A counterexample to Premet’s and Joseph’s conjectures

O. YAKIMOVA

Institut des Hautes Etudes´ Scientifiques 35, route de Chartres 91440 – Bures-sur-Yvette (France)

Janvier 2007

IHES/M/07/01 January 12, 2007 A COUNTEREXAMPLE TO PREMET’S AND JOSEPH’S CONJECTURES

O. YAKIMOVA

INTRODUCTION

Let g be a finite-dimensional reductive of rank l over an algebraically closed field K of characteristic zero, and let G be the adjoint group of g. Given x g, we denote ∈ by gx the centraliser of x in g.

Conjecture 1 (Premet). For any x g the algebra S(g )gx of g -invariants is a graded polyno- ∈ x x mial algebra in l variables.

In some particular cases the problem is simple. For example, for regular nilpotent ele-

gx ments the algebra S(gx) is known to be free. In [5], Conjecture 1 is shown to be true in types A and C. It is also verified for some nilpotent elements of orthogonal Lie algebra

and for the minimal nilpotent orbits in simple Lie algebras except of type . Later, by a different method, Brown and Brundan [1] proved that Conjecture 1 holds in type A.

Suppose that p+ and p are opposite parabolic subalgebras of g, i.e., g = p+ + p . Then − − the intersection q := p+ p is called a biparabolic or, in other terminology, seaweed subal- ∩ − gebra. Since g itself is a parabolic subalgebra, we see that parabolics are particular cases of seaweeds.

For any Lie algebra q let q" := [q, q] denote its derived algebra. In [3, Section 7.7], the following conjecture was made.

Conjecture 2 (Joseph). For any seaweed subalgebra q g the semi-invariants S(q)q! form a ⊂ polynomial algebra.

A formula for tr.deg S(q)q! is given in [3]. It is rather complicated and we are not go- ing to use it in full generality. In [2] and [3], it is proved that Conjecture 2 holds for all

2000 Mathematics Subject Classification. 17B25. Key words and phrases. Nilpotent orbits, centralisers, symmetric invariants. The author is supported by the Humboldt Foundation and RFFR Grant 05-01-00988. 1 2 O. YAKIMOVA

parabolics and seaweeds in simple Lie algebras of types A and C. As was noticed in [5, Section 4.9], minimal nilpotent orbits provide a testing site for Joseph’s conjecture as well as Premet’s one. If g is simple, then for each minimal nilpotent element e g there exists ∈ a parabolic subalgebra p g such that S(g )ge = S(p)p! . The detailed explanation of this ⊂ e ∼ p! ge construction is given below. We note only that naturally tr.deg S(p) = tr.deg S(ge) and

ge tr.deg S(ge) = l by [4].

In this note, we show that Conjecture 1 does not hold for the minimal nilpotent orbit

in the of type E8. As a consequence, a conjecture of Joseph on the semi-invariants of (bi)parabolics is not true either. Acknowledgements. This paper was written during my stay at the IHES. I thank the Institute for warm hospitality and support.

1. THEORY

Let us say a few words about the general method of [5], which, unfortunately, does not work for the minimal nilpotent orbit in E . Let g be a simple Lie algebra and e g 8 ∈ a nilpotent element. Suppose that e, h, f g is an sl -triple containing e. We identify % & ⊂ 2 G e g and g∗ by means of the . For each F S(g) let F stand for the minimal ∈ degree component of the restriction F . As was shown in [5], eF S(g )ge . A set of ho- |e+gf ∈ e mogeneous generators F , . . . , F S(g)G is said to be good if the eF ’s are algebraically { 1 l} ⊂ i independent.

Given a linear function γ on ge we denote by (ge)γ the stabiliser of γ in ge and set

(g∗) := γ g∗ dim(g ) > l . e sing { ∈ e | e γ }

G Theorem 1. [5] Suppose e admits a good generating system F1, . . . , Fl in S(g) and assume fur-

ge e e ther that (ge∗)sing has codimension ! 2 in ge∗. Then S(ge) is a polynomial algebra in F1, . . . , Fl.

Suppose now that e is a minimal nilpotent element. Then dim(ge)γ = l for generic

γ g∗, see [4]; and (g∗) is of codimension 2, see [5, Section 3.10.]. If g is of type E , ∈ e e sing ! 8 then there is no good generating system, [5, Remark 4.2.]. For that reason in Section 4.8 of [5] another approach was developed. As was proved there, Conjecture 1 holds if and only if there is a certain system of generating invariants in . COUNTEREXAMPLE 3

Since e is a minimal nilpotent element, the Z-grading defined by h is

g = g( 2) g( 1) g(0) g(1) g(2), − ⊕ − ⊕ ⊕ ⊕ with g(2) = Ke and g(1) g(2) being a Heisenberg Lie algebra. Set l := g(0)e = ge g(0). ⊕ ∩ Then g(0) = l Kh. Clearly p := g(0) g(1) g(2) is a parabolic subalgebra of g and ⊕ ⊕ ⊕ p" = g(0)" g(1) g(2). Since p = Kh ge and [h, e] = 2e, we have ⊕ ⊕ ⊕ S(p)p! S(p)e S(g ). ⊂ ⊂ e

p! ge If g is not of type A, then l = g(0)" and p" = ge. Hence S(p) = S(ge) .

p! g! Remark 1. If g is of type A, then, so far, we can only say that S(p) = S(ge) e . Set n := g(1) g(2). Then there is an ismorphism of l-modules (S(g )[1/e])n = S(l)[e, 1/e], see [5, ⊕ e ∼ Section 4.8.] or [7, Lemma 3.]. Therefore the centre of l acts on n-invariants trivially and

p! g! ge S(p) = S(ge) e = S(ge) .

From now on assume that g is of type E . Then l is of type E . For generic v g(1) the 8 7 ∈ stabiliser l is a simple Lie algebra of type E . Fix such v g(1). Let t l and ˆt l be v 6 ∈ ⊂ ⊂ v maximal tori such thatˆt t. Then there is a unique orthogonal decomposition t = ˆt Kh0. ⊂ ⊕ W Let W and W " denote the Weyl groups of l and l , respectively. Each ϕ S(t) can be v i ∈ presented uniquely as ν (j) (j) (ν) ϕ = ϕ hj ϕ S(t)W ! , ϕ = 0, ν = ν(i) . i i 0 i ∈ i ) j=0 ! " # ge˜ Theorem 2. [5, Theorem 4.14.] The algebra S(ge˜) is free if and only if there is a homoge- W (ν) ν(1) (ν) ν(7) neous generating system ϕ1, . . . , ϕ7 in S(t) such that the elements ϕ1 h0 , . . . , ϕ7 h0 are algebraically independent.

The main technical result of this paper is the following:

W Proposition 1. Suppose that ϕ1, . . . , ϕ7 is a system of homogeneous generators of S(t) with (ν) ν(1) (ν) ν(2) (ν) ν(3) deg ϕi < deg ϕj for i < j. Then the elements ϕ1 h0 , ϕ2 h0 , ϕ3 h0 are algebraically dependent.

Combining Theorem 2 and Proposition 1, we conclude that Conjectures 1 and 2 are false. 4 O. YAKIMOVA

2. CALCULATIONS

Since it is difficult to deal with E directly, we first consider a regular subalgebra sl 7 8 ⊂ E such that t sl . Let $ and $" denote the fundamental weights of E and SL , 7 ⊂ 8 i i 7 8 respectively. We use the Vinberg–Onishchik numbering of simple roots and fundamental weights, see [6, Tables]. We may (and will) assume that the simple roots of sl8 are the

first six simple roots of E7 and the lowest root δ. On the extended of E7, which is given below, the simple roots of sl8 form the upper line. Recall thatˆt is a maximal torus in a regular subalgebra E E . Without loss of generality, we may assume that ˆt 6 ⊂ 7 coincides with the annihilator of the weight $1. Expressing δ as a linear combination of the simple roots one can see that $ (δ) = 1. Hence the subtorus ˆt is also the annihilator 1 − of $" $" . 1 − 7 1 2 3 4 5 6 ...... E7 : . . . . ◦ ◦ ◦ ◦. ◦ ◦ ◦ . δ . 7 $ ◦ Without loss of generality, we may assume that t is the subspace of diagonal matrices of sl . The dual space ˜t∗ is spanned by ε , ε , . . . , ε subject to the relation ε +ε + +ε = 0 8 1 2 8 1 2 · · · 8 and the of SL8 permutes the εi’s. Since the fundamental weights $1" , $7" can be expressed as $ = ε and $ = ε , we conclude that ˆt is the annihilator of ε + ε . 1 1 7 − 8 1 8 Therefore ˆt can be presented as a linear space of diagonal matrices:

6 (1) ˆt = diag(b, b , b , . . . , b , b) b = 0 . { 1 2 6 − | i } i=1 ! Then

(2) Kh0 = diag(a, a/3, a/3, a/3, a/3, a/3, a/3, a) a K . { − − − − − − | ∈ }

Let us identify t with t∗ by means of the Killing form. Then Weyl group invariants of (ν) SL8 can be expressed in terms of variables a, b, b1, . . . , b6; the ϕi ’s will be polynomials in b, b1, . . . , b6 and h0 proportional to a. Since sl8 is a maximal rank subalgebra of E7, one can write invariants of E7 as polynomials in invariants of SL8 with unknown coefficients. (ν) ν(1) (ν) ν(2) (ν) ν(3) This will give us certain constrains on ϕ1 h0 , ϕ2 h0 , and ϕ3 h0 . COUNTEREXAMPLE 5

First we concentrate on SL8-invariants. According to formulas (1) and (2), any diagonal matrix in sl8 is of the form:

(a + b, a/3 + b , a/3 + b , a/3 + b , a/3 + b , a/3 + b , a/3 + b , a b) − 1 − 2 − 3 − 4 − 5 − 6 − 6 6 k with i=1 bi = 0. Set τk := i=1 bi , and let Si be the trace of the i + 1 power of a diagonal matrix.% Then % 6 6 2 8 S = 2a2 + 2b2 + a2 + ( ab + b2) = a2 + 2b2 + τ . 2 9 −3 i i 3 2 i=1 ! In the same way 16 S = a3 + 6ab2 aτ + τ ; 3 9 − 2 3 2 2 4 S = (2 + )a4 + 12a2b2 + a2τ aτ + 2b4 + τ ; 4 27 3 2 − 3 3 4 2 10 10 5 S = (2 )a5 + 20a3b2 a3τ + a2τ + 10ab4 aτ + τ ; 5 − 81 − 27 2 9 3 − 3 4 5 2 5 20 5 S = (2 + )a6 + 30a4b2 + a4τ a3τ + 30a2b4 + a2τ + . . . ; 6 35 27 2 − 27 3 3 4 2 28 56 70 S = (2 + )a8 + 56a6b2 + a6τ a5τ + 140a4b4 + a4τ + . . . . 8 37 36 2 − 35 3 81 4 In principle, it is possible to calculate all of them either by hand or using computer. The expression for S7 is of no importance for us. Also, the coefficients of smaller degrees of a

play no roleˆ in the following calculations. Therefore they are not written down in S6 and

S8.

Let ϕ1, ϕ2, and ϕ3 be Weyl group invariants of E7 of degrees 2, 6, and 8, respec- (ν) tively. Then deg(ϕ1 ) = 0. Since sl8 is a maximal rank subalgebra of E7, the E7- (ν) invariants are polynomials in SL8-invariants. Using this, we will show that deg(ϕ2 ) " 2 (ν) and deg(ϕ3 ) " 4. Since both these polynomials are invariants of the Weyl group of

E6, they must be algebraically dependent (recall that the degrees of -invariants are 2, 5, 6, 8, 9, 12).

(ν) Lemma 1. For the Weyl group invariant ϕ2 with deg ϕ2 = 6, we have deg(ϕ2 ) " 2.

Proof. The invariant ϕ2 is a linear combination of SL8-invariants of degree 6. One can express this as follows:

3 2 ϕ2 = x1S2 + x2S3 + x3S2S4 + x4S6, where xi K. ∈ 6 O. YAKIMOVA

(ν) (ν) Assume that deg(ϕ2 ) > 2. Since ϕ2 is an invariant of E6, it cannot be of degree 3. Hence (ν) 6 4 3 deg(ϕ2 ) ! 4 and the coefficients of a , a , and a in ϕ2 are zeros. This condition gives us four linear equations on xi. Let us write down the polynomials in question: 512 64 S3 = a6 + a4(2b2 + τ ) + . . . ; 2 27 3 2 256 32 32 S2 = a6 + a4(6b2 τ ) + a3τ + . . . ; 3 81 9 − 2 9 3 448 4 50 32 S S = a6 + (36 + )a4b2 + (2 + )a4τ a3τ + . . . ; 2 4 81 27 27 2 − 9 3 2 5 20 S = (2 + )a6 + 30a4b2 + a4τ a3τ + . . . . 6 35 27 2 − 27 3

Again we calculate only whose coefficients, which will be used. Since b and all τi are algebraically independent, we indeed obtain four linear equations.

512 256 448 2 27 x1 + 81 x2 + 81 x3 + (2 + 35 )x4 = 0 128 64 4  3 x1 + 3 x2 + (36 + 27 )x3 + 30x4 = 0   64 x 32 x + (2 + 50 )x + 5 x = 0  3 1 − 9 2 27 3 27 4 32 32 20 9 x2 9 x3 27 x4 = 0  − −  The determinant of thissystem is non-zero. Hence the only solution is trivial. Since ϕ = 0, we have proved that deg(ϕ(ν)) 2. 2 ) 2 " #

(ν) Lemma 2. For the Weyl group invariant ϕ3 with deg ϕ3 = 8, we have deg(ϕ3 ) " 4.

Proof. Argument for this invariant is essentially the same as in Lemma 1, but here calcu- lations are more involved. Again

4 2 2 2 ϕ3 = y1S2 + y2S2S3 + y3S2 S4 + y4S2S6 + y5S3S5 + y6S4 + y7S8, where yi K. ∈ We need coefficients of this seven polynomials up to a4. Here they are:

4096 4096 2048 128 S4 = a8 + a6b2 + a6τ + a4(4b4 + 4b2τ + τ 2) + . . . ; 2 81 27 27 2 3 2 2 2048 5120 512 256 416 160 8 S S2 = a8 + a6b2 a6τ + a5τ + a4b4 a4b2τ a4τ 2 + . . . ; 2 3 243 81 − 81 2 27 3 3 − 9 2 − 9 2 COUNTEREXAMPLE 7 3584 8704 1280 256 S2S = a8 + a6b2 + a6τ a5τ + 2 4 243 81 81 2 − 27 3 4064 2144 152 64 + a4b4 a4b2τ + a4τ 2 + a4τ + . . . ; 3 − 27 2 27 2 9 4 3904 20416 488 160 S S = a8 + a6b2 + a6τ a5τ + 2 6 729 243 243 2 − 81 3 820 5 40 +140a4b4 + a4b2τ + a4τ 2 + a4τ + . . . ; 27 2 27 2 9 4 2560 1280 640 320 S S = a8 + a6b2 a6τ + a5τ + 3 5 729 27 − 243 2 81 3 1240 200 10 80 + a4b4 a4b2τ + a4τ 2 a4τ + . . . ; 9 − 9 2 27 2 − 27 4 3136 448 224 448 4112 4 112 S2 = a8 + a6b2 + a6τ a5τ + a4b4 + 16a4b2τ + a4τ 2 + a4τ + . . . ; 4 729 9 81 2 − 81 3 27 2 9 2 27 4 2 28 56 70 S = (2 + )a8 + 56a6b2 + a6τ a5τ + 140a4b4 + a4τ + . . . . 8 37 36 2 − 35 3 81 4

I calculated these expansions on the computer in “Maple”. It is quite possible to check any of the coefficients by hand, but getting them all is rather tiresome. (ν) 8 6 5 4 Assume that deg(ϕ3 ) > 4. Then the coefficients of a , a , a , and a in ϕ3 are zeros. Therefore there are eight linear equations, corresponding to the summands

8 6 2 6 5 4 4 4 2 4 2 4 a , a b , a τ2, a τ3, a b , a b τ2, a τ2 , a τ4, depending on seven variables y . Since at least one 7 7 minor of this matrix is non-zero i × (it was checked on the computer), the only possible solution is zero. Thus if ϕ = 0, then 3 ) (ν) deg(ϕ3 ) " 4. #

Proof of Proposition 1. Suppose that ϕ1, . . . , ϕ7 is a system of homogeneous generators of W S(t) with deg ϕi < deg ϕj for i < j. Then deg ϕ1 = 2, deg ϕ2 = 6, and deg ϕ3 = 8. Clearly (ν) ν(1) 2 (ν) ν(1) (ν) ν(2) (ν) ν(3) ϕ1 h0 is proportional to h0. Hence the polynomials ϕ1 h0 , ϕ2 h0 , and ϕ3 h0 are (ν) (ν) algebraically independent if and only if ϕ2 and ϕ3 are. By Lemmas 1 and 2, we have (ν) (ν) (ν) (ν) deg ϕ2 , deg ϕ3 " 4. Recall that ϕ2 and ϕ3 are invariants of E6. Since the Weyl group of type E6 has no basic invariants of degrees 1, 3, and 4; and only one of degree 2, these polynomials are algebraically dependent. 2 8 O. YAKIMOVA

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