On the locus of p-Characters defining simple reduced enveloping algebras Текст научной статьи по специальности «Математика»

Tom 154, kii. 2

УЧЕНЫЕ ЗАПИСКИ КАЗАНСКОГО УНИВЕРСИТЕТА

Физико-математические пауки

2012

UDK 512.5

ON THE LOCUS OF p-CHARACTERS DEFINING SIMPLE REDUCED ENVELOPING ALGEBRAS

S.M. Skryabin

Abstract

We confirm in two cases the conjecture stating that the reduced enveloping algebra U^(g) of a restricted Lie algebra g is simple if and only if the alternating bilinear form associated with the given p-character £ G @* is nondegenerate.

Key words: restricted Lie algebras, solvable Lie algebras, Probenius Lie algebras, reduced enveloping algebras.

In the representation theory of a finite dimensional p-Lie algebra g over an algebraically closed field k of characteristic p > 0 one is naturally led to consider the family of reduced enveloping algebras U(g) associated with linear functions £ G g* (see [1]). The algebra U^(g) is defined as the factor algebra of the universal enveloping algebra U(g) % its ideal generated by central elements xp — — £(x)p • 1 with x G g, and £ is called the p-character of any g-module which can be realized as a module over U^(g). There is a certain, still far from fully understood, relation between generic properties of the family of reduced enveloping algebras and generic properties of the family of stabilizers of linear functions. The stabilizer j(£) of £ G g* coincides with the radical of the alternating bilinear form ^ : g x g ^ k defined by the rule

(x,y)= £([x,y]) forx,y G g.

The Lie algebra g is called Frobenius if ^ is nondegenerate for at least one £.

In general one cannot determine the type of one particular algebra U^(g) just knowing }(£)■ It is quite interesting and surprising that sometimes this can be done. In [2] it was conjectured that U(g) is simple if and only if $(£) = 0, that is, if and only if ^ is nondegenerate. The purpose of the present article is to verify this conjecture in

two cases. When g is solvable and p > 2 we do this using the description of irreducible g

g

gg

morpliism group. Here we apply geometric arguments to the extension of the family of reduced enveloping algebras constructed in [4].

An example at the end of the paper shows that semisimplicity of the algebra U^(g) cannot be recognized in terms of $(£) by means of a possible generalization of the above conjecture.

1. Solvable Lie algebras

It is assumed in this section that g is solvable and p > 2. Recall that a polarization of g at £ G g* is a Lie subalgebra which is simultaneously a maximal totally isotropic subspace with respect to the alternating bilinear form ^ [5].

Denote by P the set of all triples (p, a, A) such that a C p C g are vector subspaces, A G a* is a linear function and there exists a chain of subspaces

0 = ao C ai C ... C an = a c p = pn C ... C pi C po = g (1)

with the property that

[pi-i, a»] C a» and pi = {x G pi-i | A([x, ai]) = 0} (2)

for all i = 1,... ,n. As one checks by induction on i, each p» is a p-subalgebra of g, and ai is an Weal of pi-1. In particular, p is a p-subalgebra of ^^d a is an ideal of p A [p, a] [a, a]

Lemma 1. Suppose that (p, a, A) G P. // £ G g* is a linear function such that A(x)p — A(x[p]) = £(x)p for all x G a and W is an irreducible Uç(p) -module such that xw = A(x)w for all x G a and w G W, then the induced g-module Uç(g) ®Uf(P) W is irreducible.

Here Uç (p) stands for the reduced enveloping algebra of p corresponding to the restriction of £ to p. The proof is obtained by a repeated application of the characteristic p analog of Blattner's irreducibility criterion [6, Theorem 3].

We will need additional conditions on triples. Denote by P' the set of all triples (p, a, A) a C p C g A G a*

there exists a chain of snbspaces

0 = ao C ai C ... C an = a c p C pn C ... C pi C po = g (3) with the property that

[pi-i, ai] C ai, (4)

pi = {x G pi-i | A([x, ai])=0}, where ai = {y G ai | A(y) = 0}, (5)

p = {x G Pn | A([x, a]) = 0} (6)

for all i = 1,..., n. We will say that chain (3) is (p, a, A)-admissible in this case.

Lemma 2. In a (p, a, A) -admissible chain each pi is a p-subalgebra, ai is an ideal of p^i, and ai is an ideal of pi. Furthermore, p is an ideal of pn.

Proof. Since [pi, ai] C ai % (4) and A vanishes on [pi, ai] by (5), we deduce that [pi, ai] C ai. Since the normalizer of ai in g is a p-subalgebra, an induction on i shows that so too is p^^w [p, a] C a Mid A vanishes on [p, a] by (4) and (6), whence [p, a] C a'n. It follows [[pn, p], a] C [pn, a'n\ + [p, a] c a'n, and so [pn, p] c p. □

Lemma 3. It holds P' C P.

Proof. Let (p, a, A) G P'. Consider a (p, a, A)-admissible chain (3) and for each i define pi = {x G pi | A([x, ai]) = 0}. We obtain then a chain (1) with pi C pi5 and it is checked straightforwardly that (2) is fulfilled. Tims (p, a, A) G P. □

Lemma 4. Suppose that a is a one-dimensional ideal of a solvable Lie algebra h, and b is an ideal of h, minimal with respect to the property that a C b, a = b and [a, b] = 0. Then b is abelian.

Proof. Put c = {x G b | [x, b] = 0}. Then c is an Meal of ^d a C c C b. By the minimality of b we have either c = b or c = a. In the first case [b, b] = 0, and we

c = a b

alternating bilinear form b/a x b/a ^ a. In particular, b/a has even dimension. On the

b/a h b

dim b/a is a power of p. hence odd, by [3, Satz 3]. We arrive at a contradiction. □

Lemma 5. Suppose that (p, a, A) ePIf a = p, then there exists a vector subspace b С p such that a is contained, in b properly, [b, b] С a' = ker A, and, for every linear function / e b* extending A there exists q satisfying (q, b,/) e P'.

(p, a, A) a p

of pn. Let us choose an ideal b of pn such that a С b С p, a = b, and b is minimal with respect to these properties. Then [b, b] С a since pn is solvable and [b, a] С a' by (6). If a = a', then dim a/a' = 1. Lemma 4 applied to the Lie algebra pn/a' and its one-dimensional ideal a/a' shows th at b/a' is abelian in this case. Thus we have [b, b] С a' in any case. If / e b* extends A, then put

Pn+i = {x e p„ | /([x, b']) = 0} and q = {x e p„+i | /([x, b]) = 0},

where b' = {y e b | /(y) = 0}. Note th at b С q sine e / is zero on [b, b] С a'. Obviously q С pn+1 С pn. Setting an+1 = b, we obtain an extension of (3) to a (q, b, /)-admissible chain. Tims (q, b,/n) G V. □

We say that (p, a, A) e P is maximal if p = a. Denote by Pmax С P the subset of all maximal triples and put Pmax = Pmax П P'. All conclusions of the next proposition with P in place of P' were obtained by Strade [3] in a somewhat different language.

Proposition 1. (i) Given £ e g*, there exists (p, a, A) e Pmax зис^ ^hat A = £|p. In this case p is a polarization of g at £.

(ii) Given an irreducible g-module V, there exi sts (p, a, A) e Pmax such ^hat the subspace V\ = {v e V | xv = A(x)v for all x e a} is nonzero.

Proof. Denote by P' С P' the subset of those triples (p, a, A) for which A = £|p. This subset is nonempty as we may take a = 0, p = g. Suppose that (p, a, A) e P' and p = a. Find b as in Lemma 5 an d set / = £|b. There exi sts (q, b, /) e P' which belongs to P' by the choice of We have here dim b > dim a. This argument shows that P' П Pmax ^s nonvoid. Indeed, it suffices to pick out (p, a, A) e P' for which dim a is maximal possible. By Lemma 3 (p, a, A) e P. There exists then a chain (1) satisfying (2). It follows by induction on г that pj = {x e g | £([x, a®]) = 0}. Hence p = a ^s a maximal totally isotropic subspace of g with respect to в? •

Denote by PV С P' the subset of those triples (p, a, A) for which V\ =0. The triple (g, 0,0) is again in PV • Suppose that (p, a, A) e PV and p = a. Let b be as in Lemma 5. Since [b, a] С a', the subspасе is stable under b. Hence the abelian Lie algebra b/a' operates in It follows that V\ ^^^^^^^s a one-dimensional b^^^^^^ule, say kv. The equality xv = /(x)v d^nes a linear function / e b* which extends A. We have v e V^ % ^^e construction. Lemma 5 provides a triple (q, b, /) e P' which belongs to Vy. The intersection Vy П is therefore nonvoid, similarly as in case (i). □

Proposition 2. Suppose that £ e g* and (p, a, A) e Pmax A = £|p .If £ vanishes on j(£), then £(p[p]) = 0. In this case the one-dimensional p -module k\ on which p operates via A has p-character A, and so U?(g) ®Ua(P) k\ is an irreducible g -module of dimension piHdim0-dim3(5)).

Proof. For each subspace h С g denote by (j^ С g its orthogonal complement with respect to в?- One has then (h^)^ = (j + z(£). Consider a (p, a, A)-admissible chain (3). Put p = pn rnd p' = an • Note th at aj_1 С aj for all г = 1,..., n. It follows then from (5) by induction on г that pj = aj-1 for each г. For г = n we obtain p'^ = p. Hence p^ = p' + z(£). Note th at j(£) С p since p is a maximal totally isotropic subspace of g with respect to в? - Under the hypotheses of Proposition 2 j(£) С p П ker £ = p'. Thus p1 = p'.

Observe that [p, p[p]] C [p, p] sinee p is an Meal of p by Lemma 2. Henee £ vanishes on [p, p[p]], and so p[p] C p^ = p'. This shows that £(p[p]) = 0. The elaim about irredueibility follows from Lemma 1, and the dimension formula follows from the equality dimg — dimp = i(dimg — dim3(£)). □

Proposition 3. Suppose that £ G 0* and, (p, a, A) G Pmax with A = £|p. Then every maximal torus of z(£) is a maximal tor us of p.

Proof. Consider a chain (1) satisfying (2). We have a^ = pi, and therefore p^ = = z(£) + ai. As ai is an Meal of pi-i, we get [pi-i, aip]] C [ai, ai] for i > 0, which is contained in the kernel of £. This shows that aip] C p^—i = z(£) + ai-i.

Denote by bi the [p]-closure of a^en bi is an Meal of p since so is ai. Hence z(£) + b^a p-subalgebra for each i, and it follows that bip] C z(£) + bi—i.

Suppose that t is a maximal torus of s G p is a [p]-semisimple element

which centralizes t. We will prove that s G t + bi by the downward induction on i = 0,..., n. For i = n the assertion is clear since t + bn = p. Suppose that s G t + bi for some i > ^^en s = t + x, where t G t, x G bi and [t, x] = 0. By the above sM = tip] +x[p] G z(£)+bi-i. Since s is a linear combination of elements sp^th r > 0, we get s G z(£) + bi—i. The p-Lie algebra hi = (z(£) + bi—i)/bi—i is a homomorphic image of j(£), and therefore the im age of t in hi is a maximal tor us of hi by [7, Theorem 2.16]. It follows that s G t + bi—i, providing the induction step. We can now conclude that sGt+bn=t. and the proof is complete. □

Corollary 1. If z(£) is [p] -nilpotent, then so too is p.

We come to the main result of this section:

Theorem 1. Let g be a solvable finite dimensional p-Lie algebra over an algebraically closed field of characteristic p > 2, and let £ G 0*.

(i) The algebra is simple if and only if fa is nondegenerate.

(ii) If fa is nondegenerate, then £ admits a [p] -nilpotent polarization p such that £(p[p]) = 0, and the single irreducible U (g) -module is induced from the one-dimensional U,e(p)-module on which p operates via £.

Proof. Suppose that ^ is nondegenerate so that j(£) = 0. By Proposition 1 there exists (p, p, A) G Pmax such that A = £|p. Then p is [p]-nilpotent by Corollary 1. By Proposition 2 ®ux(p) is an irreducible g-modulo of dimension p2dlm0.

Since U^ (g) is of dimension pdim it has to be simple. This proves (ii) and also one implication in (i).

Suppose now that U^(0) is simple, and let V be its irreducible module. In view of Proposition 1, there exists (p, p, A) G Pmax such that V\ = 0. Let 0 = v G V> so that kv C is a one-dimensional irreducible U^(p)-submodule. By Lemma 1 the 0-modulo (H'[/£(P) kv is irredncible, hence of dimension p2 dlms. Therefore dimp =

= idimfl. Let 77 G 0* be any linear function such that ?7|p = A. By Proposition 1 p is a maximal totally isotropic subspace of 0 with respect to . The well-known formula dim0 + dimj(n) = 2dimp now yields j(n) =0. By Proposition 2 applied to the linear function n in place of £ the p-character of the p-module kv equals A. Hence A = £|p. We may thus use n = £ in the argument above to conclude that j(£) =0. The proof is complete. □

2. Frobenius Lie algebras with exponentiable adjoint derivations gp

k

X = {£ G g* | U5(g) is simple}, Y = {£ G g* | ^5 is nondegenerate}.

Lemma 6. There exists a homogeneous polynomial function f on the vector space V = g* © k such that

X = {£ G g* | f (£, 1) =0}, Y = {£ G g* | f (£, 0) = 0}.

Proof. Let n = dimg. We will exploit the algebraic family of pn-dimensional associative algebras U5,A = U5,A(g) parameterized by points (£, A) G V (see [4]). The algebra U5,A contains g as a generating subspace and has defining relations

xy — yx = A[x, y], xp = Ap-1x[p] + £(x)p • 1 (x, y G g).

In particular, two special cases of these algebras are U5= U5(g) and U5,0 = S5(g), the factor algebra of the symmetric algebra S(g) by its ideal generated by all elements xp — £(x)p • 1 with x G g.

There is a p-representation ad^,A : g ^ Der U5,A such that ad^,A(x)(y) = [x, y] for x, y G g. In this way U5,A may be regarded as a module algebra over the restricted universal enveloping algebra U0(g) and as a module over the smash product algebra R,a = U5,A#Uo(g). Let

<5,a : R,a ^ T5,a = Endfc U5,a

denote the corresponding representation. Note that dim R5,A = dim T5,a = p2n- Hence the map <5,A is bijective if and only if U5,A is a simple R5,A-module. Now the R5,A-submodules of U5,A are precisely those left ideals that are stable under the action ad^,A . A=0

U,A as a R5,A-module is equivalent to the simplicity as an algebra. In particular,

X = {£ G g* | <5is bijective}.

On the other hand, according to [4, Proposition 3.4] the algebra S5 (g) has a unique maximal g-invariant ideal /, and the codimension of this ideal is pcodimo z(5), in order that S5 (g) be a simp le R5 ,0-module, it is necessary and sufR cient that I = 0, which amounts to j(£) = 0, that is, to £ G Y- It follows that

Y = {£ G g* | <5,0 is bijective}.

It remains to show that the bijectivity of <5,A can be expressed by means of the f(£, A) = 0 f V

view R5,A and T5,A as fibers of two algebraic vector bundles R and T over V. Let e1,..., en be any basis for g. The monomials e^1 • • • e^" with 0 < a» < p form a basis for each U5,A. These monomials give rise to a basis for each R5,A and a basis for each T5,A , yielding trivializations of R and T. The entries of the matrix of <5,A in the above

(£, A) f(£, A)

matrix, we see that <5,A is bijective if and only if f (£, A) = 0.

As explained in [4], for each 0 = t G k there is a g-equivariant algebra isomorphism : U5 ,A ^ Ut5,tA(g). Hence the algebra U5 ,A has no nontri vial g-invariant ideals if and only if so does Ut5,tA(g). In other words, bijectivity of <5,A is equivalent to bijectivity of <t5,tA • It follows that the zero locus of the polynomial function f is a conical subset of V, whence / is homogeneous. □

Remark. It is possible to compute the degree of the polynomial function f in Lemma 6 proceeding as follows. The isomorphisms 0t induce actions of the one-dimensional torus <Gm on R and T. Taking quotients modulo these actions we pass to a morphism of vector bundles R —► T over the projective space P(V) associated with V. Let also U = U/Gm, where U is the vector bundle over V \ {0} with fibers U^,^ • Each line bundle over P(V) is isomorphic to some L(s), defined as the quotient of (V \ {0}) x k by the action of Gm such that t • (v, c) = (tv, tsc), where s G Z. The scalar multiples of any monomial e^1 • • • e^" produce a Gm-stable line subbundie of U. This leads to a decomposition

U = (J) L(-ai-----a„).

{(ai ") 10<a^<p}

The bundle R is isomorphic to a direct sum of pn copies of U, while T = U 0 U\

2 n _

As a result, /\p R = L{—d)1 where

{(ai J...Ja")|0<ai<p}

2»i _

while /\p T = L{0) is trivial. Now / can be identified with a section of the line bundle Hom(L(-d), L(0)) = L(d). ^^^s means that deg f = d.

Corollary 2. If 0 is Frobemus, th at is, Y = 0, the n f = 0, and therefore X = 0.

Whether X = 0 implies Y = 0 is a special case of the still open Kac-Weisfeiler conjecture from [8].

Proposition 4. If 0 is Frobenius and Y C X, then X = Y-

Proof. By Lemma 6 the complements Xc = 0* \ X and Yc = 0* \ Y are hy-persurfaces in 0*. ^^e inclusion Y C X entails Xc C Yc • Therefore each irreducible component of Xc is an irreducible component of Yc. Since Yc is a conical subset of 0*, so too is each irreducible component of Yc • It follows that Xc is a conical subset as well. Hence the polynomial function £ ^ f (£, 1) defining Xc is homogeneous. We can write

d

f (£,A) = 53 fi(£)Ai,

i=0

fi d - i 0* 0

is Frobenius, we have Y = 0, whence f0 = 0. But then we must have fi =0 for all i > 0, that is, /(£, A) does not depend on A. □

Theorem 2. Let 0 be a Frobenius p-Lie algebra with the automorphism group G. Suppose that ad 0 C Lie G. Then X = Y •

Proof. Both X Mid Y are stable under the coadjoint action of G. For any £ G Y the nondegeneracy of ^^ yields 0 • £ = 0*. Hence the tangent space at £ to the G-orbit G£ 0* G£ 0*

0* Y G X

0* X Y = 0 Y C X

4 applies. □

3. The semisimple locus: an example

g*

X = {£ e g* | U?(g) is semisimple}, Y = {£ e g* | z(£) is toral}.

It was proved in [4, Section 4] that both of them are open in g* and that Y = 0 implies X = 0. Moreover, the stabilizers j(£) of all linear functions £ e Y have equal dimensions. If s denotes their common dimension, then for each £ e X the semisimple algebra U?(g) has precisely ps nonisomorphic simple modules, all of equal dimension.

One may ask what are those p-Lie algebras for which X = Y- For instance, if g is the Lie algebra of a simply connected semisimple algebraic group G and p is good for the root system of G, then X consists precisely of the regular semisimple linear

X=Y

provide examples of nilpotent p-Lie algebras for which X = Y-

Consider a p-Lie algebra g whose center t is a toral subalgebra of codimension 2 in g and [g, g] С t. Let u, v e g span a subspace complementary to t in g. There is an element 0 = t e t such that [u,v] = t. Then [g,g] = kt is a one-dimensional snbspace.

g

cides with t. Now t С z(£) for all £ e g*. Hence j(£) is toral if and only if j(£) = t. If z(£) = t, then j(£) = g, which occurs preciselу when £ vanishes on [g, g]. It follows that

Y = {£ e g* | £(t) = 0}.

Denote by t*(1) the vector space of all p-semilinear maps t ^ k, that is, t*(1) is the Frobenius twist of the dual space t*. The mар p : t* ^ t*(1) defined by the rule

p(A)(x) = A(x)p - A(x[p]) for A e t* and x e t

is a finite snrjective morphism of algebraic varieties. There is also a bijective morphism t* ^ t*(1) given by A ^ Ap, where Ap(x) = A(x)p.

With any simple g-module V one can associate a linear function A e t* such that each element x e t acts in V as a scalar multiplication by A(x). If £ is the p-character of V, then p(A) = £p|t. Conversely, for an у pair A e t* and £ e g* satisfying the previous equality there is precisely one simple U?(g)-module V which has A as the associated function. If A(t) = 0, then [g, g] annihilates V, whence dim V = 1. V

g dim V = p p

N = pdim % for each £ e g* there are precisely N nonisomorphic simple U? (g)-modules. In order that U?(g) be semisimple, it is necessary and sufficient that its dimension pdim 0 be equal to J^(dim V)2, the sum over all those modules. This happens precisely when all simple U?(g)-modules have dimension p. We conclude that

X = {£ e g* | A(t) = 0 for each A e p-1(£p|t)}.

t tip / kt. Then neither X С Y nor Y С X.

To see this let A and £ be as above. If A(t) = 0, but A(t[p) = 0, then the equality A(t)p - A(tip) = £(t)p yields £(t) case £ e Y, but £ / X. Now the subspace

S = {A e t* | A(t) = A(t[p]) =0}

has codimension 2 in t^. Hence p(S) is a closed subvariety of codimension 2 in t*(1), and it follows that there exists £ e g* such th at £(t) = ^^ut £p|t / p(S). In this case £ / ^ut £ eX.

This work was Supported by the Russian Foundation for Basic Research (Grant No. 10-01-00431) and the Presidential Grant for Support of Leading Scientific Schools (Grant No. 5383.2012.1).

Резюме

C.M. Скрябин. О локусе р-характеров, определяющих простые редуцированные обертывающие алгебры.

В двух случаях подтверждена гипотеза, утверждающая, что редуцированная обёртывающая алгебра Щ (g) ограниченной алгебры Ли g является простой тогда и только тогда, когда альтернирующая билинейная форма, ассоциированная с заданным р -характером £ G g*, невырождена.

Ключевые слова: ограниченные алгебры Ли, разрешимые алгебры Ли, фробепиу-совы алгебры Ли, редуцированные обертывающие алгебры.

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Поступила в редакцию 03.02.12

Skryabin, Sergei Markovich Doctor of Physics and Mathematics, Professor, Department of Algebra and Mathematical Logic, Kazan Federal University, Kazan, Russia.

Скрябин Сергей Маркович доктор физико-математических паук, профессор кафедры алгебры и математической логики Казанского (Приволжского) федерального университета, г. Казань, Россия. E-mail: Serge.SkryabinQksu.ru