PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 5, May 2008, Pages 1515–1522 S 0002-9939(08)09234-4 Article electronically published on January 9, 2008

ON THE SUM OF THE INDEX OF A PARABOLIC SUBALGEBRA AND OF ITS NILPOTENT RADICAL

RUPERT W. T. YU

(Communicated by Dan M. Barbasch)

Abstract. In this short note, we investigate the following question of Panyu- shev stated in 2003: “Is the sum of the index of a parabolic subalgebra of a semisimple g and the index of its nilpotent radical always greater than or equal to the rank of g?” Using the formula for the index of parabolic subalgebras conjectured by Tauvel and the author and proved by Fauquant- Millet and Joseph in 2005 and Joseph in 2006, we give a positive answer to this question. Moreover, we also obtain a necessary and sufficient condition for this sum to be equal to the rank of g. This provides new examples of direct sum decomposition of a verifying the “index additivity condition” as stated by Ra¨ıs.

1. Introduction Let g be a Lie algebra over an algebraically closed field k of characteristic zero. For f ∈ g∗,wedenotebygf = {X ∈ g; f([X, Y ]) = 0 for all Y ∈ g}, the annihilator of f for the of g.Theindex of g, denoted by χ(g), is defined to be χ(g)=mindim gf . f∈g∗ It is well known that if g is an algebraic Lie algebra and G denotes its algebraic adjoint group, then χ(g) is the transcendence degree of the field of G-invariant rational functions on g∗. The index of a semisimple Lie algebra g is equal to the rank of g.Thiscan be obtained easily from the isomorphism between g and g∗ via the . There has been quite a lot of recent work on the determination of the index of certain subalgebras of a semisimple Lie algebra: parabolic subalgebras and related subalgebras ([2], [9], [13], [8]), centralizers of elements and related subalgebras ([10], [1], [15], [7]). Let g be a semisimple Lie algebra, p a parabolic subalgebra of g and u (resp. l) the nilpotent radical (resp. a Levi factor) of p. In [10, Corollary 1.5 (i)], Panyushev showed that (1) χ(p)+χ(u) ≤ dim l.

Received by the editors August 18, 2006. 2000 Mathematics Subject Classification. Primary 17B20.

c 2008 American Mathematical Society Reverts to public domain 28 years from publication 1515

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He then suggested [10, Remark (ii) of Section 6] that (2) χ(p)+χ(u) ≥ rk g. For example, it is well known that if b is a Borel subalgebra of g and n is its nilpotent radical, then χ(b)+χ(n)=rkg (see for example [12], [14, Chapter 40]). It is therefore also interesting to characterise parabolic subalgebras where equality holds in (2). Indeed, in [11], Ra¨ıs looked for examples of direct sum decompositions g = m ⊕ n verifying the “index additivity condition”, namely m and n are Lie subalgebras of g and χ(g)=χ(m)+χ(n).

If u− denotes the nilpotent radical of the opposite parabolic subalgebra p− of p,theng = p ⊕ u− and the Lie algebras u and u− are isomorphic. Thus parabolic subalgebras such that equality holds in (2) would provide examples of direct sum decompositions verifying the index additivity condition. Using the formula conjectured in [13] and proved in [3, 6] for the index of par- abolic subalgebras, we obtain a formula for the sum χ(p)+χ(u). By a careful analysis of root systems, we prove inequality (2) and give a necessary and sufficient condition for the equality to hold in (2) (see Theorem 2.2). To describe the index of a parabolic subalgebra and the index of its nilpotent radical, we need to recall Kostant’s cascade construction of pairwise strongly or- thogonal roots ([4], [5], [14]). Let us fix a h of g and a Borel subalgebra b of g containing + h.DenotebyR, R and Π = {α1,...,α} respectively the set of roots, positive roots and simple roots with respect to h and b. For any α ∈ R,letgα be the root subspace associated to α.ChooseXα ∈ gα such that α([Xα,X−α]) = 2. We shall ∨ ∗ ∨ ∨ write α =[Xα,X−α] ∈ h,andforλ ∈ h , λ, α  = λ(α ). For S ⊂ Π, we denote ∩ Z + ∩ + by RS = R S, RS = RS R .IfS is connected, then we shall denote by εS the highest root of RS. Let S ⊂ Π. We define inductively a set K(S) whose elements are subsets of Π as follows: a) K(∅)=∅. b) If S1,...,Sr are the connected components of S,thenK(S)=K(S1) ∪···∪ K(Sr). K { }∪K { ∈  ∨ } c) If S is connected, then (S)= S (S)whereS = α S; α, εS =0 . It is well known that (see for example [14, Chapter 40]) elements of K(S)are connected subsets of S. Moreover, if we denote by R(S)={εK ; K ∈K(S)},then R(S) is a maximal set of pairwise strongly orthogonal roots in RS. Let us also recall the following properties of K(S) which are easy consequences from the definition (see for example [14, Chapter 40]): Lemma 1.1. Let S be a subset of Π, K, K ∈K(S) and set K { ∈  ∨  } Γ = α RK ; α, εK > 0 { ∈ + ∈ \ } = α = β∈K nββ RK ; nβ > 0 for some β K K . i) We have either K ⊂ K or K ⊂ K or K and K are connected components of K ∪ K. K + \{ ∈ +  ∨  } + ii) Γ = RK β RK ; β,εK =0 .Inparticular,RS is the disjoint union of the ΓK ’s, K ∈K(S).

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SUM OF THE INDEX 1517 Table 1

Type A ,≥ 1 B ,C ,≥ 2 D ,≥ 4 E E E F G 6 7 8 4 2  +1  K(Π)  2 4 7 8 4 2 2 2

∈ K iii) α∈ΓK gα is a Heisenberg Lie algebra whose centre is gεK .Thusifα, β Γ verify α + β ∈ R,thenα + β = εK . iv) Suppose that α ∈ ΓK and β ∈ ΓK verify α + β ∈ R, then either K ⊂ K and α + β ∈ ΓK or K ⊂ K and α + β ∈ ΓK . The cardinality of K(Π) is listed in Table 1 for an irreducible R, where for any x ∈ Q,[x] is a unique integer such that [x] ≤ x<[x]+1. Examples 1.2. Let R be an irreducible root system. We shall use the numbering K of simple roots in [14, Chapter 18]. Set k = (Π). K k { } { } 1) Let R be of type A.Then (Π) = i=1 Ki where Ki = αi,...,α+1−i . For 1 ≤ i ≤ k, Ki { ··· ··· ≤ ≤ − }∪{ } Γ = αi + + αi+r,α+1−i−r + + α+1−i;0 r  2i εKi .

2) Let R be of type D2n+1.Thenk =2n, n K(Π) = {Ki}∪{Li}, i=1

where Ki = {α2i−1,...,α2n+1} and Li = {α2i−1}.For1≤ i ≤ n, Ki Li Γ = mjαj; m2i =0 , Γ = {α2i−1}. j=2i−1

3) Let R be of type .Then

K(Π) = {Π}∪{{α1,α3,α4,α5,α6}} ∪ {{α3,α4,α5}} ∪ {{α4}}. 2. Main result Recall that for any subset S ⊂ Π, pS = h ⊕ gα + α∈RS ∪R is a (standard) parabolic subalgebra of g. Any parabolic subalgebra of g is conju- gated to a standard parabolic subalgebra. The Lie subalgebra uS = gα + α∈R \RS

is the nilpotent radical of pS. ∗ For S ⊂ Π, denote by VS the vector subspace of h spanned by the elements of R(S)andR(Π). Set E { ∈K ∈ } { ∈K ∈ } S = K (Π); XεK uS = K (Π); εK RS and Q = ΓK ∩ R+. S K∈ES S

The subset ES has the following simple characterisation.

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Lemma 2.1. Let TS be the union of the subsets K ∈K(Π) verifying K ⊂ S.Then ES = K(Π) \K(TS). Proof. This is straightforward. 

Our main result is the following theorem. Theorem 2.2. Let S ⊂ Π.Then

(3) χ(pS)+χ(uS)=rkg + K(S) − K(TS)+2(K(Π) − dim VS)+QS.

We have χ(pS)+χ(uS) ≥ rk g, and the equality holds if and only if the following conditions are satisfied:

i) (K(S) ∪K(Π)) = dim VS.    ii) For any connected component S of S, we have either S ∈K(Π) or (S \TS)= 1. Proof. We may clearly assume that g is simple. The formula for the sum χ(pS)+χ(uS) is a direct consequence of the formula of the index of parabolic subalgebras conjectured in [13] and proved in [3, 6]:

(4) χ(pS)=rkg + K(Π) + K(S) − 2dimVS,

and the formula for the index of uS (see for example [14, Chapter 40]), which, in view of Lemma 1.1, can be expressed in the following way: K (5) χ(uS)=ES + Γ − dim uS = ES + QS.

K∈ES Observe that

(6) K(S) − K(TS) ≥ 0.

1) Let S1,...,Sr be the connected components of S.Foreachi, there is a unique ∈K ∈ Ki Ki (Π) (see Lemma 1.1) such that εSi Γ .IfKi = Si,thenSi is a connected component of TS .OtherwiseKi ∈ES, and we have

∈ Ki ∩ Si ⊂Q εSi Γ Γ S.

2) It follows from Point 1) that QS = ∅ if and only if K(S) ⊂K(Π) (or equiva- lently S = TS). 3) Note that the connected components of TS are the connected components of TS ∩ Si. It follows again from Point 1) that r K(S) − K(TS)= (K(Si) − K(TS ∩ Si)) i=1 = (K(Si) − K(TS ∩ Si)).

Ki∈ES

4) From Table 1, we have dim VS = K(Π) = rk g in the cases where g is of type B, C, D2n, , , and . The inequality follows immediately from (3) and (6), and by Point 2) the equality holds if and only if K(S) ⊂K(Π). On the other hand, since dim VS = K(Π) in these cases, condition i) is equivalent to K(S) ⊂K(Π). Finally, if condition i) is verified, then S = TS, and condition ii) is automatically verified. So we have the result in these cases. 5) Type A.

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For any i verifying Si = Ki, by Point 1), Lemma 1.1 and Examples 1.2, half of Si \{ } Q Si − Γ εSi belongs to S.Since(Γ )=2(Si) 1 (Examples 1.2), such an Si contributes (Si)elementsofQS. Again, since we are in type A, K(Π) is totally ordered by inclusion. It follows that TS is connected. Without loss of generality, we may assume that TS ⊂ S1. Suppose that S1 = TS. Then from the previous discussion, we deduce that

χ(pS)+χ(uS) ≥ rk g + K(S \ S1)+2(K(Π) − dim VS)+(S \ S1). But our hypothesis implies that

(7) dim VS ≤ K(Π) + K(S \ S1)=(K(Π) ∪K(S)), so χ(pS)+χ(uS) ≥ rk g + (S \ S1) − K(S \ S1).

Hence χ(pS)+χ(uS) ≥ rk g. For the equality to hold, we must have the equality in (7) and (S \ S1)=K(S \ S1).

This latter is only possible if (Si)=1fori ≥ 2, so we have conditions (i) and (ii). Conversely, suppose that conditions (i) and (ii) are verified; then (Si)=1for i ≥ 2. Consequently QS = S \ Si by Point 1) and the definition of QS. Suppose that S1 TS (this includes the case TS = ∅). Then K(S) ∩K(Π) = ∅. Thus

(8) dim VS ≤ K(Π) + K(S)=(K(Π) ∪K(S)). We deduce from Point 1) and the remark in the first paragraph of Point 5) that

χ(pS)+χ(uS) ≥ rk g + K(S) − K(TS)+2(K(Π) − dim VS)+(S) ≥ rk g + (S) − K(S) − K(TS). Hence r (S )+1 (T )+1 χ(p )+χ(u ) ≥ rk g + (S ) − i − S . S S i 2 2 i=1 Since T S , we deduce from Table 1 that S 1 (S )+1 (T )+1 (S ) − 1 − S ≥ 0. 1 2 2

So we have our inequality χ(pS)+χ(uS) ≥ rk g. Now for the equality χ(pS)+χ(uS)=rkg to hold, we must have the equality in (8) and Q = (S), S (S )+1 (T )+1 (S )+1 (S ) − 1 − S =0and(S ) − i =0 1 2 2 i 2

for i ≥ 2. This implies that (S1)=(TS) + 1, and (Si)=1fori ≥ 2. So we have conditions (i) and (ii). Conversely, if conditions (i) and (ii) are verified, then (Si)=1fori ≥ 2, and (S1)=(TS) + 1. In view of the above, to show that χ(p )+χ(u )=rkg, it suffices to check that (Q ∩ R+ )=(S ), which is a S S S S1 1 straightforward verification. 6) Type D2n+1. In this case, K(Π) = rk g − 1. Let us use the numbering of simple roots in [14, Chapter 18]. We check easily that α1,...,α−2 ∈ VΠ.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1520 RUPERT W. T. YU Table 2

∩ K K ∩ K1 ∩ S1 S1 dim VS1 TS S1  (S1)  (TS S1) K1 (Γ Γ ) A1 5 ∅ 1 0 A3 or A5 1 A2 5 ∅ 1 0 A5 2 A2 5 A1 1 1 A3 2 A3 6 A1 2 1 A5 3 A3 5 A1 2 1 E6 3 A4 6 A1 2 1 E6 4 A4 6 A3 2 2 A5 4 D4 5 A3 4 2 E6 4 D5 6 A3 4 2 E6 9

If dim VS = K(Π), then the inequality follows from (3) and (6), and the condition for equality follows from Point 2). Suppose now that dim VS =rkg and α−1 ∈ S (the case α ∈ S being analogous). Then

(9) χ(pS)+χ(uS)=rkg + K(S) − K(TS) − 2+(QS),

and the connected component S1 of S containing α−1 is not in K(Π); otherwise, we would have dim VS = K(Π). ∈Q By Point 1), εS1 S. By examining the possibilities for S1 and K1 (Examples 1.2), we verify that

K1 S1 (10) K(S1) − K(TS ∩ S1)+(Γ ∩ Γ ) ≥ 2

with equality if and only if S1 is of type A1 or A2. Therefore, we have obtained the inequality. In fact, we showed in the previous paragraph that already we have

K1 S1 rk g + K(S1) − K(TS ∩ S1) − 2+(Γ ∩ Γ ) ≥ rk g.

So if χ(pS)+χ(uS)=rkg, then from (9) and the above inequality, we must have K(S \ S1) ⊂K(Π), and also the equality in (10). Hence conditions (i) and (ii). Conversely, suppose that conditions (i) and (ii) are verified; then the fact that α1,...,α−2 ∈ VΠ implies that K(S \ S1) ⊂K(Π) and K(S1)=1.HenceS1 is of type A1, A2. It is then easy to check that χ(pS)+χ(uS)=rkg. 7) Type E6. Here, we have K(Π) = 4 and α2,α4 ∈ VΠ.LetS1 be a connected component

of S such that dim VS1 > 4. Under these conditions, the possibilities are shown in Table 2. Thus,

S1 K1 K(S1) − K(TS ∩ S1)+(Γ ∩ Γ ) ≥ 2(dim VS − K(Π)).

A direct verification gives the result. Note that as in the case of type A, K(Π) is totally ordered by inclusion, so TS is connected. 

Remark 2.3. Theorem 2.2 says that if K(S) ⊂K(Π) or, equivalently, S ∈K(Π) for  any connected component S of S,thenχ(pS)+χ(uS)=rkg.

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SUM OF THE INDEX 1521 Table 3

Type K(S) ⊂K(Π) K(S) ⊂K(Π)  +1  +1 A i = i = 2 2

B none i odd C none i =  D2n+1 i =2n, 2n +1 i<2n odd D2n none i odd or i =2n E6 i =2 , 4 i =4 E7 none i =2, 3, 5, 7 E8 none i =2, 3, 5, 7 F4 none i =2 G2 none i =1

Example 2.4. Let us consider the case of minimal parabolic subalgebras. So S = {α} and it follows that ⎧ ⎨ rk g if {α}∈K(Π), { } ∈K K χ(p)+χ(u)=⎩ rk g if α (Π) and dim VS =  (Π) + 1, rk g +2 if{α} ∈K(Π) and dim VS = K(Π).

Thus the minimal parabolic subalgebras pS verifying χ(pS)+χ(uS)=rkg are (in the simple roots numbering of [14, Chapter 18]) as shown in Table 3. Example 2.5. In the other extreme, it is easy to check that maximal parabolic subalgebras of g =sl+1 verifying χ(p)+χ(u)=rkg are exactly the ones associated to simple roots at the extremities of the . Acknowledgment The author would like to thank the anonymous referee for his or her remarks. References

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