Rend. Sem. Mat. Univ. Poi. Torino Voi. 54, 2 (1996)

L. Saal

THE AUTOMORPHISM GROUP OF A LIE ALGEBRA OF HEISENBERG TYPE

Abstract. Let rj = £ © # be a Lie algebra of Heisenberg type, where ( denotes the centre of r/ and d its orthogonal complement.

Here, it is shown that the group of automorphisms of 77 is the semidirect product of an abelian subgroup by the group of automorphisms of 77 that preserve •&, which, in tura, is described in terms of classical groups.

1. Introduction and some preliminaries

Algebras of type H were introduced by A. Kaplan in [3] as follows: Let 77 be a réal two-step nilpotent Lie algebra, endowed with an inner product (, ). Let ( denote the centre of rj and let d denote its orthogonal complement.

Let J : C —* End($) be the linear mapping defined by

(1.1) (Jzv,v') = (z,[vy]}, tvt/'Gtf, ze(.

We say that 77 is an algebra of Heisenberg type (or of H-type) if, for every unit element z € C > J% — — Id. Since Jz is skew-symmetric, this is equivalent to ask Jz to be an orthogonal transformation of fl.

We denote by Aut(^) the group of automorphisms of 77 and by A{rj) the subgroup of those automorphisms that act as orthogonal transformations on 77.

The structure of A(rj) has been described by Riehm in [5]. The aim of this note is to describe Aut(?7).

Indeed, we will see that Aut(ry) is the semidirect product of an abelian subgroup by the group of automorphisms of 77 that preserve d. We denote the last one by Aut<?(77).

Now, from (1.1), it follows by linearity and polarization that

(1.2) ' j * = -\z\2id, zec

and

(1.3) hh> + Jz<J, = -2 (z, z')Id, z, z' e C

102 L. Saal

In particular JzJzi ~f Jz'Jz = 0 if (2, z') = 0.

Let m = dim(C) and let C(m) be the Clifford algebra C(C, - | |2). By (1.2), (1.3) and the universal property of Clifford algebras, the action J of C on d extends to an unitary representation of C(m) that we also denote by J. So i? is a C(m) -module and, via the theory of orthogonal multiplication, it is possible to show that every Clifford module arises in this way [3].

So, the notions "irreducible" and "isotypic", when attributed to d, refer to its Clifford module structure.

It is known that C(m) is either simple or a direct sum of two simple algebras, (see, for example, [2]) and an explicit realization of Table 1 is given in Proposition 3 in [6].

Table 1

m, (mod 8) 0 1 2 3

C(m) ™2p C2P M 2 p- i M2P-I(8> H 2 P - I

dimM

(simple module) 2p 2p+l 2 P + I 2 P + I

m (mod 8) 4 5 6 7

C(m) IHLgp-i C 2 P M2p M2P e M2P

dimM

(simple module) 2 P + I . 2p+1 2P 2 P

We denote by M , C , M the real and complex numbers and the quaternions respectively, and by Mn , Cn , Mn, the algebra of matrices of order ri with coefficients in IR, C or H. Also p is such that m = 2p if m is even, m = 2p + 1 otherwise.

We denote by C+(m) the even Clifford algebra. Let {^1,..., zm} be an orthonormal basis of (. By the universal property of C(ra), the map given by

(1.4) ZJ -+ ZJ zm , for 1 < j < m , and zm —»• 2:m

extends to an automorphism of C(m) which tak.es C(m — 1) onto C+(m). As stated in Remark 1 in [6], this automorphism enables us to. extend the realization of C+(m) to C(m) in the same way that Prop. 3 extends it from C(m — 1) to C(m) .

The automorphism group of a Lie algebra 103

For z G C ) \z\ — 1» let pz : ( -+ ( be the reflection through the hiperplane orthogonal to z. It is easy to see that Jz : fl —• ^ is extended to an orthogonal automorphism of 77, acting on ( by —^. In this way, we denote by Pin(?7z) the subgroup of A(r)) generated by {(—pz,Jz), \z\ = 1} and by Spin(?n) the subgroup of A(rj) generated by {(pz pz*,Jz Jz>), \z\ = \z'\ = 1}. Let Cliff(ra) be the group generated

*y{(-\*\3p„J.),*e<-{0}}. Finally, we denote by Autori) the subgroup of Aut#(r7) of those automorphisms

acting trivially on the center.

In the next section we determine Aut^(rj)° in terms of classical groups and then,

we show that Aut^(r?) is the semidirect product of Aut^r/)0 by Cliff(ra), or the index

Aut^r?) : Auttf(!7)°XCliff(m)]•= 2.

Furthermore Cliff(m) fi Aut (?7)° has two or four elements. The precise situation is in Proposition 2.2.

2. Determination of Aut^r?)0

Let, as in 1, 77 = ( © ti be a Lie algebra of iJ-type, with inner product (, ). Let

ÌJJ : d x d —Ì ( be the antisymmetric form such that [z + v, z' + v'} = ^(v,i/), for

z,z' <E(, V,V' £$ From now on, we flx an orthonormal basis {zi,... ,zm} of ( and set J?; = JZi. We can

m

write \P(u, v) — 2_J i>i(u,v) zi w i m

*=i

(2.1) ,-(w, v) = (^ , -0(w, v)) = (^ , [«, v]> = («7,- w , v)

Thus

Auttf(iy) = {(h,g) G G1(C) x Gl(tf) : V(flf«,H = M«,«)Vti , t ; G <?}

where GÌ denotes the general linear group.

Also

Aut4?7)° = {g G Gl(#) : ip{gu,gv) = ^(w,v) V«,v é 0} •

From (2.1) it follows that

(2.2) # G Aut^r?)0 if and only if g% Jig = J-i for ali i = 1 , . . . , m .

where #* denotes the transpose of g.

It is easy to see that every element of A(rj) that acts on £ by the identity, commutes with the action J of C(m) on tf.

104 L. Saal

For Aut,?(77) ° we have the following

LEMMA 2.1. Every element ofAut& (n)° intertwinesthe representation J ofC+ (m) on ti.

Proof . For 1 < i, k < m,

ìpi(u, v) = (Ji u, v) = -{JlJi u, v) = .{JkJiJk u , v) = iJ)k(JiJk u,v).

Since g in A u t ^ ) 0 preserves each ipi, we have

^k(gJiJk u,gv) = ipk(JiJkgu,gv).

Since tpk is not degenerated, g(JiJk) — (JiJk)d-

REMARK 2.1. We denote by End^+/m\(ti) the algebra of linear maps on $

which intertwines the action of C+(ra). By lemma 2.1,

Aut,?(77)0 = jflf e End^+z x(^) : gi Ji g-Ji for a single Ji J .

Indeed, for k ^ z, we have

•gtJk9'= -g^JkJiJig = gtJiJhJ%g — g% hghJi = h

and so (2.2) holds.

REMARK 2.2. Set K — J\ • - • Jm. Then K commutes with C+(m). Furthermore if m = 1,2( mod 4), K2 = — 1 and since K is orthogonal A'* =' —A'. If rn = 0,3( mod 4), A 2 = 1 and K* = A.

We are now able to compute Aut^(^) . We remark that proposition 9 in [6] enable us to make a systematic use of the explicit representations of C(m) as they are given in [6j.

•m = l(mod 8).

According to proposition 4 in [6] there exists an isomorphism between C+(m) and

M2P that can be extended to one between C(m) and C2P in such a way that K = J\... Jm

acts by z'Id.

We first assume 1? irreducible. Thus there exists an isomorphism a : d —»• C as C(m) modules. Moreover since d is also an unitary irreducible Pin (m)-module we have that (u , v) = e Re [a(u) a(v)) for some positive Constant e, where ab denotes the

standard hermitian inner product for a , b E C2 .

The automorphism group of a Lie algebra 105

On the other hand, if w is an irreducible C+(m) -module, so is J;W for ali i — 1,2,..., ra. Thus w fi J;VV is invariant under C(m) and therefore trivial. In particular •d == w © Jmw and also Jmw = Kw.

We now fìx w = a - 1 (IR2p). Then for w, w' e w , (w, Kw') = e Re (a(t/;)a(it/)J = 0 since aAr = za.

So (w, Jmwf) = 0 Vtu,u/ € w. We fìx an orthonormal basis of i?, {wi,... ,W2P,Jmwi,... ,JmW2v}. Since A' intertwines w with Jmw as C+(.?TÌ) -

/ 0 - Q * \ ' modules we have that the matrix of A' in this basis is I n n ) • ^ e P r oPe r t 'e s

A = -Kf, A2 = - 1 and KJm = J m A imply that Q*Q = QQ* = 1 and Q = Ql

( • • Q2 = 1). So if # € End^+zm\(#). </ has a matrix of the form f ,n ,.. 1,

where 1 denotes the 2P x 2P identity. Since the matrix of Jm is ( . 1 we have that

^Jm f i r = Jm if and only if K ~ c J is Sp(l,M). Thus A u t ^ ) 0 ~ Sp(l,M).

If d — (d®)1 with tfo C(m) -irreducible the above argument shows that # E End^r+/ \($) has a matrix

/ a n i « i / l • —ai,/+iQ • • - « 1 , 2 / Q \

a / , i l à/,/1 • —a/.j+iQ • • ~a\,2lQ

a /+i , iO ••• a /+ i , /0 • • a /+ i f / + i l . . • a/+i,2/l

\ a2i)\Q . . . d2ì,lQ •' «2/,/+ll • 021,2/1 /

and thus Aut<?( 77)°~Sp( / ,M) .

REMARE. We can look at Q : w —»• w such that A = JmQ. Moreover Q =. Ji ... Jm_i . It is easy to see that the eigenspaces of Q associated to the eigenyalues ±1 have the same dimension and one can take a basis of eigenvectors for w.

• m = 2 (mod 8).

By proposition 3 and 4 in [6], we can extend an isomorphism between C+(m) and • C2P-i to one between C(m) and H2J>-i by sending zm to j\é. (Here we denote by i,j, K the uqits of-U).

Assume t9 C'(??2) -irreducible. Once again we can assume that (u, v) = Re (m7) where - denotes conjugation in M. By Remark 2.2 K = J\ . . . Jm commutes with C+(m) and K2 = —1. So we look at (i9, A) as a complex representation of C+(m) of (complex) dimension 2P. We can decompose it with an argument similar to the above case d = w © Jm-w, where w and J m w are irreducible, equivalent, C+(m) -modules. We

106 L. S'aal

have that Jm acts on d by ,?Id and thus (w , Jmw') = 0 Vti>, u/ € vv. We take a real basis of $ of the form

{ W1, . . . , U/ 2 P-I , I<Wi, . . . , KlU2P-i, JmWi,..., JmW2P-i, Jm KwX, . . . ,

JmKw2P-i}.

The operator "multiplication by j on the right" intertwines w and J m w as C+(ra) -modules. We denote it by <p. We know that <p commutes with A and that <p is skew-symmetric and (p2 = — 1.

Moreover the matrix of p in the above basis is Moreover the matrix of <p in the above basis is

0 " -pi

n o

/ - ì o \ 0

0 1

1 0 0

\o - 1 /

/o 1

Also the matrix of K is

- 1

0

0 \

\ 0

0 1

A'i 0 0 -A'i

Thus

- 1 0 ) g € End^.+ / \($) if and only if the matrix of r is

«i l + & n # i -¥>i(ai2 - &i2#i)

<Pi(a2i + 621 A i ) a22 - 6 2 2 Ai

Then gt Jmg = J m if and only if ana22 - bnb22 + ai2«2i - b21b12 = 1 and

«21^12 + «12621 + «11&22 + &n«22 = 0 if and only if

«i i4-&i i« —fli2 — &12*

«21 + 621« 022.+ &22« isinSp (1,C)

If tf = (^0)' with i?o C(m) -irreducible, a similar computation shows that Aut (7?) ~

Sp(/,C).

• m = 4 (mod 8).

In this case, C+(m) ~ H2r-2 © H2P-2 and it has two irreducible, inequivalent modules. (we consider H" as a right H-module, Hn acting on it on the left).

The automorphism group of a Lie algebra 107

Let us consider ti irreducible. Since K2 = 1, ti splits as ti — ti* © ti~ where ti* are the eigenspaces of K associated to the eigenvalues ±1. We have that J(K = -KJi for ali i, so Ji : ti* —• ^^ and dim $+ = dim t?~. Since A' is in the center of C*(m) , ti* are inequivalent C+(m) -modules and, by dimension, they must be irreducible. Thus we identify End^+z \(tf) with M x l (acting by right multiplication).

If ti — (tio)1 with tio irreducible, we have ti* ~ ( H 2 " j , that is, the 2P~2 x / matrix over

Hand so End^+z \(ti) ~Wi x Hj. Since ti = ti+ 0 Jiti+, there exists an orthonormal

basis of ti such thàt Ji — { 1. If we write g 6 End^+/ \ (ti) as a pair g = (gì ,g2),

the condition g% Jig — Ji implies that g[ = g^1.

Thus Alitai;)0 ~G1/(!Q.

• m = 0 (mod 8).

This case is completely analogous to the above: C*(m) ~ M2P-i ®M.2P-I , K2 — 1, and the eigenspaces of K are the two inequivalent, irreducibles modules of C*(m).

For ti —- (tio)1 with ti0 irreducible, we have End^f-wm\(ti) ~ IR/ x Mj and A u t ^ ) 0 ~

G1.(/,M).

• m = 3, 7 (mod 8).

We remark that in these cases C(m) has two inequivalent, irreducible modules.

Assume, first, rj isotypical, that is, ti = (^o)' with tio an irreducible C(m)-module. Since A' commutes with C(m) and K2 = 1, K acts on tio by K — 1 or by K = — 1. Therefore J\ = J2- - Jm or J\ — -J2 • • • Jm- Since J2 • • • Jm is in C*(m) , every element of Autd(rj)° commutes with J\. Thus, every element of Aut (?y) is an orthogonal automorphism (see [5]).

We assume m = 3(8) (resp. m = 7(8)) and we now compute Aut,j(?7) in the case that i] is not isotypical.

According to proposition 3 in [6] we can extend the isomorphism between C(m— 1) and M2P-i (resp. M.2P ) which sends Zi -+ j Z i , for 1 < i < m - 1, to an isomorphism beetwen C(rn) and H 2 P-I © M 2 P - I (resp M2P ^R2P) by the rule

Zi -*Ji = (jZiJ2i), 1 <* < m - 1

Zm y Jm — \Jzi • • • Jzm-i J ~~3z\ • • • Jzm.-i)

Furthermore, as is stated in Remark 1 in [5], the automorphism from C(m - 1) to C+(m) which sends Zi —> ZiZmì 1 < i < m - 1, enables us to apply Prop. 3 to C(m)

108 L. Sciai

extending the representation from C+(m) to C(m) in the same way that Prop. 3 extends it from C(m — 1) to C(m).

So the two irreducible -up to isomorphism- inequivalent representations of C(m), say tfj and tfjj", are equivalent as C+(m) -modules, being the identity an intertwining operator.

Assume, now, ti = d+ © d~, where K = J\... Jm acts by ±1 on #* respectively. Thus é+ = tf J 8 . . . 0 $£, p times, and i?~ = !?Q © .. . © I?Q > <Z times;

By the above, End^+z M) =" G7(p -f g,H) for ?n = 3(8) and End^+z N(^) S*

G/(p + g,lR) forra = 7(8).

We remember that Aut#(r))° = Ig £ End^+z x(#) : flf*Jmflf = J m | . If </ G

Aut#(?7) , g commutes with J\... Jm_i and so giJmg = Jm if and only if sr*A'(/ = A\

We set IPtg = ( 1 I , lp x p matrix, — lqxq matrix.

So, for m = 3(8), Aiit*(»/)0 * {g e Gl(p + q,M) : g*IPtqg = IPt9}

= U(p, f,H) where - denotes conjugation in M and for m = 7(8), Aut (?7) = o(P,q,m).

• m = 5 (mod 8).

Assume, first, $ irreducible. Set A'i = J\ - • • Jm-\. Then K\ = 1 and ?? = ??+©*/"", where i?* are the eigenspaces of K\ associated to ±1, respectively. Furthermore Jm — ±K on i?±. We also observe that End^+z A($) commutes with Ki and so it preserves tì±.

Bydimension, $ is C,+ (?n)-irreducibleand wecan identify End^+z \($) withM(by

right multiplication). Now K commutes with C,+ (77?.).and, by Remark 2.2, Kt — —K and A'2 = —1. Thus K acts on d by a quaternion, say q\, which is imaginary pure and it has norm one. Moreover we can assume that q\ — b\i + cij + diK, with bi = c\ — 0 , &\ — 1. Associating to each g in End/^+/ \(i?) a quaternion </(#), we have that the condition

giJmg — Jm is equivalent to q{g)q\q{g) = q\, when q(g) denotes the conjugate of q(g). This implies that the norm of q(g) must be one and so q(g) commutes with q\. Thus Aut^??)0 ~ U(l). v

Assume i? = (tfo)' with tf0 C(ra) -irreducible. Then End^+/ \(i9) ~ H/ and we

need to determine \ q E H/ : qt{q\I)q — q\I k

Now, let <£> : H/ —• C2/ be the canonical homomorphism given by g —s> f ^ _ J,

with a = A + Bi, (3 = D + Ciifq = A + Bi + Cj+DK. Then y(</i) = f j "l1 J. Since

«KtfiMtfMtfi)-1 = ¥>(<?), we nave that tf'M = qi if and only ify>(tf) <p(qi)<p(q) = y?(tfi)

The automorphism group of a Lie algebra 109

if and only if <p(q) <p(q) = 1. Thus Aut^??)0 = Gl(/,M)n 0(2J,C) = SO* (2/).

• m = 6 (mod 8).

Here A'* = —A', K anticommutes with Ji and K2 = — 1. Assume, first, t? irreducible. Thus (i9, K) is a complex vector space and, by dimension, it must be irreducible as a (7+(?72)-module. Thus g 6 EIHL~»+/ \(i?) if and only if g = a + 6A', a, 6 6 R.

Now, (7*«7»<7 = «/,• if and only if (a + bK*) Ji(a + bK) = /; if and only if a2Ji + abK*Ji + abJjK -f- b2K%JiK = Ji iff (a2 - 62) J; + 2abJiI< = «/;. Since J and AT are linearly independent, a2 — b2 — 1, and aò = 0 which implies Aut# (??) = {±1}. We assume ti = (^o)' with i?o irreducible. We identify Q with End^+/m\(??) via the mie that maps the matrix A+iB = [asj -f ib8j].1<8 ? < / to the element g = (asj -f bsjK0)1<s -<{

where A'o is the restriction of K on $o- Also let Jm the restriction of Jm on $o-

Now, gtJmg — Jm if and only if

/ J (aji + hi^oì Jm (ajr + bjrK0) ~ SirJm .

3

But this implies

/ _j ydjittjr ~~ bjiVjr) Jm • \O,jr0jr -\- Ojì<ljr) J m A o — ^ir^rn

3

and so A'A - B*B = 1, A<5 + Bl A = 0, ie. A + K£ <E O (/,C). Thus Aut^T/)0- O (/,€).

We have, just, proved the following

PROPOSITION 2.1. Let rj = ((Bti be an algebra of H-type.

Let dim^d = n. Then Aut#(rf) is isomorphic to

Sp ( n 2 ~ ( ^ ) , M ) if m = \{mod 8)

Sp (nT^), CÌ i/ m = 2{mod 8)

£/ (?212_(i2^i"),n_i2-(22^i),IHl) if m = Z(mod8)

Gì ( n 2 - ( ^ ) , H ) iym = 4(/iwrf 8)

110 L. Saal

SO* (2n2-(s^1)\ ìf m = b(mod 8)

O ( n 2 " ( ^ ) , c ) ifrn = Q(modS)

O (ni2"( i r^ i) ,n_i2~( i !^ i) ,]R) ifm = 7(modS)

Gì (n2~W,R\ ìfm = 0(modS)

He re we kept the notation ofTheorem 6 in [5] and ni and n_ i = n — n\ are the dimerisions of the eigenspaces of fì with respect to K = J\- • • Jm.

Our purpose, now, is to describe Aut^(n).

We denote by 6d , d > 0, the automorphism given by

6d(z, v) = (d2z, dv) , ze( , v é'tf •

We recali that {zi,... ,zm} denotes an orthonormal basis of (. Thus, for h G GÌ (£), there exist orthogonal matrices &i and k2 such that

(2.3) h = k1h0k2

where ho is diagonal with positive entries di,..., dm.

Set dimtf = n.

We first assume dimC = m is even.

In this case Pin(m) acts on £ as the full non-connected O (m) since det(—pz) = - 1 . Thus, by (2.3), we can write each element (h,g) G A u t ^ ) as

(hìg) = (k1,gi)(h0}go){k2ìg2), with (&<,#') G Pin(m) for i = 1, 2 .

Since-(/io, </o) £ Aut#(?7), we have that 4>i(goii,gov) = diijji(u,v) foru,v G i?, i = l , . . . , m ; that is ggJigo = diJi. This implies d? = (det(/o) so that /i0 = di and (ho, do) — Sd1/2 ' (lì 9')- Thus we conclude that

(h,g) = Sdi/2(k1,g1)(ì, ^^2^2)

So

(2.4) ( M = W*3,03)(l ,f lf ' ' )

with (k3ig3) e Pin(?7i), (l,g") G A u t ^ ) 0 .

We now assume m = dim( is odd.

The automorphism group of a Lie algebra 111

In this case, Pin(m) acts on ( as SO {m).

Case m = 3 , 7 (mod 8). If (~1,#) is in kui^rj), we have ìjji(gu,gv) = -ipi(u,v) so

\halipi(gJkJiU}gv) = -il>i(JkJiU,v) = -ipk(u,v) - ^{gu.gv) = ipi(JkJi9u,gv).

So ( - 1 , g) G Auttf (»;) if and only if

(2.5) 9eEndC+(m)W a n d 9ÌJm9 = -Jm

As before, we decompose t? = tf+ 0 i9~, where A' = Ji . . . Jm acts by ± 1 on tì^

respectively and tì+ = tff 0 . . . 0 tfj, p times, ?9~ == tfg 0 . . . 0 tfj", q times.

We have that g satisfaces (2.5) if and only ìf'g G End£.+ / \(i9) and giKg = —K.

So if p = q, there exists such g: indeed, for m = 3(8), g G Gì (p + q,M) and

gflp^g — —IPtq. For example g = I ) . The same holds for m = 7(8).

On the other hand, we have that if dim?9+ = dimz9~ and <j>(z) — -z V z G (

and we set Jz = J(f>(z) then Jz \#+ is equivalent to J<j,(z) U- a s C(?72)-modules. So there

exists an orthonormal g : d —»• $ such that gJz = J$(z)g. So (/ G End^+/ \(i?) and

gK =-Kg.

Now if p ^ q, there is not such # since if 5 is a non degenerated symmetric bilinear

form on a real vector space V, diml^ = n, then there exists a basis of V {{3j} such that

B({3j ,/3j) = l for 1 < j < r and B(/3j, /?j) = — 1 for r + 1 < j < n and the number r is

independent of the choice of the basis (see, for example, [1]).

Thus for (/*,#) G A u t ^ ) , either det/i > 0 and we can write (h,g) as in (2.4), or

(h,g) = (-l,<7o)(/A</') with deth' > 0.

Case m = 1 , 5 (mod 8). Here, there exists ( - l ,#o) € *4(??) (see [4]).

Then, we have the following

PROPOSITION 2.2. tóN = Aut0(»y)onClifT(m) and G = Aut#(r))°><ì Cliff(m).

- Ifm = 0,2 (mod 4) A u t ^ ) ~ G/JV « ^ W = {±1}.

- Ifm = 1 (mod 4), the index [Aut^(?y) : G/N] = 2 and N = { ± 1 , ± A } .

- //??! = 3 (mod 4) Auto (77) ~ G/N for dimt;+ ^ dimu" A/M/-[Aut^ (r;) : G/JV] = 2

/or dimu+ = dimt; - , where N = {±Ì,±K} if v is not isotypical and N = {±1} if v is

isotypical.

Proof. We only need to compute N.

We have that {u G C(m) : u commutes with C+(m)} = MI 0 MA'. Furthermore, if

g G A u t ^ ) 0 , g must commute with C+(m) and (det#) = 1. Thus N C {±l,±K}.

But now, Remark 2.2 implies that for m = l ,3 (mod4) , K%3iK = Ji and for

112 L Saal

m = 0,2(mod 4) , IVJÌK = -«/,-.

PROPOSITION 2.3. Aui(rj) is the semidìrectproduct ofa normal, abelian subgroup by Autd(ri).

Proof. We fìx {i>i,..., vn} an orthonormal basis of d. We take T £ Aut(?7) and we write

3 = 1

m

TVÌ = ^2 ajiZj + ] T 9JivJ • 3=1 3=1

We take a E Mm x n , h E Mm and g E Mn given by

a = [atij] , h = [hij] and g = [^j] .

Now, T is an automorphism if and only if

(ZJ , T[vi,vk]) = {Zj , [Tvi,Tvk]) ,l<j<m;l<i,k<n.

ni

Since [vi,vk] = ^^/(v.',Vjb)^., ^K:,n-] = ^ *Pi(vi,vk) hsi zs and [T^, Ti>fc] = f= l l < * , K m

y ^ flfr»^fcK-,Wt] = X ] 5 - 1 9 r i 9 t k ll>s(Vr,Vt)zs.

1<» ' )<<W l < r , t < n s = l

ni

Therefore T is an automorphism if and only if V\ V>/(w», vk)hji — YJ </™#ifc V"j (vfc, v*) /=1 l < r , i < «

i.e. if and only if (h,g) E Aut (7?). m

For (3 = [fa] E Mmxn , we defìne Tp(zi) = z>, Tp(v{) = ^ + ^ f t ^ j .

Then J5 = {Tp , /3 E Mm x n} is a normal abelian subgroup of Aut(?;) and by the above Aut(?7) ~ B x Auttf (»;).

REMARK 2.3. It is known that every automorphism of a classical Heisenberg algebra (dimC = 1) can be obtained as the product of an inner automorphism by an element in Aut^(?;) .

We see here that, in general, this is not the case. Indeed, we take x = z + v , z E C , v 6 ti. Since i] is two step nilpotent, Ad (exp x) = I + a óv and the matrix of

The automorphism group of a Lie algebra 113

Ad (exp x) with respect to the basis {z\,..., zm, v1,..., vn} is

I with a — :

Furthermore, the rows of a are orthogonals since Ji(v) = (ipi{v,vi), . . . , ipi(v, vn)) i =

l , . . ." ,m.

If m = 1 the above assertion holds since for each &G l n there exists a unique

v E i? such that J i (v) = 6.

If m / 1, take T a with rank(a) < m. This Ta can not be written in this form.

Acknowledgements. I am indebted to Aroldo Kaplan with whom I had the pleasure to work

on this problem. I am also grateful to Fulvio Ricci for several invaluable conversations.

Research partially supported by CONICET, CONICOR and SecytUNC.

REFERENCES

[1] HOFFMAN K., KUNZE R.., Linear Algebra. Prentice-Hall, 1973. [2] HUSSEMOLLER D., Fibre bundles, 2. ed., Springer 1975. [3] KAPLAN A., Fundamental solutions far a class of hipoelliptic P D E generated by

composition of quadratic forms. Trans. Amer. Math. Soc. 259 (1980), 145-153. [4] KAPLAN A., RICCI F., Harmonic analysis on groups of Heisenberg type. Lecture Notes in

Mathematics 992, 416-435 (Springer, 1983). [5] RIEHM, C , The automorphism group of a composition of quadratic forms. Trans. Amer.

Math. Soc. 269 (1982), 403-415. [6] R.IEHM, C , Explicit spin representations and Lie algebras of Heisenberg type. J. London

Math. Soc. 29 (1984), 49-62.

Linda SAAL

Facultad de Matematica, Astronomia y Fisica

Universidad Nacional de Cordoba, Ciudad Universitaria, 5000 Cordoba, Argentina.

Lavoro pervenuto in redazione il 15.9.1995 e, informa definitiva, il 20.4.1996.