Algebra with a symplectic form

ANNALES SOC 1 ET AT IS MATHE MAT1CAE POLONAE Series I: CO M MENTATIONES MATHEMATICAL XXVI (1986) ROCZNIKI POLSK1EGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXVI (1986)

T

W

(Lôdz)

Algebra with a symplectic form

Abstract.

Introduction. In this note we generalize the concept of a Lie symplectic group considered by Chu [3]. We are interested in purely algebraic structures related to this subject. We consider an algebra A equipped with a skew-symmetric non-degenerate 2-form со satisfying the condition

(1) — co(xy, z) + co(xz, y) — co(yz, x) = 0.

Theorem 1 can be considered as our main result. It says that any 2n- dimensional, nilpotent symplectic algebra A over a field of characteristic Ф 2 contains and n-dimensional subalgebra В such that со (В, В) = 0.

In the case of a Lie group condition (1) says that the form со is closed, i.e., dco = 0.

1. Definitions and examples.

D

1 (Sternberg [7]). A symplectic manifold is a pair (M, со), where M is a 2n-dimensional differentiable manifold and со is a closed differential 2-form which is of maximal rank everywhere. Such a form is called a symplectic form on the manifold M.

Let G be a Lie group and H a closed subgroup of G.

D

2 (Chu [3]). We-say that a homogeneous space M = G/H is symplectic if there exists a symplectic form со on M which is invariant under diffeomorphisms Lg, geG, given by Lg{g' H) = gg' H (i.e., Lg* со = со for all

g g

G).

D

3 (Chu [3]). A symplectic group is a pair (G, со), where G is a Lie group and со is a closed left-invariant 2-form of maximal rank.

If со is a left-invariant symplectic form on G, then со defines a skew- symmetric bilinear form coe of maximal rank on the Lie algebra 51 of G satisfying the condition

— coe{\X, Y\, Z) + coe(\X, Z|, У) —сое(|У, Z|, X) = 0 for all X, У, Z.

Conversely, every 2-cocycle of maximal rank on ЗД defines a left-invariant symplectic form on G.

D

4. An algebra with symplectic form (or symplectic algebra) is a pair (A, со), where A is an algebra over a field К and со: A x A - * K is an alternating bilinear non-degenerated form satisfying the condition

(1) — co(xy, z) + co(xz, y) — co(yz, x) = 0 for all x, y, z e A.

A subalgebra of a symplectic algebra {A, со ) is a pair (A', со'), where A' is a subalgebra of A and со' = со/A is a non-degenerated form. A weekly- symplectic algebra is a pair (A, со), where A is an algebra over a field К and со is an alternating bilinear form satisfying condition (

).

P

1. Let (A, со) be a symplectic algebra. Then the multiplication on A satisfies the condition

(

) a

=

for all a e A .

P roof. Since co(a2, x) = co(ax, a) — co(ax, a) = 0 for all x e A , we have a

=

.

E

1. Let A = ( R \ + , *, o) be a 4-dimensional algebra over R with a base (e

e2, еъ, e4), where + ; R

x R4 -> R

and •; R x R 4 -* R

are given by natural rules and the multiplication o; R

x fl4-> R

is defined as follows:

0 e l e"l2 ee33 e"44

ei ⁰ e3 e t e4

e2 ~ e3 ⁰ ei e3

e3 -< ? i - e x 0 ~ e2

<-'4 ^{-< ? 4} ~ e3 e2 ⁰

Let со be the form R

x R

((cc1,

, a 3, a

fi2, Ръ-> /*

)) ->(Xi fi

— fii a

+ a

fi

— fi

ot

)eR. The pair (A, со) is a symplectic algebra.

R em ark 1. Example 1 shows that a symplectic algebra need not be a Lie algebra.

P

2. Let К be a field of characteristic # 2. Let (A, со) be a finite-dimensional symplectic algebra over K. Then dim A = 2n and A

=

J---L E„, where E{ (i = 1, n) are hyperbolic planes.

C

1. I f the hyperbolic planes E{ satisfy conditions

at • bi e £; for all at ,b ieE i (i =

, ..., n),

then (£,, co\E.) are symplectic algebras.

Algebra with a symplectic form 363

C

2. Let I be an ideal of A. Then I

is a subalgebra of A. I f I is a non-degenerated ideal of A, then (I1, J is a symplectic algebra.

Proof. Since co(xx x 2, a) — a, x2) — co(x2 а, x y) = 0 for all x l , x

e l \ a e l , then we have x

x

e l 1. If / is a non-degenerated ideal, then I

is also a non-degenerated algebra (Lang [4], p. 384).

2. Nilpotent and solvable symplectic algebras. Polarization of symplectic algebras.

P

3. An anticommutative algebra A over afield К is nilpotent if and only if the algebra A is associative-nilpotent.

P ro o f. See Schafer [6], p. 18, Theorem 1.

C

3. I f a symplectic algebra {A, со) is nilpotent of the length n, n

^ 4, then it is solvable of the length less than n.

P

4. I f a symplectic algebra (A , со) over a field К is nilpotent, then any 2-dimensional symplectic subalgebra (В , cot) of {A, со) is abelian.

P ro o f. Let (e

e2) be the base of В such that co{ex, e2) = 1. Let n be a length of nilpotence. It is easy to see that

(3) 0 = со (

?! , et (<?,, (... e

(e2 e2)))) = jun~ 1,

where et e

= Àei + pe2, А, р е К and the product in (3) consists of n factors.

Thus /i = 0. Similarly, it is easy to verify 2 = 0.

R em ark 2. The sequence A x —A, А г = А -А, А ъ = A- A 2, ... denotes the lower central series of A. The sequence A(1) = A, A(2) = A ’A, A(3) = A(

) -A(2), ... denotes the derived series of A.

P

5. Any non-zero symplectic algebra has no unit element. I f (A, со) is an associative symplectic algebra over a field К of characteristic ф 2, then A

= 0 and со (A2, A 2) = 0.

P ro o f. Suppose that l e A. Since

cd

(1, a) = co(a, 1) = 0 for all a e A ,

then we have 1 = 0, which proves the first part of our proposition.

Since 0 = |x, Iy, z|| + |y|z, x|| + |z, |x, y\\ = 4 xyz for all x, y , z e A (|x, y\

= xy — yx), then we have A

= 0 and co{A2, A2) = 0, which proves the second part of our proposition.

D

5. A polarization of a finite-dimensional symplectic algebra {А, ш) is a subalgebra В of A satisfying the conditions

(4) d im B = id im A ,

(5) co(B, B) = 0 (i.e^ В is an isotropic subalgebra of A).

We are going to prove that every nilpotent symplectic algebra has a

polarization.

R em ark 3. Let Л be a nilpotent algebra. If V is subspace of the vector space A and Ak a V с Ak^ x, к < m, where m is the length of nilpotence A, then V is an ideal of A.

In fact A • V c A • x = Ак a V.

Pr o p o s it io n 6.

Let

A , со) be a finite-dimensional nilpotent, symplectic algebra over a field К o f characteristic Ф 2. Let dim A = 2n and m be the length of nilpotence. Then

(a) co(A2, Am- 1) = 0, (b) dim A

^ 2(n— 1),

(c) if dim A

= 2(n — 1), n Ф 1, then dim A

= 2n — 3,

(d) if A is a Lie algebra, then co{As, At) = 0 for all s , t e N such that s + t = m+ 1,

(e) if either m = 2 or m — 3, then (A , со) has a polarization which is an ideal o f A.

P ro o f. Suppose that dim A

= 2n—1. Let (et , e2, ..., e2n- 1, e2„) be a base of A such that {ex, e2, ..., e2n-\) is a base of A 2. The ideal A

is generated by the set В = [е^у i, j = 1 ,..., 2n, i < j}. Since В <=. A3, we have A

— A

contradicting the fact that A is nilpotent. Now, let dim A

= 2(n— 1) (и > 1). Let (ei , .... e2n) be a base of A such that (el5 ..., ^(«-n)

is a base of A 2. The ideal A

is generated by the set B = \е{еу i ,j

= 1, ..., 2n, i < j}. We have e{ ej e A

for i = 1, ..., 2n — 2 and j — 1, ..., 2n, and c2i,-i ' е

пфАз- Thus dim A

— 2n — 3. The proof of conditions (b) and (c) is finished.

Suppose that A is a Lie algebra. Since As- A, cz As+t for all s, t, we have

oj ( f l ! ( a 2 ( . . . ( a s _ , as) . . . ) ) , b t ( b 2 ( . . . (b t _ t bt) . . . ) ) )

= со (a

(a

(... (as_ , as) ...)), ax (bx (... (b, _ , bt) ...)))

= ... = m (us as_ ! (as_ 2 (... (ax (bl (...(bt_ l bt) ...))}))) = 0 for all generators ax (a

a j...)) of As and bl (b

(...{bt- l bt)...)) of At.

This proves (d).

For m = 2 Proposition 6 (e) is clear. If U is maximal isotropic subspace

of A ([1], Theorem 3.11, p. 122), then it is also an ideal of A. Let m = 3. We

have by (a) that A

is an isotropic subalgebra (ideal) of A. If dim A

= n,

then A

is a polarization of (A, cu). Let dim A

= к < n. There exists a

hyperbolic extension В of A

such that Л = В_Ц31 (i.e., A = В ф В

and

co(B, B) = 0). Taking into account Corollary 2 we conclude that the pair

(B1,

|

_

) is a 2(w — k)-dimensional symplectic algebra with degree of

nilpotence m = 2. Let U be a maximal isotropic subspace of B \ It is easy to

check that C = A 2@U is a polarization of (A, со). This completes the proof

of Proposition 6.

Algebra with a symplectic form 365

We shall deal exclusively with a nilpotent finite-dimensional symplectic algebra (A, со) over field К of characteristic Ф 2. Let dim A = 2n and m denote the length of nilpotence.

Le m m a

1. Let (A , of) be a symplectic algebra. I f a sequence A

= B0,

Bi , Bk satisfies the conditions

1° B{_ t с: В^ and dim B, = dim B,_ x + 1 for i = 1, . . k, 2° Bt is ideal of B f t ,

3° B fr \ A

— Bk,

4° if index r{ (i = 1, . . k) is the greatest possible such that Bi a Ar., then B f n Ar.+

cz Bi,

then B0, Bl5 Bk satisfy also the conditions (a) Bf cz A 2, i = 1, . . k,

(Й B f cz В f„ l CZ...C B f, (y) aj(Bi, B() = 0, i = 0, 1, k, (<5) dim Bk ^ n.

R em ark. Condition 2° of Lemma 1 makes sense. Indeed, since B

= A

is an ideal of A, then by Corollary 2, B f is a subalgebra of A.

Similarly, it is easy to verify that B f, i = 1, ..., k, are subalgebras of A.

P ro o f. Conditions (a) and (/?) are clear. From Proposition 6 (a) it follows that B

is an isotropic ideal of A. Assume that condition (y) is satisfied for some B(, i < k. From conditions 1° and 2° we gather that there exists an element es + 1e B f \B t such that Bi + l = _L Lin(es+1). Hence we get that oj(Bi + 1, Bi+1) = 0. From (y) and Theorem 3.11 of [1] it follows that (<5) is satisfied.

Le m m a

2. Let (A, со) be a symplectic algebra. There exists a sequence

,4m_ i — B0, Bl , . . . , B k of subalgebras of A satisfying conditions l°-4° of Lemma 1.

P ro o f. We shall construct element Bx of our sequence satisfying

conditions 1°, 2°, 4°. By Proposition 6 (a), we have A

c: B f — A f - l . We

can assume (Proposition 6 (e)) that m > 3. Hence we see that there

exists an element ex e Am_ 2\B 0 c B f. Let Bl = B

l L i n ^ ) . Since

Am^ l cr B x cz Am_2, then (Remark 3) Bk is an ideal of A. It is easy to see

that Bj satisfies conditions 1°, 2°, 4°. Assume then that we have a sequence

B0, Bl5 ..., Bs satisfying conditions 1°, 2°, 4° and such that B f п А

Ф Bs. We

shall construct an algebra Bs+1 satisfying conditions 1°, 2°, 4°. Let r be

the greatest possible index such that there exists an element

es+ie (B f n Ar)\B s ( B f n A

Ф Bf). Let Bs+1 = Bs _L Lin(es+ j). It is clear that

Bs+1 satisfies condition 1°. Since Bs Bf_

czBs then (Lemma 1) we have

Bs B fcz Bs. Since B f n A fs+l cz Bs, so r ^ rs. Since B f n A r+l cz Bs, then we

deduce that condition 2° is satisfied. Indeed

B f 'B

= B f(B s 1 Lin (es+ j)) = B f Bs + B f Lin (es+1) cz Bs + B f Ar+ i c Bs c Bs+1.

Since Bxn A

c Bs, then B f+1n Ar+l cz Bs cz Bs+l (i.e., condition 4° is satisfied). This completes the proof of Lemma 2.

Th e o r e m 1.

Every finite-dimensional, nilpotent symplectic algebra (A, со)

over a field К of characteristic Ф 2, has a polarization.

P ro o f. From Lemma 2 it follows that there exists a sequence A

= B0, ..., Bk satisfying conditions l°-4° of Lemma 1. If dim Bk = n, then Bk is a polarization of (A, со). Assume that dim Bk = r < n. Since by Teorem 3.5 of [1] we have dim Bk > n, there exists an element ak+ie B k \B k. Let Bk+1

= Bk 1 Lin(ak+1). Of course Bk+l is an isotropic vector space. Similarly we define B

= Bk+Ï 1 Lin(ak+1), ..., £ k+(n_r) = fîk+(„_r)_ 11 Lin(ak+n_r), where ak+le B k+i- l \B k+i- i for i = 1, . . . , n — r. Bk+(n_r) is a «dimensional isotropic vector space. It is also a subalgebra of A. In fact, since Bk + („_r) = Bk± L in (ak+1, ..., ak + n- r) c Bk , then

Bk2+n_r cz (Bkx)2 c: B jfn A

= Bk c: Bk + n_r.

From Proposition 6 (d) and the proof of Theorem 1, we get the following

Co r o l l a r y 4.

Let (A, со) be a symplectic Lie algebra satisfying

assumptions o f Theorem 1. Let

2

m

j ( w + l )

for m even, for m odd,

r = dim Ak — dim Ak+ s = n — dim Ak+l. I f r ^ 2 s , then (A, со) has a polarization which is an ideal of A.

3. Connections of symplectic and invariant forms. In this section we shall deal exclusively with vector spaces and algebras over a field К of characteristic Ф 2.

Let F be a linear space over К equipped with a non-degenerated symmetric form Q. We denote by Alt(F) (see [4]) the set of all alternating endomorphisms f : V - * V with respect to the form Q (i.e., endomorphisms / satisfying the condition Q (f{x), y) = G(x, —f(y)) for all x, у of V).

It is known that the map

(6) Alt ( V ) 3 fi- > a ( f( - ) ,- ) e I l( V ) ,

where L

(V) denotes the linear space of all bilinear alternating forms on V, is

an isomorphism.

Algebra with a symplectic form 367

The bracket [/,<?] = f g —gj makes Alt(K) into a Lie algebra of the form Q (or the orthogonal Lie algebra of Æ) which is denoted by Lc .

Let A be an algebra and denote the Lie algebra of all derivations of A by Da.

Th eo r e m 2.

Let A be an anticommutative algebra with a symmetric non­

singular invariant bilinear form Q. I f there exists a 1-1 derivation D e L a, then the pair (A, со), where со = Q (/)(•), •), is a symplectic algebra. The map D (£>(•), •) is one-to-one. Conversely, if (A, со) is a symplectic algebra with a non-singular invariant symmetric bilinear form Q, then there exists 1-1 derivation D e L Q satisfying со = Q (£)(• ), •).

P ro o f. We have

(o(x, y) = Q(D(x), y) = -Q (D (y), x) = -co(x, y) and

— co(xy, z) + co(xz, y) — oo(yz, x) = Q(x, yD(z)) + Q(x, D(y)z) — Q(x, D{yz))

= Q (x, yD (z) + D(y)z — D (yz)) = 0

for all x, y, z e A . Since co(a, y) = Q(D(a), у) = 0 for all y e A implies a = 0, then a; is a symplectic form.

Conversely, since the map defined by (6) is an isomorphism, there exists an alternating endomorphism Z)eAlt(F) such that со = Q(D( •), •). Similarly the first part of the proof, it is easy to verify that D e L a.

Coro llary 5.

Any semisimple (real) Lie algebra L has no

derivation.

P ro o f. Since L is semisimple, the Killing’s form Q of L is non- degenerated and every derivation of L is inner. Suppose that there exists an element x 0e L such that the inner derivation adx

is one-to-one. It is easy to see that adx

e L n. By Theorem 2, we have that the pair (L, со), where (7) cu(x, y) = Q(adx

(x), y) = tr ad(x0, x)-ady for all x, y e L ,

is a semisimple symplectic real Lie algebra. This contradicts Theorem 8 of [3].

Co ro llary 6.

Let A be an algebra satisfying the assumptions of Theorem 2. The set S of all weekly-symplectic forms on A is a linear space. The map h defined by

(8) DAn L ae fe + Q ( f( - ) , )e S

is a linear isomorphism. The law of composition

(9) S

S

(

<и2)|-+ 0([Л "1(о 1), / Г 1 (a>2)]( • ), *)eS makes S into a Lie algebra.

P ro o f. Since the map defined by (6) is an isomorphism, then h is also

an isomorphism. Therefore, S is a Lie algebra.

Exa m ple 2.

Let L = (Æ6, + , -, [% •]) be a 6-dimensional algebra over R with a base (el5 e2, e3, e4, e5, e6), where + ; R6 x R 6-+ R6 and •; R x R 6 R6 are given by natural rules and the multiplication [•»•]; R6 x R 6-+R6 is

L is a nilpotent Lie algebra (see [2], p. 107). There exists a bilinear non­

degenerate symmetric invariant form Q on L such that

&(^i, e6) = &(e6, *i) = 1* 0 (e3, e4) = fi(e4, e3) = 1,

&(e2, e5)

Q(e

and the other values of Q at ordered pairs of the basis of L in question are 0.

Let D: R

-* R

be an isomorphism (of linear spaces) defined as follows:

D{ex) = eu D(e2) = e3, D(e3) = eu D(e4) = es, D(es) = e6, D(e6) = - e 4. It is easy to check that D e L a and D is a derivation of L. From Theorem 2, it follows that the pair (L, со), where со = (£)(•), •), is a symplectic Lie algebra.

References

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[3] M o n -Y a o C hu, Symplectic homogeneous spaces, Trans. Amer. Math. Soc. (1974), 145-159.

[4] S. Lang, Algebra, Warszawa 1973.

[5 ] Y. M a ts u s h im a , Differentiable manifolds, New York 1972.

[6 ] R. D. S c h a fe r , An introduction to nonassociative algebras, New York and London 1966.

[7 ] S. S te r n b e r g , Symplectic homogeneous spaces, Trans. Amer. Math. Soc. (1975), 113-130.

[8] V. S. V a r a d a r a ja n , Lie groups, Lie algebras and their representations, New Jersey 1974.

[9] A. W e in s te in , Lectures on symplectic manifolds. Regional Conference Series in Math.

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