One of the main motivations for studying Lie algebras is the connection with groups. Starting off with a group a Lie algebra is constructed that reflects (parts of) the structure of the group. In many cases it can be shown that questions about the group carry over to questions about the Lie algebra, which are (usually) easier to solve. Various constructions of this kind have been used, each for a specific type of group. For instance, if G is a Lie group, then it can be shown that the tangent space at the identity is a Lie algebra. This connection is for example used to study symmetry groups of differential equations (see [10], [67]). Also if G is an algebraic group, then the tangent space at the identity has the structure of a Lie algebra (see, e.g., [12]). Here we sketch how a Lie algebra can be attached to a p-group.
In this case the Lie algebra is not a tangent space. But it is quite clear how the Lie algebra structure relates to the group structure.
Let G be a group. Then for g, h E G we define their commutator to be the element
(g, h) = g - l h - l gh.
Furthermore, if H is a subgroup of G then (G, H) is the subgroup generated by all elements (g, h) for g C G and h E H.
1.4 Lie algebras from p-groups 11 We set ")'1 ( a ) - G, and for k >__ 1, 3'k+ 1 ( a ) - ( a , 7k ( a ) ) . T h e n ")'1 (G) >__
")'2(G) ___ . . . is called the lower central series of G. If 7n(G) - 1 for some n > 0, then G is said to be nilpotent.
A group G is said to be a finite p-group if G has pk elements, where p > 0 is a prime. For the basic facts on p-groups we refer to [40],
[43]. We
recall t h a t a p-group is necessarily nilpotent.
For a group H we denote by H p the subgroup generated by all h p for h c H . Let G be a p-group. We define a series G - N 1
(G)
_> t~2(G) >__ . . . byi~n(a) - ( ~ n - l ( G ) , a ) n m ( a ) p,
where m is the smallest integer such t h a t p m > n. This series is called the Jennings series of G. The next theorem is [44], C h a p t e r VIII, T h e o r e m 1.13.
T h e o r e m 1.4.1 We have
1. ( n m ( a ) , n n ( G ) ) <__ nn+m(a), and 2. nn(G) p < t~pn(G).
Set
Gm
= ~ m ( G ) / ~ m + l ( G ) . T h e n by T h e o r e m 1.4.1, all elements of Gm have order p, and Gm is Abelian. Now since any Abelian group is the direct p r o d u c t of cyclic groups we see t h a t Gm is the direct p r o d u c t of cyclic groups of order p, i.e., Gm = H1 x ... x Hk where Hk ~- Z / p Z . Now fix an k
generator hi of Hi. T h e n any element in Gm can be w r i t t e n as h ~ l . . . h k where 0 < ni _< p - 1. Set Vm - F~p. T h e n we have an i s o m o r p h i s m am : Gm ~ Vm of Gm onto the additive group of Vm. It is given by am(h~ ~ ''" h k ) = ( n l , . . . ,nk). rtk For j > 1 we let Tj be the composition
9 j(v) - %
Gj -% 5
where 7rj is the projection map.
Set L = V1 @ 1/2 | T h e n L is a vector space over Fp. We fix a basis B of L t h a t is the union of bases of the c o m p o n e n t s Vm. Let x, y C B be such t h a t x C ~ and y C I/). Furthermore, let g c hi(G) and h E nj(G) be such t h a t Ti(g) = x and 7-j(h) = y respectively. T h e n we define
y] - h ) ) .
This p r o d u c t is well-defined, i.e., [x, y] does not depend on the choice of g, h. This follows from the identity (g, f h ) - (g, h)(g, f ) ( ( g , f ) , h), which holds for all elements f, g, h in any group.
12 Basic c o n s t r u c t i o n s L e m m a 1 . 4 . 2 F o r x, y e B we have that [x, x] - 0 a n d [x, y] + [y, x] - O.
P r o o f . Let x E ~ a n d y C Vj. Let g C a i ( G ) a n d h E a j ( G ) be p r e - i m a g e s of respectively x u n d e r Ti a n d y u n d e r 7). T h e n
[x,x] - ~-2i((g,g)) - T2i(1) -- 0.
A n d
[x,y] + [y,x] - T i + j ( ( g , h ) ) + ~'i+j((h, 9))
= Ti+j((g, h ) ( h , g))
= vi+j(1) -- 0.
[3
L e m m a 1 . 4 . 3 F o r X l , X 2 , X 3
IX3, [Xl, X2]] -- 0.
e t3 we have
[Xl, [X2,X3]] %" IX2, [X3,Xl]] %"
P r o o f . S u p p o s e t h a t xi E Vm~ for i - 1, 2, 3.
image of xi u n d e r ~-i for i - 1, 2, 3. T h e n
Let gi E am~ (G) be a pre-
[Xl, IX2, X3]] %"Ix2, Ix3, Xl]] %" [X3, [Xl, X2]]-
TmlWm2-k-m3
((gl, (g2, g3))(g2, (g3, gl))(g3, (gl, g2))).
Now the result follows from the fact t h a t
(gl,
(g2, g3))(g2, (g3, gl))(g3, (gl, g2)) C I'g, lltl.-~m2-I-m3.-~-l (G ) (see [44], C h a p t e r VIII, L e m m a 9.2).Now we e x t e n d the p r o d u c t [ , ] to L • L by bilinearity.
C o r o l l a r y 1 . 4 . 4 W i t h the p r o d u c t [ , ] 9 L • L ~ L the v e c t o r space L becomes a L i e algebra.
P r o o f . T h i s follows i m m e d i a t e l y from the above l e m m a s (cf. L e m m a 1.3.1).
D
E x a m p l e 1 . 4 . 5 Let G be the g r o u p g e n e r a t e d by t h r e e elements gl, g2, g3 s u b j e c t to the relations (g2, g l ) - g3, (gn,gl) - (g3,g2) - 1 a n d g~ - g2 2 -- g3 a n d g~ - 1. T h e first relation is the same as g2gl - glg2g3,
1.5 On algorithms 13 whereas the second and t h i r d relations can be w r i t t e n as g a g 1 - g t g a and
g 3 9 2 - g 2 g 3 . These relations allow us to rewrite any word in the generators
to an expression of the form
9 i2 ia ( 1 . 8 )
g~l g2 g3 9
Using the r e m a i n i n g relations we can rewrite this to a word of the form (1.8) where 0 _< ik _< 1. This rewriting process is called c o l l e c t i o n . We do not discuss this process here as it is clear how it works in the example.
For a more e l a b o r a t e t r e a t m e n t of the collection process we refer to, e.g., [40], [80]. Every element of G has a unique r e p r e s e n t a t i o n as a word (1.8) such t h a t ik - 0, 1. Hence G contains 2 a - 8 elements. We have 71(G) - G, ~2(G) - (g3} (where (9a) denotes the subgroup g e n e r a t e d by ga) and
7 a ( a ) - 1. T h e Jennings series of G is ~I(G) - G, t~2(G) - (g3) and
~3(G) - 1. So G1 - G / ( g a ) - (01,02), where g 2 g l - g i g 2 . Therefore G1 =
{ 1, ~]1, (72,9192}.
Let V1 be a 2-dimensional vector space over I72 s p a n n e d by {el, e2}. Let ~1 "G1 --4 V1 be the m o r p h i s m given by cr~ (0~) - e~ for i - 1, 2 (so cr l ( g l g 2 ) -- e l + e2). A l s o w e have t h a t G2 - (ga)/1 - {1, ga}. Let V2 be a 1-dimensional vector space over F2 s p a n n e d by e3. T h e n a2 " G2 --+ V2 is given by cr2(g3) - e3. Now set L - V1 (9 V2. We calculate the Lie p r o d u c t of el and e2"[el, e2] -- T 2 ( ( g l , g 2 ) ) -- 7 2 ( g 3 ) -- e3.
Similarly it can be seen t h a t [el, e3] - [e2, e3] - 0.
1.5 On algorithms
Here we will not give a precise definition of the notion of "algorithm" (for this the reader is referred to the literature on this subject, e.g., [52]). For us
"algorithm" roughly means "a list of consecutive steps, each of which can be p e r f o r m e d effectively, t h a t , given the input, produces the o u t p u t in finite time". So there is an a l g o r i t h m known for adding n a t u r a l numbers, b u t not for proving m a t h e m a t i c a l theorems.
We are concerned with algorithms t h a t calculate with Lie algebras. So we need to represent Lie algebras, and subalgebras, ideals, and elements thereof in such a way t h a t t h e y can be dealt with by a c o m p u t e r , e.g., as lists of numbers. For Lie algebras there are two solutions to this p r o b l e m t h a t i m m e d i a t e l y come to mind: we can represent a Lie algebra as a linear Lie algebra, or by a table of s t r u c t u r e constants. We briefly describe b o t h approaches.
14 Basic constructions available for performing the elementary arithmetical operations (addition and multiplication) for the fields that are input to our algorithms. Also we assume that elements of these fields can be represented on a computer.
This does not pose any problems for the field of rational numbers Q, nor for number fields, nor for finite fields.
1.5 On algorithms 15 randomized algorithms called Las Vegas algorithms. An algorithm for com- puting a function f ( x ) is called Las Vegas if on input a it either computes as Gaussian elimination~ which works over all fields for which we can perform the elementary arithmetical operations. The basic procedure is described in many monographs on linear algebra, see, e.g., [60].
Sometimes we use the connections between associative algebras and Lie algebras to study the structure of a Lie algebra. Doing so we also need algorithms for calculating various objects related to an associative algebra.
We have described some of them in Appendix A.
16 Basic constructions represented as coefficient vectors. So the Gaussian elimination procedure allows us to perform Step 1 and Step 2.
1.6 Centralizers and normalizers 17 It follows t h a t the coefficients of the m a t r i x of adx are ~-~in=l o~ickij. So we have an a l g o r i t h m AdjointMatrix t h a t for a Lie algebra L a n d an element x E L constructs the m a t r i x of adx relative to the input basis of L. We note t h a t the coefficients cq of x are i m m e d i a t e l y available since x is i n p u t as a coefficient vector.
1.6 C e n t r a l i z e r s a n d n o r m a l i z e r s
In this section we construct centralizers and normalizers in a Lie algebra L.
We also give algorithms for calculating bases of these spaces. For t h a t we assume t h r o u g h o u t t h a t L has basis { x l , . . . ,Xn} and s t r u c t u r e constants ci jk relative to this basis (see Section 1.5)
Let S be a subset of L. T h e n the set
CL(S) - {x e L ] [x,s] - 0 for all s e S )
is called the centralizer of S in L. We prove t h a t CL(S) is a s u b a l g e b r a of L. Let x, y C CL(S) and s C S. T h e n by the Jacobi identity
y], - - [ [ y , x] - x], y] - 0, (1.9) so t h a t [x, y] e CL (S) and CL (S) is a s u b a l g e b r a of n.
It is s t r a i g h t f o r w a r d to see t h a t CL (S) is equal to the centralizer of K in L, where K is the subspace s p a n n e d by S. Therefore, in the a l g o r i t h m for c o n s t r u c t i n g the centralizer we assume t h a t the input is a basis { y l , . . . , yt) of a subspace K of L, where
n
Y l - E / ~ l j X j . (1.10)
j----1
T h e n x - }-~i aixi lies in CL (K) if and only if [x, Yl] - 0 for 1 < l < t. This is equivalent to
i~l /~ljCfj a i -- 0 for 1 < k < n and 1 < l < t.
9 j = l
It follows t h a t we have nt equations for the n u n k n o w n a l , . . . , an. By a G a u s s i a n elimination we can solve these; and therefore we find an a l g o r i t h m Centralizer for calculating the centralizer of a subspace K of L.
T h e subset
C(L) - {x e L ] [x,
y]-
0 for all y e L}18 Basic constructions is called the centre of L. We have t h a t the centre of L is the centralizer of L in itself, i.e., C ( L ) - CL(L). As [x, y] - 0 for all x C C ( L ) a n d y C L, it is i m m e d i a t e t h a t C ( L ) is an ideal in L. T h e centre is the kernel of the m a p a d " L ~ E n d ( L ) , i.e.,
C ( L ) - {x e L I a d z - 0}.
So if we s t u d y the s t r u c t u r e of L via its adjoint map, t h e n we lose "sight"
of the centre.
If C ( L ) - L, t h e n L is said to be Abelian or commutative.
T h e a l g o r i t h m for calculating the centralizer also yields an a l g o r i t h m for calculating the centre. In this case the requirement for an element x - }-~i a i x i to belong to C ( L ) is [x, xj] - 0 for 1 _< j <_ n, which boils down to
Z
n CijO~ i - - 0 for 1 _< j, k <_ n.i = 1
So we have n 2 equations for the n unknowns a l , . . . , an, which can be solved by G a u s s i a n elimination. This gives us an algorithm Centre.
Let V be a subspace of L. T h e n the set
N L ( V ) - {z e L I[x, v] e V for all v E V}
is called the normalizer of V in L. In the same way as for the centralizer we can prove t h a t the normalizer of V in L is a subalgebra of L. If V h a p p e n s to be a subalgebra of L, then V is an ideal in the Lie algebra N L ( V ) .
Now we describe an algorithm for calculating the normalizer. Let V be the subspace of L s p a n n e d y l , . . . ,Yt, where the yl are as in (1.10).
T h e n x - Y'~i c~ixi is an element of NL(V) if and only if there a r e ~ l m for 1 _< l, rn _< t such t h a t
[x, yl] -- ~llyl + . . . + ~itYt f o r l - - 1 , . . . , t .
This a m o u n t s to the following linear equations in the variables c~i a n d flzm:
~ljeij O t i - /~mk~lm - - 0 for 1 _< k <_ n and 1 _< 1 _< t.
9 j = l m = l
Again by a Gaussian elimination we can solve these equations. However, we are not interested in the values of the film, so we throw the p a r t of the solution t h a t corresponds to these variables away, and we find a basis of N L ( V ) . As a consequence we have an algorithm Normalizer for calculating the normalizer of a subspace V of L.
1.7 C h a i n s of ideals 19 E x a m p l e 1.6.1 Let L be the Lie algebra with basis { x l , . . . ,x5} a n d mul- t i p l i c a t i o n table
[x~, x~] - x~, [ ~ , x~] - - x ~ , [x~, x4] - ~ , [ ~ , ~ ] - ~1, [~4, ~ ] - x3.
(As usual we only list p r o d u c t s [xi,xj] for i < j; a n d we omit those t h a t are 0.) We calculate a basis of the centre of L. Let x = ~ 5 i = l O~iXi be an a r b i t r a r y element of L. T h e n x C C ( L ) if a n d only if [xi, x] - 0 for 1 <_ i <_ 5. So 0 - [xl,x] - a 4 x l - ( ~ 5 x 2 , from which it follows t h a t c~4 - c~5 - 0. T h e n also [x2, x] - 0. It is easily seen t h a t Ix3, x] - 0. F r o m 0 - [x4, x] = -c~lXl - ~ 2 x 2 + c~5x3 we infer t h a t c~1 - c~2 - c~5 - 0. F i n a l l y [x5, x] = c~1x2 - c ~ 2 x l -c~4x3, from which c~1 - c~2 - c~4 - 0. It is seen t h a t only O~3 can be non-zero. It follows t h a t C ( L ) is s p a n n e d by x3.
Let V be the subspace of L s p a n n e d by x l , x 5 . Let x = }--~.i=1 a i x i be an 5
element of L. T h e n x C N L ( V ) if a n d only if Ix, Xl] ---- a x l + bx5
[X, X5] ~--- CXl -~- dxs.
T h e first of these r e q u i r e m e n t s p r o d u c e s a5 - 0 a n d a + c~4 - 0. T h e second boils d o w n to c - c~2 = 0 a n d a l = o~4 - - 0. Hence N L ( V ) is s p a n n e d by
X2 ~ X 3 .
1 . 7 C h a i n s o f i d e a l s
Here we c o n s t r u c t chains of ideals of L. First we note t h a t , in order to check t h a t a subspace I of L is an ideal, it suffices to prove t h a t Ix, y] E I for all x C L a n d y C I. T h i s follows from (1.1).
Let L be a Lie algebra a n d let V, W be subspaces of L. T h e n the linear s p a n of the elements [v, w] for v ~ V a n d w C W is called the product space of V a n d W. It is d e n o t e d by IV, W].
L e m m a 1 . 7 . 1 Let I and J be ideals of L. Then also [I, J] is an ideal of L.
P r o o f . Let x C I, y C J , a n d z E L t h e n by the Jacobi identity [~, [~, y]] - - [ ~ , [y, z]] - [y, [~, ~]],
which lies in [I, J]. Now if u e [I, J] is a linear c o m b i n a t i o n of elements of the form Ix, y] for x C I a n d y C J , t h e n also, by linearity of the p r o d u c t , for all z C L we have t h a t [z, u] is a linear c o m b i n a t i o n of elements of the
20 Basic constructions
form [a, b] for a E I and b C J. So [I, J] is an ideal of L. v1 The following is an algorithm for calculating the product space [K1,/42]
of two subspaces K1 a n d / ( 2 of L.