Пополнения алгебр Ли Текст научной статьи по специальности «Математика»

МАТЕМАТИКА

Вестн. Ом. ун-та. 2013. № 4. С. 67-70.

УДК 512.554.3 М. Шахрияри

ПОПОЛНЕНИЯ АЛГЕБР ЛИ

Алгебра Ли над полем характеристики нуль называется пополнением рациональной алгебры Ли, если она содержит ее в качестве рациональной подалгебры и совпадает с линейным подпространством, порожденным этой подалгеброй. Изучается класс всех пополнений фиксированной рациональной алгебры Ли.

Ключевые слова: алгебра Ли, пополнение.

Let L be a Lie algebra overQ , the field of rational numbers. A pair (K,E) with E a field characteristic zero and K a Lie algebra over E, is a completion of L, if L <Q K (a Q -Lie subalgebra), andK = SpanE (L). In manysituations,

dimg L is infinite, while dimg K is finite. Therefore, we can applysome properties of finite dimensional Lie algebras in characteristic zero to L.For example, we will show that, if L is nilpotent, then it can be embeddedin the algebra of strictly upper triangular n x n matrices with entries fromE, for some n. In

some cases (L,E) is a completion of L, for example, let L = , the infinite di-

mensional Q -algebra of three dimensional real vectorswith the wedge product as the Lie bracket. Although L is infinite dimensional, its completion (L,R) has dimension 3. This observations motivateus to study the class of completions of rational Lie algebras. Most of thework here can be done for other fields, but we do it just for Q .

The type of problems we are going to deal with in this article, are notnew; at least in two situations, similar problems are studied. The first oneis the study of centroid of Lie algebras. Recall that for a Lie algebra L, itscentroid is defined by

r(L) = {a e End(L): Vx ,y eL,[a(x),y ] = a[x,y ]}.

The centroid of L is the largest ring in which L is a Lie algebra over it. So if(L,E) is a completion of L, then E <T(L) . If L is simple, then its centroidis a field (see [1]), so it is the largest field E, with (L,E)a completion.

The second similar situation can be seen in the theory of exponential-groups, where the concept of completion for groups is defined. The conceptof centroid can be also defined for groups and it is the largest ring in whichthe group admits as a ring of scalars. For a complete description of exponential groups see [2] and [3].

This article is arranged in three sections; in the first one, we show thatany abelian rational Lie algebra (a vector space over Q ) is potentially

onedimensional. In the second section, we introduce the class XL of all completions of L and we show that this class is elementary in the case dimL < <x>.

In the section three, we introduce entangled ideals of E ®qL and we

studytheir connections to completions. This section contains also some results onthe completions in some finite dimensional cases.

1. Abelian Lie algebras

In this section, we show by an elementary argument that any abelian Liealgebra (i.e. any vector space over Q ), is potentially one dimensional,

i.e.for any vector space L over Q , there exists a field E of characteristic zero, such that dimE L = 1. Then we show that this cannot be generalized fornilpo-tent Lie algebras of class 2.

© M. Шахрияри, 2013

Proposition 1.1. Let L be a vector space over Q . Then there exists a field Eof characteristic zero, such that L is a vector space over E and dimE L = 1.

Proof. First, let dim^ L = n and x 1,...,xn be a basis of L. Suppose

q(t) = a0 + at + • •' + an-1tn-1 + antn e Q[t] is an irreducible polynomial. Let X e C be any root of q (t) and E = Q(X) . Define an action of E on L by the Q -bilinear extension of following rules

Xx 1 = x2

2 — x 3

X^n =~a0x 1 -a1x 2--------an -1xn .

It is easy to check that L is a vector space over E of dimension 1. Now, suppose dimQ L — a is infinite. Let X be any set with the cardinality a and suppose Q(X) is the field of rational functions with variables in the set X.Note that

Q(X) = |J Q(Y),

Y c X

where Y varies in the set of all finite subsets of X. Since every Q(Y) iscountable, so we have

|Q(X)| = |X|. Hence as Q -spaces, we have L = Q(X) .Let f: L ^ Q(X) be a Q -isomorphism. Now for q e Q(X) and x e L , define q -x =f \qf (x )).

So, by assuming E = Q(X), L is an Espace with dimE L =1 .

One may wants to generalize this proposition for the case of nilpotent Liealgebras of higher nilpotency classes. But this is not true in general, sincethe three dimensional Heisenberg algebra H = (x, y, z ^ with [x, y ] = z and z e Z (H) is a nilpotent rational Lie algebra of class two but there is no field E with dimE H = 2 . In general, the extension degree

[E: Q] is restricted bythe numbers

dimQ CL (x), x e L . To see this, let L be any Lie

algebra over E. Then clearly for any x e L , we have SpanE (x) c CL (x), so

[E : Q] = dimQ Spane (x) < dimQ Cl (x) -

Therefore, we have

Lemma 1.2. Let L be a rational Lie algebra and E be any field such that(L,E) is a completion for L. Then

[E: Q]< mindimQ Cl(x).

By a similar argument, we can prove the following proposition.

Proposition 1.3. Let L be an infinite dimensional rational Lie algebrawhich has a finite dimensional completion of the form (L,E). Then i )either L is solvable or for any

ordinal a we have dimQ L(a) = Ki.

ii) either L is nilpotent or for any ordinal a we have dimQ ya (L) = rn .

2. The class of completions

In this section, we show that the class of completions of any finite dimensional rational Lie algebra is elementary (axiomatizable in some firstorder language). We also show that this is not true for the class of all fieldsE in which L is a Lie algebra over E. Remember that a pair (K,E) is acompletion of L, if

i) E is a field of characteristic zero,

ii) K is a Lie algebra over E,

iii) L <q K (a Q -Lie subalgebra),

iv) K = SpanE (L).

Let XL be the class of all completions of L. Theorem 2.1. If dim^ (L) is finite, then the class XL is elementary.

Proof. Let v 1,...,vn be aQ -basis of L. We introduce a first order language

L = (0,1, +,x,q,(a*,a eL))

which contains constant symbols 0, 1 and a for all a e L . It also has twobinary functional symbols + and x, and q is a unary predicate symbol whichwill be interpreted as "to be scalar1’. Let E be the following set of axioms:

1) VxVy(q(x) aq(y))^(q(x + y) aq(xy)).

2) VxVy (q( x) a q( y))

^ (x + y = y + x a x x y = y x x).

3) VxVyVz (q(x) a q(y) a q(z)) ^

^( (x + y) + z = x + (y + z) a (x x y) x z =

= x x (y x z) )

4) Vx q(x) ^ x + 0 = x.

5) Vx 1 x x = x.

6) Vx q(x) ^(3y q(y) ax + y = 0).

7) VxVyVz (q(x) a q(y) a q(z)

^ (x x (y + z) = x x y + x x z).

8) Vx (q(x) a x ^ 0)) ^ (3y q(y) a x x y = 1).

9) VxVy (~ q(x) a - q(y))

^ (- q(x + y)a ~ q(x x y)).

10) VxVy (~ q(x)a~ q(y)x + y = y + x.

Пополнения алгебр Ли

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11) Vx ~ q(x) = x x x = 0*.

12) VxVyVz (- q(x)a ~ q(y)a ~ q(z)) =>

= x + (y + z) = x + (y + z).

13) Vx - q(x) = x + 0* = x.

14) Vx - q(x) = (y - q(y) ax + y = 0*).

15) VxVyVz q(x)a - q(y)a - q(z)) =

= x x (y x z) + y x (z x x) + z x (x x y) = 0*.

16) VxVyVz (- q(y)a - q(z)) =

= x x (y + z) = x x y + x x z.

17) VxVy (q(x)a - q(y)) =- q(x x y).

18) VxVyVz (q(x)a - q(y) a - q(z))=

= x x (y x z) = (x x y) x z = y x (x x z).

19) VxVyVz (q(x) a q(y)a - q(z)) =

= x x (y x z) = (x x y) x z.

20) VxVyVz (q(x)a - q(y)a - q(z)) =

= (x + y) x z = x x z + y x z.

21) for all a e L the axiom: - q(a ).

22) for alldistinct a,b e L the axiom: a* ^ b*.

23) for all a, b e L the axiom: (a + b) = a + b and [a,b]* = a* x b*.

24) for all natural number m the axiom:

Vx (q( x) a mx = 0) = x = 0.

25) Vx - q(x) = (3x13x2 ..3xn(a”=1 q(xi)) a

a(x = E”=1 xt xv*)).

Now, let M e Mod(E) be a model of E. Define

K = {x e M : ~ q (x)}

and

E ={x e M : q (x)}

Then it is easy to verify that (K,E) is a completion of L. Conversely if(K,E) is a completion of L then we let M = K U E . For all x , y e M , define x + y to be their sum in E, if x, y e E , their sum in K if, x , y e K , and 0 otherwise. Similarly, define x x y to be xy , if x, y e E, [x, y ], if x, y e K ,and the scalar product otherwise. It can be verified that M is a model for E and so E axiomatizes the class XL .

It is clear that the class of all Lie algebras over a fixed field E is elementary.We show that the class of all fields for a fixed rational Lie algebra is notelementary. Let L be such a Lie algebra. By FL we denote the class of all fields E such that (L,E) is a completion of L. Note

that every element of FL is a sub-ring of the centroid r(L ) .

Proposition 2.2. Let L be a finite dimensional Lie algebra over Q . Then FL is not elementary.

Proof. Let L be a first order language and E be a set of axioms of FL . Since L is countable, we may assume that L is also countable. Now Q is a modelfor E and Q| > |L|. So by the Lowenheim-Skolm theorem, for any in-finitecardinal a , the set E has a model E of size a . Now

dimQ L = [ E: Q] dimEL,

and the left hand side is finite. So [E: Q] is finite, contradicting |E | = a .

Similarly, one can prove that the class FL is not elementary for any abelianLie algebra L.

3. Entangled ideals

In this section, we introduce the concept of entangled ideals in the tensorpro-duct E ®q L and we show that they have close

connections with the completions of L. For a rational Lie algebra L and any field E of char-acteristiczero, we say that a Lie algebra K is an E-completion of L, if (K,E) is acompletion for L.

Definition 3.1. An ideal N of the E-algebra E ®qL is called entangled, if

N f| (1 ® L) = 0.

Proposition 3.2. Let K be an E-completion of L. Then there exists anentangled ideal N <1E ®q L such that

N

Conversely, for any entangled ideal N, the Lie algebraKdefined by theabove equality is an E-completion.

Proof. Define a map <p :E x L ^ K by (p(x,a) = xa . This map is Q -bilinearand so there is a Q -linear map p : E ®qL ^K with

p (x ® a) = xa . Since K = SpanE (L), so p is surjective. Also for any Xe E , we have p (X(x ® a)) = p> ((Xx) ® a) = Xp> (x ® a) hence p is E-linear. Further

p ([x ®a,y ®a]) = p ((xy) ® [a,6 ]) = xy [a,b ] = [p(x ® a), p (y ® a)]

So, p is an epimorphism of E-Lie alge-bras.Let N = kerp . Then

E ®qL

K =-----

N

Note that

N = {Z x ® a e E ®ql : Z xa = 0}

i i

So, if 1 ® a e N , then a = 0 and therefore N is an entangled ideal.

Conversely, let N < E ®qL be an entangled ideal. Let K be defined asabove. Then the map a ^1 ® a + N is a Q -embedding of L in K. So, we may assume that L <q K. Now

clearly K = SpanE (L) and so K is an E-completion of L.

Example 3.3. Let L be a Lie algebra over E. Then in the same time L isa Lie algebra over Q . Let I <qL, f e AutQ (L) , and <J e Aut(E) . Define

N (I, f ,a) =

= {Zxi ® a> e E ®qI : Za(x) f (a,) = 0}.

i i

Note that the map p:E xI ^L defined by p(x ,a) = a(x f (a) is Q -bilinearand so

N (I ,f ,a) is well-defined. It can be easily seen that N (I ,f ,a) is anentangled ideal. So we have a completion

E ®qL

K (I, f ,a) =------Q----

N (I, f,a).

Proposition 3.4. Let E be a finite extension of Q and L be a finite dimensional Lie algebra over E. With the above notations, we have dimE K (I f ,a) = dimE K f (I),1,1), and in the special case,

dimE K(L f ,a) = dimE L Proof. Define a Q -linear map

<j® f: E ®qL ^ L by (a® f)(x ® a) =

= a(x) f (a) . This is an invertible map and we have

(a®f )(N (I f ,a)) = N f (I ),1,1).

Therefore,

dimQ N (I f ,a) = dimQ N f (I ),1,1). Now, since [E: Q] is finite, so

dimE K(I f ,a) = dimE K f (I),1,1).

As an special case, when I = L , we have dimE K(L f ,a) = dimE K(L,1,1).

But by an easy argument, one can see

that

dimE N(L,1,1) = ([E: Q] -1) dimE L ,

so we have

dimE K (L f, a) = dimE L .

We can say more about the structure of K (L f ,a) . Let a ^ a be theembedding of L in K (L f ,a) . We have,

Proposition 3.5. By the above notations,

K(L f ,a) = {a :aeL}

and for allx e E, x .a =f '(a(x f (a)) .

PToof. Let N = N (L f ,a) and

X =Zx, ®at + N eK(L f ,a) . We knowthat

Za(xi)f (ai) e L , so for some a e L , we have

f (a)=Z a(x; f (ai ) . Therefore 1® a —

—Z xt ® at e N and hence X = a . Now,

ifx e E , then x a = x ® a + N = 1 ® b + N , for some b. Clearly, f (b) = a(x )f (a), so

b =f \a(x f (a)).

Hence

x a =f — (a(x f (a)).

Now, we show that nilpotency (solvablity) of L implies nilpotency (solvablity) of its completions.

Proposition 3.6. Let L be a rational Lie algebra and K be a completionfor L over some field E. If L is nilpotent, then K is nilpotent of the sameclass. If L is solvable, then K is also solvable of the same derived length.

Proof. We know that there is an entangled ideal N such that

E ®n L K = -

yQ-

N

So, for any n we have

Yn ( K ) =

E ®Q Yn (L) + N

() E ®QLn ) + N

K (n ) = Q_________

N N

This shows that, if Yn (L) = 0 (L(n) = 0 ), then Yn (K ) = 0 ( K(n ) = 0 ). Onthe other hand, if Yd (K ) = 0 ( K(d ) = 0 ) , then since L œ K , we

must have Yd (L ) = 0 ( L(d ) = 0 ).

A Lie algebra L overQ is potentially finite dimensional, if it has a finitedimensional completion. Using a theorem of Ado (see [1]), it is easy toprove that L is potentially finite dimensional, if and only if L <q gln (E) forsome field

E of characteristic zero and a natural number n. By the aboveproposition, we have

Corollary 3.7. Let L be a potentially finite dimensional nilpotent rationalLie alge-

bra.ThenL <q n+ (E), for some field E

and n > 1, where n + (E ) is the Lie algebra of strictly upper triangular n x n matrices over E.

REFERENCES

[1] Jacobson N. Lie algebras. Dover Publications, 1971.

[2] Myasnikov A.G., Remeslennikov V. N. Exponential groups I: Foundations of thetheory and tensor completions // Siberian Math. Journal. 1994. Vol. 35. № 5. Р. 986-996.

[3] Myasnikov A. G., Remeslennikov V. N. Exponential groups II: Extensions of centralizers and tensor completions of CSA-groups // Int. J. algebra and computations. 1996. Vol. 6. № 6. Р. 687-711.