ) = rk(Λ) and def(Λ t ) = − def(Λ).We define a mapping Υ from symbols to bi-partitions by(2.1) Υ : (cid:18) a , . . . , a m b , . . . , b m (cid:19) (cid:20) a − ( m − , a − ( m − , . . . , a m − − , a m b − ( m − , b − ( m − , . . . , b m − − , b m (cid:21) . The image Υ(Λ) in fact depends only on the equivalence class of Λ. Moreover, it iseasy to check thatrk(Λ) = | Υ(Λ) | + (def(Λ) − , if def(Λ) is odd; | Υ(Λ) | + (def(Λ)) , if def(Λ) is even . For a symplectic group or an orthogonal group G , we define the following setsof (similar classes of) symbols: S O +2 n = { Λ ∈ S | rk(Λ) = n, def(Λ) ≡ } ; S Sp n = { Λ ∈ S | rk(Λ) = n, def(Λ) ≡ } ; S SO n +1 = S O n +1 = { Λ ∈ S | rk(Λ) = n, def(Λ) ≡ } ; S O − n = { Λ ∈ S | rk(Λ) = n, def(Λ) ≡ } . The following is modified from the result proved by Lusztig ( cf . [Lus77]):(1) Suppose that G = Sp n , O ǫ n , SO n +1 . There is a parametrization E ( G ) by S G , so a unipotent character ρ of G is written as ρ = ρ Λ for Λ ∈ S G . If G = O ǫ n , we know that ρ Λ · sgn = ρ Λ t .(2) Suppose that G = O n +1 . Then a unipotent character ρ ∈ E ( G ) can bewritten as ρ = ρ Λ or ρ = ρ Λ · sgn for Λ ∈ S SO n +1 Unipotent characters of a unitary group.
In this subsection, supposethat G is a unitary group. For a symbol Λ, we define(2.2) rk U (Λ) = 2 | Υ(Λ) | + | def(Λ) | ( | def(Λ) | + 1) . SHU-YEN PAN
It is easy to check that(2.3)rk U (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) = m X i =1 a i + m X j =1 b j + | m − m | − ( m + m )( m + m − . Then we let S U n be the set of (similar classes of) symbols Λ ∈ S such that • def(Λ) is either even and non-negative, or odd and negative; • rk U (Λ) = n .Again, there is a parametrization E (U n ( q )) by S U n ( cf . [FS90] or [Pan20b]), i.e.,for ρ ∈ E (U n ( q )) , we can write ρ = ρ Λ for a unique Λ ∈ S U n .2.4. Lusztig correspondence.
The following result is proved by Lusztig ( cf . [DM91]theorem 13.23, remark 13.24): Proposition 2.4 (Lusztig) . There is a bijection L s : E ( G ) s → E ( C G ∗ ( s )) such that R GT ∗ ,s ε G ε C G ∗ ( s ) R C G ∗ ( s ) T ∗ , where C G ∗ ( s ) denotes the centralizer in G ∗ of s , ε G = ( − κ and κ is the relative rank of G . The bijection L s called a Lusztig correspondence . Remark 2.5.
Clearly, def(Λ) ≡ t ) ≡ L : E (Sp n ( q )) → E (SO n +1 ( q )) is a bijectiongiven by ρ Λ ρ Λ t .For s ∈ ( G ∗ ) , we define G (0) = G (0) ( s ) = Y h λ i⊂{ λ ,...,λ n } , λ = ± G [ λ ] ( s ); G ( − ) = G ( − ) ( s ) = G [ − ( s ); G (+) = G (+) ( s ) = ( G [1] ( s )) ∗ , if G is symplectic; G [1] ( s ) , otherwise , (2.6)where G [ λ ] ( s ) is given in [AMR96] subsection 1.B (see also [Pan19b] subsection 2.2).Hence we have a bijection E ( C G ∗ ( s )) ≃ E ( G (0) × G ( − ) × G (+) ) × {± } , if G is odd-orthogonal; E ( G (0) × G ( − ) × G (+) ) , otherwise . It is known that G (0) is a product of unitary groups or general linear groups, and G ( − ) , G (+) are given as follows: • if G = O ǫ n , then G ( − ) = O ǫ ′ n ( − ) , and G (+) = O ǫ ′ ǫ n (+) for some ǫ ′ = ± ; • if G = Sp n , then G ( − ) = O ǫ ′ n ( − ) , and G (+) = Sp n (+) for some ǫ ′ = ± ; • if G = O n +1 , then G ( − ) = Sp n ( − ) , and G (+) = Sp n (+) ; • if G = U n , then G ( − ) = U n ( − ) , and G (+) = U n (+) , HETA RANKS 7 where n (+) = n (+) ( s ) (resp. n ( − ) = n ( − ) ( s )) denotes the multiplicity of +1 (resp. − { λ , . . . , λ n } of “eigenvalues” of s .So now a Lusztig correspondence L s induces a mappingΞ s : E ( G ) s → E ( G (0) ( s ) × G ( − ) ( s ) × G (+) ( s )) . The mapping Ξ s is two-to-one and Ξ s ( ρ ) = Ξ s ( ρ · sgn) when G is an odd-orthogonalgroup; and Ξ s is a bijection otherwise. We can write Ξ s ( ρ ) = ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) .Because now ρ ( ε ) where ε = ± is a unipotent character of a symplectic group,an even-orthogonal group, or a unitary group, we can write ρ ( ε ) = ρ Λ ( ε ) for somesymbol Λ ( ε ) ∈ S G ( ε ) . Now ρ Λ ( ε ) is a mapping from E ( G ) s to S G ( ε ) and Λ ( ε ) isalso denoted by Λ ( ε ) ( ρ ), and hence we can writeΞ s ( ρ ) = ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Note that when G is an orthogonal group, ρ ∈ E ( G ) s if and only if ρ · sgn ∈ E ( G ) s . Lemma 2.7.
Let G be an orthogonal group and ρ ∈ E ( G ) s . Then Λ ( ε ) ( ρ · sgn) = Λ ( ε ) ( ρ ) t , if G is an even orthogonal group ;Λ ( ε ) ( ρ ) , if G is an odd orthogonal group . Proof.
This is obvious from the definition in (2.6). (cid:3)
Recall that a linear character χ G of order two is defined when G is an orthogonalgroup or a unitary group ( cf . [Pan16] subsection 3.3). Then there exists a bijection E ( G ) s → E ( G ) − s given by ρ ρχ G . Lemma 2.8.
Let G be an orthogonal group or a unitary group, and let s be asemisimple element in ( G ∗ ) . Then G (0) ( − s ) = G (0) ( s ) , G ( − ) ( − s ) = G (+) ( s ) and G (+) ( − s ) = G ( − ) ( s ) .Proof. This is obvious from the definition in (2.6). (cid:3)
Lemma 2.9.
Let G be an orthogonal group, and let ρ ∈ E ( G ) . Then Λ ( ε ) ( ρχ G ) =Λ ( − ε ) ( ρ ) for ε = ± .Proof. Let ρ ∈ E ( G ) s for some semisimple s ∈ ( G ∗ ) . It is known that χ G R GT ∗ ,s = R GT ∗ , − s for any rational maximal torus T ∗ in G ∗ containing s . Then from Lemma 2.8we have G ( − ) ( − s ) = G (+) ( s ) and G (+) ( − s ) = G ( − ) ( s ). If we write Ξ s ( ρ ) = ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) and Ξ − s ( ρχ G ) = ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) , then we have ρ ′ ( − ) = ρ (+) and ρ ′ (+) = ρ ( − ) . This means that Λ ( − ) ( ρχ G ) = Λ (+) ( ρ ) and Λ (+) ( ρχ G ) = Λ ( − ) ( ρ ). (cid:3) Lusztig correspondence and parabolic induction.
Let G n denote thegroup of split rank n in its Witt series, i.e., G n is one of Sp n , O +2 n , O − n +2 , O n +1 ,U n , or U n +1 . For m ≥ n , let E G m G n ( ρ ) denote the set of irreducible componentsof Ind G m P n ( ρ ⊗ ) where P n is a parabolic subgroup of G m whose Levi factor isisomorphic to G n × T m − n and T m − n is the direct product of m − n copies of GL ( q ) SHU-YEN PAN if G n is a symplectic group or an orthogonal group; T m − n is the direct product of m − n copies of GL ( q ) if G n is a unitary group. For m < n and ρ ∈ E ( G n ), let E G m G n ( ρ ) denote the set of ρ ′ ∈ E ( G m ) such that ρ ∈ E G n G m ( ρ ′ ).Let ρ ∈ E ( G n ) and ρ m ∈ E G m G n ( ρ ) for some m ≥ n . BecauseInd G m P n ( R G n T ∗ ,s ⊗ ) = R G m T ∗ × T m − n ,s m where s m = ( s, ∈ ( G ∗ n ) × T m − n ⊂ ( G ∗ m ) , we see that h ρ m , R G m T ∗ × T m − n ,s m i 6 = 0if and only if h ρ, R G n T ∗ ,s i 6 = 0. This means that if ρ ∈ E ( G n ) s for some s , then ρ m ∈ E ( G m ) s m . In particular, if ρ is unipotent, then every element in E G m G n ( ρ ) isunipotent. Lemma 2.10.
Suppose that m ≥ n , ρ ∈ E ( G n ) s for some s , and Ξ s ( ρ ) = ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) . Then E G m G n ( ρ ) = { ρ m ∈ E ( G m ) s m | Ξ s m ( ρ m ) = ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) m , ρ (+) m ∈ E G m (+) G n (+) ( ρ (+) ) } where m (+) = n (+) + ( m − n ) .Proof. Let ρ ∈ E ( G n ) s . If ρ m ∈ E G m G n ( ρ ), then from above we see that ρ m ∈E ( G m ) s m where s m = ( s, ∈ ( G ∗ n ) × T m − n ⊂ ( G ∗ m ) . Then G (0) ( s m ) = G (0) ( s ), G ( − ) ( s m ) = G ( − ) ( s ), and G (+) ( s m ) = G m (+) where m (+) = n (+) + ( m − n ). It isknown that R G m T ∗ × T m − n ,s m = Ind G m P n ( R G n T ∗ ,s ⊗ ) where P n is a parabolic subgroupof G m whose Levi factor is isomorphic to G n × T m − n . Then the lemma followseasily. (cid:3) Suppose that ρ ∈ E ( G n ) . We write ρ = ρ Λ ( ρ = ρ Λ or ρ = ρ Λ · sgn if G n =O n +1 ) for Λ ∈ S G n . From the Littlewood-Richardson rule, we know that anelement ρ ′ ∈ E G n +1 G n ( ρ ) is of the form ρ Λ ′ ( ρ Λ ′ or ρ Λ ′ · sgn if G n = O n +1 ) whereΛ ′ ∈ S G n +1 such that def(Λ ′ ) = def(Λ) and Υ(Λ ′ ) is obtained from Υ(Λ) = (cid:2) µν (cid:3) byadding a box to some row (including an empty row) of µ or ν . Example 2.11.
Suppose that ρ ∈ E (Sp n ) s and Ξ s ( ρ ) = ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) suchthat ρ Λ (+) is cuspidal. Hence n (+) = t ( t + 1) for some non-negative integer t andΛ (+) = (cid:0) t,t − ,..., , − (cid:1) , if t is even; (cid:0) − t,t − ,..., , (cid:1) , if t is odd . Hence the parabolic induced character Ind Sp n +1) P n ( ρ ⊗ ) is the sum of two irre-ducible charactersΞ − s, (( ρ (0) ⊗ ρ ( − ) ⊗ ρ Λ ) + ( ρ (0) ⊗ ρ ( − ) ⊗ ρ Λ ))where Ξ ( s, is the bijectionΞ ( s, : E (Sp n +1) ( q )) ( s, → E ( G (0) × G ( − ) × Sp t ( t +1)+1) ( q )) and { Λ , Λ } = (cid:8)(cid:0) t +1 ,t − ,t − ,..., , − (cid:1) , (cid:0) t +1 ,t,t − ,..., , (cid:1)(cid:9) , if t is even; (cid:8)(cid:0) − t +1 ,t − ,t − ,..., , (cid:1) , (cid:0) t +1 ,t,t − ,..., , (cid:1)(cid:9) , if t is odd . Finite Howe correspondences on unipotent characters.
First we recallthe concept of finite reductive dual pairs from [How]. Let ( G ′ , G ) be a finitereductive dual pair of one of the following types:(I) (U n ′ , U n );(II) (O ǫ n ′ , Sp n ) or (Sp n ′ , O ǫ n );(III) (O n ′ +1 , Sp n ) or (Sp n ′ , O n +1 )where n ′ , n are non-negative integers. By restricting the Weil character to the dualpair, we obtain a decomposition ω ψ G ′ , G = X ρ ′ ∈E ( G ′ ) , ρ ∈E ( G ) m ρ ′ ,ρ ρ ′ ⊗ ρ where m ρ ′ ,ρ = 0 ,
1. The we define a relation between E ( G ′ ) and E ( G ):Θ G ′ , G = { ( ρ ′ , ρ ) ∈ E ( G ′ ) × E ( G ) | m ρ ′ ,ρ = 0 } , called the Θ -correspondence for the dual pair ( G ′ , G ). For ρ ′ ∈ E ( G ′ ), we writeΘ( ρ ′ ) = Θ G ( ρ ′ ) = { ρ ∈ E ( G ) | ( ρ ′ , ρ ) ∈ Θ G ′ , G } . It is well known that for cases (I) and (II) above, the unipotent characters arepreserved by Θ-correspondence, i.e., when ( ρ ′ , ρ ) ∈ Θ G ′ , G , then ρ ′ ∈ E ( G ′ ) if andonly if ρ ∈ E ( G ) . For these two cases, the unipotent part of the Weil characterhas the decomposition ω G ′ , G , = X ρ ′ ∈E ( G ′ ) , ρ ∈E ( G ) m ρ ′ ,ρ ρ ′ ⊗ ρ = X (Λ ′ , Λ) ∈B G ′ , G ρ Λ ′ ⊗ ρ Λ where the relation B G ′ , G are given in [Pan19a] and [Pan19b].2.7. Howe correspondence and Lusztig correspondence.
The following re-sult on the commutativity between the Lusztig correspondence and the Howe cor-respondence is from [Pan19b]:
Proposition 2.12.
Let ( G ′ , G ) be a finite reductive dual pair, and let ρ ′ ∈ E ( G ′ ) s ′ , ρ ∈ E ( G ) s for some s ′ , s . (i) If ( G ′ , G ) = (O ǫ n ′ , Sp n ) or (U n ′ , U n ) , then G ′ (0) ≃ G (0) , G ′ ( − ) ≃ G ( − ) , ( G ′ (+) , G (+) ) is a dual pair of types (I) or (II) in Subsection 2.6, and Ξ s ′ , Ξ s can be chosen so that the following diagram ρ ′ Θ G ′ , G −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ id ⊗ Θ G ′ (+) , G (+) −−−−−−−−−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) commutes. (ii) If ( G ′ , G ) = (O n ′ +1 , Sp n ) , then G ′ (0) ≃ G (0) , G ′ (+) ≃ G (+) , ( G ′ ( − ) , G ( − ) ) is a dual pair of types (I) or (II) in Subsection 2.6, and Ξ s ′ , Ξ s can be cho-sen so that the following diagram ρ ′ Θ G ′ , G −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ Θ G ′ ( − ) , G ( − ) ⊗ id −−−−−−−−−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) commutes. Recall that an irreducible character ρ of G = Sp n ( q ) is called quadratic unipotent if ρ ∈ E ( G ) s such that C SO n +1 ( s ) = O ǫ n for some ǫ = ± . Now from the propositionwe have several lemmas concerning the first occurrences of unipotent or quadraticunipotent characters. Lemma 2.13.
Let ρ be an irreducible unipotent character of Sp n ( q ) . Suppose that ρ occurs in the Θ -correspondence for the dual pair (O n ′ +1 , Sp n ) . Then n ′ ≥ n .Proof. Consider the dual pair (O n ′ +1 , Sp n ) and the following commutative dia-gram: ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ Θ ⊗ id −−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) . Because ρ is a unipotent character of Sp n ( q ), we see that G (0) = U , G ( − ) = O +0 ,and G (+) = Sp n . This means by Proposition 2.12 that G ′ (+) = Sp n and hence n ′ ≥ n . (cid:3) Lemma 2.14.
Let ρ be an irreducible quadratic unipotent character of Sp n ( q ) .Suppose that ρ occurs in the Θ -correspondence for the dual pair (O ǫ n ′ , Sp n ) . Then n ′ ≥ n .Proof. Consider the dual pair (O ǫ n ′ , Sp n ) and the following commutative diagram: ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ id ⊗ Θ −−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) Because ρ is a quadratic unipotent character of Sp n ( q ), we see that G (0) = U , G ( − ) = O ǫ ′ n for some ǫ ′ = ± , and G (+) = Sp . This means by Proposition 2.12that G ′ ( − ) = O ǫ ′ n and hence n ′ ≥ n . (cid:3) Lemma 2.15.
Let ρ be an irreducible unipotent character of O n +1 ( q ) . Supposethat ρ occurs in the correspondence for the dual pair (Sp n ′ , O n +1 ) . Then n ′ ≥ n . HETA RANKS 11
Proof.
Consider the dual pair (Sp n ′ , O n +1 ) and the following commutative dia-gram: ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ Θ ⊗ id −−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) Because ρ is a unipotent character of O n +1 ( q ), we see that G (0) = U , G ( − ) = Sp ,and G (+) = Sp n . This means by Proposition 2.12 that G ′ (+) = Sp n and hence n ′ ≥ n . (cid:3) Lemma 2.16.
Let ρ be an irreducible unipotent character of O ǫ n ( q ) . Suppose that ρχ G occurs in the correspondence for the dual pair (Sp n ′ , O ǫ n ) . Then n ′ ≥ n .Proof. Consider the dual pair (Sp n ′ , O ǫ n ) and the following commutative diagram: ρ ′ Θ −−−−→ ρχ G Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ id ⊗ Θ −−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) Now ρχ G is in E (O ǫ n ( q )) − , we see that G (0) = U , G ( − ) = O ǫ n , and G (+) = O +0 .This means by Proposition 2.12 that G ′ ( − ) = O ǫ n and hence n ′ ≥ n . (cid:3) Lemma 2.17.
Let ρ be an irreducible unipotent character of U n ( q ) , and let χ be anon-trivial linear character of U n ( q ) . Suppose that ρχ occurs in the correspondencefor the dual pair (U n ′ , U n ) . Then n ′ ≥ n .Proof. Consider the dual pair (U n ′ , U n ) and the following commutative diagram: ρ ′ Θ −−−−→ ρχ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ ′ ( − ) ⊗ ρ ′ (+) id ⊗ id ⊗ Θ −−−−−−→ ρ (0) ⊗ ρ ( − ) ⊗ ρ (+) Now ρ is unipotent and χ is a nontrivial linear character, we see that ρχ is in E ( G ) s where s does not have eigenvalue 1, i.e., G (+) = U . Because now G ′ (0) = G (0) and G ′ ( − ) = G ( − ) by Proposition 2.12, we conclude that n ′ ≥ n . (cid:3)
3. Θ -ranks of Irreducible Characters
Definition of Θ -ranks. Let G be a symplectic group, an orthogonal groupor a unitary group. For ρ ∈ E ( G ), the Θ -rank of ρ , denoted by Θ-rk( ρ ), is definedas follows:(1) If G is a symplectic group, we consider dual pairs (O ǫk , G ) and defineΘ-rk( ρ ) = min { k | ρ ∈ Θ( ρ ′ ) for some ρ ′ ∈ E (O ǫk ( q )) , ǫ = ± } . (2) If G is an orthogonal groups, we consider dual pairs (Sp k , G ) and defineΘ-rk( ρ ) = min { k | ρχ ∈ Θ( ρ ′ ) for some ρ ′ ∈ E (Sp k ( q ))and some linear character χ ∈ E ( G ) } . So the Θ-rank of an irreducible character of an orthogonal group is alwaysan even number. Note that for a non-trivial orthogonal group, there arefour linear characters: , sgn, χ G , and sgn χ G .(3) If G is a unitary group, we consider dual pairs (U k , G ) and defineΘ-rk( ρ ) = min { k | ρχ ∈ Θ( ρ ′ ) for some ρ ′ ∈ E (U k ( q )) and linear character χ ∈ E ( G ) } . From the above definition, we see that Θ-rk( ρ ) = Θ-rk( ρχ ) for any linear character χ of G . Example 3.1.
By convention, we have ω ψ G ′ , G = G ′ ⊗ G when ( G ′ , G ) =(Sp , O ǫn ), (O +0 , Sp n ) or (U , U n ). Therefore linear characters are the only irre-ducible characters of G of Θ-rank 0. Remark 3.2.
Let ρ be an irreducible character of G .(1) Suppose that G = Sp n . Let n ǫ ( ρ ) denote the minimum of 2 n ′ such that ρ occurs in the Θ-correspondence for the dual pair (O ǫ n ′ , Sp n ). It is knownfrom [Pan19b] that n +0 ( ρ ) + n − ( ρ ) ≤ n + 2 . Because now both n +0 ( ρ ) , n − ( ρ ) are even, we see that at least one of n +0 ( ρ ) , n − ( ρ )is less than or equal to 2 n . Moreover, Θ-rk( ρ ) ≤ min { n +0 ( ρ ) , n − ( ρ ) } by thedefinition, so we have 0 ≤ Θ-rk( ρ ) ≤ n .(2) Suppose that G = O ǫ n or O n +1 . Let n ( ρ ) denote the minimum of 2 n ′ such that ρ occurs in the Θ-correspondence for the dual pair (Sp n ′ , G ). Itis known that n ( ρ ) + n ( ρ · sgn) ≤ n, if G = O ǫ n ;4 n + 2 , if G = O n +1 . Now both n ( ρ ) , n ( ρ · sgn) are even. It again implies that 0 ≤ Θ-rk( ρ ) ≤ n .(3) Suppose that G = U n . Let n +0 ( ρ ) (resp. n − ( ρ )) denote the minimum of2 n ′ (resp. 2 n ′ + 1) such that ρ occurs in the Θ-correspondence for the dualpair (U n ′ , U n ) (resp. (U n ′ +1 , U n )). It is known that n +0 ( ρ ) + n − ( ρ ) ≤ n + 1 . This implies that 0 ≤ Θ-rk( ρ ) ≤ n . HETA RANKS 13 -ranks of unipotent or quadratic-unipotent characters.
Let Λ = (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) ∈S G where G is a symplectic group, an orthogonal group, or a unitary group. Be-cause (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) ∼ (cid:0) a +1 ,a +1 ,...,a m +1 , b +1 ,b +1 ,...,b m +1 , (cid:1) and rk(Λ) = rk(Λ ′ ), rk U (Λ) = rk U (Λ ′ )if Λ ∼ Λ ′ , we may assume that both rows { a i } , { b i } are non-empty. Then we define(3.3) Θ-rk(Λ) = min { r (cid:0) a − (cid:1) ) , r (cid:0) − b (cid:1) ) } , if G is Sp or O;min { rk U (Λ r (cid:0) a − (cid:1) ) , rk U (Λ r (cid:0) − b (cid:1) ) } if G is Uwhere Λ r (cid:0) a − (cid:1) = (cid:0) a ,...,a m b ,b ,...,b m (cid:1) , etc. Note that Θ-rk(Λ) depends only on the equiv-alence class of Λ. Proposition 3.4.
Let G be a symplectic group, an orthogonal group, or a unitarygroup, and let ρ ∈ E ( G ) . Write ρ = ρ Λ ( ρ = ρ Λ or ρ = ρ Λ · sgn if G = O n +1 )for some symbol Λ ∈ S G . Then Θ-rk( ρ Λ ) = Θ-rk(Λ) .Proof. Let Λ = (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) ∈ S G .(1) Suppose that G = Sp n . By Lemma 2.13, to compute Θ-rk( ρ ) we only needto consider the dual pairs (O ǫ n ′ , Sp n ) for ǫ = ± , i.e., we do not need toconsider the first occurrence of ρ Λ in the correspondence for the dual pair(O n ′ +1 , Sp n ). From [Pan19b] section 8, we know that the minimum of 2 n ′ such that ρ Λ occurs in the Θ-correspondence for the dual pair (O +2 n ′ , Sp n )is equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) . Similarly, the minimum of 2 n ′ such that ρ Λ occurs in the Θ-correspondence for the dual pair (O − n ′ , Sp n ) is equal to2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) .(2) Suppose that G = O ǫ n . By Lemma 2.16, if ρ Λ χ G or ρ Λ · sgn χ G occurs in theΘ-correspondence for the dual pair (Sp n ′ , O ǫ n ), then n ′ ≥ n . So we needonly to consider the first occurrences of ρ Λ and ρ Λ · sgn in the correspondencefor the dual pair (Sp n ′ , O ǫ n ). We know that the minimum of 2 n ′ suchthat ρ Λ occurs in the Θ-correspondence for the dual pair (Sp n ′ , O ǫ n ) isequal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) when ǫ = +; and equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) when ǫ = − . Similarly, the minimum of 2 n ′ such that ρ Λ · sgn occurs in theΘ-correspondence for the dual pair (Sp n ′ , O ǫ n ) is equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) when ǫ = +; and equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) when ǫ = − .(3) Suppose that G = O n +1 . By Lemma 2.15, we know that if ρ Λ or ρ Λ · sgnoccurs in the Θ-correspondence for the dual pair (Sp n ′ , O n +1 ), then n ′ ≥ n . So we need only to consider the first occurrences of ρ Λ χ G and ρ Λ · sgn χ G in the correspondence for the dual pair (Sp n ′ , O ǫ n ).Now ρ Λ χ G ∈ E ( G ) − and Λ − ( ρ Λ χ G ) = Λ t . So we have a commutativediagram ρ ′ Θ G ′ , G −−−−→ ρ Λ χ G Ξ y y Ξ − ⊗ ρ ′ ( − ) ⊗ id ⊗ Θ G ′ ( − ) , G ( − ) ⊗ id −−−−−−−−−−−−−→ ⊗ ρ Λ t ⊗ Now G ( − ) = Sp n and G ′ ( − ) = O +2 n ′ , so the minimum of 2 n ′ such that ρ Λ χ G occurs in the Θ-correspondence for the dual pair (Sp n ′ , O n +1 ) is equal tothe minimum of 2 n ′′ such that ρ Λ t occurs in the Θ-correspondence forthe dual pair (O +2 n ′′ , Sp n ), and hence equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) . Similarly, ρ Λ · sgn χ G ∈ E ( G ) − and Λ − ( ρ Λ · sgn χ G ) = Λ t . Now G ( − ) = Sp n and G ′ ( − ) = O − n ′ , so the minimum of 2 n ′ such that ρ Λ · sgn χ G occurs in theΘ-correspondence for the dual pair (Sp n ′ , O n +1 ) is equal to the minimumof 2 n ′′ such that ρ Λ t occurs in the Θ-correspondence for the dual pair(O − n ′′ , Sp n ), and hence equal to 2rk (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) .(4) Suppose that G = U n . By Lemma 2.17, to compute Θ-rk( ρ ), we do not needto consider the first occurrences of ρ Λ χ for any nontrivial linear character χ of G . Now the minimum of n ′ such that ρ Λ occurs in the Θ-correspondencefor the dual pair (U n ′ , U n ) is equal to rk U (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) if n ′ + n is even; andis equal to rk U (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) if n ′ + n is odd. (cid:3) The following two corollaries follow from the proof of the above propositionimmediately.
Corollary 3.5.
Then Θ -rank of an irreducible unipotent character of Sp n ( q ) iseven. Note that if ρ Λ ∈ E (O n +1 ( q )), then ρ Λ t ∈ E (Sp n ( q )). Corollary 3.6. If ρ Λ ∈ E (O n +1 ( q )) , then Θ-rk( ρ Λ ) = Θ-rk( ρ Λ · sgn) = Θ-rk( ρ Λ t ) . Example 3.7.
It is clear from the result in [AM93] and Lemma 2.16 that theΘ-ranks of unipotent cuspidal characters are described as follows:(1) The unique unipotent cuspidal character of Sp d ( d +1) ( q ) is 2 d .(2) The two unipotent cuspidal characters of O ǫ d ( q ) (for d ≥
1) is 2 d ( d − ǫ = + if d is even, and t = − if d is odd.(3) The two unipotent cuspidal characters of O d ( d +1)+1 ( q ) is 2 d .(4) The unique unipotent cuspidal character of U d ( d +1) ( q ) is d ( d − Example 3.8.
Now we want to compute the Θ-rank of a Steinberg character.(1) Let G = Sp n . The Steinberg character St Sp n is the unipotent characterassociated to the symbol (cid:0) n,n − ,..., n,n − ,..., (cid:1) . Therefore,Θ-rk(St Sp n ) = min { (cid:0) n − ,n − ,..., n,n − ,..., (cid:1) , (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) } = 2 n, i.e., St Sp n is a character of maximal Θ-rank.(2) Suppose that G = O +2 n and n ≥
1. The two Steinberg characters ρ, ρ · sgnare the unipotent characters associated to the symbols (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) or (cid:0) n − ,n − ,..., n,n − ,..., (cid:1) . ThereforeΘ-rk( ρ ) = Θ-rk( ρ · sgn) = min { (cid:0) n − ,n − ,..., n − ,n − ,..., (cid:1) , (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) } = 2 n − . HETA RANKS 15
Suppose that G = O − n and n ≥
1. The Steinberg characters are the unipo-tent characters associated to the symbols (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) or (cid:0) n − ,n − ,..., n,n − ,..., (cid:1) .Therefore,Θ-rk( ρ ) = Θ-rk( ρ · sgn) = min { (cid:0) n − ,n − ,..., n − ,n − ,..., (cid:1) , (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) } = 2 n − . (3) Suppose that G = O n +1 . The two Steinberg characters ρ, ρ · sgn of G arethe two unipotent characters associated to the symbol (cid:0) n,n − ,..., n,n − ,..., (cid:1) . There-fore, by the same argument in the symplectic case, we obtain Θ-rk( ρ ) =Θ-rk( ρ · sgn) = 2 n .(4) Suppose that G = U n and n ≥ n is even. The Steinberg character of G is the unipotentcharacter associated to the symbol Λ = (cid:0) n − , n − ,..., n , n − ,..., (cid:1) and Υ(Λ) = (cid:2) − , ,..., (cid:3) ( n copies of “1”). ThenΘ-rk(St U n ) = rk U (cid:18) n − , n − , . . . , n − , n − , . . . , (cid:19) = 2 (cid:16) n − (cid:17) + 1 ·
22 = n − . (b) Suppose that n is odd. The Steinberg character of G is the unipotentcharacter associated to the symbol Λ = (cid:0) n − , n − ,..., n − , n − ,..., (cid:1) and Υ(Λ) = (cid:2) , ,..., − (cid:3) ( n − copies of “1”). ThenΘ-rk(St U n ) = rk U (cid:18) n − , n − , . . . , n − , n − , . . . , (cid:19) = 2 (cid:18) n − (cid:19) + 0 ·
12 = n − . Lemma 3.9.
Let G be Sp n , O +2 n , or O n +1 , and let ρ ∈ E ( G ) . If Θ-rk( ρ ) = 2 n ,then ρ is a Steinberg character , if G = Sp n , O n +1 ; n = 0 , if G = O +2 n . Moreover, O − n does not have any unipotent character of Θ -rank n .Proof. Let ρ = ρ Λ ∈ E ( G ) ( ρ = ρ Λ or ρ = ρ Λ · sgn if G = O n +1 ) such thatΘ-rk( ρ ) = 2 n , and writeΛ = (cid:18) a , a , . . . , a m b , b , . . . , b m (cid:19) ∈ S G , Υ(Λ) = (cid:20) µ , µ , . . . , µ m ν , ν , . . . , ν m (cid:21) . (1) Suppose that G = Sp n . By Proposition 3.4, we need both rk (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) = n and rk (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) = n , and then (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) ∈ S O +2 n and (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) ∈S O − n . So we need µ = 0 and ν = 1, thus Υ(Λ) = (cid:2) − , ,..., (cid:3) ( n copies of“1”). That is, ρ Λ is the Steinberg character of Sp n ( q ).(2) Suppose that G = O n +1 . By Corollary 3.6, we know that Θ-rk( ρ Λ ) = 2 n if and only if Θ-rk( ρ Λ t ) = 2 n . Now by (1), ρ Λ t is the Steinberg characterof Sp n ( q ), hence ρ Λ is a Steinberg character of O n +1 ( q ).(3) Suppose that G = O +2 n . By Proposition 3.4, we need both rk (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) = n and rk (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) = n , and then (cid:0) b ,b ,...,b m a ,a ,...,a m (cid:1) ∈ S Sp n and (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) ∈ S Sp n . So we need µ = 0 and ν = 0, thus Υ(Λ) = (cid:2) −− (cid:3) , i.e., ρ Λ is unipotentcuspidal. But from Example 3.7, we must have n = 0.(4) Suppose that G = O − n (where n ≥ | Υ(Λ) | = n − r (cid:0) a − (cid:1) ) and rk(Λ r (cid:0) − b (cid:1) ) is less than or equal to n −
1. This means that the maximum value of Θ-rk( ρ ) for ρ ∈ E (O − n ( q )) is at most 2( n − (cid:3) Lemma 3.10.
Let ρ be an irreducible quadratic unipotent character of Sp n ( q ) .Then Θ-rk( ρ ) = Θ-rk( ρ ( − ) ) + 1 , in particular, Θ-rk( ρ ) is odd.Proof. By Lemma 2.14, we only need to consider the first occurrence of ρ for thedual pair (O n ′ +1 , Sp n ). Now we have ρ ∈ E ( G ) s such that G (0) = U , G ( − ) = O ǫ ′ n for some ǫ ′ , and G (+) = Sp . Hence ρ ′ ∈ E ( G ′ ) s ′ such that G ′ (0) = U , G ′ ( − ) =Sp n ′ and G (+) = Sp . Now ρ of Sp n ( q ) occurs in the correspondence for the dualpair (O n ′ +1 , Sp n ) if and only if ρ ( − ) of O ǫ ′ n ( q ) occurs in the correspondence forthe dual pair (Sp n ′ , O ǫ ′ n ). Thus the lemma is proved. (cid:3) Lemma 3.11.
Let ρ ∈ E (U n ( q )) . If Θ-rk( ρ ) = n , then n = 0 .Proof. Suppose that ρ = ρ Λ ∈ E ( G ) . By the same argument in the proof ofLemma 3.9, we must have Υ(Λ) = (cid:2) −− (cid:3) , i.e., ρ Λ is cuspidal. This means thatrk U (Λ) = d ( d + 1) and Θ-rk(Λ) = d ( d −
1) for some non-negative integer d . Theequalities n = d ( d + 1) = d ( d −
1) imply that n = d = 0. (cid:3) -ranks for symplectic groups or orthogonal groups. Now we can de-scribe the Θ-rank of a general irreducible character of a symplectic group or anorthogonal group in terms of its data under the Lusztig correspondence.
Proposition 3.12.
Let G be a symplectic group or an orthogonal group, and let ρ ∈ E ( G ) . Suppose that ρ ∈ E ( G ) s and write Ξ s ( ρ ) = ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Then Θ-rk( ρ ) is equal to min { n − n (+) + Θ-rk(Λ (+) ) , n − n ( − ) + Θ-rk(Λ ( − ) ) + 1 } , if G = Sp n ;min { n − n (+) + Θ-rk(Λ (+) ) , n − n ( − ) + Θ-rk(Λ ( − ) ) } , if G = O ǫ n , O n +1 where n ( − ) , n (+) are given in Subsection 2.4.Proof. First suppose that ρ ′ ⊗ ρ occurs in the correspondence for the dual pair( G ′ , G ) for some ρ ′ ∈ E ( G ′ ) s ′ and write Ξ s ′ ( ρ ′ ) = ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) . HETA RANKS 17 (1) Suppose that G = Sp n . First we consider the dual pair (O ǫ n ′ , Sp n ) forsome n ′ . Then we have a commutative diagram ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ id ⊗ Θ −−−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Now we have G ′ ( − ) = O ǫ ′ ǫ n ′ ( − ) and G ′ (+) = O ǫ ′ n ′ (+) ; G ( − ) = O ǫ ′ ǫ n ( − ) and G (+) = Sp n (+) for some ǫ ′ depending on s . Because now G ′ (0) = G (0) and G ′ ( − ) = G ( − ) , we have 2 n ′ − n ′ (+) = 2 n − n (+) , i.e., 2 n ′ = 2 n − n (+) +2 n ′ (+) . Now n and n (+) are fixed and the minimal value of 2 n ′ (+) (for bothpossible ǫ = ± ) is equal to Θ-rk(Λ (+) ).Next we consider the dual pair (O ǫ n ′ +1 , Sp n ). Then we have a commu-tative diagram ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ Θ ⊗ id −−−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Now we have G ′ ( − ) = Sp n ′ ( − ) and G ′ (+) = Sp n ′ (+) ; G ( − ) = O ǫ ′ ǫ n ( − ) and G (+) = Sp n (+) for some ǫ ′ . Because now G ′ (0) = G (0) and G ′ (+) = G (+) ,we have 2 n ′ − n ′ ( − ) = 2 n − n ( − ) , i.e., 2 n ′ + 1 = 2 n − n ( − ) + 2 n ′ ( − ) + 1.Now n and n ( − ) are fixed and the minimal value of 2 n ′ ( − ) (for both possible ǫ = ± ) is equal to Θ-rk(Λ ( − ) ).From the definition of Θ-rank and Proposition 3.4, we see thatΘ-rk( ρ ) = min { n − n (+) + Θ-rk( ρ Λ (+) ) , n − n ( − ) + Θ-rk( ρ Λ ( − ) ) + 1 } = min { n − n (+) + Θ-rk(Λ (+) ) , n − n ( − ) + Θ-rk(Λ ( − ) ) + 1 } . (2) Suppose that G = O ǫ n . Now we consider the dual pair (Sp n ′ , O ǫ n ) andwe have a commutative diagram ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ id ⊗ Θ −−−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Now we have G ′ ( − ) ( s ′ ) = O ǫ ′ ǫ n ′ ( − ) and G ′ (+) ( s ′ ) = Sp n ′ (+) ; G ( − ) ( s ) =O ǫ ′ ǫ n ( − ) and G (+) ( s ) = O ǫ ′ n (+) for some ǫ ′ .Now we need consider the occurrences for four irreducible characters ρ , ρ · sgn, ρχ G and ρχ G · sgn. Note that by Lemma 2.9, we have Λ ( ε ) ( ρ · sgn) =(Λ ( ε ) ( ρ )) t and Λ ( ε ) ( ρχ G ) = Λ ( − ε ) ( ρ ) for ε = ± . By the similar argumentin (1), we conclude thatΘ-rk( ρ ) = min { n − n (+) + Θ-rk(Λ (+) ) , n − n ( − ) + Θ-rk(Λ ( − ) ) } . (3) Suppose that G = O n +1 . Now consider the dual pair (Sp n ′ , O n +1 ) andwe have a commutative diagram ρ ′ Θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ Θ ⊗ id −−−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) . Now we have G ′ ( − ) ( s ′ ) = O ǫ ′ ǫ n ′ ( − ) and G ′ (+) ( s ′ ) = Sp n ′ (+) ; G ( − ) ( s ) =Sp n ( − ) and G (+) ( s ) = Sp n (+) for some ǫ ′ . Then by the similar argumentin (2), we conclude thatΘ-rk( ρ ) = min { n − n (+) + Θ-rk(Λ (+) ) , n − n ( − ) + Θ-rk(Λ ( − ) ) } . (cid:3) Example 3.13.
There are two irreducible characters ρ , ρ of degree ( q n + 1)of Sp n ( q ) for n ≥
1. They are in the Lusztig series E ( G ) s such that G (0) = U , G ( − ) = O +2 n , G (+) = Sp , i.e., n ( − ) = n and n (+) = 0. Moreover, we know thatΛ ( − ) = (cid:0) n (cid:1) or (cid:0) n (cid:1) , and Λ (+) = (cid:0) − (cid:1) . ThereforeΘ-rk( ρ ) = Θ-rk( ρ ) = min { n − (cid:0) − (cid:1) , n − n + Θ-rk (cid:0) n (cid:1) + 1 } = min { n, } = 1 . In fact, both ρ , ρ are quadratic unipotent and both ρ ( − )1 , ρ ( − )2 are trivial or sgncharacters of O +2 n ( q ). So the above result is also a direct consequence of Lemma 3.10. Corollary 3.14.
Let G = Sp n , O n +1 , O ǫ n , and let ρ ∈ E ( G ) s for some s . Then Θ-rk( ρ ) = 2 n if and only if n ( − ) = 0 and ρ (+) is a Steinberg character , if G = Sp n ; both ρ ( − ) , ρ (+) are Steinberg characters , if G = O n +1 ; both n ( − ) = n (+) = 0 , if G = O ǫ n . Proof.
Suppose that G = Sp n . By Proposition 3.12, we know that Θ-rk( ρ ) = 2 n if and only if Θ-rk( ρ ( − ) ) = 2 n ( − ) and Θ-rk( ρ (+) ) = 2 n (+) . Now G ( − ) = O n ( − ) and G (+) = Sp n (+) , so we need n ( − ) = 0 and ρ (+) to be the Steinberg character byLemma 3.9. The proofs for two other cases are similar. (cid:3) We provide some necessary condition on the possible value of Θ-rk( ρ ) in Re-mark 3.2. Now we have the following sufficient condition: Proposition 3.15. (i)
Suppose that G = O ǫ n . (a) For each even integer k such that ≤ k < n , there exists an irre-ducible character ρ ∈ E (O ǫ n ( q )) such that Θ-rk( ρ ) = k . (b) If n ≥ , or q > , or ǫ = − , then there exists an irreducible character ρ ∈ E (O ǫ n ( q )) such that Θ-rk( ρ ) = 2 n . (c) There is no irreducible character of O +2 (3) of Θ -rank . HETA RANKS 19 (ii)
For each integer k such that ≤ k ≤ n , there exists an irreducible char-acter ρ ∈ E (Sp n ( q )) such that Θ-rk( ρ ) = k . (iii) For each even integer k such that ≤ k ≤ n , there exists an irreduciblecharacter ρ ∈ E (O n +1 ( q )) such that Θ-rk( ρ ) = k .Proof. (1) Suppose that G = O − n . Because it is trivial when k = 0, we willassume that k is an even integer such that 2 ≤ k ≤ n .(a) Suppose that 2 ≤ k < n . Let Λ k = (cid:0) k − , k − ,..., n − , k , k − ,..., (cid:1) . Then we haveΥ(Λ k ) = (cid:2) − n − k − , ,..., (cid:3) ∈ P ( n − (cid:0) k , k − ,..., k − , k − ,..., (cid:1) ∈ S Sp k and (cid:0) − n, , (cid:1) ∈ S Sp n +2 if k = 2; (cid:0) k − , k − ,..., n − , k , k − ,..., (cid:1) ∈ S Sp n +2 if k ≥
4. ThenΘ-rk( ρ Λ k ) = min { k, n + 2 } = k .(b) It is known that there exists a semisimple element s ∈ G ∗ = O − n ( q )such that C G ∗ ( s ) ≃ U ( q n ). That is, n ( − ) = n (+) = 0, and then any ρ ∈ E ( G ) s has Θ-rk( ρ ) = 2 n by Proposition 3.12.(2) Suppose that G = O +2 n and k is an even integer such that 0 ≤ k ≤ n .(a) Suppose that k < n . Let Λ k = (cid:0) n, k , k − ,..., k , k − ,..., (cid:1) ∈ S O +2 n . Then we haveΥ(Λ k ) = (cid:2) n − k , ,..., − (cid:3) ∈ P ( n ), (cid:0) k , k − ,..., k , k − ,..., (cid:1) ∈ S Sp k , and (cid:0) n, k , k − ,..., k − , k − ,..., (cid:1) ∈S Sp n . Hence by Proposition 3.4, Θ-rk( ρ Λ k ) = min { k, n } = k .(b) Suppose that q > n ≥
2. It is known that there exists a semisimpleelement s ∈ G ∗ = O − n ( q ) such that C G ∗ ( s ) ≃ GL ( q n ). That is, n ( − ) = n (+) = 0, and then any ρ ∈ E ( G ) s has Θ-rk( ρ ) = 2 n byProposition 3.12.(c) Now O +2 (3) is a finite group of 4 element. Each irreducible characterof O +2 (3) is a linear character and hence has Θ-rank 0.(3) Suppose that G = Sp n and k is an integer such that 0 ≤ k ≤ n .(a) Suppose that k is even and 0 ≤ k < n . Let Λ k = (cid:0) n, k , k − ,..., k − , k − ,..., (cid:1) andthen Υ(Λ k ) = (cid:2) n − k , ,..., − (cid:3) ∈ P ( n ). Then (cid:0) k − , k − ,..., k , k − ,..., (cid:1) ∈ S O +2 k and (cid:0) k − , k − ,..., n, k , k − ,..., (cid:1) ∈ S O − n +2 . Thus Θ-rk( ρ Λ k ) = min { k, n + 2 } = k .(b) Suppose that k = 2 n . Let Λ k = (cid:0) n,n − ,..., n,n − ,..., (cid:1) ∈ S Sp n . Then Υ(Λ k ) = (cid:2) − , ,..., (cid:3) ∈ P ( n ). Now (cid:0) n,n − ,..., n − ,n − ,..., (cid:1) ∈ S O +2 n and (cid:0) n − ,n − ,..., n,n − ,..., (cid:1) ∈S O − n . Thus Θ-rk( ρ Λ k ) = 2 n .(c) Suppose that k is odd. Then k − ≤ k − < n Let ρ ∈ E ( G ) s such that G (0) = U , G ( − ) = O +2 n and G (+) = Sp , i.e., ρ is quadratic unipotent. Let Ξ s ( ρ ) = ⊗ ρ Λ ( − ) ⊗ where Λ ( − ) ∈ S O +2 n is the symbol Λ k − given in (2.a). Then by Lemma 3.10, Θ-rk( ρ ) =Θ-rk( ρ Λ k − ) + 1 = k − k .(4) Suppose that G = O n +1 and k is an even integer such that 0 ≤ k ≤ n .Let ρ = ρ Λ where Λ t is equal to the symbol Λ k ∈ S Sp n given in (3.a) and(3.b). Then by Corollary 3.6, we have Θ-rk( ρ ) = Θ-rk( ρ Λ t ) = k . (cid:3) -ranks for unitary groups. In this subsection, we consider G = U n . For s ∈ G ∗ , we can write C G ∗ ( s ) = r Y i =1 t i Y j =1 GL ( − i n ij ( q i )for some non-negative integers r, t i , n ij such that P ri =1 P t i j =1 in ij = n where GL +1 n ij =GL + n ij = GL n ij and GL − n ij = GL − n ij = U n ij . For i = 1, we can let t = q + 1 andeach j = 1 , . . . , t corresponds an eigenvalue λ j ∈ f q of s such that λ q +1 j = 1.Let ρ ∈ E ( G ) s for some s ∈ G ∗ . Then for the Lusztig correspondence L s : E ( G ) →E ( C G ∗ ( s )) we can write(3.16) L s ( ρ ) = r O i =1 t i O j =1 ρ ( ij ) for some ρ ( ij ) ∈ E (GL ( − i n ij ( q i )) . Note that ρ (1 j ) is a unipotent character of U n j ( q ). Proposition 3.17.
Keep the above notation. Let ρ ∈ E (U n ( q )) s for some s . Then Θ-rk( ρ ) = min { n − n j + Θ-rk( ρ (1 j ) ) | j = 1 , . . . , q + 1 } . Proof.
Write L s ( ρ ) as in (3.16). Suppose that G (+) ( s ) = U n , . Then the minimalvalue of n ′ such that ρ occurs in the Θ-correspondence for the pair (U n ′ , U n ) isequal to n − n , + Θ-rk( ρ (1 , ). Suppose that χ is a linear character of U n ( q ) andwrite L s ( ρχ ) = r ′ O i =1 t ′ i O j =1 ρ ′ ( ij ) . Then ρ ′ (1 , , . . . , ρ ′ (1 t ) are a permutation of ρ (1 , , . . . , ρ (1 t ) . Hence the propositionfollows from the definition of the Θ-rank. (cid:3) Remark 3.18.
The same result of this proposition can be found in [GLT20] the-orem 3.9.
Corollary 3.19.
Let ρ ∈ E (U n ( q )) . Then Θ-rk( ρ ) = n if and only if n j = 0 for j = 1 , . . . , q + 1 .Proof. By Proposition 3.17, Θ-rk( ρ ) = n implies that Θ-rk( ρ (1 j ) ) = n j for j =1 , . . . , q +1. Then by Lemma 3.11, we conclude that n j = 0 for j = 1 , . . . , q +1. (cid:3) Proposition 3.20.
For each integer k such that k = 0 if n = 0 , ; and ≤ k ≤ n if n ≥ , there exists an irreducible character ρ ∈ E (U n ( q )) such that Θ-rk( ρ ) = k .Proof. The case for n = 0 , n ≥ ≤ k < n .(a) Suppose that n + k is even. LetΛ k = (cid:0) n , k , k − ,..., k , k − ,..., (cid:1) ; (cid:0) n − , k − , k − ,..., k +12 , k − ,..., (cid:1) , Υ(Λ k ) = (cid:2) n − k , , ,..., − (cid:3) , if n is even; (cid:2) n − k , , ,..., − (cid:3) , if n is odd. HETA RANKS 21 (i) Suppose that n is even. Now (cid:0) k , k − ,..., k , k − ,..., (cid:1) ∈ S U k and (cid:0) k − , k − ,..., n , k , k − ,..., (cid:1) ∈S U n +1 . Hence Θ-rk( ρ Λ k ) = min { k, n + 1 } = k .(ii) Suppose that n is odd. Now (cid:0) k +12 , k − ,..., k − , k − ,..., (cid:1) ∈ S U k and (cid:0) k − , k − ,..., n − , k − , k − ,..., (cid:1) ∈S U n − . Hence Θ-rk( ρ Λ k ) = min { k, n − } = k .(b) Suppose that n + k is odd. LetΛ k = (cid:0) k − , k − ,..., n , k − , k − ,..., (cid:1) ; (cid:0) k − , k − ,..., n − , k , k − ,..., (cid:1) , Υ(Λ k ) = (cid:2) − n − k +12 , , ,..., (cid:3) , if n is even; (cid:2) − n − k − , , ,..., (cid:3) , if n is odd.(i) Suppose that n is even. Now (cid:0) k − , k − ,..., k − , k − ,..., (cid:1) ∈ S U k and (cid:0) n , k − , k − ,..., k − , k − ,..., (cid:1) ∈S U n . Hence Θ-rk( ρ Λ k ) = min { k, n } = k .(ii) Suppose that n is odd. Now (cid:0) k , k − ,..., k − , k − ,..., (cid:1) ∈ S U k and (cid:0) n − , k , k − ,..., k − , k − ,..., (cid:1) ∈S U n +2 . Hence Θ-rk( ρ Λ k ) = min { k, n + 2 } = k .(2) Suppose that k = n .(a) If n is even, let ρ ∈ E ( G ) s such that C G ∗ ( s ) = GL ( q n ).(b) If n is odd and n ≥
3, let ρ ∈ E ( G ) s such that C G ∗ ( s ) = U ( q n ).For each case, such a semisimple element s exists, and by Corollary 3.19,we have Θ-rk( ρ ) = n . (cid:3) -ranks and parabolic induction. Let G n be one of Sp n , O +2 n , O − n +2 ,O n +1 , U n , or U n +1 . For ρ ∈ E ( G n ) and an integer m ≥
0, recall that a subset E G m G n ( ρ ) ⊂ E ( G m ) is defined in Subsection 2.5. Lemma 3.21.
Let ρ ∈ E ( G n ) and m ≥ n . Then Θ-rk( ρ ) = min { Θ-rk( ρ ′ ) | ρ ′ ∈ E G m G n ( ρ ) } Proof.
Let ρ be a unipotent character of G n . It is clear that we need only to showthat Θ-rk( ρ ′ ) ≥ Θ-rk( ρ ) for any ρ ′ ∈ E G n +1 G n ( ρ ), and there exists ρ ∈ E G n +1 G n ( ρ ) suchthat Θ-rk( ρ ) = Θ-rk( ρ ).Now we can write ρ = ρ Λ ( ρ = ρ Λ or ρ = ρ Λ · sgn if G = O n +1 ) for some symbolΛ = (cid:0) a ,a ,...,a m b ,b ,...,b m (cid:1) ∈ S G n . Consider the following reductive dual pairs ( G ′ n ′ , G n ):(a) (O +2 n ′ , Sp n ), (Sp n ′ , O +2 n ), (U n ′ , U n ), (U n ′ +1 , U n +1 );(b) (O − n ′ , Sp n ), (Sp n ′ , O − n +2 ), (U n ′ +1 , U n ), (U n ′ , U n +1 ).Let ρ ′ ∈ E G n +1 G n ( ρ ). Then ρ ′ is also unipotent and we write ρ ′ = ρ Λ ′ ( ρ ′ = ρ Λ ′ or ρ ′ = ρ Λ ′ · sgn if G = O n +1 ) where Λ ′ is equal to one of the following: (cid:0) a ,...,a i − ,a i +1 ,a i +1 ,...,a m b ,b ,...,b m (cid:1) , (cid:0) a +1 ,a +1 ,...,a m +1 , b +1 ,b +1 ,...,b m +1 , (cid:1) , (cid:0) a ,a ,...,a m b ,...,b i − ,b i +1 ,b i +1 ,...,b m (cid:1) , (cid:0) a +1 ,a +1 ,...,a m +1 , b +1 ,b +1 ,...,b m +1 , (cid:1) for some index i .By Proposition 3.4, it is clear thatΘ-rk( ρ Λ ′ ) = Θ-rk(Λ ′ ) ≥ Θ-rk(Λ) = Θ-rk( ρ Λ ) . Now define Λ ∈ S G n +1 byΛ = (cid:0) a +1 ,a ,...,a m b ,b ,...,b m (cid:1) , if ( G ′ n ′ , G n ) is in case (a); (cid:0) a ,a ,...,a m b +1 ,b ,...,b m (cid:1) , if ( G ′ n ′ , G n ) is in case (b) . Let ρ = ρ Λ . Then ρ ∈ E G n +1 G n ( ρ ) and by Proposition 3.4, we know that Θ-rk( ρ Λ ) =Θ-rk( ρ Λ ). (cid:3) From the proof of the lemma, we have the following result:
Corollary 3.22.
Let ρ ∈ E ( G n ) and ρ ′ ∈ E G n +1 G n ( ρ ) . Then either Θ-rk( ρ ′ ) =Θ-rk( ρ ) or Θ-rk( ρ ′ ) = Θ-rk( ρ ) + 2 . Then we generalize Lemma 3.21 to any irreducible characters.
Lemma 3.23.
Let ρ ∈ E ( G n ) and m ≥ n . Then Θ-rk( ρ ) = min { Θ-rk( ρ ′ ) | ρ ′ ∈ E G m G n ( ρχ ) , χ linear character of G n } Proof.
As in the proof of Lemma 3.21, we need only to show that Θ-rk( ρ ′ ) ≥ Θ-rk( ρ )for any ρ ′ ∈ E G n +1 G n ( ρχ ) for any linear character χ , and there exists ρ ∈ E G n +1 G n ( ρχ )for some linear character χ such that Θ-rk( ρ ) = Θ-rk( ρ ). Now we suppose that ρ ∈ E ( G n ) s for some s . When G n is a symplectic group or an orthogonal group,we write Ξ s ( ρ ) = ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) for some Λ ( ε ) ∈ S G ( ε ) where ε = ± .(1) Suppose that G n = Sp n and Θ-rk( ρ ) is even. By Proposition 3.12, we seethat Θ-rk( ρ ) = 2 n − n (+) + Θ( ρ Λ (+) ). Let ρ ′ ∈ E G n +1 G n ( ρ ). Then we knowthat ρ ′ ∈ E ( G n +1 ) s n +1 where s n +1 = ( s, ∈ ( G ∗ n ) × T ⊂ ( G ∗ n +1 ) . NowΞ s n +1 ( ρ ′ ) = ρ (0) ⊗ ρ ( − ) ⊗ ρ ′ (+) where ρ ′ (+) ∈ E G (+) n (+)+1 G (+) n (+) ( ρ (+) ) by Lemma 2.10.Then Θ-rk( ρ ′ ) = 2( n + 1) − n (+) + 1) + Θ-rk( ρ ′ (+) ) ≥ n − n (+) + Θ-rk( ρ (+) ) = Θ-rk( ρ ) . Next let Λ (+)0 be defined as Λ in the proof of Lemma 3.21, and let ρ = Ξ − s n +1 ( ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+)0 ) . Then we have ρ ∈ E G n +1 G n ( ρ ) and Θ-rk( ρ ) = Θ-rk( ρ )(2) Suppose that G n = Sp n and Θ-rk( ρ ) is odd. Then by Proposition 3.12, wehave Θ-rk( ρ ) = 2 n − n ( − ) + Θ( ρ Λ ( − ) ) + 1. The remaining proof is similarto case (1) with only “+” replaced by “ − ”.(3) Suppose that G n is an orthogonal group. By twisting a linear character χ of G n , by Lemma 2.9 and Proposition 3.12, we may assume Θ-rk( ρ ) =2 n − n (+) + Θ( ρ Λ (+) ). Then the remaining proof is similar to case (1). HETA RANKS 23 (4) Suppose that G n is a unitary group. Write L s ( ρ ) = N ri =1 N t i j =1 ρ ( ij ) asin (3.16). By twisting a linear character χ of G n , by Proposition 3.17, wemay assume that Θ-rk( ρ ) = n − n , + Θ( ρ (1 , ) where the notations are asin Subsection 3.4. Then the remaining proof is similar to case (1), again. (cid:3)
4. Θ -ranks and Other Ranks U -rank, A -rank, A -rank and tensor ranks. First we recall the definitionof the U -rank and the asymptotic rank of an irreducible character ρ of a classicalgroup G = G ( V ) given in [GH17] and [GH20] where V is the standard vector spacewith an associated form over f q (if G is a symplectic group or an orthogonal group)or f q (if G is a unitary group) on which G acts. Suppose that we have a Wittdecomposition V = X ⊕ V ⊕ Y where X, Y are totally isotropic and dual to eachother, and V is anisotropic. The group U = { g ∈ G ( V ) | ( g − X ⊕ V ) = 0 , ( g − Y ) ⊂ X } is an abelian unipotent subgroup of G . Then we have a decomposition ρ | U = X φ ∈E ( U ) m φ φ for m φ ∈ N ∪ { } . We say that ρ | U contains φ if m φ = 0. Now each character φ of U is naturally assigned a rank.The irreducible character ρ is said to be of U -rank k , denoted by U -rk( ρ ) = k ,if the restriction ρ | U contains a character φ of rank k , but contains no character ofhigher rank. From the definition, we see that 0 ≤ U -rk( ρ ) ≤ n where n = dim( X ) =dim( Y ). Moreover, it is clear that U -rk( ρ ) = U -rk( ρχ ) for any linear character χ of G because χ | U is always trivial. An irreducible ρ ∈ E ( G n ) is said to be of low U -rank if U -rk( ρ ) < n, if G n = Sp n , U n or U n +1 ; n − , if G n = O +2 n , O − n +2 , or O n +1 . An irreducible character ρ of G n is said to have asymptotic rank less than orequal to k , if for all sufficiently large N , there is a character ρ N ∈ E ( G N ) such that U -rk( ρ N ) ≤ k < N and ρ occurs in ρ N | G n . The asymptotic rank of ρ is denoted by A -rk( ρ ). As indicated in [GH20], it is known that U -rk( ρ ) ≤ A -rk( ρ ) for ρ ∈ E ( G ).To study the representation theory of reductive groups, parabolic induction isusually more convenient than usual induction, so we modified the definition ofasymptotic rank to define the parabolic asymptotic rank as follows. An irreduciblecharacter ρ of G n is said to have parabolic asymptotic rank less than or equalto k , if for all sufficiently large N , there is a character ρ N ∈ E G N G n ( ρ ) such that U -rk( ρ N ) ≤ k < N . The parabolic asymptotic rank of ρ is denoted by A -rk( ρ ). Itis clear from the definition that A -rk( ρ ) ≤ A -rk( ρ ) for ρ ∈ E ( G ). Lemma 4.1.
Let ρ be an irreducible character of G n . Then A -rk( ρ ) = min { A -rk( ρ ′ ) | ρ ′ ∈ E G m G n ( ρ ) } for any m ≥ n .Proof. Let ρ ′ ∈ E G m G n ( ρ ) and suppose that A -rk( ρ ′ ) = k . This means for all suffi-ciently large N , there exists ρ N ∈ E G N G m ( ρ ′ ) such that U -rk( ρ N ) = k < N . Clearly, ρ N is in E G N G n ( ρ ), so we have A -rk( ρ ) ≤ k .On the other hand, suppose that A -rk( ρ ) = k . Then for all sufficiently large N ,there exists ρ N ∈ E G N G n ( ρ ) such that U -rk( ρ N ) = k < N . For m ≥ n , we know that E G N G n ( ρ ) = [ ρ ′ ∈E GmGn ( ρ ) E G N G m ( ρ ′ ) , i.e., ρ N ∈ E G N G m ( ρ ′ ) for some ρ ′ ∈ E G m G n ( ρ ). Then we have A -rk( ρ ′ ) ≤ k for some ρ ′ ∈ E G m G n ( ρ ). Thus the lemma is proved. (cid:3) According to [GH20] definition 6.4.1, the tensor rank of an irreducible character ρ ∈ E ( G ), denoted by ⊗ -rk( ρ ), is defined as follows:(1) Suppose that G is a symplectic group. An irreducible character ρ ∈ E ( G )has tensor rank less than or equal to k if ρ occurs in( ω O ǫ , G ⊗ ω O ǫ , G ⊗ · · · ⊗ ω O ǫk , G ) | G for ǫ i = ± .(2) Suppose that G is an orthogonal group. An irreducible character ρ ∈ E ( G )has tensor rank less than or equal to 2 k if ρχ occurs in ( ω Sp , G ) ⊗ k | G forsome linear character χ of G .(3) Suppose that G is a unitary group. An irreducible character ρ ∈ E ( G ) hastensor rank less than or equal to k if ρχ occurs in ( ω U , G ) ⊗ k | G for somelinear character χ of G .Here ω ⊗ k means the tensor product ω ⊗ · · · ⊗ ω of k copies of ω . Because( ω O ǫ , G ⊗ ω O ǫ , G ⊗ · · · ⊗ ω O ǫk , G ) | G = ( ω O ǫk , G ) | G ( ω Sp , G ) ⊗ k | G = ( ω Sp k , G ) | G ( ω U , G ) ⊗ k | G = ( ω U k , G ) | G for some ǫ = ± , we see that ⊗ -rk( ρ ) = Θ-rk( ρ ) for ρ ∈ E ( G ).4.2. η -correspondence and θ -correspondence. The following fundamental the-orem is from [GH20]:
Proposition 4.2 (Gurevich-Howe) . Let ( G ′ , G ) be a dual pair in stable range, i.e.,it is of one of the following types: (i) (Sp k , O +2 n ) or (Sp k , O − n +2 ) with k even and k ≤ n ; (ii) (O ǫk , Sp n ) with k ≤ n ; HETA RANKS 25 (iii) (U k , U n ) or (U k , U n +1 ) with k ≤ n .Then for ρ ′ ∈ E ( G ′ ) , there is a unique irreducible character η ( ρ ′ ) ∈ Θ G ( ρ ′ ) of U -rank k , and all other elements in Θ G ( ρ ′ ) have U -rank less than k . Furthermore,the mapping η : E ( G ′ ) → E ( G ) is injective. The mapping η : E ( G ′ ) → E ( G ) is called the η -correspondence for the dual pair( G ′ , G ) in stable range. In [Pan20a] and [Pan20b], a sub-relation of Θ, called the θ -correspondence , is defined as follows. Let ( G ′ , G ) be one of the following typesof dual pairs:(I) (O +2 n ′ , Sp n ) or (Sp n ′ , O +2 n )(II) (O − n ′ , Sp n ) or (Sp n ′ , O − n )(III) (U n ′ , U n ) or (U n ′ +1 , U n +1 )(IV) (U n ′ , U n +1 ) or (U n ′ +1 , U n )Let Λ ′ ∈ S G ′ and Λ ∈ S G such that (Λ ′ , Λ) ∈ B G ′ , G . Then we know that | Υ(Λ) | = | Υ(Λ ′ ) | + τ where τ is given by τ = n − n ′ + j def(Λ)2 k , for case (I); n − n ′ − j def(Λ)2 k , for case (II); n − n ′ + | def(Λ) | , for case (III); n − n ′ − | def(Λ) | − , for case (IV).For Λ ′ ∈ S G ′ n ′ such that τ ≥
0, we define θ (Υ(Λ ′ )) by θ (cid:18)(cid:20) µ , µ , . . . , µ m ν , ν , . . . , ν m (cid:21)(cid:19) = (cid:2) ν ,ν ,...,ν m µ ,µ ,...,µ m (cid:3) ∪ (cid:2) τ − (cid:3) , for cases (I),(III) ; (cid:2) ν ,ν ,...,ν m µ ,µ ,...,µ m (cid:3) ∪ (cid:2) − τ (cid:3) , for cases (II),(IV).Then we define θ (Λ ′ ) to be the symbol Λ such that Υ(Λ) = θ (Υ(Λ ′ )). Then weobtain a relation θ between E ( G ′ ) and E ( G ) by θ ( ρ Λ ′ ) = ρ θ (Λ ′ ) .Next we extend the domain of θ by letting it compatible with the Lusztig corre-spondence, i.e.,(1) for cases (I),(II),(III),(IV) above and ρ ′ ∈ E ( G ′ ) s ′ and ρ ∈ E ( G ) s , we havethe following commutative diagram: ρ ′ θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ id ⊗ θ −−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) (2) for dual pairs (O n ′ +1 , Sp n ) or (Sp n ′ , O n +1 ), we have the following com-mutative diagram ρ ′ θ −−−−→ ρ Ξ s ′ y y Ξ s ρ ′ (0) ⊗ ρ Λ ′ ( − ) ⊗ ρ Λ ′ (+) id ⊗ θ ⊗ id −−−−−→ ρ (0) ⊗ ρ Λ ( − ) ⊗ ρ Λ (+) We know that θ is a sub-correspondence of Θ, and we can also define the θ -rank of ρ ∈ E ( G ) by the same way as Θ-rank. It is know that(4.3) θ -rk( ρ ′ ) = Θ-rk( ρ ′ )for any ρ ∈ E ( G ′ ). Proposition 4.4.
Let ( G ′ , G ) be a reductive dual pair in stable range. Then η = θ .Proof. In [Pan20a], we show that for the dual pair ( G ′ , G ) = (O ǫk , Sp n ) and ρ ′ ∈E ( G ′ ), if ρ ∈ Θ G ( ρ ′ ) r { θ ( ρ ′ ) } , then the occurrence of ρ in Θ G ′ , G is not the firstoccurrence, and hence ρ = η ( ρ ′ ) by Proposition 4.2. Therefore we must have θ ( ρ ′ ) = η ( ρ ′ ). It is not difficult to see that the above argument holds for other dualpairs. (cid:3) Lemma 4.5.
Let ρ ∈ E ( G n ) . If Θ-rk( ρ ) ≤ n , then U -rk( ρ ) = Θ-rk( ρ ) .Proof. Suppose that Θ-rk( ρ ) = k ≤ n . So we have θ -rk( ρ ) = k ≤ n from Re-mark 3.2, i.e., ρ = θ ( ρ ′ ) for some ρ ′ ∈ E (O ǫk ( q )) for some ǫ = ± . Now the dual pair(O ǫk , Sp n ) is in stable range, so the two mappings η and θ coincides by Proposi-tion 4.4, i.e., ρ = η ( ρ ′ ) and hence U -rk( ρ ) = k by Proposition 4.2. (cid:3) Corollary 4.6.
We have A -rk( ρ ) ≤ Θ-rk( ρ ) for ρ ∈ E ( G n ) .Proof. Let ρ ∈ E ( G ) and suppose that Θ-rk( ρ ) = k . So now ρ ′ ⊗ ρ occurs inthe Θ-correspondence for some dual pair ( G ′ , G n ). Then we know that, for any N ≥ n , ρ ′ ⊗ ρ N occurs in the Θ-correspondence for the dual pair ( G ′ , G N ) for some ρ N ∈ E G N G n ( ρ ) such that Θ-rk( ρ N ) = k . When N ≥ k , by Lemma 4.5, we see that U -rk( ρ N ) = k . Then by definition, we conclude that A -rk( ρ ) ≤ U -rk( ρ N ) = k . (cid:3) Proof of Theorem 1.5.
Now we can summarize the relation between these variousranks as U -rk( ρ ) ≤ A -rk( ρ ) ≤ A -rk( ρ ) ≤ Θ-rk( ρ ) = ⊗ -rk( ρ )for any ρ ∈ E ( G ). Then Theorem 1.5 follows from Lemma 4.5 immediately. (cid:3) Agreement of A -ranks and Θ -ranks. It is proved by Howe-Gurevich in[GH20] theorem 11.4 that A -rk( ρ ) = ⊗ -rk( ρ ) for any ρ ∈ E (GL n ( q )). Now we wantto apply their method to other classical groups.Let G = G n = G ( V ) and V = X n ⊕ V ⊕ Y n be a complete polarization of V where V is anisotropic. Let { x , x , . . . , x n } be a basis of X n , and let { y , y , . . . , y n } bethe corresponding dual basis of Y n , i.e., h x i , y j i = δ ij where δ ij is the Kroneckerdelta symbol. For 1 ≤ k ≤ n , let X k (resp. Y k ) denote the space spanned by { x , x , . . . , x k } (resp. { y , y , . . . , y k } ). So we have the decomposition V = X k ⊕ V n − k ⊕ Y k where V n − k is a space of Witt index n − k . Define H k = { g ∈ G ( V ) | ( g − X k ) = 0 , ( g − Y k ) = 0 } . Then H k is a subgroup of G ( V ), H k ⊂ H k − for each k , and H = G n . HETA RANKS 27
Lemma 4.7.
Let ρ ∈ E ( G n ) be of low U -rank, i.e., U -rk( ρ ) = k < n, if G n is Sp n , U n or U n +1 ; n − , if G n is O +2 n , O − n +2 or O n +1 . Then ρ | H k contains a linear character.Proof. Suppose that ρ ∈ E ( G n ) is of low U -rank. Define the following unipotentsubgroups of G ( V ): U = { g ∈ G ( V ) | ( g − X k ) = 0 , ( g − V n − k ⊕ Y k ) ⊂ X k } ,U = { g ∈ G ( V ) | ( g − X k ⊕ V n − k ) = 0 , ( g − Y k ) ⊂ X k } ,U = U ∩ H k . It is not difficult to see that U ≃ U × U , and U is normalized by H k . Define thesubgroup U of G ( V ) as follows:(1) If G ( V ) is a symplectic group, then k < n , and let U be the subgroup ofelement g ∈ G ( V ) such that • ( g − y k +1 = ax k +1 , a ∈ f q , and • ( g − X k +1 ⊕ V n − k − ⊕ Y k ) = 0.(2) If G ( V ) is an orthogonal group, then k < n −
1, and let U be the subgroupof element g ∈ G ( V ) such that • ( g − y k +2 = ax k +1 and ( g − y k +1 = − ax k +2 , a ∈ f q , and • ( g − X k +2 ⊕ V n − k − ⊕ Y k ) = 0.(3) If G ( V ) is a unitary group, then k < n , and let U be the subgroup ofelement g ∈ G ( V ) such that • ( g − y k +1 = ax k +1 , a ∈ f q , a = − a , and • ( g − X k +1 ⊕ V n − k − ⊕ Y k ) = 0.Here a a is the nontrivial automorphism of f q over f q .Finally we define the subgroup U of G ( V ) as follows:(1) If G ( V ) is a symplectic group or a unitary group, let U be the subgroupof element g ∈ G ( V ) such that • ( g − X k +1 ⊕ V n − k − ) = 0 and • ( g − Y k +1 ) ⊂ X k +1 .(2) If G ( V ) is an orthogonal group, let U be the subgroup of element g ∈ G ( V )such that • ( g − X k +2 ⊕ V n − k − ) = 0 and • ( g − Y k +2 ) ⊂ X k +2 .There exists a linear character ψ of U of rank k and trivial on U , so ψ | U stillhas rank k . It is easy to see that h ( u , u ) h − = ( u , u ′ ) for h ∈ H k , u ∈ U and u ∈ U . Because now ψ is trivial on U , ψ is invariant under the conjugation by H k . Because now U -rk( ρ ) = k , ρ | U contains a nontrivial ψ -eigenspace, denoted by ρ ψ . Because ψ is invariant under the conjugation of H k , the space ρ ψ is invariantunder the action of ρ | H k . Let σ denote the representation of H k on the space ρ ψ .Now we claim that the representation σ of H k is trivial on the commutatorsubgroup of H k . Suppose that the claim is not true, i.e., σ is not trivial on thecommutator subgroup of H k Then σ is not trivial on U , because the normal sub-group of H k generated by U is the full commutator subgroup of H k . The space ρ ψ is decomposed into eigenspaces for U . Let ρ ψ be the eigenspace correspondingto a non-trivial linear character λ of U . The group U is commutative and con-tains both U and U . So the space ρ ψ is decomposed into eigenspaces of U . Thecharacter of U which corresponding a nontrivial eigenspace in ρ ψ of U has rankgreater than k and we get a contradiction.Now the representation σ of H k is trivial on the commutator subgroup of H k ,we see that σ is a sum of linear characters of H k . (cid:3) Recall that under the
Shr¨odinger model of the Weil representation ω , the group G ( V ) acts on the space L (Hom( X k , V )) of functions on Hom( X k , V ) by( ω ( g ) f )( T ) = χ k ( g ) f ( g − ◦ T )for g ∈ G ( V ) and T ∈ Hom( X k , V ) where χ k is a linear character of G ( V ) of orderat most 2. Recall that L ( G/H k , χ − k ) denotes the space of functions f on G suchthat f ( gh ) = χ − k ( h ) f ( g ) for g ∈ G and h ∈ H k . Then G ( V ) acts on the space L ( G/H k , χ k ) by ( π ( g ) f )( x ) = f ( g − x )for g, x ∈ G ( V ).Fix an element T ∈ Hom( X k , V ) such that the image of T is equal to X k . Notethat for h ∈ H k , we have h ◦ T = T from the definition of H k . Then it is notdifficult to see that the mapping φ : L (Hom( X k , V )) → L ( G/H k , χ k )given by φ ( f )( x ) = χ k ( x − ) f ( x ◦ T ) is a surjective linear transformation. Moreover,we have π ( g ′ )( φ ( f ))( g ) = φ ( f )( g ′− g ) = χ k ( g − g ′ ) f ( g ′− g ◦ T ) φ ( ω ( g ′ ) f )( g ) = χ k ( g − )( ω ( g ′ ) f )( g ◦ T ) = χ k ( g − g ′ ) f ( g ′− g ◦ T ) , i.e., φ commutes the two actions ω and π of G . Therefore π can be regarded as asubrepresentation of ω . Lemma 4.8.
Let ρ ∈ E ( G n ) and ≤ k ≤ n . If ρ | H k contains a linear character,then Θ-rk( ρ ) ≤ k .Proof. Suppose that ρ ∈ E ( G n ) and ρ | H k contains a linear character. Since alinear character of H k is the restriction of a linear character of G n , after twisting alinear character of G n , we may assume that ρ | H k contains the character χ − k . By HETA RANKS 29
Frobenius reciprocity, ρ is contained in L ( G n /H k , χ − k ) under the action π of lefttranslation of G n . Then by the above discussion, we see that ρ is contained in theSch¨ondinger model L (Hom( X k , V )) of the Weil representation ω . This means that ρ occurs in the Θ-correspondence for the dual pair ( G ′ , G ) where G ′ is the groupacting on a space V ′ which has a Witt decomposition with a trivial anisotropickernel and a maximal totally isotropic space isomorphic to X k , i.e., dim( V ′ ) = 2 k .This concludes that Θ-rk( ρ ) ≤ k . (cid:3) Now we have an analogous result of [GH20] proposition 11.2.
Corollary 4.9. If ρ ∈ E ( G n ) is of U -rank k ≤ n , then U -rk( ρ ) = Θ-rk( ρ ) .Proof. Suppose that U -rk( ρ ) = k ≤ n . Then by Lemma 4.7 and Lemma 4.8 we haveΘ-rk( ρ ) ≤ k ≤ n . Then by Lemma 4.5, we conclude that Θ-rk( ρ ) = U -rk( ρ ) = k . (cid:3) Theorem 4.10.
Let ρ ∈ E ( G ) where G is a symplectic group, an orthogonal group,or a unitary group. Then A -rk( ρ ) = Θ-rk( ρ ) .Proof. Let ρ ∈ E ( G n ) where G n is Sp n , O +2 n , O − n +2 , O n +1 , U n , or U n +1 . Sup-pose that A -rk( ρ ) = k and Θ-rk( ρ ) = k . We know that k ≤ k by Corollary 4.6and k ≤ n + 2 from the definition of Θ-ranks. Then by definition we have(4.11) k = min { U -rk( ρ N ) | ρ N ∈ E G N G n ( ρχ ) , χ linear , N >> n } , and by Lemma 3.23 we know that(4.12) k = min { Θ-rk( ρ N ) | ρ N ∈ E G N G n ( ρχ ) , χ linear , N >> n } . Let N be sufficiently larger than n , in particular, we assume that N ≥ n + 4.Then by (4.11) there is ρ ∈ E G N G n ( ρχ ) for some linear character χ of G n such that U -rk( ρ ) = k ≤ n + 2 ≤ N . Then by Lemma 4.8, we have Θ-rk( ρ ) ≤ k ≤ N .Then by Lemma 4.5, we have Θ-rk( ρ ) = U -rk( ρ ) = k . Then (4.12) implies that k ≤ k .Because now N is sufficiently large, by (4.12) there is a ρ ∈ E G N G n ( ρχ ) for somelinear character χ of G n such that Θ-rk( ρ ) = k < N . Then Lemma 4.5 impliesthat U -rk( ρ ) = k and we have k ≤ k by (4.11). (cid:3) References [AM93] J. Adams and A. Moy,
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Department of Mathematics, National Tsing Hua University, Hsinchu 300, Taiwan
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