A Partial Order on Bipartitions From the Generalized Springer Correspondence
aa r X i v : . [ m a t h . R T ] F e b A Partial Order on Bipartitions From the Generalized SpringerCorrespondence
Jianqiao XiaMarch 2, 2018
Abstract
In [1], Lusztig gives an explicit formula for the bijection between the set of bipartitions and theset N of unipotent classes in a spin group which carry irreducible local systems equivariant for thespin group but not equivariant for the special orthogonal group. The set N has a natural partialorder and therefore induces a partial order on bipartitions. We use the explicit formula given in[1] to prove that this partial order on bipartitions is the same as the dominance order appeared inDipper-James-Murphy’s work ([2]). For group G = Spin n ( k ), where k is a field of characteristic not equal to 2, let N be the set of unipotentclasses in G which carry irreducible local systems, equivariant for the conjugation action of G , butnot equivariant for the conjugation action of the special orthogonal group. Then N has a one-to-one correspondence with a certain set of partitions X n (see [1], section 14). X n consists of partitions λ = ( λ ≥ λ ≥ · · · λ m ) of n , such that each λ i ∈ N + and1. for each integer n ∈ Z + 1, the set { i ; λ i = n } has at most one element;2. for each integer n ∈ Z , the set { i ; λ i = n } has even number of elements.Let Irr W s be the set of all bipartitions of s . Then the Generalized Springer Correspondence for Spingroup gives a bijection X n ←→ G t ∈ Z + n Irr W ( n − t + t ) . (1)In [1], Lusztig gives an explicit formula for this bijection. Specifically, let λ = ( λ ≥ · · · ≥ λ m ) ∈ X n .Define t i = X j ≥ i +1 d ( λ j ) . (2)and t = X j ≥ d ( λ j ) . (3)Here d ( λ j ) = ( λ j is even.( − λj ( λj − If λ j is odd. (4)Then the image of λ under the bijection can be constructed in the following way:1. If λ i ∈ Z + 1, then lable this entry by a , and replace this entry by ( λ i − − t i .2. If λ i ∈ Z + 3, then lable this entry by b , and replace this entry by ( λ i −
3) + t i .3. If λ i = e ∈ Z + 2, then by definition it appears 2 p times. Replace these entries by14 ( e −
2) + t i ,
14 ( e + 2) − t i , · · · ,
14 ( e + 2) − t i (5)respectively, and label them as b, a, b, · · · , a, b, a .1. If λ i = e ∈ Z , then by definition it appears 2 p times. Replace these entries by14 e + t i , e − t i , · · · , e − t i (6)respectively. Label them as b, a, b, · · · , a, b, a .The modified entries with lable a form an decreasing sequence α . The entries with lable b form andecreasing sequence β . If t >
0, then λ corresponds to ( α, β ) in the bijection. If t ≤
0, then λ corresponds to ( β, α ). Moreover, the bipartion ( α, β ) (when t ≥
0) or ( β, α ) (when t ≤
0) is an elementin Irr W ( n − t + t ) . Remark.
In Lusztig’s paper [1], he gives the formula for partitions in increasing order. Here I simplytranslated everything in decreasing order, for convenience of the following proof. Moreover, for a partitionin decreasing order, we can view it as an infinite sequence, by adding 0’s.There is a natural partial order on N : c ≤ c ′ if c is contained in the closure of c ′ . This partial orderis given below, in terms of elements in X n : Definition 1.1.
For λ, µ ∈ X n and each is in decreasing order. We say λ ≤ µ if and only if for all i ∈ N X j ≤ i λ j ≤ X j ≤ i µ j . (7)From the bijection (1), we have an induced partial order on the set of bipartions Irr W m , for each t . This partial order is closely related to that found in Dipper-James-Murphy’s paper ([2]), and alsoappears in Geck and Iancu’s paper ([3]) as the aymptotic case for their pre-order relation on Irr W ,indexed by two parameters a, b . In the aymptotic case b > ( n − a , their pre-order is a partial order,and is defined by Definition 1.2. (Dipper-James-Murphy) The dominance order between ( λ, µ ) , ( λ ′ , µ ′ ) ∈ Irr W , each indecreasing order, is ( λ, µ ) ≤ ( λ ′ , µ ′ ) ⇔ (P j ≤ k λ j ≤ P j ≤ k λ ′ j for all k | λ | + P j ≤ k µ j ≤ | λ ′ | + P j ≤ k µ ′ j for all k. (8)The main result of this paper is Theorem 1.
For t ≥ m , the induced partial order on Irr W m from the inclusion Irr W m ֒ → X t − t +4 m ,is the dominance order. Let f m,t : Irr W m ֒ → X t − t +4 m be the inclusion from the Generalized Springer Correspondence. Wefirst make the following observation: Lemma 1. If t ≥ m , and λ ∈ f m,t (Irr W m ), then λ i ∈ Z ∪ (4 Z + 1). Proof.
Suppose on the contrary there is an i such that λ i ∈ Z + 3. By definition, t = P i d ( λ i ). Each λ i ∈ Z + 1 contributes +1, and each λ i ∈ Z + 3 contributes −
1. By definition of X n , each odd integerappears at most once. So t = |{ i ; λ i ∈ Z + 1 }| − |{ i ; λ i ∈ Z + 3 }| . (9)And then |{ i ; λ i ∈ Z + 1 }| ≥ t + 1. So2 t − t + 4 m = | λ | = X i λ i ≥ X i,λ i ∈ Z +1 λ i ≥ t X j =0 (4 j + 1)= 2 t + 3 t + 1 ≥ t − t + 4 m + 1 . (10)2his is a contradiction! This lemma also proves that there are exactly t odd integers in λ , each is in4 Z + 1.Now the picture is clear for t ≥ m . In fact, if ( α, β ) corresponds to λ , then α represents the deviationof odd integers of λ from (4 t − , t − , · · · , β is the even integers of λ , up to scalar. We havethe following lemma: Lemma 2.
Suppose t ≥ m , and ( α, β ) ∈ Irr W m corresponds to λ ∈ X t − t +4 m . Let α ′ = ( α ′ , · · · , α ′ t )be the decreasing sequence of odd integers in λ , and β ′ = ( β ′ , · · · , β ′ k ) be the decreasing sequence ofeven integers of λ . Then α i = 14 ( α ′ i − (4( t − i ) + 1)) (11)and β i = 12 β ′ i . (12) Proof.
Suppose β ′ s = e ∈ Z , and 4 l − < e < l + 1. We prove that 1 , , , · · · , l − α ′ . Otherwise, there are at least t − l + 1 odd integers greater than 4 l −
3, and then2 t − t + 4 m = | λ | ≥ | α ′ | + 2 e ≥ · · · + (4 l −
7) + (4 l + 1) + · · · + (4 t + 1) + 2 · (4 l − t + 3 t + 4 l. (13)So t ≤ m − l. (14)This is only possible when l = 0.Now suppose e = β ′ = β ′ = λ k and l the same as above. Then t k = l . If l = 0, then the lemma isautomatically true. Otherwise, from Lusztig’s formula, suppose there are 2 p such e in λ . There are twocases:1. e = 4 l −
2. Then those 2 p numbers are replaced by ( e −
2) + l, ( e + 2) − l, · · · alternatively. Theseare exactly 2 l − , , l − · · · ,
0. So β = 2 l − β ′ .2. e = 4 l . Similar as above, β = e + l = 2 l = β ′ .In either cases, β i = β ′ i .For α , notice that the above calculation shows that all a labels from even integers gives modifiednumber 0. Since α is in decreasing order, we only need to consider a lables from odd integers. For a lables from elements λ i ≤ β , we replaced it by ( λ i − − t i . Notice that the odd integers below β areexactly 1 , , , · · · , l −
3. So λ i is exactly the ( t i + 1)-th odd integer. They contributes to 0 in α .For λ i ≥ β , there are exactly i − λ i . So λ i = α ′ i and there are t − i oddintegers λ j with index j greater than i . By definition, t i = t − i . Therefore, α i = 14 ( λ i − − t i = 14 ( α ′ i − t − i ) − . (15)Now we use the above observation to prove the main theorem. Let ( α, β ),( α ′ , β ′ ) be bipartitions withorder m . They correspond to λ, λ ′ from the inclusion f m,t : Irr W m ֒ → X t − t +4 m . Here t ≥ m is afixed integer. Proof of main theorem: (a) If ( α, β ) ≥ ( α ′ , β ′ ) in the dominance order, then λ ≥ λ ′ . Proof.
Suppose λ = ( λ , λ , · · · ) in decreasing order, and λ ′ = ( λ ′ , λ ′ , · · · ) also in decreasing order.3otice that 2 t − t + 4 m = X i λ i = X λ i odd λ i + X λ i even λ i ≥ t − X i =0 (4 i + 1) + X λ i even λ i = 2 t − t + X λ i even λ i . (16)Since even numbers appear in pairs, we conclude that λ i ≤ m , for even entries.If there is an s such that 4 s + 1 does not appear in λ , suppose s is the smallest one. Then2 t − t + 4 m ≥ t − X i =0 (4 i + 1) − (4 s + 1) + (4 t + 1) = 2 t − t + 4( t − s ) . (17)So s ≥ t − m , and 4 s + 1 > m . This means the part that contributes to α and the part that contributesto β are separated. This separation is independent of λ . In particular, we know λ m = λ ′ m = 4( t − m ) + 1,and the odd integers after them form an arithemetic sequence with common difference 4.Therefore, for k ≤ m −
1, according to lemma 2, the conditions λ + λ + · · · + λ k ≥ λ ′ + · · · + λ ′ k (18)is equilvalent to α + · · · + α k ≥ α ′ + · · · + α ′ k (19)If λ ≥ λ ′ does not hold, then there is a smallest integer k , such that λ + · · · + λ k < λ ′ + · · · + λ ′ k . (20)And we know from definition of k that λ k < λ ′ k and k > m . Suppose the remaining odd integers of λ are 1 , , , · · · , u + 1, and the remaining odd integers for λ ′ are 1 , , , · · · , u ′ + 1. Then 4 u + 2 ≤ λ k <λ ′ k ≤ u ′ + 5. So u ≤ u ′ . There are two cases.(1) u ′ − u is even. The number of even integers appeared in { λ , · · · , λ k } is k − ( t − ( u +1)) = k − t + u +1.By assumption, k − t + u + 1, k − t + u ′ + 1 are both odd integers or both even integers. If they areboth odd, then we consider k ′ = k + 1. In this case, λ k +1 = λ k . So by adding one term, we still have λ + λ + · · · + λ k ′ ≤ λ ′ + · · · + λ ′ k ′ . (21)So we will only deal with the case k − t + u + 1 , k − t + u ′ + 1 both even. In this case, since | λ | = | λ ′ | ,we have X i>k λ i > X i>k λ ′ i . (22)So from lemma 2, u X i =0 (4 i + 1) + 4( m − | α | − β − · · · − β k − t + u +12 ) > u ′ X i =0 (4 i + 1) + 4( m − | α ′ | − β ′ − · · · − β ′ k − t + u ′ +12 ) . (23)Let S ( x ) = | α | + β + · · · + β x . Similarly define S ′ . Then the above can be written as0 > ( u ′ − u )(2 u + 2 u ′ + 3) + 4 S ( k − t + u + 12 ) − S ′ ( k − t + u ′ + 12 ) . (24)4otice that S ( k − t + u + 12 ) − S ′ ( k − t + u ′ + 12 ) = S ( k − t + u ′ + 12 ) − S ′ ( k − t + u ′ + 12 ) − k − t + u ′ +12 X i = k − t + u +12 +1 β i ≥ − k − t + u ′ +12 X i = k − t + u +12 +1 β i ≥ − u ′ − u u + 42 . (25)The last inequality is from the definition of u , which implies 4 u + 5 is some λ i , i ≤ k . So all the eveninteger after that must be less than or equal to 4 u + 4. Therefore the corresponding β is less thanor equal to (4 u + 4) /
2. Combine these inequality, we get0 > ( u ′ − u )(2 u + 2 u ′ + 3) − ( u ′ − u )(4 u + 4) = ( u ′ − u )(2( u ′ − u ) − ≥ . (26)This is a contradiction, since u ′ − u is an integer.(2) u ′ − u is odd. Then u ′ − u ≥
1. Let A ( x ) = λ + · · · + λ x , and similarly define A ′ ( x ). Then B ( k ) := A ( k ) − A ′ ( k ) <
0. Consider B ( k + 1). Since λ k +1 ≤ u + 4 < u ′ + 1 ≤ λ ′ k , so we still have B ( k + 1) < u ′ ≤ u , at some point. Notice that u ′ − u changes by at most 1 ateach step. So the first time this process ends is when u ′ = u . It must stop at some point, since bothsequences λ, λ ′ are eventually 0, and B ( x ) = 0 for large x . Now we can apply same method in case(1), or simply notice that in the case u ′ = u , we must have B ( x ) = A ( x ) − A ( x ) ′ = 4( | α | + β + · · · + β k − t + u + l ) − | α ′ | + β ′ + · · · + β ′ k − t + u + l ) ≥ . (27)This is a contradiction.(b) Suppose λ ≥ λ ′ , then ( α, β ) ≥ ( α ′ , β ′ ). Clearly α ≥ α ′ from lemma 2 and the discussion at thebeginning of the proof above. So if ( α, β ) ≥ ( α ′ , β ′ ) does not hold, then there is a smallest k , such that | α | + β + · · · + β k < | α ′ | + β ′ + · · · + β ′ k . (28)We still use the notation A ( x ) for the sum of first x terms of λ , and S ( x ) for the sum of first x termsof β and | α | . So S ( k ) < S ( k ) ′ , k ≥
1. By assumption of k , β k < β ′ k . If β k = 0, then it is automaticallya contradiction, since the left side is then m = | α ′ | + | β ′ | . So 0 < β k < β ′ k . They both comes fromsome even integers 2 β k , β ′ k in the corresponding partition. Suppose they correspond to λ x − , λ x and λ ′ x ′ − , λ ′ x ′ , respectively. Suppose 4 u + 5 > β k > u + 1 and 4 u ′ + 5 > β ′ k > u ′ + 1. Then u ′ ≥ u . So x = 2 k + t − u − ≥ x ′ = 2 k + t − u ′ −
1. Now A ( x ) − A ′ ( x ′ ) = A ( x ′ ) − A ′ ( x ) + x X i = x ′ +1 λ ′ i ≥ x X i = x ′ +1 λ ′ i . (29)Also notice that A ( x ) − A ′ ( x ′ ) = u ′ X i = u +1 (4 i + 1) + 4( S ( k ) − S ′ ( k )) < u ′ X i = u +1 (4 i + 1) . (30)5his means u ′ X i = u +1 (4 i + 1) > x X i = x ′ +1 λ ′ i . (31)However, this is a contradiction, since { u + 5 , u + 9 , · · · , u ′ + 1 } ⊂ { λ x ′ +1 , λ x ′ +2 , · · · } , and x − x ′ = u ′ − u .We now give an example that violates the above partial order for t = m −
1. The partition λ =(4 t + 1 , t − , · · · , , , ,
1) corresponds to ( α, β ), where α = (1 , , · · · ,
1) ( t “1”s.) and β = (1).The partition λ ′ = (4 t + 1 , t − , · · · , , , ,
2) corresponds to ( α ′ , β ′ ), where α ′ = (1 , , · · · , , t + 1 “1”s.) and β ′ = (0).Then λ > λ ′ , but ( α, β ) < ( α ′ , β ′ ). Acknowledgement
The author would like to thank prof. George Lusztig for suggesting this problem and providing usefulcomments and directions. This project is funded by the Undergraduate Research Opportunities Programof MIT.
References [1] George Lusztig,
Character sheaves on disconnected groups, II , Representation Theory (2004), no. 4, 72–124.[2] Richard Dipper and Gordon James and Eugene Murphy, Hecke Algebras of Type Bn at Roots of Unity , Proceedings ofthe London Mathematical Society s3-70 (1995), no. 3, 505–528.[3] Meinolf Geck and Lacrimioara Iancu,
Ordering Lusztig’s families in type B , Journal of Algebraic Combinatorics (2013), no. 2, 457–489.(2013), no. 2, 457–489.