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Z2 -contractions of classical Lie algebras and symmetric polynomials
Journal of Algebra 489 (2017) 25–37
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Journal of Algebra www.elsevier.com/locate/jalgebra
Z2 -contractions of classical Lie algebras and symmetric polynomials Oksana Yakimova 1 Institut für Mathematik, Friedrich-Schiller-Universität Jena, Jena, D-07737, Germany
a r t i c l e
i n f o
Article history: Received 5 November 2016 Available online 5 July 2017 Communicated by Shrawan Kumar MSC: 17B20 17B70
a b s t r a c t We consider Z2 -contractions of classical Lie algebras and the behaviour of the symmetric invariants under these contractions. It is demonstrated on three different examples how the theory of symmetric polynomials works in this invariant-theoretic problem. © 2017 Elsevier Inc. All rights reserved.
Keywords: Coadjoint representation Symmetric invariants Classical Lie algebras Symmetric polynomials
Contents 1. Introduction . . . . . . . . . . . . . . . . . . . . . 2. Some properties of symmetric polynomials 3. The symmetric pair (sp2n+2l , sp2n ⊕ sp2l ) . 4. The symmetric pair (sl2n , sp2n ) . . . . . . . . 5. The symmetric pair (so2N , glN ) . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .
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E-mail address: [email protected]. The author is supported by the Graduiertenkolleg GRK 1523 “Quanten- und Gravitationsfelder”.
http://dx.doi.org/10.1016/j.jalgebra.2017.06.026 0021-8693/© 2017 Elsevier Inc. All rights reserved.
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O. Yakimova / Journal of Algebra 489 (2017) 25–37
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1. Introduction The goal of this note is to demonstrate how the theory of symmetric polynomials works in some invariant-theoretic problems related to Z2 -contractions of simple Lie algebras. Let g = g0 ⊕ g1 be a Z2 -grading of a simple Lie algebra g of rank r. Then the ab ab ˜ = g0 gab semi-direct product g 1 , where [g , g ] = 0, is a Z2 -contraction of g, see [1] for generalities and the results mentioned in this paragraph. The ring of symmetric g-invariants S(g)g is graded polynomial, of Krull dimension r, and there is a procedure ˜ as vector that takes a homogeneous H ∈ S(g)g to an element of S(˜ g)g˜ (we identify g and g spaces). Namely, let H • be the bi-homogeneous component of H that has maximal degree w.r.t. g1 . Then H • ∈ S(˜ g)g˜ . This is related to the structure of S(˜ g)g˜ as follows. Suppose g that there are homogeneous generators H1 , . . . , Hr ∈ S(g) such that H1• , . . . , Hr• are algebraically independent. Then S(˜ g)g˜ is a polynomial ring generated by H1• , . . . , Hr• . Such a family {H1 , . . . , Hr } is called a good generating system (= g.g.s.) in S(g)g with ˜. respect to the contraction g ; g It is known that a g.g.s. does not always exist [1, Remark 4.3]. There are exactly four Z2 -gradings of exceptional algebras such that S(˜ g)g˜ is not a polynomial ring [4]. Moreover, a g.g.s. is a subtle object, i.e., usually an initial choice of H1 , . . . , Hr does not provide a g.g.s., and further adjusting is required. If g is classical, then (the possibility of) such a “fine tuning” is closely related to some properties of symmetric polynomials. For all but three classical symmetric pairs g.g.s. are constructed in [1]. The remaining ones are treated in [3] making use of Poison tensors and differential forms. Here we present an elementary approach to these cases. The existence of a g.g.s. for each of these three pairs follows from a certain general observation, Theorem 2, on symmetric polynomials. Thereby we obtain another proof of the fact that for these three symmetric pairs, S(˜ g)g˜ ˜ g is a polynomial ring. The ring S(˜ g) is always bi-graded and we indicate the bi-degrees of the basic invariants. Theorem 2 is in itself a new and amusing statement. A useful trick is that in place of S(g)g one can work with the isomorphic algebra S(t)W , where t is a suitable Cartan subalgebra (= CSA) of g and W is the Weyl group of (g, t). For generalities on Cartan subalgebras, Weyl groups, and Cartan subspaces, see e.g. [2]. Let c be a Cartan subspace of g1 . For a generic x ∈ c, let l be the centraliser of x in g0 and m be a CSA of l. Then c ⊕ m =: t is a CSA of g, and this yields a bi-graded structure in S(t)W that is compatible with the bi-graded structure of S(g)g under the Chevalley isomorphism. This means that in place of the basic invariants H1 , . . . , Hr ∈ S(g)g , one can work with the basic invariants hi = Hi |t∗ ∈ S(t)W . For a homogeneous h ∈ S(t)W , let h• denote its bi-homogeneous component of maximal degree w.r.t. c. Then we have: if h•1 , . . . , h•r are algebraically independent, then H1 , . . . , Hr is a g.g.s. If g is classical, then W is an extension of the symmetric group Sr and therefore basic W -invariants are related to some symmetric functions. The three pairs that are not considered in [1] are: I. (g, g0 ) = (sp2n+2l , sp2n ⊕ sp2l ) with n l,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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II. (g, g0 ) = (sl2(n+1) , sp2(n+1) ), III. (g, g0 ) = (so2N , glN ). For each of them, we construct polynomials H1 , . . . , Hr with the following properties. If i n, then h•i ∈ S(c) and these h•i are algebraically independent. r−n If i > n, then h•i = j=1 ui−n,j fj , where k[f1 , . . . , fr−n ] = S(m)Wl with Wl being the Weyl group of (l, m), and ui−n,j ∈ S(c). We prove then that the determinant of (ui,j ) is a nonzero polynomial. This shows that ˜∗ the polynomials h•1 , . . . , h•r are algebraically independent. Moreover, let us identify g ∗ ∗ ∗ ∗ • • ∗ ˜ of H to g + y is an with g0 ⊕g1 and take a generic y ∈ c ⊂ g1 . The restriction H 0 i i element of S(l) by [4, Proposition 2.7]. The argument with the determinant shows also ˜ • with i > n form a generating set of S(l)l . Thus, in these three that the polynomials H i particular cases we obtain an elementary proof of [4, Theorem 2.8]. The explicit description of the polynomials h•i , which is presented here, helps to deal ˜-invariants and will be used in forthcoming papers. with the symmetric g Warning. We use adjective “symmetric” in several different senses. The term “symmetric invariants” refers to the q-invariants in the symmetric algebra S(q) of a Lie algebra q. The term “symmetric polynomials” refers to the invariants of a symmetric group Sr and the term “symmetric pair” refers to (g, g0 ). Notation. N tk (x1 , . . . , xN ) = tk (x) = i=1 xki is the k-th power sum. σk (x1 , . . . , xN ) = σk (x) = i1 n.
(2.1) (2.2)
Let Qm be the unique polynomial such that tm = Qm (t1 , . . . , tn ), m n + 1. Let Sk be the unique polynomial such that σk = Sk (t1 , . . . , tn ). Newton’s formulae imply that Sk depends only on t1 , . . . , tk . For instance, σ1 = t1 , σ2 = 12 (t21 − t2 ), and if n = 2, then
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O. Yakimova / Journal of Algebra 489 (2017) 25–37
t3 = Q3 (t1 , t2 ) = t1 t2 − t1 12 (t21 − t2 ) = 12 (3t1 t2 − t31 ),
(2.3)
t4 = Q4 (t1 , t2 ) = t22 − 2( 12 (t21 − t2 ))2 = 12 t22 − 12 t41 + t21 t2 .
We are interested in certain coefficients of Qm and Sk . To this end, we need a special subset of (b1 , . . . , bn ) ∈ kn . More precisely, consider B = {(b1 , . . . , bn ) | σ1 (b) = . . . = σn−1 (b) = 0}.
(2.4)
By Newton’s formulae, this is equivalent to that t1 (b) = . . . = tn−1 (b) = 0. The coordinates of n-tuples in B are roots of the polynomial Y n + (−1)n σn (b). Hence bni = (−1)n−1 σn (b) for each i and n1 tn (b) = (−1)n−1 σn (b) whenever b ∈ B. Lemma 1. [tk ]Sk (t1 , . . . , tk ) =
(−1)k−1 . k
Proof. Substituting bk+1 = . . . = bn = 0, we reduce the problem to the case k = n. Let ˜ 1 , . . . , tn−1 ). Assuming that b ∈ B, we obtain σn (b) = σn = Sn (t1 , . . . , tn ) = αtn + S(t (−1)n−1 . 2 αtn (b). Hence α = n Proposition 1. For m = n + k with 1 k n, we have [tn tk ]Qm (t1 , . . . , tn ) =
1 1 k + n, 1 n,
k
.
Proof. It follows from (2.2) and the definition of the polynomials Sk that tm = tm−1 S1 (t1 ) − tm−2 S2 (t1 , t2 ) + . . . + (−1)n−1 tm−n Sn (t1 , . . . , tn ). This readily implies that the monomial tn tk occurs only in the summands tn Sk and tk Sn . Thus, if k < n, then using Lemma 1, we obtain [tn tk ]Qm (t1 , . . . , tn ) = (−1)k−1 [tk ]Sk (t1 , . . . , tk ) + (−1)n−1 [tn ]Sn (t1 , . . . , tn ) = If k = n, then [t2n ]Q2n (t1 , . . . , tn ) = (−1)n−1 [tn ]Sn (t1 , . . . , tn ) =
1 n.
1 k
+ n1 .
2
Motivated by examples discussed in Sections 3–5, we consider some polynomials that depend on two groups of variables a = (a1 , . . . , an ) and b = (b1 , . . . , bn ). Set pm = pm (a, b) = tm (b) + γm
n
bm−d adi , where m = 1, . . . , 2n. i
i=1
Here d is fixed and 1 d n. Since we wish to have polynomials, we also assume that γm = 0 for m < d. For our applications, we only need the case with d = 1, 2. Hence the only practical constraint is that γ1 = 0 if d = 2.
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The polynomials pm are homogeneous, with deg pm = m. We are interested in their bi-homogeneous components of maximal degree w.r.t. the variables {bj } (= highest b-components). Denoting such a component by p•m , we see that p•m = tm (b). Hence p•1 , . . . , p•n are algebraically independent, while p•m = Qm (p•1 , . . . , p•n ) for m > n. Our goal is to modify the last n polynomials in such a way that the obtained 2n polynomials have algebraically independent highest b-components. The transformation pm ; pm − Qm (p1 , . . . , pn ), m > n, kills the bad highest b-component of pm , and we are going to prove that under certain mild constrains this yields a desired family of polynomials. Let ψm−n,j = [adj ] pm − Qm (p1 , . . . , pn ) ∈ k[b1 , . . . , bn ] for n + 1 m 2n. Set Ψ := (ψk,j ). m−d Theorem 2. Suppose that b ∈ B. Then letting k = m−n, we have ψk,j = (γm − m . k γk )bj γm γk Therefore, Ψ(b) is nondegenerate if and only if m = k for k = 1, . . . , n and then
det Ψ(b) =
2n γk (2n)! γm − σn (b)n−d+1 det V(b1 , . . . , bn ), n! m=n+1 m k
where V is a Vandermonde matrix, i.e., νk,j = bk−1 . In particular, det Ψ ∈ k[b] is a j nonzero polynomial and the 2n polynomials n • p•1 = t1 (b), . . . , p•n = tn (b), pm − Qm (p1 , . . . , pn ) = ψm−n,j adj , m = n + 1, . . . , 2n, j=1
are algebraically independent. Proof. Let us look at the coefficient [adj ]Qm (p1 , . . . , pn ). If b ∈ B and k = n, then tk (b) = 0 and hence pk (a, b) lies in the ideal generated by the adj ’s. Therefore, summands of degree d with respect to {aj } occur only in the monomial pn pk in Qn+k (p1 , . . . , pn ). Then using Proposition 1, we see that if k < n, then ψk,j = [adj ] pm − Qm (p1 , . . . , pn ) = [adj ]pm − [adj ] Qm (p1 , . . . , pn ) = = [adj ]pm − (
1 1 1 1 + )[ad ]pn pk = γm bm−d − ( + )tn (b)γk bk−d = j j n k j n k n m (γm − γk − γk )bm−d = (γm − γk )bm−d . j j k k
And for k = n, i.e., m = 2n, we similarly obtain 1 [adj ] p2n − Q2n (p1 , . . . , pn ) = [adj ] p2n − [adj ] p2n = n γ2n bj2n−d −
2 tn (b)γn bn−d = (γ2n − 2γn )bj2n−d , j n
O. Yakimova / Journal of Algebra 489 (2017) 25–37
30
as required. It follows that det Ψ(b) =
2n m=n+1
(k−1)+(n−d+1)
bj
γm m
. Hence det Ψ =
m
γ
−
m n γ
γ (2n)! m k − n! m=n+1 m k 2n
m
γk ˜ det Ψ(b), where ψ˜k,j = k bn−d+1 · det V(b1 , . . . , bn ). 2 j
j=1
In the subsequent sections, γm /m is going to be an affine-linear form αm + β. Then − γkk = α(m − k) and ψk,j = mα(m − k)bm−d = αmnbm−d . j j
3. The symmetric pair (sp2n+2l , sp2n ⊕ sp2l ) For g = sp2n+2l , W Sn+l (Z2 )n+l and a CSA t ⊂ g can be chosen as the set of diagonal matrices t = {diag(x1 , . . . , xn+l , −xn+l , . . . , −x1 ) | xi ∈ k}. Then S(t)W consists of the symmetric polynomials in x21 , . . . , x2n+l . Next, l = (sl2 )n × sp2l−2n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , a1 , . . . , an , c1 , . . . , cl−n , −cl−n , . . . , −c1 , −an , . . . , −a1 , −an , . . . , −a1 }. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , −b1 , . . . , −bn , 0, . . . , 0, bn , . . . , b1 , −bn , . . . , −b1 ) | bi ∈ k}. Thus, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , a1 − b1 , . . . , an − bn , c1 , . . . , cl−n ) and S(t)W consists of the symmetric polynomials in (bi + ai )2 , (bi − ai )2 , and c2j . Then we need their bi-homogeneous components of highest degree w.r.t. b = (b1 , . . . , bn ). Let t k be the half of the k-th power sum and σ k be the k-th elementary symmetric polynomial for the l + n indeterminates (variables) (ai ± bi )2 and c2j , where i = 1, . . . , n and j = 1, . . . , l − n. That is, 1 2k 1 t k = (ai + bi )2k + (bi − ai )2k + c . 2 i=1 2 j=1 j n
l−n
We begin with the set of basic W -invariants that consists of t k (1 k 2n) and σ k
(2n + 1 k l + n). Note that the power sums tk contain no terms of odd degree w.r.t. the ai ’s and bj ’s. In particular,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
t k =
n i=1
31
b2k i +
l−n 1 2k 2k 2k−2 2 ai + (terms of degree 4 w.r.t. {ai }) + c . bi 2 j=1 j 2
(3.1)
n • • Then t k = tk (b21 , . . . , b2n ) = t2k (b) for k 2n and σ k = σk−2n (c21 , . . . , c2l−n ) i=1 b4i for • • • • k > 2n. Therefore, t 1 , . . . , t n , σ
2n+1 ,...,σ
n+l are algebraically independent, whereas • • •
tm each tm (n + 1 m 2n) is a polynomial in t 1 , . . . , t n . Our task is to modify
tm , will have “suitable” highest b-components. To such that the new W -invariants, say achieve this goal, we use the above polynomial Qm (see Section 2) and set tm =
tm − Qm (t 1 , . . . , t n ). • • • • tm tn+1 , . . . , We are going to prove that is a polynomial in {ai , bj } and that t 1 , . . . , t n , • • t2n are algebraically independent. Then the above expressions for σ k for k > 2n (which include all the elementary symmetric polynomials in {c2i }) will guarantee us • • • • • • that t 1 , . . . , t n , ,σ
2n+1 ,...,σ
n+l are algebraically independent. tn+1 , . . . , t2n • More precisely, we are going to prove that is a bi-homogeneous polynomial of tn+i
degree n + i − 2 w.r.t. b. This allows us to make the following simplification of notation. • • • • First, since t 1 , . . . , t n , do not include the indeterminates {ci }, we may tn+1 , . . . , t2n forget about the cj ’s and W -invariants σ k (2n + 1 k l + n). In other words, we may assume w.l.o.g. that n = l and there are no variables c1 , . . . , cl−n at all. Second, since our considerations involve only terms of degree 2k − 2 w.r.t. b in 2k 2k−2 2 n each t k , it suffices to deal with the truncated power sums p˜k := i=1 b2k ai i + 2 bi for 1 k 2n (cf. (3.1)) and the corresponding modifications p˜m − Qm (˜ p1 , . . . , p˜n ) for n + 1 m 2n. Third, one can get rid of the squares in the above formula for p˜k and work with the polynomials pk =
n
bki +
i=1
n 2k k−1 bi ai = tk (b) + k(2k − 1) bk−1 ai i 2 i=1
(3.2)
and the modifications pm − Qm (p1 , . . . , pn ) for n + 1 m 2n. This reduces our problem to the situation of Section 2 and yields the main technical assertion related to the symmetric pair (sp4n , sp2n ⊕ sp2n ) and thereby to all (sp2n+2l , sp2n ⊕ sp2l ) with l n. Let ψm−n,j ∈ k[b1 , . . . , bn ] be the coefficient of aj in pm − Qm (p1 , . . . , pn ) for n + 1 m 2n. Consider the n×n matrix Ψ = Ψ(b) = (ψi,j ). Theorem 3. The determinant of Ψ is a nonzero polynomial in the bi ’s. Equivalently, the 2n polynomials p•j = tj (b), (1 j n),
n • pm − Qm (p1 , . . . , pn ) = ψm−n,j aj , (n + 1 m 2n) j=1
are algebraically independent.
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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Proof. The assertion on det Ψ is a special case of Theorem 2, with d = 1 and γm = m(2m − 1). The matrix entries here are ψm−n,j = 2mnbm−1 . As already mentioned in j Section 2, p•j = tj (b) (1 j n),
n • pm − Qm (p) = ψm−n,j aj (n + 1 m 2n) j=1
are algebraically independent if and only if det Ψ = 0. 2 Example 1. If n = 2, then p1 = b1 + b2 + a1 + a2 and p2 = b21 + b22 + 6(b1 a1 + b2 a2 ). Using (2.3), we may compute ψk,j = [aj ] p2+k − Q2+k (p1 , p2 ) as follows: ψ1,1 ψ1,2 ψ2,1 ψ2,2
= 6b21 − 6b1 b2 , = 6b22 − 6b1 b2 , = 28b31 − 6b1 (b21 + b22 ) + 2(b1 + b2 )3 − 2(b1 + b2 )(b21 + b22 ) − 6b1 (b1 + b2 )2 , = 28b32 − 6b2 (b21 + b22 ) + 2(b1 + b2 )3 − 2(b1 + b2 )(b21 + b22 ) − 6b2 (b1 + b2 )2 .
It is readily seen that det Ψ is a non-zero polynomial if b1 +b2 = 0. For, then ψ1,1 = 12b21 , ψ1,2 = 12b22 = 12b21 , ψ2,1 = 16b31 , and ψ2,2 = 16b32 = −16b31 . The bi-degrees of the basic invariants in S(˜ g)g˜ are (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), . . . , (2, 4n − 2); (2, 4n), . . . , (2(l − n), 4n) . n
n
l−n
4. The symmetric pair (sl2n , sp2n ) It is more convenient to work with g = gl2n and the same g0 . This adds a trivial 1-dimensional g0 -module to g1 . Then a CSA t ⊂ g can be chosen as the set of all diagonal matrices t = {diag(x1 , . . . , x2n ) | xi ∈ k}, W S2n , and S(t)W consists of the symmetric polynomials in x1 , . . . , x2n . Next, l = (sl2 )n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 | ai ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 ) | bj ∈ k}. Thus,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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t = m ⊕ c = {diag(a1 + b1 , . . . , an + bn , bn − an , . . . , b1 − a1 ) | ai , bj ∈ k} and S(t)W consists of the symmetric polynomials in (ai +bi ) and (bi −ai ). Here everything is similar to the previous case, but some details are different. Let t k be the half of the k-th power sum for the 2n variables (bi ± ai ). That is, 1 t k = (ai + bi )k + (bi − ai )k 2 i=1 n
Since t k is stable under the substitution ai → −ai , its expansion contains no monomials of odd degree w.r.t a. We begin with the set of basic W -invariants that consists of t k • (1 k 2n). Since t k = tk (b) for k 2n, it is not a g.g.s. As in Section 3, we then tm = t consider the modifications m − Qm (t 1 , . . . , t
n ) for n + 1 m 2n. Since the passage t ; t kills the highest b-component in t m m m (i.e., the component of degree zero w.r.t. {ai }) and we are going to prove that the new highest b-components (i.e., the component of degree 2 w.r.t. ai ), together with t1 (b), . . . , tn (b), are algebraically independent, it suffices to deal with the truncated power sums qk :=
n n k k k−2 2 k(k − 1) k−2 2 bi ai bi + bi ai = tk (b) + 2 2 i=1 i=1
(4.1)
for 1 k 2n and corresponding modifications qm − Qm (q1 , . . . , qn ) for n + 1 m 2n. This leads to the main technical result related to the symmetric pair (gl2n , sp2n ). Let φm−n,j ∈ k[b1 , . . . , bn ] be the coefficient of a2j in qm − Qm (q1 , . . . , qn ) for n + 1 m 2n. Here deg φm−n,j = m − 2. Consider the n×n matrix Φ = Φ(b) = (φi,j ). Theorem 4. The determinant of Φ is a nonzero polynomial in the bi ’s. Equivalently, the 2n polynomials qj• = tj (b), (1 j n),
•
qm − Qm (q1 , . . . , qn )
=
n
φm−n,j a2j , (n + 1 m 2n)
j=1
are algebraically independent. Proof. The assertion on det Φ is a special case of Theorem 2, with d = 2 and γm = m(m − 1)/2. (Note that the necessary condition γ1 = 0 is satisfied!) The matrix entries here are φm−n,j = 12 mnbm−2 . j We leave it to the reader to verify that the polynomials qj• = tj (b) (1 j n),
•
qm − Qm (q)
=
n
φm−n,j a2j (n + 1 m 2n)
j=1
are algebraically independent if and only if det Φ = 0. 2
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34
The conclusion is that t 1 , . . . , t n , t m − Qm (t 1 , . . . , t
n ) for n + 1 m 2n is a g.g.s. ˜ g here. The bi-degrees of the basic invariants in S(˜ g) are (0, 1), (0, 2), . . . , (0, n); (2, n − 1), . . . , (2, 2n − 2) . n
n
Remark 5. These bi-degrees refer to the case of g = gl2n . For g = sl2n , one has to remove the invariant of bi-degree (0, 1) that comes from the trace of gl2n . 5. The symmetric pair (so2N , glN ) We assume that N 4. For g = so2N , W SN (Z2 )N −1 and a CSA t ⊂ g can be chosen as the set of diagonal matrices t = {diag(x1 , . . . , xN , −xN , . . . , −x1 ) | xi ∈ k}. Then S(t)W is generated by the symmetric polynomials in x21 , . . . , x2N and σN (x) = x1 · · · xN . Our preferred set of the basic W -invariants consists of 12 tk (x21 , . . . , x2N ), k = 1, . . . , N − 1, and σN (x). Here one has to distinguish the case of even and odd N . 5.1. N = 2n If N = 2n, then l (sl2 )n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 , a1 , . . . , an , −an , . . . , −a1 ) | ai ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 , −b1 , . . . , −bn , −bn , . . . , −b1 ) | bj ∈ k}. Hence, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , bn − an , . . . , b1 − a1 ). Using the previous notation, our preferred set of the basic W -invariants is 1 (ai + bi )2k + (bi − ai )2k = t k = 2 i=1 n 2k 2 2 tk (b1 , . . . , bn ) + b2k−2 a2j + (terms of degree 4 w.r.t. {ai }) 2 j=1 j n
O. Yakimova / Journal of Algebra 489 (2017) 25–37
35
for k = 1, . . . , 2n − 1 and also σ2n =
n i=1
(bi + ai )(bi − ai ) =
n
b2i −
i=1
n ( b2j )a2i + (terms of degree 4 w.r.t. {ai }). i=1 j=i
n • • Then t k = tk (b21 , . . . , b2n ) for k 2n − 1 and σ2n = i=1 b2i = σn (b21 , . . . , b2n ). Therefore, • • • • • = Qm (t 1 , . . . , t n ), t 1 , . . . , t n are algebraically independent. For m n +1, we have
tm • • • also σ2n = Sn (t 1 , . . . , t n ). To kill the “bad” highest b-components, we make the modifications tm − Qm (t 1 , . . . , t n ), and σ 2n = σ2n − Sn (t 1 , . . . , t n ). tm =
Again, as in the previous sections, we can replace t k and σ2n with the truncated invariants: n 2k 2 2
b2k−2 a2j tk ; p k = tk (b1 , . . . , bn ) + 2 j=1 j σ2n ; q˜2n =
n
b2i −
i=1
n ( b2j )a2i . i=1 j=i
Then we get rid of the squares in p˜k , q˜2n and consider n n n 2k pk = tk (b1 , . . . , bn ) + bk−1 a and q = b − ( bj )ai . j 2n i 2 j=1 j i=1 i=1 j=i
Then our task is to prove that the polynomials p•1 , . . . , p•n , (pm − Qm (p1 , . . . , pn ))• , m = n+1, . . . , 2n−1, and (q2n − Sn (p1 . . . , pn ))• (5.1) are algebraically independent in k[a, b]. Actually, it suffices to prove that the n by n matrix Ξ formed by the coefficients of the aj ’s in pm −Qm (p1 , . . . , pn ) and q2n −Sn (p1 . . . , pn ) is nonsingular. More precisely, set ξm−n,j = [aj ](pm − Qm (p1 , . . . , pn )) for n + 1 m 2n − 1; and ξn,j = [aj ](q2n − Sn (p1 . . . , pn )). Theorem 6. The matrix Ξ is nonsingular and hence pm − Qm (p1 , . . . , pn ))• =
n
ξm−n,j aj ,
(q2n − Sn (p1 . . . , pn ))• =
j=1
Therefore, all the polynomials in (5.1) are algebraically independent.
n j=1
ξn,j aj .
O. Yakimova / Journal of Algebra 489 (2017) 25–37
36
Proof. Assume that b ∈ B. By Theorem 2 with d = 1 and γm = m(2m − 1), we then have ξm−n,j = 2nmbm−1 if m < 2n. To compute the last row of Ξ, we notice that pi j does not contain the terms of degree zero w.r.t. {aj } if i < n and b ∈ B. Therefore, the only monomial of Sn (p1 , . . . , pn ) that contains terms of degree 1 w.r.t. {aj } is pn . Hence ξn,i = [ai ]q2n − [ai ]Sn (p1 , . . . , pn ) = [ai ]q2n − [ai ]pn ·[pn ]Sn (p1 , . . . , pn ) = =−
bj −
j=i
(−1)n−1 n(2n − 1)bn−1 = (−1)n bn−1 + (−1)n (2n − 1)bn−1 . i i i n
One then easily verifies that here det Ξ = −(2n)n σn (b)n−1 (2n−1)! V(b1 , . . . , bn ) = 0. The n! rest is standard. 2 The bi-degrees of the basic invariants in S(˜ g)g˜ are (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), . . . , (2, 4n − 4); (2, 2n − 2). n
n−1
5.2. N = 2n + 1 If N = 2n + 1, then l (sl2 )n × t1 and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 , c, a1 , . . . , an , −an , . . . , −a1 ) | ai , c ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 , 0, −b1 , . . . , −bn , −bn , . . . , −b1 ) | bj ∈ k}. Hence, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , c, bn − an , . . . , b1 − a1 ). Using the previous notation, our preferred set of the basic W -invariants is 1 1 (ai + bi )2k + (bi − ai )2k + c2k = t k = 2 i=1 2 n 1 2k tk (b21 , . . . , b2n ) + c2k + b2k−2 a2j + (terms of degree 4 w.r.t. ai ’s) 2 k j=1 j n
for k = 1, . . . , 2n and also
O. Yakimova / Journal of Algebra 489 (2017) 25–37
σ2n+1 = c
n
37
n (bi + ai )(bi − ai ) = c (b2i − a2i ).
i=1
i=1
n • • Then t k = tk (b21 , . . . , b2n ) for k 2n and σ2n+1 = c i=1 b2i = cσn (b21 , . . . , b2n ). Therefore, • • • • • • are algebraically independent, while
= Qm (t 1 , . . . , t n ) for m t 1 , . . . , t n , σ2n+1 tm n + 1. To kill the “bad” highest b-components, we make the modifications tm =
tm − Qm (t 1 , . . . , t n ), for m > n. The situation with t k , k = 1, 2, . . . , 2n, is the same as in Section 3. We make similar truncations and simplifications, and then we obtain the same n by n matrix Ψ. The • presence of the additional invariant σ2n+1 does not create any problem, since σ2n+1 con• • • • • tm+1 , . . . , t2n , σ2n+1 tains a new variable c. The resulting conclusion is that t 1 , . . . , t n , are algebraically independent. Hence a g.g.s. exists in this case, too. g)g˜ are The bi-degrees of the basic invariants in k(˜ (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), (2, 2n + 2), . . . , (2, 4n − 2); (1, 2n). n
n
References [1] D. Panyushev, On the coadjoint representation of Z2 -contractions of reductive Lie algebras, Adv. Math. 213 (2007) 380–404. [2] .B. Vinberg , A.L. Oniw ik , Seminar po gruppam Li i algebraiqeskim gruppam , Nauka, Moskva, 1988 (in Russian); English translation: A.L. Onishchik, E.B. Vinberg, Lie Groups and Algebraic Groups, Springer, Berlin, 1990. [3] O. Yakimova, One-parameter contractions of Lie-Poisson brackets, J. Eur. Math. Soc. 16 (2014) 387–407. [4] O. Yakimova, Symmetric invariants of Z2 -contractions and other semi-direct products, Int. Math. Res. Not. 2017 (2017) 1674–1716.
Contents lists available at ScienceDirect
Journal of Algebra www.elsevier.com/locate/jalgebra
Z2 -contractions of classical Lie algebras and symmetric polynomials Oksana Yakimova 1 Institut für Mathematik, Friedrich-Schiller-Universität Jena, Jena, D-07737, Germany
a r t i c l e
i n f o
Article history: Received 5 November 2016 Available online 5 July 2017 Communicated by Shrawan Kumar MSC: 17B20 17B70
a b s t r a c t We consider Z2 -contractions of classical Lie algebras and the behaviour of the symmetric invariants under these contractions. It is demonstrated on three different examples how the theory of symmetric polynomials works in this invariant-theoretic problem. © 2017 Elsevier Inc. All rights reserved.
Keywords: Coadjoint representation Symmetric invariants Classical Lie algebras Symmetric polynomials
Contents 1. Introduction . . . . . . . . . . . . . . . . . . . . . 2. Some properties of symmetric polynomials 3. The symmetric pair (sp2n+2l , sp2n ⊕ sp2l ) . 4. The symmetric pair (sl2n , sp2n ) . . . . . . . . 5. The symmetric pair (so2N , glN ) . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .
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E-mail address: [email protected]. The author is supported by the Graduiertenkolleg GRK 1523 “Quanten- und Gravitationsfelder”.
http://dx.doi.org/10.1016/j.jalgebra.2017.06.026 0021-8693/© 2017 Elsevier Inc. All rights reserved.
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O. Yakimova / Journal of Algebra 489 (2017) 25–37
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1. Introduction The goal of this note is to demonstrate how the theory of symmetric polynomials works in some invariant-theoretic problems related to Z2 -contractions of simple Lie algebras. Let g = g0 ⊕ g1 be a Z2 -grading of a simple Lie algebra g of rank r. Then the ab ab ˜ = g0 gab semi-direct product g 1 , where [g , g ] = 0, is a Z2 -contraction of g, see [1] for generalities and the results mentioned in this paragraph. The ring of symmetric g-invariants S(g)g is graded polynomial, of Krull dimension r, and there is a procedure ˜ as vector that takes a homogeneous H ∈ S(g)g to an element of S(˜ g)g˜ (we identify g and g spaces). Namely, let H • be the bi-homogeneous component of H that has maximal degree w.r.t. g1 . Then H • ∈ S(˜ g)g˜ . This is related to the structure of S(˜ g)g˜ as follows. Suppose g that there are homogeneous generators H1 , . . . , Hr ∈ S(g) such that H1• , . . . , Hr• are algebraically independent. Then S(˜ g)g˜ is a polynomial ring generated by H1• , . . . , Hr• . Such a family {H1 , . . . , Hr } is called a good generating system (= g.g.s.) in S(g)g with ˜. respect to the contraction g ; g It is known that a g.g.s. does not always exist [1, Remark 4.3]. There are exactly four Z2 -gradings of exceptional algebras such that S(˜ g)g˜ is not a polynomial ring [4]. Moreover, a g.g.s. is a subtle object, i.e., usually an initial choice of H1 , . . . , Hr does not provide a g.g.s., and further adjusting is required. If g is classical, then (the possibility of) such a “fine tuning” is closely related to some properties of symmetric polynomials. For all but three classical symmetric pairs g.g.s. are constructed in [1]. The remaining ones are treated in [3] making use of Poison tensors and differential forms. Here we present an elementary approach to these cases. The existence of a g.g.s. for each of these three pairs follows from a certain general observation, Theorem 2, on symmetric polynomials. Thereby we obtain another proof of the fact that for these three symmetric pairs, S(˜ g)g˜ ˜ g is a polynomial ring. The ring S(˜ g) is always bi-graded and we indicate the bi-degrees of the basic invariants. Theorem 2 is in itself a new and amusing statement. A useful trick is that in place of S(g)g one can work with the isomorphic algebra S(t)W , where t is a suitable Cartan subalgebra (= CSA) of g and W is the Weyl group of (g, t). For generalities on Cartan subalgebras, Weyl groups, and Cartan subspaces, see e.g. [2]. Let c be a Cartan subspace of g1 . For a generic x ∈ c, let l be the centraliser of x in g0 and m be a CSA of l. Then c ⊕ m =: t is a CSA of g, and this yields a bi-graded structure in S(t)W that is compatible with the bi-graded structure of S(g)g under the Chevalley isomorphism. This means that in place of the basic invariants H1 , . . . , Hr ∈ S(g)g , one can work with the basic invariants hi = Hi |t∗ ∈ S(t)W . For a homogeneous h ∈ S(t)W , let h• denote its bi-homogeneous component of maximal degree w.r.t. c. Then we have: if h•1 , . . . , h•r are algebraically independent, then H1 , . . . , Hr is a g.g.s. If g is classical, then W is an extension of the symmetric group Sr and therefore basic W -invariants are related to some symmetric functions. The three pairs that are not considered in [1] are: I. (g, g0 ) = (sp2n+2l , sp2n ⊕ sp2l ) with n l,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
27
II. (g, g0 ) = (sl2(n+1) , sp2(n+1) ), III. (g, g0 ) = (so2N , glN ). For each of them, we construct polynomials H1 , . . . , Hr with the following properties. If i n, then h•i ∈ S(c) and these h•i are algebraically independent. r−n If i > n, then h•i = j=1 ui−n,j fj , where k[f1 , . . . , fr−n ] = S(m)Wl with Wl being the Weyl group of (l, m), and ui−n,j ∈ S(c). We prove then that the determinant of (ui,j ) is a nonzero polynomial. This shows that ˜∗ the polynomials h•1 , . . . , h•r are algebraically independent. Moreover, let us identify g ∗ ∗ ∗ ∗ • • ∗ ˜ of H to g + y is an with g0 ⊕g1 and take a generic y ∈ c ⊂ g1 . The restriction H 0 i i element of S(l) by [4, Proposition 2.7]. The argument with the determinant shows also ˜ • with i > n form a generating set of S(l)l . Thus, in these three that the polynomials H i particular cases we obtain an elementary proof of [4, Theorem 2.8]. The explicit description of the polynomials h•i , which is presented here, helps to deal ˜-invariants and will be used in forthcoming papers. with the symmetric g Warning. We use adjective “symmetric” in several different senses. The term “symmetric invariants” refers to the q-invariants in the symmetric algebra S(q) of a Lie algebra q. The term “symmetric polynomials” refers to the invariants of a symmetric group Sr and the term “symmetric pair” refers to (g, g0 ). Notation. N tk (x1 , . . . , xN ) = tk (x) = i=1 xki is the k-th power sum. σk (x1 , . . . , xN ) = σk (x) = i1
(2.1) (2.2)
Let Qm be the unique polynomial such that tm = Qm (t1 , . . . , tn ), m n + 1. Let Sk be the unique polynomial such that σk = Sk (t1 , . . . , tn ). Newton’s formulae imply that Sk depends only on t1 , . . . , tk . For instance, σ1 = t1 , σ2 = 12 (t21 − t2 ), and if n = 2, then
28
O. Yakimova / Journal of Algebra 489 (2017) 25–37
t3 = Q3 (t1 , t2 ) = t1 t2 − t1 12 (t21 − t2 ) = 12 (3t1 t2 − t31 ),
(2.3)
t4 = Q4 (t1 , t2 ) = t22 − 2( 12 (t21 − t2 ))2 = 12 t22 − 12 t41 + t21 t2 .
We are interested in certain coefficients of Qm and Sk . To this end, we need a special subset of (b1 , . . . , bn ) ∈ kn . More precisely, consider B = {(b1 , . . . , bn ) | σ1 (b) = . . . = σn−1 (b) = 0}.
(2.4)
By Newton’s formulae, this is equivalent to that t1 (b) = . . . = tn−1 (b) = 0. The coordinates of n-tuples in B are roots of the polynomial Y n + (−1)n σn (b). Hence bni = (−1)n−1 σn (b) for each i and n1 tn (b) = (−1)n−1 σn (b) whenever b ∈ B. Lemma 1. [tk ]Sk (t1 , . . . , tk ) =
(−1)k−1 . k
Proof. Substituting bk+1 = . . . = bn = 0, we reduce the problem to the case k = n. Let ˜ 1 , . . . , tn−1 ). Assuming that b ∈ B, we obtain σn (b) = σn = Sn (t1 , . . . , tn ) = αtn + S(t (−1)n−1 . 2 αtn (b). Hence α = n Proposition 1. For m = n + k with 1 k n, we have [tn tk ]Qm (t1 , . . . , tn ) =
1 1 k + n, 1 n,
k
.
Proof. It follows from (2.2) and the definition of the polynomials Sk that tm = tm−1 S1 (t1 ) − tm−2 S2 (t1 , t2 ) + . . . + (−1)n−1 tm−n Sn (t1 , . . . , tn ). This readily implies that the monomial tn tk occurs only in the summands tn Sk and tk Sn . Thus, if k < n, then using Lemma 1, we obtain [tn tk ]Qm (t1 , . . . , tn ) = (−1)k−1 [tk ]Sk (t1 , . . . , tk ) + (−1)n−1 [tn ]Sn (t1 , . . . , tn ) = If k = n, then [t2n ]Q2n (t1 , . . . , tn ) = (−1)n−1 [tn ]Sn (t1 , . . . , tn ) =
1 n.
1 k
+ n1 .
2
Motivated by examples discussed in Sections 3–5, we consider some polynomials that depend on two groups of variables a = (a1 , . . . , an ) and b = (b1 , . . . , bn ). Set pm = pm (a, b) = tm (b) + γm
n
bm−d adi , where m = 1, . . . , 2n. i
i=1
Here d is fixed and 1 d n. Since we wish to have polynomials, we also assume that γm = 0 for m < d. For our applications, we only need the case with d = 1, 2. Hence the only practical constraint is that γ1 = 0 if d = 2.
O. Yakimova / Journal of Algebra 489 (2017) 25–37
29
The polynomials pm are homogeneous, with deg pm = m. We are interested in their bi-homogeneous components of maximal degree w.r.t. the variables {bj } (= highest b-components). Denoting such a component by p•m , we see that p•m = tm (b). Hence p•1 , . . . , p•n are algebraically independent, while p•m = Qm (p•1 , . . . , p•n ) for m > n. Our goal is to modify the last n polynomials in such a way that the obtained 2n polynomials have algebraically independent highest b-components. The transformation pm ; pm − Qm (p1 , . . . , pn ), m > n, kills the bad highest b-component of pm , and we are going to prove that under certain mild constrains this yields a desired family of polynomials. Let ψm−n,j = [adj ] pm − Qm (p1 , . . . , pn ) ∈ k[b1 , . . . , bn ] for n + 1 m 2n. Set Ψ := (ψk,j ). m−d Theorem 2. Suppose that b ∈ B. Then letting k = m−n, we have ψk,j = (γm − m . k γk )bj γm γk Therefore, Ψ(b) is nondegenerate if and only if m = k for k = 1, . . . , n and then
det Ψ(b) =
2n γk (2n)! γm − σn (b)n−d+1 det V(b1 , . . . , bn ), n! m=n+1 m k
where V is a Vandermonde matrix, i.e., νk,j = bk−1 . In particular, det Ψ ∈ k[b] is a j nonzero polynomial and the 2n polynomials n • p•1 = t1 (b), . . . , p•n = tn (b), pm − Qm (p1 , . . . , pn ) = ψm−n,j adj , m = n + 1, . . . , 2n, j=1
are algebraically independent. Proof. Let us look at the coefficient [adj ]Qm (p1 , . . . , pn ). If b ∈ B and k = n, then tk (b) = 0 and hence pk (a, b) lies in the ideal generated by the adj ’s. Therefore, summands of degree d with respect to {aj } occur only in the monomial pn pk in Qn+k (p1 , . . . , pn ). Then using Proposition 1, we see that if k < n, then ψk,j = [adj ] pm − Qm (p1 , . . . , pn ) = [adj ]pm − [adj ] Qm (p1 , . . . , pn ) = = [adj ]pm − (
1 1 1 1 + )[ad ]pn pk = γm bm−d − ( + )tn (b)γk bk−d = j j n k j n k n m (γm − γk − γk )bm−d = (γm − γk )bm−d . j j k k
And for k = n, i.e., m = 2n, we similarly obtain 1 [adj ] p2n − Q2n (p1 , . . . , pn ) = [adj ] p2n − [adj ] p2n = n γ2n bj2n−d −
2 tn (b)γn bn−d = (γ2n − 2γn )bj2n−d , j n
O. Yakimova / Journal of Algebra 489 (2017) 25–37
30
as required. It follows that det Ψ(b) =
2n m=n+1
(k−1)+(n−d+1)
bj
γm m
. Hence det Ψ =
m
γ
−
m n γ
γ (2n)! m k − n! m=n+1 m k 2n
m
γk ˜ det Ψ(b), where ψ˜k,j = k bn−d+1 · det V(b1 , . . . , bn ). 2 j
j=1
In the subsequent sections, γm /m is going to be an affine-linear form αm + β. Then − γkk = α(m − k) and ψk,j = mα(m − k)bm−d = αmnbm−d . j j
3. The symmetric pair (sp2n+2l , sp2n ⊕ sp2l ) For g = sp2n+2l , W Sn+l (Z2 )n+l and a CSA t ⊂ g can be chosen as the set of diagonal matrices t = {diag(x1 , . . . , xn+l , −xn+l , . . . , −x1 ) | xi ∈ k}. Then S(t)W consists of the symmetric polynomials in x21 , . . . , x2n+l . Next, l = (sl2 )n × sp2l−2n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , a1 , . . . , an , c1 , . . . , cl−n , −cl−n , . . . , −c1 , −an , . . . , −a1 , −an , . . . , −a1 }. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , −b1 , . . . , −bn , 0, . . . , 0, bn , . . . , b1 , −bn , . . . , −b1 ) | bi ∈ k}. Thus, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , a1 − b1 , . . . , an − bn , c1 , . . . , cl−n ) and S(t)W consists of the symmetric polynomials in (bi + ai )2 , (bi − ai )2 , and c2j . Then we need their bi-homogeneous components of highest degree w.r.t. b = (b1 , . . . , bn ). Let t k be the half of the k-th power sum and σ k be the k-th elementary symmetric polynomial for the l + n indeterminates (variables) (ai ± bi )2 and c2j , where i = 1, . . . , n and j = 1, . . . , l − n. That is, 1 2k 1 t k = (ai + bi )2k + (bi − ai )2k + c . 2 i=1 2 j=1 j n
l−n
We begin with the set of basic W -invariants that consists of t k (1 k 2n) and σ k
(2n + 1 k l + n). Note that the power sums tk contain no terms of odd degree w.r.t. the ai ’s and bj ’s. In particular,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
t k =
n i=1
31
b2k i +
l−n 1 2k 2k 2k−2 2 ai + (terms of degree 4 w.r.t. {ai }) + c . bi 2 j=1 j 2
(3.1)
n • • Then t k = tk (b21 , . . . , b2n ) = t2k (b) for k 2n and σ k = σk−2n (c21 , . . . , c2l−n ) i=1 b4i for • • • • k > 2n. Therefore, t 1 , . . . , t n , σ
2n+1 ,...,σ
n+l are algebraically independent, whereas • • •
tm each tm (n + 1 m 2n) is a polynomial in t 1 , . . . , t n . Our task is to modify
tm , will have “suitable” highest b-components. To such that the new W -invariants, say achieve this goal, we use the above polynomial Qm (see Section 2) and set tm =
tm − Qm (t 1 , . . . , t n ). • • • • tm tn+1 , . . . , We are going to prove that is a polynomial in {ai , bj } and that t 1 , . . . , t n , • • t2n are algebraically independent. Then the above expressions for σ k for k > 2n (which include all the elementary symmetric polynomials in {c2i }) will guarantee us • • • • • • that t 1 , . . . , t n , ,σ
2n+1 ,...,σ
n+l are algebraically independent. tn+1 , . . . , t2n • More precisely, we are going to prove that is a bi-homogeneous polynomial of tn+i
degree n + i − 2 w.r.t. b. This allows us to make the following simplification of notation. • • • • First, since t 1 , . . . , t n , do not include the indeterminates {ci }, we may tn+1 , . . . , t2n forget about the cj ’s and W -invariants σ k (2n + 1 k l + n). In other words, we may assume w.l.o.g. that n = l and there are no variables c1 , . . . , cl−n at all. Second, since our considerations involve only terms of degree 2k − 2 w.r.t. b in 2k 2k−2 2 n each t k , it suffices to deal with the truncated power sums p˜k := i=1 b2k ai i + 2 bi for 1 k 2n (cf. (3.1)) and the corresponding modifications p˜m − Qm (˜ p1 , . . . , p˜n ) for n + 1 m 2n. Third, one can get rid of the squares in the above formula for p˜k and work with the polynomials pk =
n
bki +
i=1
n 2k k−1 bi ai = tk (b) + k(2k − 1) bk−1 ai i 2 i=1
(3.2)
and the modifications pm − Qm (p1 , . . . , pn ) for n + 1 m 2n. This reduces our problem to the situation of Section 2 and yields the main technical assertion related to the symmetric pair (sp4n , sp2n ⊕ sp2n ) and thereby to all (sp2n+2l , sp2n ⊕ sp2l ) with l n. Let ψm−n,j ∈ k[b1 , . . . , bn ] be the coefficient of aj in pm − Qm (p1 , . . . , pn ) for n + 1 m 2n. Consider the n×n matrix Ψ = Ψ(b) = (ψi,j ). Theorem 3. The determinant of Ψ is a nonzero polynomial in the bi ’s. Equivalently, the 2n polynomials p•j = tj (b), (1 j n),
n • pm − Qm (p1 , . . . , pn ) = ψm−n,j aj , (n + 1 m 2n) j=1
are algebraically independent.
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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Proof. The assertion on det Ψ is a special case of Theorem 2, with d = 1 and γm = m(2m − 1). The matrix entries here are ψm−n,j = 2mnbm−1 . As already mentioned in j Section 2, p•j = tj (b) (1 j n),
n • pm − Qm (p) = ψm−n,j aj (n + 1 m 2n) j=1
are algebraically independent if and only if det Ψ = 0. 2 Example 1. If n = 2, then p1 = b1 + b2 + a1 + a2 and p2 = b21 + b22 + 6(b1 a1 + b2 a2 ). Using (2.3), we may compute ψk,j = [aj ] p2+k − Q2+k (p1 , p2 ) as follows: ψ1,1 ψ1,2 ψ2,1 ψ2,2
= 6b21 − 6b1 b2 , = 6b22 − 6b1 b2 , = 28b31 − 6b1 (b21 + b22 ) + 2(b1 + b2 )3 − 2(b1 + b2 )(b21 + b22 ) − 6b1 (b1 + b2 )2 , = 28b32 − 6b2 (b21 + b22 ) + 2(b1 + b2 )3 − 2(b1 + b2 )(b21 + b22 ) − 6b2 (b1 + b2 )2 .
It is readily seen that det Ψ is a non-zero polynomial if b1 +b2 = 0. For, then ψ1,1 = 12b21 , ψ1,2 = 12b22 = 12b21 , ψ2,1 = 16b31 , and ψ2,2 = 16b32 = −16b31 . The bi-degrees of the basic invariants in S(˜ g)g˜ are (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), . . . , (2, 4n − 2); (2, 4n), . . . , (2(l − n), 4n) . n
n
l−n
4. The symmetric pair (sl2n , sp2n ) It is more convenient to work with g = gl2n and the same g0 . This adds a trivial 1-dimensional g0 -module to g1 . Then a CSA t ⊂ g can be chosen as the set of all diagonal matrices t = {diag(x1 , . . . , x2n ) | xi ∈ k}, W S2n , and S(t)W consists of the symmetric polynomials in x1 , . . . , x2n . Next, l = (sl2 )n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 | ai ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 ) | bj ∈ k}. Thus,
O. Yakimova / Journal of Algebra 489 (2017) 25–37
33
t = m ⊕ c = {diag(a1 + b1 , . . . , an + bn , bn − an , . . . , b1 − a1 ) | ai , bj ∈ k} and S(t)W consists of the symmetric polynomials in (ai +bi ) and (bi −ai ). Here everything is similar to the previous case, but some details are different. Let t k be the half of the k-th power sum for the 2n variables (bi ± ai ). That is, 1 t k = (ai + bi )k + (bi − ai )k 2 i=1 n
Since t k is stable under the substitution ai → −ai , its expansion contains no monomials of odd degree w.r.t a. We begin with the set of basic W -invariants that consists of t k • (1 k 2n). Since t k = tk (b) for k 2n, it is not a g.g.s. As in Section 3, we then tm = t consider the modifications m − Qm (t 1 , . . . , t
n ) for n + 1 m 2n. Since the passage t ; t kills the highest b-component in t m m m (i.e., the component of degree zero w.r.t. {ai }) and we are going to prove that the new highest b-components (i.e., the component of degree 2 w.r.t. ai ), together with t1 (b), . . . , tn (b), are algebraically independent, it suffices to deal with the truncated power sums qk :=
n n k k k−2 2 k(k − 1) k−2 2 bi ai bi + bi ai = tk (b) + 2 2 i=1 i=1
(4.1)
for 1 k 2n and corresponding modifications qm − Qm (q1 , . . . , qn ) for n + 1 m 2n. This leads to the main technical result related to the symmetric pair (gl2n , sp2n ). Let φm−n,j ∈ k[b1 , . . . , bn ] be the coefficient of a2j in qm − Qm (q1 , . . . , qn ) for n + 1 m 2n. Here deg φm−n,j = m − 2. Consider the n×n matrix Φ = Φ(b) = (φi,j ). Theorem 4. The determinant of Φ is a nonzero polynomial in the bi ’s. Equivalently, the 2n polynomials qj• = tj (b), (1 j n),
•
qm − Qm (q1 , . . . , qn )
=
n
φm−n,j a2j , (n + 1 m 2n)
j=1
are algebraically independent. Proof. The assertion on det Φ is a special case of Theorem 2, with d = 2 and γm = m(m − 1)/2. (Note that the necessary condition γ1 = 0 is satisfied!) The matrix entries here are φm−n,j = 12 mnbm−2 . j We leave it to the reader to verify that the polynomials qj• = tj (b) (1 j n),
•
qm − Qm (q)
=
n
φm−n,j a2j (n + 1 m 2n)
j=1
are algebraically independent if and only if det Φ = 0. 2
O. Yakimova / Journal of Algebra 489 (2017) 25–37
34
The conclusion is that t 1 , . . . , t n , t m − Qm (t 1 , . . . , t
n ) for n + 1 m 2n is a g.g.s. ˜ g here. The bi-degrees of the basic invariants in S(˜ g) are (0, 1), (0, 2), . . . , (0, n); (2, n − 1), . . . , (2, 2n − 2) . n
n
Remark 5. These bi-degrees refer to the case of g = gl2n . For g = sl2n , one has to remove the invariant of bi-degree (0, 1) that comes from the trace of gl2n . 5. The symmetric pair (so2N , glN ) We assume that N 4. For g = so2N , W SN (Z2 )N −1 and a CSA t ⊂ g can be chosen as the set of diagonal matrices t = {diag(x1 , . . . , xN , −xN , . . . , −x1 ) | xi ∈ k}. Then S(t)W is generated by the symmetric polynomials in x21 , . . . , x2N and σN (x) = x1 · · · xN . Our preferred set of the basic W -invariants consists of 12 tk (x21 , . . . , x2N ), k = 1, . . . , N − 1, and σN (x). Here one has to distinguish the case of even and odd N . 5.1. N = 2n If N = 2n, then l (sl2 )n and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 , a1 , . . . , an , −an , . . . , −a1 ) | ai ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 , −b1 , . . . , −bn , −bn , . . . , −b1 ) | bj ∈ k}. Hence, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , bn − an , . . . , b1 − a1 ). Using the previous notation, our preferred set of the basic W -invariants is 1 (ai + bi )2k + (bi − ai )2k = t k = 2 i=1 n 2k 2 2 tk (b1 , . . . , bn ) + b2k−2 a2j + (terms of degree 4 w.r.t. {ai }) 2 j=1 j n
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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for k = 1, . . . , 2n − 1 and also σ2n =
n i=1
(bi + ai )(bi − ai ) =
n
b2i −
i=1
n ( b2j )a2i + (terms of degree 4 w.r.t. {ai }). i=1 j=i
n • • Then t k = tk (b21 , . . . , b2n ) for k 2n − 1 and σ2n = i=1 b2i = σn (b21 , . . . , b2n ). Therefore, • • • • • = Qm (t 1 , . . . , t n ), t 1 , . . . , t n are algebraically independent. For m n +1, we have
tm • • • also σ2n = Sn (t 1 , . . . , t n ). To kill the “bad” highest b-components, we make the modifications tm − Qm (t 1 , . . . , t n ), and σ 2n = σ2n − Sn (t 1 , . . . , t n ). tm =
Again, as in the previous sections, we can replace t k and σ2n with the truncated invariants: n 2k 2 2
b2k−2 a2j tk ; p k = tk (b1 , . . . , bn ) + 2 j=1 j σ2n ; q˜2n =
n
b2i −
i=1
n ( b2j )a2i . i=1 j=i
Then we get rid of the squares in p˜k , q˜2n and consider n n n 2k pk = tk (b1 , . . . , bn ) + bk−1 a and q = b − ( bj )ai . j 2n i 2 j=1 j i=1 i=1 j=i
Then our task is to prove that the polynomials p•1 , . . . , p•n , (pm − Qm (p1 , . . . , pn ))• , m = n+1, . . . , 2n−1, and (q2n − Sn (p1 . . . , pn ))• (5.1) are algebraically independent in k[a, b]. Actually, it suffices to prove that the n by n matrix Ξ formed by the coefficients of the aj ’s in pm −Qm (p1 , . . . , pn ) and q2n −Sn (p1 . . . , pn ) is nonsingular. More precisely, set ξm−n,j = [aj ](pm − Qm (p1 , . . . , pn )) for n + 1 m 2n − 1; and ξn,j = [aj ](q2n − Sn (p1 . . . , pn )). Theorem 6. The matrix Ξ is nonsingular and hence pm − Qm (p1 , . . . , pn ))• =
n
ξm−n,j aj ,
(q2n − Sn (p1 . . . , pn ))• =
j=1
Therefore, all the polynomials in (5.1) are algebraically independent.
n j=1
ξn,j aj .
O. Yakimova / Journal of Algebra 489 (2017) 25–37
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Proof. Assume that b ∈ B. By Theorem 2 with d = 1 and γm = m(2m − 1), we then have ξm−n,j = 2nmbm−1 if m < 2n. To compute the last row of Ξ, we notice that pi j does not contain the terms of degree zero w.r.t. {aj } if i < n and b ∈ B. Therefore, the only monomial of Sn (p1 , . . . , pn ) that contains terms of degree 1 w.r.t. {aj } is pn . Hence ξn,i = [ai ]q2n − [ai ]Sn (p1 , . . . , pn ) = [ai ]q2n − [ai ]pn ·[pn ]Sn (p1 , . . . , pn ) = =−
bj −
j=i
(−1)n−1 n(2n − 1)bn−1 = (−1)n bn−1 + (−1)n (2n − 1)bn−1 . i i i n
One then easily verifies that here det Ξ = −(2n)n σn (b)n−1 (2n−1)! V(b1 , . . . , bn ) = 0. The n! rest is standard. 2 The bi-degrees of the basic invariants in S(˜ g)g˜ are (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), . . . , (2, 4n − 4); (2, 2n − 2). n
n−1
5.2. N = 2n + 1 If N = 2n + 1, then l (sl2 )n × t1 and a CSA in l can be chosen as the set of diagonal matrices m = {diag(a1 , . . . , an , −an , . . . , −a1 , c, a1 , . . . , an , −an , . . . , −a1 ) | ai , c ∈ k}. A Cartan subspace in g1 can be identified with the orthogonal complement of m in t, i.e., c = {diag(b1 , . . . , bn , bn , . . . , b1 , 0, −b1 , . . . , −bn , −bn , . . . , −b1 ) | bj ∈ k}. Hence, the first half of the diagonal entries of the elements of t = m ⊕ c is (a1 + b1 , . . . , an + bn , c, bn − an , . . . , b1 − a1 ). Using the previous notation, our preferred set of the basic W -invariants is 1 1 (ai + bi )2k + (bi − ai )2k + c2k = t k = 2 i=1 2 n 1 2k tk (b21 , . . . , b2n ) + c2k + b2k−2 a2j + (terms of degree 4 w.r.t. ai ’s) 2 k j=1 j n
for k = 1, . . . , 2n and also
O. Yakimova / Journal of Algebra 489 (2017) 25–37
σ2n+1 = c
n
37
n (bi + ai )(bi − ai ) = c (b2i − a2i ).
i=1
i=1
n • • Then t k = tk (b21 , . . . , b2n ) for k 2n and σ2n+1 = c i=1 b2i = cσn (b21 , . . . , b2n ). Therefore, • • • • • • are algebraically independent, while
= Qm (t 1 , . . . , t n ) for m t 1 , . . . , t n , σ2n+1 tm n + 1. To kill the “bad” highest b-components, we make the modifications tm =
tm − Qm (t 1 , . . . , t n ), for m > n. The situation with t k , k = 1, 2, . . . , 2n, is the same as in Section 3. We make similar truncations and simplifications, and then we obtain the same n by n matrix Ψ. The • presence of the additional invariant σ2n+1 does not create any problem, since σ2n+1 con• • • • • tm+1 , . . . , t2n , σ2n+1 tains a new variable c. The resulting conclusion is that t 1 , . . . , t n , are algebraically independent. Hence a g.g.s. exists in this case, too. g)g˜ are The bi-degrees of the basic invariants in k(˜ (0, 2), (0, 4), . . . , (0, 2n); (2, 2n), (2, 2n + 2), . . . , (2, 4n − 2); (1, 2n). n
n
References [1] D. Panyushev, On the coadjoint representation of Z2 -contractions of reductive Lie algebras, Adv. Math. 213 (2007) 380–404. [2] .B. Vinberg , A.L. Oniw ik , Seminar po gruppam Li i algebraiqeskim gruppam , Nauka, Moskva, 1988 (in Russian); English translation: A.L. Onishchik, E.B. Vinberg, Lie Groups and Algebraic Groups, Springer, Berlin, 1990. [3] O. Yakimova, One-parameter contractions of Lie-Poisson brackets, J. Eur. Math. Soc. 16 (2014) 387–407. [4] O. Yakimova, Symmetric invariants of Z2 -contractions and other semi-direct products, Int. Math. Res. Not. 2017 (2017) 1674–1716.