Generalizations of Steinberg groups

= e, a = b = c = d = e, aa2 — =-:b2u — = c\ <* , 2 a = b2 = c 2 , cd = dc~ dc~ \ab \ab = ba~ ba~-■l1 x cd — dc~ ,ab = ba~l, 4

4

2

=■ d

4

2

= [a ,d\ == [b,c]= [b,d[ =- e. [«,c] [a, c] = [a,d] = [b, c] = [b,d] = e. Now let a : N —► N be defined by Now let a : N —► N be defined by

0 1—> ab2cd

o 1—> ab2cd a

b1 \-^> c b 1—► c aa C H 8

d 1\—> —► c6. cb.

50

Linkage Graphs and Linkage Groups

By checking the relations of TV it can be shown that a is an automorphism of TV of order 5. (This can be checked by hand or by computer. In a later section we will actually check this with CAYLEY.) Let G be the split extension of TV by (a). Now let R = {0,2,4,6} be the ring of even integers modulo 8. Define cf>:R-+G

% ■-> c{ Since (R, +) is cyclic of order 4 it follows that is injective and that (r + »)* = r V , for all r, s e R. Now consider the word w(x,y) = [xa,ya~ ). We have io((2i)*,(2i)*)=u,(d ! ,e'') = = [(c«y,(c°-1y}

= [a\»} = {a,bYJ =

= (a~ 2 )^' = c 2i J = (2»2j)*. Thus, ^ is a verbal embedding. The constant group (a) is of order 5. This is in contrast with Theorem 2.2.4 in which it was shown that the constant group has to have order at least 6 when the ring being embedded has an identity. Futhermore, if we identify R with its image in G, we have RnRa
={a2) =

{e,a2}.

This contrasts with Theorem 1.2.1 and Lemma 2.4.2. Finally, normal closure of R in G is the group TV which has order 32 = is nilpotent and not perfect as required by Theorem 1.2.1. Now suppose that /? is any automorphism of TV such that if H is the split extension of TV by (/?), it may be checked that R-+H 2i h+c'

note that the 2 5 . Hence, TV cP = ad. Then the mapping

2.7 Counter

Examples

51

is a verbal embedding using the encoding word [x13, y] with constant group (/3). There are many candidates for /?. For example a

H->

ab2

b t—y a2bcd ct-y ad d t—> ac

The automorphism /? has order 2. Other such automorphisms exist of orders 3 and 4 as will be shown in a later section dealing with computer calculations. Note that a word of the form [x@, y] can never embed a ring with identity since this would give r = [l^r] = (l^)-lr-1l"r. This implies that ( l ^ ) _ 1 r _ 1 l ' ? = e, and hence r = e for all r which is not possible. Thus, we have here an example of a ring R that can be embedded with a given constant group C = (/3). However, R is contained in the ring S of integers modulo 8, and S cannot be embedded using the same constant group and encoding word.

52

Linkage Graphs and Linkage

Figure 2

Groups

Linkage Graph for ST(3,R)

2 . 7 Counter

Figure 3

Examples

Linkage Graph for ST(4,R)

53

54

Linkage Graphs and Linkage

Figure 4

Truncated Cube

Groups

2.7 Counter

Figure 5

Examples

Heptagon

55

56

Linkage Graphs and Linkage Groups

Figure 6

Hexagon 1

2.7 Counter

Figure 7

Examples

Hexagon 2

57

58

Linkage Graphs and Linkage Groups

Figure 8

Subgraph 1

2.7 Counter

Figure 9

Examples

Subgraph 2

59

60

Linkage Graphs and Linkage

Figure 10

Subgraph 3

Groups

2.7 Counter

Figure 11

Examples

Subgraph 4

61

62

Linkage Graphs and Linkage Groups

Figure 12

Subgraph 5

Chapter 3 COMBINATORICS IN LINKAGE GROUPS 3.1

Free Products With Amalgamation

In the previous chapter we discussed the manner in which linkage graphs arise from verbal embeddings. These linkage graphs have the property that the ring in question is embedded at each vertex in the linkage graph. In this chapter we will approach linkage graphs from another direction. We will construct linkage graphs using other means and then show that the ring is embedded at each vertex. Then we will use the linkage groups constructed to demonstrate the existence of verbal embeddings with various constant groups. First note that we can construct a linkage group merely by specifying a linkage graph and a ring. However, this is not what we mean by constructing a linkage group. For example, if R is a ring with identity then L{R, II5) is the trivial group by Lemma 2.4.1. We are looking for linkage groups in which the ring embeds at each vertex and this requires more than merely specifying a linkage graph and a ring. The first tool we will use in our construction is the free product with amalgamation. For complete details on this construction see for example [16]. In order to be able to refer to subgroups of linkage groups more easily we make the following definition. Definition 1. Let S be a non-empty subset of the set of vertices of a linkage graph T. The vertex subgroup corresponding to S is the following subgroup of L(R,T): R 1 E S). s=(= ri (n \r£R,: \r£R,ieS). Rs-If S consists of a single vertex i then Rs will be denoted by Ri as before. Consider first the case of two linkage graphs each consisting of a basic

63

64

Combinatorics

in Linkage

Groups

linkage, say, Ti consists of the linkage (1,2,3) and T2 consists of the linkage (4,5,6). Then we know that for any ring R the linkage groups G\ = L(R, Ti) and Gi = L(R, T2) are each isomorphic to the group of 3 x 3 upper unitriangular matrices over R. The subgroups Hi = =

<£(fl,r,)

R{2„3}
and Hi Hi == -ft{4,5} #{4,5}


are each isomorphic to R © R, the direct sum of two copies of the additive group of the ring. We identify these groups as follows Hi —►Hi Hi ^ -* h-» r44 r 2 H-> . H-> r& r3 i-> r5

We now form the following free product with amalgamation G\ ** G G22.. G =--Gt Hi HI

Now consider the group G3 — L(R,T3) where T3 has vertices {1,2,3,4} and linkages (1,2,3) and (2,3,4). The group G and G3 can be presented with the same generators and relations and are therefore isomorphic. The theory of free products with amalgamation shows that Giand G2 are both embedded as subgroups of G and hence in G3. Thus, the ring R is embedded at each vertex of G3 = L{R,T3). The construction presented above was first used by Leo Comerford in an (as far as we know) unpublished note. He continued with the technique to construct L(R, II m ) as follows. Construct an n-chain Cn and an m-chain Cm for m, n > 6. Consider the vertices of C m to be {1,2,3, ■ • •, m) and the linkages to be (1,2,3), (2,3,4),- ■ ■. Consider the vertices of Cn to be{m + l , m + 2 , • ■ • ,m + n) and the linkages to be (m + 1, m + 2, m + 3), (m + 4, m + 4, m + 6), • • ■. Consider the following subgroups

Hr #1 — = R{l,2,m-l,m.} R{l,2,m-l,m} <<

L(R,Cmm) L(R,C

and Hi-H2 =-

R{m+l,m.+2,m+n-l,m+n} R{m+1, m+2, 7n+n--l,m+ra} <

L(R,Cn).

65

3.1 Free Products with Amalgamation

Since m,n > 6 the subgroups Hi and H2 are both isomorphic to the free product (.R © R) * (R® R) and may be identified by the isomorphism *2 H2 Hi — ^H ri i—> ^m+7i—1 T2 H-V rm+n

r

m

.

*-* >-» r mm ++ i

rmm~i -i

l—

*

r

m+2

Now we identify # 1 and #2 by this isomorphism and form the following free product with amalgamation G = L(R,C G--L(R,C L(R,C m)* n). m) * L(R,C n). Hi Hi

As in the previous example it is clear that G = L(R, Cm+n-i). Clearly the above constructions can be generalized to include cases where the arrows in the linkages have different orientations. For example, we could have begun with the groups Gi and G2 above but amalgamated H% and H^, — R{i,2} where these subgroups are identified as follows —>H H3 H22 Hz —* i"i H-> rs r 5 ■. ri t~* h - > rr 4 r 22 H-> 4

Then the group L(R,Ti)

* L(R,T2)

is isomorphic to L(R,T)

where T has

vertices {1,2,3,4} and linkages (3,2,1) and (2,3,4). Similarly we can construct an n-chain Cn, n > 3 in which the orientation of the arrows in the linkages is arbitrary. We can also construct II„, n > 8. We now have the following result. Proposition 3.1.1. (Comerford) (i) For n > 3, R is embedded at each vertex of L(R, Cn) for every ring R. (ii) For n > 8, R is embedded at each vertex of L(R, Y[n) for every ring R. (iii) For n > 8, the cyclic group of order n serves as the constant group for a verbal embedding for any ring R. (iv) For n > 4, the dihedral group Dn of order 2n serves as the constant group for a verbal embedding for any ring R.

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Proof. T h e proofs of the first two parts of this result are contained in the previous paragraphs. For the third part we merely construct L(R, Un) as above. Then the cyclic group C = (c) of order n clearly acts as a group of linkage graph au­ tomorphisms on I I n by sending i t o i + l(mod n). Thus C acts as a group of automorphisms on L(R,Un). Let G be t h e split extension of L(R,Un) by C. Define t h e following mapping 4>:R^G > rir\ Clearly <j> is a monomorphism from (R, +) into G. Now r\ = r 2 and r\ T h u s we have = (rs)i. [rC

=

rn-\.

,41 =

Thus <j> is a verbal embedding with constant group C. For t h e last part of the theorem let n > 4 and construct polygonal linkage graph n 2 n with vertices { 1 , 2 , 3 , • • •, 2n} and linkages ( 1 , 2 , 3 ) , ( 4 , 3 , 2 ) , ( 3 , 4 , 5 ) , ( 6 , 5 , 4 ) , - - - , (2rc - l , 2 n , l ) , (2, l , 2 n ) . Let a b e t h e linkage graph automorphism given by 1H->2

1l 3 H-» 2n 4 i(-> - > 22nn --- l1 ■■ i-> 2n — - 2 5 H-» 2H>

: T h u s , a corresponds to the geometric reflection of Il2n about t h e line joining the midpoint of t h e edge {1,2} and the edge {n + 1, n + 2 } . Similarly let /3 be the linkage graph automorphism given by 1 i->2n ► 2n 2n — 2 i— (-»■ - 11 2n -— 2 . 3 I— H->>• 2n 4 t-» !-»• 3n — - 3

Thus, /3 corresponds to the geometric reflection of Il2 n about the line joining the midpoint of the edge { l , 2 n } and {n,n + 1}.

3.1 Free Products with Amalgamation

67

Let Dn = (a,b\ a2 — b2 — e, (ab)n = e). Then Dn acts on II2„ on each vertex i as follows i" = 1° i" = i" ■ Let G be the split extension of L(r,H2n)

by Dn. Then the mapping

cj>:R - > G r t-> rj

is a verbal embedding since [rj,s°] = [r„_i,s 2 ] = (rs)!. This completes the proof. □

The above discussion constructs linkage groups using the free product with amalgamation construction and groups corresponding to basic linkages. We will now continue in the same manner except that we need some more groups to serve as building blocks. Some linkage groups that we will construct will be groups which correspond to linkage graphs that are derived from specific verbal embeddings using Theorem 2.2.1. These graphs were constructed by computer and their origins will be discussed in a later chapter.

Let r be the linkage graph in Figure 1. Note that the two headed arrow in the diagram indicates one of two possible linkages. Thus, for example (2,1,4) is a linkage or (4,1,2) is a linkage, but not both. The group L(R, T) is easy to construct. Let Ai be the linkage graph with vertices 1,2,3 and edges {1,2}, {2,3}, {1,3} and no linkages. Let A 2 be graph with vertices l',2',4 and either (2', 1', 4) or (4,1', 2') as its sole linkage (and the appropriate edges). Then L(R, Aj) is the direct product of three copies of the additive group of the ring R and L(R, A 2 ) = U(R), the group of 3 x 3 upper unitriangular matrices over R. Let S — {1,2} and S' = {l',2'}. The Rs and Rs' are isomorphic via the map ?"! I—► Ty

r2 >-> r 2 /

68

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Groups

We can now form the free product with amalgamation L(R, Ai) * L(R.A2). Clearly this free product is isomorphic to L(R, T) and we can now say that R is embedded at each vertex (for any ring R).

Figure 1

Linkage Graph 1

Now let T be the linkage graph in Figure 2. We will now construct L(R, r ) and show that R is embedded at each vertex for any ring R. First let N denote the direct sum of three copies of the additive group of the ring.

3.1 Free Products with

Amalgamation

69

We identify Ri with elements of N of the form (r, 0,0). Similarly, elements of i?2 and i?3 are identified with elements of the form (0, r, 0) and (0,0, r), respectively. For elements y, z, s G R we define the following functions „ (,. ,\ _ J Vs pi\y,s> | _s2/

i i f ( 2 .1.4) i s a linkage ^ if ^ 1 2 ) i s a l i n k a g e

. | , , \ _ I *s ^ ' ; ~ \ -a*

linkage , if (3,1,4) is a Hi , if (4,1,3) is a lit linkage

and P2

For each a G -R wee can can define define an an operator operator A A33 :: N N —* —► N 1 by (x, y, z)A. = (x + pi(jf, a) + p 2 (z, a), y, z). Clearly, A.TA.S

If we identify the element a4 let G be split extension of N We will now do some niteness we will assume that

— A r _(_ s .

G R4 with .As then i?4 acts on N. We may now by i? 4 . sample computations in G. For the sake of defi(4,1,2) and (3,1,4) are linkages.

( r , 0 , 0 ) - = (r, 0,0)4, = (r +Pi(0,a) + p 2 (0,a),0,0) = (r,0,0), (0,r,0)'* = (0,r,0)A S = ( P l (r,a) + p 2 (0, a),r,0) = = (-ar,r,0), (0,0, r) s - = (0,0, r)As = ( P l (0, a) + p2(r, a), 0, r)) = = (ra,0,r). Thus, a4 commutes with ri = (r, 0,0). Also, [a 4 ,r 2 ] = (r2)-°*r2 = (-ar.O.Oj-^O.r.O) = (ar, 0,0) =

(sr^.

Similarly, [r 3 ,a 4 ] = K T 1 ^ ) 8 4 = ( 0 , 0 , - r ) ( r a , 0 , r ) = (ra, 0,0) =

(rs^.

Hence, the commutator relations are satisfied and G is the group whose rela­ tions are given by the linkage graph in Figure 2. Similar computations yield the same result for any of the various possible orientations of the linkages.

70

Combinatorics

in Linkage Groups

We can now construct the groups L(R, T) for T as in Figures 3 and 4, for example, if T is the linkage graph of Figure 4, by an iterated free product with amalgamation. We can similarly construct the linkage groups corresponding to Figure 5 and Figure 6. These, in turn, can be used as building blocks for many other linkage groups. Definition 2. Suppose that V is a linkage subgraph ofT and that A' is a linkage subgraph of A and that V and A' are isomorphic as linkage graphs. We define the free product of T with A amalgamating F* and A' to be the linkage graph T * A whose vertices, edges, and linkages are those of T and A where the vertices, linkages, and edges of V and A' are identified via their isomorphism. If r ' is a linkage subgraph of T then it need not be the case that L(R, T') = Rr'- For example, if R is any ring with identity, T = n 5 , and T' is a basic linkage of T,then we know from Lemma 2.4.1 that i?r' < L(R, II5) = {e}. However, L(R, V) is the group of 3 x 3 upper unitriangular matrices over JR. Note that if V is a linkage subgraph of T then there is an obvious epimorphism L{R,T')^R L(R, V) —>ri Rr> /g ^ rv 1 y rv

(3.1)

where rv is first considered to be an element of L(R, V) and then to be an element of i?r'- We will refer to 3.1 as the standard epimorphism. We are most interested in the situations in which this epimorphism is an isomorphism. This is an extension of the concept of embedding the ring at each vertex of L(R,T). The free product with amalgamation used above is very useful for showing that 3.1 is an isomorphism. In this case we will refer to 3.1 as the standard isomorphism. The following is clear. Proposition 3.1.2. Suppose that V is a linkage subgraph ofT and that A' is a linkage subgraph of A and that A and A' are isomorphic as linkage graphs. If standard epimorphisms L(R,F') —> Rr< and L(R,T) —> Rr are isomorphisms then •fc L(R,T* *A)*L(R,r) A)^L(R,r) L(R,T L}£L(R,A). Ai)L(R,A).

3.S The Collection

3.2

Process

71

The Collection Process

The method of construction of the group L(R, n m ) is valid as long as m > 8. Hexagonal linkage graphs such as Figures 2.6 and 2.7 can be constructed as in Chapter 2. This leaves heptagonal linkage graphs. We will come back to all of these in a later chapter when linkage graphs are connected to Lie Algebras. In this section we will describe a combinatorial method for constructing II m for TO > 7. This method involves a process reminiscent of Hall's collection process for nilpotent groups. Consider the polygonal linkage graph Um, m>7, with vertices 1,2,3,- • -,m. Let R be a ring (not necessarily associative but with at least 2 distinct ele­ ments). We introduce a relation -< on the vertices of n m as follows: 1 -< 2,1 -< 3 2^3,2^4 3x4,3-<5 m — \ < m,m— 1 ^ 1 m -< 1,TO -< 2 Note that the relation -< is not transitive. Although we read "~<" as "less than" each vertex is "less than" exactly two other vertices. If the relation were transitive then each vertex would be "less than" every other vertex. Let F(R, n m ) denote the free group generated by the set {r,- | r e -R\{0} and i is a vertex of n m } .

Definitions: 1. A word in F(R,Iim) is not freely reduced if it contains 0,- or if it contains an adjacent pair (ri)h(si)k for some vertex i and some integers h, k. Otherwise, a word is said to be freely reduced. The empty word is freely reduced. 2. If a, b are vertices in some linkage ofHm, we say that a and b react with each other and we denote this by a ~ 6. 3. Consider the word w = (ri) 0 ,(r 2 ) 0 2 ---(r n ) a „

72

Combinatorics

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where r; £ R. We say that vertex a,- moves in front of vertex aj if i > j , a,- ~ ak for k — j,j + !,•■• ,i — 1, and either a,- -< aj or ai = aj.

We now describe an algorithm that defines our collection process on words w = (ri)Si(r2)S2 ■ ■ ■ {rn)s„ in F(R, Um). Intuitively, in this algorithm we look for left most vertex ai in w that moves in front of a\. If there is one, then we move it in front of a\ by using steps that obey the relations of L(.ft,IIm). We continue this until nothing more moves to the front. Then we look for the first vertex that moves to the second position of the word. We continue until nothing else can be moved. The exact algorithm is as follows. COLLECTION PROCESS Step 1 Ifw is the empty word then w is called reduced and the process stops. Step 2 If w is not freely reduced then replace any occurrence of (rv)h(sv)k by (hr + ks)v for every vertex v and integers h, k. Replace any occurrence of 0V by the empty word. If w is now the empty word return to Step 1. If w is not freely reduced then began Step 2 again. Step 3 Find the first j > I such that aj moves in front of aj. If none exists, then find the first j• > 2 such that aj moves in front of a2, etc. If no such aj is ever found then we say w is reduced and the process stops. As soon as some aj is found that moves in front of some a,-, then we move (rj)aj in front o/(j"i)0l by a sequence of replacement moves of the following types: (i) Replace raSf, by str a if a and b are vertices joined by an edge. (ii) Replace raSf, by S),ra(rs)c if (a,c,b) is a linkage and a -4 c. (iii) Replace raS(, by Sb(rs)cra if (a,c,b) is a linkage and c -< a. (iv) Replace rasi, by Sbra(—sr)c if (b,c,a) is a linkage and a -< c. (v) Replace raSf, by si,{—sr)cra if (b,c,a) is a linkage and c -< a. After {rj)aj is moved in front o/(r,) a i then go to Step 2 and continue with w replaced by the newly transformed state of w.

3.2 The Collection

Process

73

In the Collection Process the replacement moves are taking place in words from the free group F(R,Um). The replacement moves correspond to relations in L(R, II m ) and so when a word is changed by a replacement move it does not change as an element of L(R, Tlm). It is necessary to prove two things as we proceed: • The Collection Process terminates after a finite number of steps and gives as an output a reduced word. • Two words represent the same element in L(R, II m ) if and only if the Collection Process yields the same reduced word when it is applied to them.

If these two things are shown then the reduced word that comes from the collection process will be a normal form, that is, we will be able to tell when two elements are equal in L(R, Um) merely by applying the Collection Process to the words. We will now give several examples of the Collection Process as applied to the group L(Z, II7). In these examples the notation w —y u will mean that the word u is derived from the word w by a finite series of replacement moves. 1. w — I1I4I1 and u = I4I1I4 are both reduced. (Note that in the symbol 1 4 the 1 represents a ring element, that is, an integer, and the 4 represents vertex number 4.) 2. w = 2 4 32li —* 32 (—6)324li = u. Here u is reduced. 3. w = l 1 l 2 2 5 (-6) 6 37 -+ l i l 2 2 5 3 7 ( - 6 ) 6 -> -+ 1x12372566 ( - 6 ) 6 -+ l 1 3 7 ( - 3 ) 1 l 2 2 5 6 6 ( - 6 ) 6 -» -+ 37li(-3)il 2 2 5 6 6 (-6) 6 -+ 37li(-3)il 2 2 5 0 6 ^ —+ 37(—2)il22s = u. Once again u is reduced.

We will now show by example that the collection process need not yield a normal form for L(R, Rm) if m < 7.

74

Combinatorics

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Groups

1. (Non Termination of Collection Process) Let m — 5, and let R be a ring with a non-zero element a such that a2 = 0. Then w = axa,2(i4 —> and-^a-i —> a2<24ai —► a\a2a4, etc. Note: This example could really have been done with any ring R but it is easier to keep track of the commutator terms with a2 = 0. 2. (Unequal Reduced Forms For Equal Elements) Let m = 6, and let R = Z2 be the ring of integers modulo 2. We will first do some calculations in the group L(R, Te). Over this ring we do not have to keep track of minus signs and this makes our calculations easier. For example, l4=[l3, Is] = I3l5l3l5- However, calculations similar to the one we are about to make can be done over any ring. I4I1U I 4 l l l 4 == l 3I3I5I3L. l 5 l 3 l 5al ilil44 == = I1I2I3I5I6I2 I1I2I3I5I6I2I3I5I6I4 I3I5L3I4 == I1I4I2I6I2I6 — = = llUll.

Now I4I1I4 and I1I4I1 are both reduced and have a different form but they are equal in the group. 3.3

Termination of the Collection Process

Let w b e a reduced word in F(R, II m ), m > 6. In studying the collection process we will initially consider the word w r,- for some vertex i and some r £ R. To simplify notation we assume without loss of generality that i = 3. The vertices 1 and 5 will be called gates for 3. Since m > 6 it follows that 3 and 5 do not lie in the same linkage. This fact separates the II5 and lie from II m for m > 6, and in fact, this is crucial for what follows. Let w = (ri)ai(r2)a2 ''' (r™)a„ be a reduced word. If 3 lies in a common linkage with each of the vertices a,i and j is minimal with respect to aj being a gate for 3, then ((ri}a r i ) a 1i ((r%}m r 2 ) a 2 -!i--'(rs)a ••(»"i)a;i is called the first factor of w with respect to 3, or just the first factor of w. Thus, if a,j = 1 and TJ = r, then the first factor of w if either nr\ ,, or or

U47"i , o r

v2u4ri , or vv22tt33U4ri. u4rx.

3.3 Termination

of the Collection Process

75

We will denote this first factor by v2t3u4ri with the understanding that v,t or u could be equal to 0, in which case the term v2 or u4 is considered to be absent. If aj = 5 then the first factor of w is of the form V r!, v22tt3U 3u44n,

r €e R\0,

u,v,te u,v,t£R. R.

If j is maximal with respect to a,j being a gate for 3, then (ri)^ •

\. r n)a„

is called the last factor of w with respect to 3,.or just the last factor of w. If we use a notation similar to that for the first factor of to, then the last factor of w is of the form riu2t3v4, r5u2, rsU2,

r G € R\0,

u,v,t € R, or

r € JR\0, R\0, «u eG R.

In a similar fashion we can define the first and last factors of w with respect to any other vertex. In some of the replacement moves in the Collection Process we obtain ring elements of the form ±rs or ± s r . We will denote any one of these four signed products by p{r,s). Elements of the form p(p(r,s),t) or p(r,p(s,t)) will be denoted by p(r, s, t). When a word u has be reduced by the collection process, then we will denote the reduced form of the word by u. The following extremely technical result describes what happens in the collection when r3 moves to the first factor of w r3. Lemma 3.2.1. Let w = (ri) a i (r 2 ) a 2 • • ■ (r n ) a „ be a reduced word in F(R,Um) where m > 6. Suppose that w has at least one gate for 3 and that 3 is in a common linkage with every vertex a,- of w. Suppose further that in applying the collection process to w r3 that r 3 moves into the first factor of w. Then (i) The collection process on w r3 terminates after a finite number of steps. (ii) The gates for 3 in w r3 are the same as the gates for 3 (and in exactly the same order) as the gates for 3 in w. (iii) In applying the collection process to (w r3)s3, the s3 moves to the first factor of w r3.

76

Combinatorics

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(iv) The first factor ofw is either a^Ste^i or £22/324*5 (depending on whether the gate for 3 in the first factor is 1 or 5), where x,y,z £ R and s,t G /?\{0}. (If x — 0, y = 0, or z — 0, then we consider x2,y3, or Z4 to be the empty word and thus not actually appearing in w. (v) The first factor ofw r 3 is either (x')2(j/')3S4 1 and that the gates for 3 are (in order) gi,gi,--- ,gicCase 1: gk-i = gk = 1. Then w = ■ • ■ s ^ ^ u i j A i where x,y G R and s,t,u G i?\{0}. Hence, w r 3 -> • • • s1x2t4u1 r3 (p(r, u))2 y4. w'

The induction hypothesis applies to w' r3, and thus it may be reduced by applying the Collection Process. Since the r4 moves into the first factor there are two subcases: (i) Si is immediately preceded in w' by v$, v ^ 0. (ii) 5i is immediately preceded in w' by V4, v / 0. Suppose that (i) holds. If p(r, u) = 0 then we have w r3 —> (w' r3) uiy4. Thus we may assume that p(r, u) ^ 0. Then we have wr3-y-

v&si (x + p(r,s) + p(r,u))2 s

(p(r,u,t))s

t4«i2/4-

*

v w"

Now w" is reduced. If p(r,u,t) hypothesis applies to

= 0 then we are done. If not, then induction w"(p(r,u,t))3.

3.3 Termination

of the Collection

77

Process

Thus, w r3 -+ ■ ■ ■ u 5 si (x + p(r, s) + p(r, u) +p(s,r,u,

t))2 <4Wii/4

and this final word is reduced. Suppose that (ii) holds. Since w is reduced it must be that x = 0. We have w r3-> ■■■ U43ir3i4ui (p(r, u))2 y4 ->■ • ■ v4S! r3t4Ui (p(r, u)\ y4. w'"

Now w'" is reduced. If p(r, u) = 0 then we are done. If not, then (p(r, u))2 moves in. The gates for 2 are 4 and m. There can be no m's in w'" since otherwise r 3 would not be able to into the first factor. Since w'" is reduced and has the same gates for 3 as the word w with the last factor removed, it follows that the number of 4's in w'" is less than the number of gates for 3 in w. Hence the induction hypothesis applies to moving in (p(r, u)) 2 , and this case is finished. Case 2: gk-i = 9k = 5. We have w = ■ • ■ S s ^ s ^ where x £ R and s,t,u £ R\{0). Hence wr3~*y-

s5t2r3 (p{r, u)) 4 u5x2. V VJ1

By induction we have wr3 ->■ • • s5t2

{p(r,u))4usx2.

reduced

Now the gates for 4 are 2 and 6, but there are no 6's in w'. If p(r, u) = 0 we are done. If not, then (p(r, u)) 4 moves into the first factor, or until it meets a vertex of 1, or until it meets another vertex of 4. As (p(r, u)) 4 moves in it cannot create any commutator except 3. By induction these 3's move to the from as the original r3 does and this case is done. Case 3: gk-i = 1, 3k = 5. In this case we have w = ■ ■ ■ Six2y4t5z2 where x,y,z € R and s.t G R\{0}. Thus i o r 3 - > ■■■s1x2r3 {y + P(r,t))4 t5z2. induction applies

If Si is immediately preceded by v& where v ^ 0, then w r3 ->- - • u 5 si (x + p(r, t))2 (y + p(r, t))415 z2, * ' w reduced by induction

78

Combinatorics in Linkage Groups

and this final word is reduced. If Si is not preceded by v5 then w r3 reduces according to the lemma. Case 4: 9k-i = 5, gu — 1. Here we have w = • • ■ s5tix2y4 where x,y £ R and s, t G # \ { 0 } . Hence, «>r3-» ■■■s5r3t1

(x + p(r,t))2yi

->

induction applies

-» ■■■s5ti

(x+p(r,t))2y4

and this last word is reduced. The proof is now complete. □ The above lemma is the starting point for the next result which will be fundamental in deriving our normal form results. Before starting we need another definition. Definition 2. For any vertex i mutator of i. The vertices i — 1 and of i. If j is an m-fold commutator of i, commutators of i. (Of course the +1

of Hm we say that i is a 0-fold com­ i + 1 are called 1-fold commutators then j ' — 1 and j + 1 are ( m + l ) - f o l d and —1 here are done modulo m.)

Lemma 3.2.2. Let w be a reduced word in L(R,Tlm) for m > 6. Suppose that w does not end in t 3 , t ^ 0. If wr3 is not reduced then (i) the collection process on w r3 terminates after a finite number of steps and yields the reduced word WT3, and (ii) the last factor of wr3 is the same as the last factor of w unless w ends in ■85^1^22/4, where s,t ^ 0. In this caseWr^ ends in s$t\a(x-\- p(r,t))2 y4. Proof. Assume that the result is false and pick a counter example w of minimal length. Suppose that in applying the Collection Process to w r3 no n-fold commutator of 3 ever gets collected to the first factor of w. If w = saw' where s ^ O , then w' is a counter example of shorter length. Hence we may assume that some n-fold commutator of 3 reaches the first factor of w. We proceed by induction on n. If n = 0 then the result is the previous lemma. Hence, we may assume that n > 0.

3.3 Termination

of the Collection

Process

79

Thus we have w = w'w" where w' is a non-empty reduced word and r3 moves all the way to the first vertex position of w" in the Collection Process. In Figure 8 we have a tree that will help us keep track of the possible vertices of w. In this tree the symbol \3J indicates the position in w to which r3 is initially collected. Note that w may or may not have an element S3 at this position. Beginning at \z\ and going to the right a finite sub-branch of this tree represents the vertices of w". A finite sub-branch going to the left represents the vertices of w'. Suppose the vertices of w" are represented by the branch of Vertex Tree 1 which begins [3]—5 . Collecting r 3 to the front of w" we may produce more vertices, but the only vertex that could be produced and move across the initial 5 of w" is a vertex of 4 or another 3. We need to know what could cause a 4 to be moved into w'. Vertex Tree 2 in Figure 9 shows what vertices must be present in w' in order to draw in a vertex of 4. After collecting r3 to the [3j position of w, a commutator S4, s ^ 0, may be produced at r3 moves across the vertex 5. The next step in the Collection Process would be applied to (w'r3)

s4.

reduced

(Refer to Figure 9.) Since an n-fold commutator of 3 reaches the first factor of w, it follows that an (n-l)-fold commutator of 4 reaches the first factor of w' 7-3. Hence, w'r3s4 collects according to the induction hypothesis. Thus, if w' = ■ ■ ■ t$U2X3Vs where t, M, V ^ 0 then we have w r3 —> • ■ • t6u2{x + r)3 S4V5 •■•-»••■ t6u2(x + r + p(s, M)) 3 V5

In collecting w'" everything proceeds essentially as it did for w"r3 except at than any 4-vertex that collects in front of v5 must also collect across the 3vertex and the ^6^2- By induction ^ 2 remains. The steps remaining in the Collection Process are those from reducing w'r3 and this collects according to Lemma 3.2.1. This ends this case. If the vertices of w" are represented by the branch of Vertex Tree 1 that begins with [§]—4, then the proof if completely analogous to the above and hence it will be omitted. This completes the proof. □

80

3.4

Combinatorics in Linkage Groups

The Normal Form Theorem

In the previous section we proved that the Collection Process (over the link­ age graph II m , m > 6) acting on words wr3 where w is a reduced word, always terminates after a finite number of steps. Now suppose that w = (n)ai(r2)a 2 ■ ■ • (r„) a „ where r^ £ .ft\{0} and each a,- is a vertex. We apply the Collection Process in a left-normed fashion, that is, we first reduce (ri) 0 1 (r2)fl2. Let w' be its reduced form. Then we reduce w'(r3)a3. Continuing in this fash­ ion we will after a finite number of steps reach a reduced word w. We may view L(R,Tlm) in several ways. For example, let N be the set of reduced words . Then L(R, II m ) may be regarded as the group whose elements are the elements of N with the group operation denned as follows. If w, u 6 N then the product of w and u is the word wu obtained from the left-normed Collection Process as applied to the concatenation of w and u.

There is another way to view L(R,Um). Let F(R,Um) defined in the previous section and let Sym(N) be the group of all permutation on N. There is a homomorphism e : F(R,Ilm) -r Sym{N) n \-r (r,)* where u;(r;)* = Wfl. Then w may be viewed as the image in N of the empty word under the permutation we : N —► N. It is this view that we will take. Note that all of the above work holds no matter what orientation the arrows in the linkages have. Thus, we may be working over II m or over II m as long as ra > 7. We can now prove the following. Theorem 3.4.1 (NORMAL FORM THEOREM).Le< m > 6. Let w,u two reduced words in F(R, II m ). Then w = w if and only ifw = u in the group L(R,nm). Proof. Let K be the set of relations in L(R, Um). Let u and w be reduced words in F(R,Um). If u = w in the group L(R,Um) then u = vw for some v £ K. Then u — vw = vw = w and both u and w have the same normal form. If two elements of F(R,Um) have the same normal form then they are clearly equal in L(R, Um). This proves the theorem. □

3.5 Consequences of the Normal Form

Theorem

81

Remarks 1. If we restrict out attention to the m-chain graph Cm rather n m than the Normal Form Theorem still holds (for m > 3). The proof is actually simpler since, for example in C$ the vertices 1 and 6 do not lie in the same linkage. 2. If Coo is the linkage graph whose set of vertices is set of Integers and whose linkages are (i,i + l,i + 2) or (i + 2,i + l,i), then the Normal Form Theorem holds for L(R, Coo). 3. In view of the fact that 0 is a homomorphism, the Collection Process need not be performed in a left normed fashion. Thus, if the replacement moves are performed in any order at all on a word w then whenever a reduced word is reached, then this reduced word is the word w yielded by the left-normed Collection Process. 3.5

Consequences of the Normal Form Theorem

There are many consequences of the Normal Form Theorem for the combina­ torial structure of the group L(R, II m ), m > 6. We will describe some of the most important in this section. For m > 7 most of these results follow from the free product with amalgamation construction given in section 3.1. For m = 7 these results need the Normal Form Theorem. Proposition 3.5.1 R is embedded at each vertex of L(R,]lm),

m > 6.

Proof. Let i be a vertex of II m . The word rt- is in reduced form. Hence the mapping R —t L(R,flm) given by r t-* r{ is thus one-to-one for each i. □ Proposition 3.5.2. Form > 6 the vertices 1,2,3, • • ■ ,m — 2 in Um are vertices of a subgraph isomorphic to Cm-2 (the orientation of the arrows in the subgraph being the same as the corresponding arrows in the graph). For m > 6 the mapping L{R,Um) ^ L(R,flm) is a monomorphism.

82

Combinatorics

in Linkage Groups

Proof. The Normal Form Theorem applies elements in L(R, Cm) and the Collection Process works in exactly the same way for these elements whether they are considered to be in L(R, Cm) or in L(R, Ylm). O Proposition 3.5.3. Let R be a ring with more than one element. (i) Every subgroup of finite index in L(R, II m ), m > 6, contains a free subgroup of rank 2. (ii) Every subgroup of finite index in L(R,Cm), group of rank 2.

m > 3, contains a free sub-

Proof. In either case (i) or (ii) we consider the subgroup #{1,3,4}. From the Normal Form Theorem we have

(R3 S© Ri). #{i,3,4} = Ri B RA). X * {Rz {i, 3,4} ^R R

It is a standard exercise (see [16], page 195, exercise 19) that the free product of two non-trivial groups with at least one of them of order more than 2 contains a free subgroup of rank 2. Since R has at least two elements, i?2 @ Ri has at least 4 elements. Hence, #{1,3,4} has a free subgroup, say F, of rank at least 2. Now let H be any subgroup of finite index k in L(R, II m ) or L(R, II m ). By the Nielsen-Schreier Theorem ([21], Theorem 6.1.1) H H F is a free group of rank 2& + 1 — k = k+1. Hence, H 0 F is a free subgroup of H of rank at least 2. D Proposition 3.5.4. If S is a subring of R then the following mappings are monomorphisms: (i)

L(S,C ) -+L{R,C L(S,Cmm)-^L(R,C m) m) t-t Si Si H->

for for m m > >33.. (ii)

L(s,nm) -*L(R,nm) L(s,nm)Si ^ L(R,nm) *-* Si for m m > 77..

Si t-* Si

3.5 Consequences of the Normal Form

Theorem

83

Proof. The Normal Form Theorem applies to elements that come from S as well as R. □ Proposition 3.5.5. Ifm > 7 then any element of finite order in L(R,U.m) is conjugate to an element in a basic linkage. Hence, if the additive group of R is torsion-free, then so is L(R,Tlm). The same is true of L(R,Cm), m > 3. Proof. In this proof we will use explicitly the iterated free product with amalgamation construction of L(R, II m ) for m > 7. Recall that in a free prod­ uct with amalgamation an element of finite order is conjugate to an element in one of the factors. This forces an element of finite order in L(R, fim) to be conjugate to an element in one of the basic linkages. These basic linkages are isomorphic to the 3 x 3 upper unitriangular matrix group U(R). Since U(R) is an extension of (R, +) by (R, +) ® (R, +), it follows that L(R, fim) is torsion-free if (R, +) is torsion-free. The same argument applies to L(R, Cm). □ Proposition 3.5.6. / / R has more than one element and m > 6, then L(R, II m ) is centerless. Proof. Let w be a non-trivial word in the center of L(R, II m ). If m > 7 then the result is true by the free product with amalgamation construction. Let m = 7. Without loss of generality we may assume that w ends in • ■ • si for some s € i?\{0}. Let r G i?\{0}. Consider a word of the form u =

r4r7r3r6r2r5ri

In u the vertices start at 4 and then each vertex is the previous vertex plus 3 (modulo 7). It is clear that the normal form of wu is just wu (the concatenation of w and u with no reduction possible). The word w may begin with any vertex. For the sake of definiteness we assume w = t\ ■ • ■ s\. We also assume that u = TA ■ ■ • r5ri- ID applying the reduction process to uw we find that the £4 of w does not move in. If some other vertex from w moves in across the ri of u, then it must be of the form xt- where i is a vertex that reacts with 1 and 4. The only vertices that react with both vertices 1 and 4 are 2 and 3 However, if a 2 or 3 moves in, it would have already moved into the front of w to begin with. Since w is reduced this is not possible. Hence, the normal form

84

Combinatorics

in Linkage Groups

of uw is uw itself. Since w is central we thus have that uw = wu. This cannot happen unless w is of the same form as u. Hence there are many elements of the group that do not commute with w. □ Proposition 3.5.7. / / R has more than one element and m > 1 then G — L(R,Hm) contains no non-trivial normal subgroups of finite order. Proof. Let x be an element of finite order in G. Then with out loss of generality we may assume that x lies in a single linkage. Thus, suppose that in normal form x = r^tz- Let c be a non-zero element of R. Now let Un =C6C 2 C6C 2 • ■ • C2Cg . " v ' 3+2n times

Then w"1 x un is in reduced form since m > 7, and is in the normal closure of a; in G. Hence this normal closure is infinite. □ Remark: The example in chapter 1 of a verbal embedding of the integers modulo 2 into the unitary group 1/3(3) gives rise to the heptagonal linkage graph H7. When it was initially studied it was not apparent whether or not L(R, H7) had order greater than 1. The unitary example shows that L(Z2,Hr) has order at least 6048 and embeds the ring at each vertex. The combinatorial results listed above show that L(R, H7) is infinite for any ring with more than one element. In a later chapter we will try to identify the kernel of the mapping

L(z2,n7)-+r/3(3). We now know that this kernel is infinite and must contain a free subgroup of rank 2 (and hence a free subgroup of countably infinite rank). We also know that L(Z 2 ,n7) is centerless and that the elements of finite order have orders 1,2, or 4. 3.6

Standard Linkage Graphs

In this section we will describe some special linkage graphs T such that the consequences of the Normal Form Theorem described in the last section all apply to L(R, T) for any ring R with identity. Let £ be one of the linkage graphs in Figure 5 or 6. Now consider the group G = L(R, S ) where R is

3.6 Standard Linkage Graphs

85

a ring with identity. The group G is a free product with amalgamation and as such its center is a subgroup of the amalgamated part of the two factors. However, this means that the center of G is trivial. Also note that G has a free subgroup of rank 2 and as in Proposition 3.4.3 ever subgroup of G of finite index has such a free subgroup. Definition 1. The linkage graphs £ and IIm, m > 7, are standard linkage graphs. Any linkage graph that is obtained from two standard linkage graphs by the free product with amalgamated subgraph is also a standard linkage graph. If T is a standard linkage graph then L{R, T) is called a standard linkage group. By using the properties of the free product with amalgamated subgroup construction we may now easily prove the following. Theorem 3.6.1. Let R be a ring with identity and let T be a standard linkage graph. Let G = L(R, T). Then (i) The ring R is embedded at each vertex of G. (ii) Let A be a linkage subgraph of T. Then the inclusion map L(R,A)^L(R,T) is a monomorphism. (iii) A subgroup of finite index in G contains a free subgroup of rank 2, and hence contains a free subgroup of countably infinite rank. (iv) Suppose that S is a subring of R. Then the inclusion map

L(s,r) -> L(R,r) is a monomorphism. (v) An element of finite order in G is conjugate to and element in a basic linkage. Thus, if the additive group of R is torsion-free, then G is torsionfree. (vi) The group G has trivial center.

86

Combinatorics in Linkage

Figure 2

Linkage Graph 2

Groups

3.6 Standard Linkage

Figure 3

Graphs

Linkage Graph 3

87

88

Combinatorics

Figure 4

in Linkage Groups

Linkage Graph 4

3.6 Standard Linkage Graphs

Figure 5

Basic Piece

89

90

Combinatorics

Figure 6

in Linkage Groups

Basic Piece

3.6 Standard Linkage Graphs

Figure 7

Heptagonal Linkage Graph

91

92

Combinatorics

Figure 8

in Linkage

Groups

Vertex Tree 1

3.6 Standard Linkage Graphs

Figure 9 Vertex Tree 2

93

Chapter 4 LINKAGE LIE ALGEBRAS AND GROUPS 4.1

The Exponential Map

The bracket operator [_, _] occurs in group theory and denotes the commutator operation. It also occurs in the theory of Lie rings and Lie algebras where it is referred to as the Lie bracket. These two bracket operators are, of course, closely related. Definition 1. Let R be a ring. A Lie ring over R is a module M over R with a binary operation [_,_] on M such that 1. [a;, x] = 0, for all x £ M. 2. [a;, y] = —[y, x], for all x,y £ M. 3. [rx + sy,z] = r [x, z] + s [y, z], for all x,y,z

£ M, and all r, s £ R.

4. (Jacobi Identity) [a;, y,z] + [y,z,x] + [z,x,y] = 0, for all x,y,x where [x,y,z] = [[x,y],z]

£ M,

If R is a field then a Lie Ring is called a Lie algebra. In most of our considerations it will not matter whether or not R is a field. Hence we will loosely speak of Lie algebras even when we should more properly speak of Lie rings. This will cause no confusion. We will now use a linkage graph to define a Lie algebra by generators and relations in the same way we used linkage graphs to define groups by generators and relations. Definition 2. Let R be a ring and Y be a linkage graph. The linkage Lie algebra defined by R and T is the Lie algebra generated by the set of

95

96

Linkage Lie Algebras and Groups

vertices ofT together with the standard Lie relations and the following linkage relations (i) [r u,su] = 0, for all r,s G R, for all vertices u, (ii) [ru,«»] = 0,for all r,s € R, if u and vare connected by an edge in T, (iii) [ru,su] = (rs)w, for all r,s G R, if (u,w,v)is algebra will be denoted by C{R,G).

a linkage ofT. This Lie

As a first example, let T be the hexagonal linkage graph of Figure 2.6. We will now construct a multiplication table for the bracket operation in C(R, T). We will temporarily denote [i,j] by ij for the sake of simplicity. We can read off many of the multiplications immediately from the linkage graph T. For example, 1 1 = 0 , 1 2 = 0 , 1 3 = 2. The rest of the multiplications we can compute using the Jacobi identity. For example, (1 3 ) 5 + (3 5 ) 1 + (5 1)3 = 0 which implies 25 + 4 1 + 6 3 = 0. Hence, 141 = - ( 4 1 ) 1 = (2 5)1 + (63)1 = = - ( 5 1 ) 2 - (12) 5 - (31) 6 - (16) 3 = = -62-0 + 26-0 = = 26 + 26 = 1 + 1. Hence, 1(14) = -(14)1 = - 1 4 1 = - 1 - 1 . We will now perform several more calculations in this algebra. 251 = 1 4 1 - 6 3 1 = 1 + 1 + 316 + 163 = = 1 + 1 + 3 1 6 = 1 + 1 - 2 6 = 1 + 1 - 1 = 1. Hence, 1(25) = - 1 . 142 = - 4 2 1 - 2 1 4 = 2 4 1 = - 3 1 = = 13 = 2.

4-1 The Exponential Map

Hence, 2(14) = - 2 . 252 = 1 4 2 - 6 3 2 = = - 4 2 1 - 2 1 4 + 326 + 263 = = - 3 1 + 13 = 2 + 2. Hence, 2(25) = - 2 5 2 = - 2 - 2. 143 = - 4 3 1 - 3 1 4 = 2 4 = - 3 . Hence, 3(14) = 3. 2 5 3 = - 5 3 2 - 3 2 5 = 352 = 42 = 3. Hence, 3 (2 5) = - 3 . 144 = 254 + 634 = = -542-425-346-463 = = 2 4 5 - 4 6 3 = - 3 5 + 53 = - 4 - 4 . Hence, 4(14) = 4 + 4. 145 = - 4 5 1 - 5 1 4 = - 6 4 = - 5 . Hence, 5(14) = 5. 256 = - 5 6 2 - 6 2 5 = 265 = - 1 5 = 6. Hence, 6(2 5) = - 6 . (14)(25) = - 4 ( 2 5 ) 1 - 2 5 1 4 = = 2541-2514 = = - 4 1 - 1 4 = 0.

97

98

Linkage Lie Algebras and Groups

[,] 1 2 3 4 5 6 14 25

1 0 0 -2 -14 6 0 1+1 1

2 0 0 0 3 -25 -1 2 2+2

3 2 0 0 0 -4 -25+14 -3 3

Table 4.1

4 14 -3 0 0 0 5 -4-4 -4

5 -6 25 4 0 0 0 -5 -5-5

6 0 1 25-14 -5 0 0 6 6

14 -1-1 -2 3 4+4 5 -6 0 0

25 -1 -2-2 -3 4 5+5 -6 0 0

Lie Multiplication Table

The above calculations allow us to create a multiplication table for the Lie alge­ bra £ = £(R, T). This algebra has as its basis the elements 1,2,3,4,5,6,1 4,2 5. The complete bracket operation is given in Table 4.1. Notice that in this bracket operation we have x i i i — 0 for ever vertex i and every x 6 C For each vertex i we let ad{i) denote the adjoint m a p p i n g ad(i) : C —► C x i-> [x,i] In the algebra we are considering we have ad(i)h = 0, for all integers k > 2. For every vertex i and every r 6 jR we define the exponential m a p °° rk exp (r ad(i)) = I + £ TfO^O* where / is the identity map. Ostensibly the definition of the exponential map is an infinite sum and requires — to be in the ring for all k. However, in our k\ present example the sum is not infinite since ad(i)k = 0 for k > 2. It would still appear that - must be in the ring in our present example. We will sidestep this issue for the moment and assume that our ring R contains the inverses of enough integers to define the exponential map, and that R is commutative. We say that the linear transformation ad(i) is nilpotent if there is some integer k such that ad(j)k = 0. We say that ad(i) is locally nilpotent, if for each x e £{R, T) there is a positive integer k such that x ad(i)k = 0. Whenever we say that t h e exponential m a p can b e defined we mean that each ad(i)

4-1 The Exponential

99

Map

is nilpotent or locally nilpotent and that the commutative ring R contains enough inverses of integers. In order to continue with this example and others we need to know more about the exponential map. The next few results are special cases of the Campbell-Baker-HausdorfF formula (see [16]). Proposition 4.1.1. Let r,s G R and i,j be vertices ofT. Suppose that R is commutative. If the exponential map can be defined then: (i) exp (r ad(i)) exp (s ad(i)) = exp ((r + s) ad(i)), (ii) If i and j are joined by an edge in T, then exp ((r ad(i)) exp (s ad(j)) — exp (s ad(j)) exp (r ad(i)).

Proof. Since the exponential map can be defined we can may assume that the infinite sums involved are convergent. We first consider the following product exp (r ad(i) ) exp (s ad(j)) = 2

2

= (I + r ad{i) + j ad(i)2 + ■ ■ ■)(/ + s ad(j) + i - ad(jf

+ ■••) =

r2 = I + rad(i) + — ad(i)2 + • ■ ■

u ^

r2s +s ad(j) + rs ad(i) ad(j) + — ad(i) 2 ad(j) + ■ ■ ■ +~ ad{jf + ^ y - ad(i) ad{j)2 + ■■■ Now suppose that i = j . Then (4.1) becomes / + (r + s) ad(i)

r 2 +

+

2 r s

+ g2

ad{i)2

+

..,

(42)

100

Linkage Lie Algebras and Groups

In this sum the coefficient of ad(i)n is n n

fc n -k r„fc„n—fc .s

E £«(»-*)!" fc!(n- ■k)V

fc=0

Here we have of course used that R is commutative. Hence the coefficient of ad(i)n is n\ A r r*k3sn-~*k = E L o C O ^ " " * _ fr + «)" n! n! £Qk\{n-k)\ n\t £ k\(n - *)! n\ n\ Thus, (4.2) becomes

^{r±sr , ad{i)n = exp ((r + s) ad[i)) I+ ad(i)).. I + E ^ ad{tT 71=1

This proves part (i). Now suppose that i and j are joined by an edge in I\ Then [i,j] = 0 in C(R, T). Thus by the Jacobi identity we have

[fc.*1.J] = = [*i* ,/] = = -[*>/•«]-!>*,*•»]--[*>ii«] - [ > , ar.i] = -[[),*]■+ i, i] == = --[o,a] [*,j,«] +-[*»j

= [[x [[*,»i]. j], »]. »1 Therefore, xad(i)ad(j) = xad(j)ad(i), that is, ad(i)ad(j) Thus, exp(r i) exp(s, j ) = exp(sj') exp(ri).

=

ad(j)ad(i).

This completes the proof. □ We will use the bracket operator [-., _] to denote the group commutator when we are dealing with a group operation, the Lie multiplication when we are dealing with the operation in £(R, Y), and we when we are dealing with linear transformations such as ad{i) and ad(j) we will us it to denote the ring commutator [ad(i),ad(j)] = ad(i)ad(j) — ad(j)ad(j) (which is, of course, a Lie multiplication).

4-1 The Exponential Map

101

Suppose that the exponential map can be defined. By part (i) of the previous result it follows that exp(r ad(i)) is an invertible linear transformation on C(R, T) and that (exp(r ad(i))) - 1 = exp (—r ad(i)).

Definition 3. If the exponential map can be defined on C(R, T) then the Lie group associated t o C(R, T) is defined to be the group G(R,T) = (exp(ri) \reR,ie

V(T)).

The group G(R, T) is also called a linkage Lie group. We need the following technical lemma. Lemma 4.1.2. Suppose that for some x € C — C (R, T) and that X and Y are linear transformations of C Suppose there is a positive integer m such that x Xm = xYm = x[X,Y]m Suppose also that [[X,Y],X]

= 0.

= 0 and [[X,Y],Y] = 0. Then (X + Y)4m'3

=0

and exp (X + Y) exp (±[X, Y]j = exp (X) exp (Y). Proof.

The first thing we do is to prove the following XYk = YkX-kYk-1[Y,X],

forfc= l,2,---.

For k = 1 this formula is

XY =

YX-[Y,X]

(4.3)

102

Linkage Lie Algebras and Groups

which is true. Assume that (4.3) is true for k. Then k k XYk+1 = {XY)Y (XY)Y = 1 k l = YkXY-kY XY k~-kY -- [Y,XY} = [Y,XY} k 1 = yYk* ( 7 I(YX-[Y,X})~ = - [Y,X]) - kY k Y^IY,XY] ~ [Y,XY] = k k 1 = Yk YX Y-kYX-Y Yk [Y,X] [Y,X}- kY k r*" ~ *[Y,X]Y [Y,X] Y = k l 1 [Y, [Y,X]-k X] -kYY**~ [Y,X]Y [Y, X)Y = =

= Yk+1 X-Yk

k+1 k = Yk+1 YX-Y X-Yk[Y,X] -kY -kYkk[Y,X} [Y,X] = k+1 = Yk+1 X-(k X-(k

+ 1) l)Y Yk k[[Y,X]. Y,X}.

Thus (4.3) is true by induction. Next we will prove

(X + Y)n n!

b

v

■xg '"^ "

(4(4.4) 4)

'

If n = 1 then k = 0 and either i = l,j = 0 or i = 0,j = 1. Thus (4.4) becomes x X

+ + yV = = xX + + y. V-

For n = 2 the left side of (4.4) becomes X*_ IF X2 XY + 2 2

_xX2 2

71 YX ' 22

+

XY XY XY + + = 2 1 2 1 2 2 2 2

+

Y2 __ + _ 2T =

\X,Y] [X,Y] Y2 —-—= 2 H 2 22 [X,Y] Y 2 2

4 + *r + 2

This last expression is the right hand side of (4.4) for n = 2.

4-1 The Exponential Map

103

We now proceed to the induction step and assume the result is true for n -1. Then fc (X (x + Y)n _ X X+Y I _ „, (-i)*yyj[.Y,y]*\ (-i) jf y j ' p w \ _ (i\){j\){k\){2*) "n!! W s U -^ i (*!)ti!)(fc!)(2*) jy \i+J+2fc=Ti-l »» U (W)(*i)(*) /"

=

i 11 k k 1 y, (~l)kkxXi+ +Y'[x,Y] Y^[X,Y] 1i (-i) _, k fc + n ., (i\)(j\)(k\)(2 ,+, i, +JLt^^_, , , " « , + J + k„-i (i0(i!)(^)(2 )) " ' n« iW

i _^

^ v

((-i)* - l f(~l) j: X''+ CkX ' +i+1i1y^[x,y] yY>[X,Y] i ^ y ] *Ak

'»

H f l L_ .,

((«)(j'-)(«)(2 W ) ( K ) ( * ))

1. i +

"

ii_

fc

k i (-l) ( - lk^ Xij(-l) Yi f '+1 yX[X,Y] ' Yi+ + ^ j1k[X,Y] f . y ] *k (i\)(j\)(M)(2k) i+i+2i=n _, (TO)(*0(2*) i+i+k.-i"

^

4 2 v

k

1

1

l{-l) u - i j ' xX^ ^ y ' i x .Y^X^]^ y ] *1*^- 11 i(-i)* x'^y^^y] (i\)(j\)(k\)(2k)

t'+j+2A:=ra-l

II1 n

kk i ki i k fc (-l)hfc(-l) X Y'[X,Y] (~l) (-i) Xkjf"y^[x,y] Yi[X,Y] X Y^[X,Y] k (i-l)l(j\)(k\)(2 ) i+i+2*;=r ,

^ ^ ^

i+h=n

{i-mmmm

ii k i 1i (-l) (-l) ( - lfsk^X(-l) ^Y'[X,Y] Y'[X,Y] yX^Y^[X,Y] x .kyk ] *k v^ t!)(j-l)!(*!)(2fc) » .■+j+2i=n . +J ^ = n ((fl)0-l)!(H)(2*) (i!){j-l)!(fc!)(2*) (i!){j-l)!(fc!)(2*) + J ^=n

+1 E "

11 +

(-lfrrpfty fc (*!)(i!)(fc!)(2 (i!)(j!)f Jb!)(2*. ) W(J0(*0(2*>

t 4 . iA-1k~n.

^

E

« HJL-J

n

;+j+2/fc=n-l

fc FJ[X,F]* (-l) fc JP F*[X,F]* (*!)(j!)(fe-l)!(2*-i) (*0(j-')(fe-l)!(2*- 1 )"

i_ y . i(-i)*.yy*[A",y]* ••(-i)*jr*y*[jr,y]* (-i)*jif*y [jf,yj* = E ~~ " « S U (i!)(i0(^)(2 ) (*f)(7l)(H)(2*) fc

n

i+i+2*:=n

i1 xx'''yy ij [ xj rk,,yy]]** ;ij((--iiik))Xfk*eiY'[X,Y] j(-l) v + nkk (i\)(j\)(k\)(2 (il)(j\)(k\)(2 )) » wJzU (W)(*0(2*)

4 "

1i_ 71

£ E

i+j+2k=n

^

:i i kA: h 2* 2fc(-i)' J>£:'y^[x,y] 2k(-l) Y>[X,Y\ _ '.(-l)kkXX Yi[X,Y] k (z\)(jl)(k\)(2 ) i+i+2*:=n

„ ^

104

bras and Gro Linkage Lie Algebras Groups

■e\

+ 2k\ 2k\

(i+j (i+j

» /)

«

(-l)kXiYi[X,Y]k (-l)*X'"i k , (t!)(j!)(*!)(2*) iM=n (^)(i!)(*!)(2fc) (maw i+J v

h=n

mmm )

k k {-l) ( - 1k)X*iYi[X,Y] * ' ' Y'[X,Y] (fl)(j!)(t!)(2») ~it +jh=n„ (fl)(J0(«)(2*)

^

"

(X + rY) )n nl •' »1

This finishes the induction and proves (4.4). Now suppose that Xm = Ym = [X,Y]m = 0. Suppose that i < m - 1, j < m — 1, and k < m — 1. Then m - l 1 + l(m - 1) = 4m - 4. i +-J-jj + 22kA<: <m m-- -1l++ m Now if i + j + 2k = 4m — 3 then either i > m, or j > m, or k > m. In any of these cases we have from (4.3) that im 3 (X + Y) y)«m-3 (X + ~ _ = 0 n!

and hence

m 33 (X + yK) ~- == 0. ) 44m

Thus, exp(X + y ) is defined. We can now finish up the rest of the proof.

exP(x (x + r) y) exp n

)n = f ; (^* Y ±^ = n=0 n=0

"■ "■

f/ (-l^yn^yM r S l U w U (W)(«)(*) )■

(4.5)

Also, X,,yr ] ) = e x p ( X ) e x pP ( y ) e x p ( - i [ X

j +. v f r

(-i)***y'ix,Y]*\

~ £ UsL mumrn ) • Comparing (4.5) and (4.6) we see that exp(X + exp(X) iexp(y) e x p (( - ■ipr,n). + F) Y) === exp(X) I[Jf, K]) . exp(K).exp

(4(4.6) 6)

-

4-1 The Exponential Map

105

We can rewrite this as exp(A: + Y) expQ[X,Y]) == exp(Ar)exp(y) exp(X + Y) exp Q [ X , Y]) = exp(X) exp(y) and this completes the proof. O Proposition 4.1.3. Suppose that X,Y and [X, Y] are locally nilpotent linear transformations and that [[X,Y],X] == [[X,Y],Y] 0. l\X,Y],X}-[[X,Y],Y}-== 0. Then [exp(X),exp(Y)]=exp{[X,Y]). [exp(X),exp(Y)\ == exp{[X,Y}).

Proof.

Our conditions on X, Y and [X, Y] imply that X[X,Y] == [X,Y)X [X,Y]X X[X,Y]--

Y[X,Y] ==*[X,Y]Y. [X,Y]Y Y[X,Y] Hence + Y) y ) exp e x p(j\X, ( i [ X ,Y]J y ] ) .. eexp x p (( ii [[ X X ,, yY]) ] ) ee*p(* xP(X + + Y)= y ) =--exp(X ^M* +

Thus we have:

106

Linkage Lie Algebras and Groups

[exppO,exp(Y)] = = exp(-X)exp(-r)exp(X)exp(F) = = exp (-(X + Y)) exp (±[X, F]) exp(X + Y) exp fyx, F]) = = exp (-(X + Y)) exp (X + Y) exp (i[X, F]) exp (i[X, F]) = = exp(i[X,y])exp(^,K]) =

= exP((I+ I)[X,r]) = = exp([X,K]). This completes the proof. □

Theorem 4.1.4. Suppose that C(R,T) is a linkage Lie algebra and that each ad(i) is locally nilpotent. Also suppose that R is a commutative ring that contains the Rational Numbers as a subring. Then (i) exp (r ad(i)) exp (s ad(i)) = exp ((r + s) ad(i)) (ii) exp(rad(i)) exp(sad(j)) = exp(sad(j))exp(r nected by an edge in V. (iii) [exp (r ad(i)) ,exp (s ad(j))] = exp(rsad(!)),

ad(i)), ifi and j are con­

where (i,l,j)

is a linkage of

r. Proof.

Parts (i) and (ii) are from Proposition 4.1.1. For part (ii) we apply Proposition 4.1.3 with X = r ad(i) and Y —

4-1 The Exponential Map

sad(j).

107

First note that for all x G C we have x ad{i) ad(l) = [x, i, I] =

= -[«\f,x]-[/,a;,t| = = —[/, x, i] = [x, I, i] = x ad(l) ad(i). Also, x [ad(i),ad(j)] = x (ad(i)ad(j) = [x,i,j]-[x,j,i]

— ad(j) ad(i)) — =

= [x,i,j] + \j,hx] + [i,x,j] = = [x,i,j] + \j,.i,x] - [x,i,j] = = —[i,j,x] = —[l,x] = xad(l). Thus, [[ad(i),ad(j)],ad(i)] = \ad(l), ad(i)] = 0. Similarly, [[ad(i), ad(j)}, ad(j)] = [ad(l), ad(j)] = 0. Therefore, [exp (r ad{i)), exp (s ad(j))] — exp ([r ad(i), s ad (j)]). Now, for all x G C we have x [r ad(i), s ad (j))] = rs [x, i, j] - sr [x,j, i] = = rs[x,i,j}

+ sr \j, i, x] + sr [i, x, j] =

= rs[x,i,j]

+ sr\j,i,x]

~sr[x,i,j]

=

= sr [j, i, x] = sr [—1, x] = x (rs ad(l)). This ends the proof. □

108

Linkage Lie Algebras and Groups

Remarks 1. The above result would also be true if there were some n > 1 such that ad(i)n = 0 for all vertices i, and (ad(i) + ad(j))n - 0 for all vertices i,j that belong to the same linkage, and if the commutative ring R contained inverses for all integers k such that 1 < k < n. For example, if ad(i)2 = 0 then m = 2 and 4m — 3 = 5 so that (ad(i) + ad(j))5 = 0. We would need the inverses of 2,3,4. These inverses exist in a field of characteristic 0, or in a field of characteristic p > 3. 2. This result says that if the linkage Lie algebra satisfies the relations of the linkage graph V as a Lie algebra, the associated Lie group satisfies the relations of V as a linkage group

Now let C = £(R,F) denote the hexagonal Lie algebra whose bracket operation is given in Table 4.1. Let R be the ring of Rational Numbers. Then the exponential map is defined and it follows from the previous theorem that the associated linkage Lie group Q — Q{R,Y) is an image of the groupG = L(R,Y) = ST(3,R). Since Q is a group of linear transformations of an 8 dimensional space we can get a concrete 8 x 8 matrix representation of Q. The matrix representing exp(r ad(l))may be easily computed. One computation needed, for example, is r2 4 exp (r ad(l)) = 4 + r [4,1] + — [4,1,1] =

= 4-r[l,4] + £ ( ~ l - l ) = = 4-r[l,4]-rl.

4-2 Generators and Relations in Lie Algebras

109

The matrix representing exp (r ad(l)) is: 1 0 0 0 0 0 0 0

0 0 -r2 1 —r 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 —r 0 0 0

0 0 0 0 1 r 0 0

0 0 0 0 0 1 0 0

2r 0 0 0 0 0 1 0

r \ 0 0 0 0 0 0 1/

(4.7)

This matrix is computed with respect to the order basis {1,2,3,4,5,6, [1,4], [2,5]}. All operators in this book act on the right so that the first row of matrix 4.7 represents (l)exp(rad(i)) = 1 - r 2 4 + 2 r [1,4] + r [2,5]. Similarly, the matrix for exp (r ad(ij) can be computed for all vertices i — 1,2,3,4,5,6. Notice that all of these matrices have entries that are integers or integer multiples of r and r 2 . In particular there are no fractional entries. Thus, these matrices can be defined over any ring R. We now have a concrete matrix representation of the linkage group L(R,T). We will use this same technique in later sections to construct concrete representations of linkage groups for more complicated linkage graphs. We will also describe exactly what group has been constructed here. 4.2

Generators and Relations in Lie Algebras

In this section we will briefly outline a description by generators and relations of the classical Lie Algebras. This complete description due to Serre, may be found in [12]. When dealing with classical Lie algebras one generally first looks at root systems and Cartan subalgebras and from these constructs Cartan matrices. We will describe all Cartan matrices (without the motivation of describing root systems) and construct Lie algebras beginning from these matrices. There are various types of Cartan matrices with four of the types be­ ing families of matrices. These are given below. The subscripts denote the dimension of the matrices.

110

Linkage Lie Algebras and Groups

Type An ( « > 1 ) : /

2 -1 0

-1 2 -1

- ]L

2

0 -1

0

0

0

0

k

Type Bn

0

0\ 0 0

0 . . -1

2,

.

0 \ 0

{n>2): /

2 -1 0 0

k

-1 0 2 -1

.

0 0

. . . .

■

0 0

■

■

-1 2 0 -1

-2 2

/

Type Cn (n > 3) : /

-1 0 2 -1 -1 2

2 -1 0

K

0 0

. -1

0

0 0

■

•

.

•

.

. .

0 \ 0

■

.

. .

•

•

- ]L

2

0

-2

o 0 -1 2 /

Type Z>„ (n > 4) : /

\

2 - 1 0 - 1 2 - 1 0 0 0 0

0 \ 0

0 0 0 0

- 1 2 - -1 . - 1 !I 0 --1 0 --1

0 -.I 2 0

Type £ 6 : (

2

0 -1 0 0

Io

0 2 0 -1 0 0

-1 0 2 -1 0 0

0 -1 -1 2 -1 0

0 0 \ 0 0 0 0 -1 0 2 -1 -1 2 )

0 -1 0 2

)

111

4-2 Generators and Relations in Lie Algebras

Type E7 : /

VK

2 0 - - 11 0 0 0 0 0 \ 0 22 0 0 - - 1 00 0 0 0 0: - 11 0 0 2 - -1 0 0 0 0 ; 0 - - 1 -- 11 22 - -11 0 0 0 0 0 0 - 1- 1 2 2 - 1 1 0 0 0 0 0 0 - -1 1 22 -- 11 0 0 0 0 0 - 1 22 ;) o0

Type E& :

22

0 - - 11 0 0 0 0 0 0 0 22 0 0 - - 11 00 0 0 0 0 0 0 2 - -1 0 0 - 11 0 0 0o 0 0 0 - - 1 -- 11 22 - -11 0 0 0 0 0 - - 11 22 - -11 0 0 0 0 0 0 0 0 0 0 - 1- 1 2 2 - 1 1 00 0 0 0 0 0 0 - -1 1 22 -- 11 0 ^ 0 0 0 0 0 - 1 22 Vo

I/

\ |

! j/

Type F4 : / 22 - 1 0 0 \ - 11 22 - -22 0 0 - - 1 2 2- 1- 1 \\ o0 0 0 - -1 1 22 J/

/

Type G2 :

J' (-3 1-3 2 n (2

-l\

In [12] the notation [a, b, c, ■ ■ ■] denotes the right normed commutator, that is, [a, [6, [c, • • •]]]. This is opposite of the notation we have been using. We will continue to use our left normed commutators for now. Let A = (ay) be an n X n Cartan matrix and let C be the Lie algebra generated by x Xi,2-15 ' ' '' >) ^71) nt 2/l) V\) '' "■,Un,hi," » Vnt **1 j * *? tl>n " j *^n

with the following relations:

112

Linkage Lie Algebras and Groups

1. [hi, hj] = 0, for all i, j 2. [x;,yi] = hi for all i, and [x;,g/j] = 0 for all i ^ j 3. [hi,Xj} = cijiXj,

4. [xhXi,---,Xi)

[hi,yj} =

=0,

-djiyj

i=f}

—Ojt+1 times

5- [Vj, Vir--,yi]

= 0, j

^j.

—aji+1 times

Serre's Theorem (see [12]) says that the algebra £ defined above is a simple Lie Algebra with Cartan subalgebra generated by hi, ■ ■ ■, hi and with a root system that yields the Cartan matrix A. The Cartan matrices listed above are for simple Lie algebras. Proposition 4.2.1. Let A he a Cartan matrix such that there exist i,j such that aij = aji — —1. Then a;,-, Xj, yi, j/j generate a subalgebra over the Complex Numbers whose generators and relations may be described by the linkage graphs in Figure 1. Proof. Let £ be the Lie algebra defined by the Cartan matrix A. In Cartan matrices the diagonal entries are always 2. Thus, i ^ j . Relation 2 of the defining relations of C imply that [xi, y,] = 0. Relations 4 and 5 say that [xi, [xh Xj]] = 0,

[XJ, [XJ, Xi]]

=0

and hfi,bi,yj]] = o, [yy,[tr,-,»t]j = o. Also the Jacobi identity and relation 3 imply 0 = [[xi,Xj],yi] + [[xj,yi],xi] + [[yi,Xi],Xj] = = [[xi,Xj],yi] + 0 + [-hi,Xj].

4-2 Generators and Relations in Lie Algebras

Figure 1

Hexagonal Linkage Graphs

113

114

Linkage Lie Algebras and Groups

Thus,

[l*h£/],«/»] */], yi\== [k,*i] [hi,Xf]==-XJ. -XJ. H«i,

(4.8) (4.8)

Similarly, 0 =--[[x;,Xj],3/y] [[xi,Xj],yj] ++[[xj,?/j],a;;] [[xj,yj],Xi] ++ [[l/j,Xi],Xj] [[l/j,Xi],Xj]== = [[*<»*/l»lfi]+0 [[*<»*/l»tfj]+0 + [fci,Xj]. |fc,Xj]. Hence,

II*i»*i], »j] == —[A*. II*i»*i]»»j] —[*«i *il *i] = X** J.

(4.9) (4-9)

Also, 0 = [[yi,yj],xi] + [[yj,xi],yi] + [[**,»], y*| = = fc,3/#],a?t]+0+[fe fc,3/#],a?t] + 0 + [fe t-,|fi]. t-,|fi]. Hence, [[Vi, Vj],Xi]==-[fc»,»il -[k, Vj]= =»i.Vj. [[|K,RJ3J*»]

(4.10) (4.10)

Finally, 0 == [[ifi, Vj], */••] Xji] ++ [fc> [bi>*j]»yj] ZiliVi)++[fc>, [[XJ,»»li Vi],fj] [[?<, &•]> wl == = [[yi,yj],Xj] + [-hi,yi] + 0. = Therefore, [[yi,yj],xj] [[Vi,yj],xj] = [hj,yi] = = Ki. yt. We may now use (4.8), (4.9), (4.10), and(4.11) to prove the result. □ The above result applies to the following Lie algebras: n>2 >2 -A ^ nT,l ) n >3 Bn, m > Cnn, , n >> 3 (s Dn, n n>4 >4 Dn, E$, E Ee, Ei7,, Eg, £ g , ^F44-. It does not apply to A\, B2, or G2. We will deal with G2 later. The Cartan matrix for A2 is (2

l-l

-I)

2J-

(4.11)

4-2 Generators and Relations in Lie Algebras

115

What we have done above shows that the Lie algebra A2 is isomorphic to the hexagonal Lie algebra £(R, V) where T is the hexagonal linkage graph of Figure 2.6. It is also isomorphic to the Lie algebra of C{R, V) where Tis the hexagonal linkage graph of Figure 2.7 as we know from example 4 in section 2.3. We now wish to pass from a Lie algebra C defined by a Cartan matrix to the associated Lie group. This is done by applying the exponential map to the set of roots of C , that is, all elements of the form x •£«1 121 ' ' '' i 3**1 [i^-iu iii xXifi

Vi,

x

ik\i lXik\i

blii,Via," *»»*]■

For example, if z is one of the above elements it is a basic result that there is an n such that ad(z)n = 0. Hence, the exponential map can be defined: r2 r2

exp(r ad(z)) ad(z)) = -=I I++r rad(z) ad(z) + — ad(z) ad(z) + + •■ •■•.•. 2

We denote exp(rad(z)) by Xz(r). If we represent Xz{r) as a matrix, and the entries in that matrix are polynomials in r with integer coefficients, then this allows Xz(r) to be defined over any field K. The group generated by all Xz(r) for all z and all r G K is the Chevalley group C{K). This group is usually simple. Suppose, for example that C = An. Then the group An(K) is the projective special linear group PSL{n + l,K), that is, the central factor group of all (n + 1) x (n + 1) matrices of determinant 1 over K. This group is simple if n + 1 > 2, or if K has more than 3 elements. Since the hexagon algebras in the previous result are isomorphic to the algebra A2, it follows that the corresponding group is PSL(3,K). We already know that this group has a verbal embedding of K. The previous result now says that any of the Lie algebras to which it applies has associated Chevalley group Q{K) and these groups contain PSL(3, K). Hence the groups Q(K) each admit a verbal embedding of K. These groups are simple and hence they are the images of linkage groups by Theorem 2.2.4. The finite non-abelian simple groups consist of the alternating group An,n > 5, the sporadic simple groups, the Chevalley groups and the twisted Chevalley groups. The twisted Chevalley groups tend to be images of linkage groups in the same way as the Chevalley groups. We have already seen in

Linkage Lie Algebras and Groups

116

section 2.6 that 25 of the 26 sporadic simple groups are image of linkage groups. The alternating group As does not admit a verbal embedding of any ring with identity as we saw in Corollary 1.2.5. In a later section we will see that groups An, n > 5 admit verbal embeddings and are thus the images of linkage groups. Thus, we may say that in some sense most finite simple groups admit verbal embeddings and are therefore images of linkage groups. 4.3

The Heptagonal Algebra

Now let II7 be the heptagonal linkage graph of Figure 3.7. We will try to construct the Lie algebra C — C{R, II7) in the same way that we constructed the hexagonal algebra above. We will first try to construct a Lie multiplication table for C. Again we will denote the Lie operation [i,j] by ij. We will attempt to construct a multiplication table for the elements 1, 2, 3, 4, 5, 6, 7, 14, 2 5, 36, 4 7, 5 1 , 6 2, 73 (which form an ordered basis). These vertices can be divided into two sets of seven vertices which are each cyclically symmetric of order 7, namely, they can be divided into 1,2,3,4,5,6,7 and 1 4 , 2 5 , 36, 47, 5 1 , 6 2 , 73. . Hence, if we know one column in the multiplication table we will be able to compute the rest by this symmetry. We will once again let ijk denote [[i,j], k]. We need a number of preliminary calculations. 361 = - 6 1 3 - 1 3 6 = - 7 3 - 2 6 =

= -73 + 62. Similarly, 621 = - 2 1 6 - 1 6 2 = 6 1 2 = = 72=1. Also, 731 = - 3 1 7 - 1 7 3 = 27 + 0 = = -1.

4-3 The Heptagonal Algebra

[,] 1 2 3 4 5 6 7 14 25 36 47 51 62 73 Table 4.2

117

1 0 0 -2 -14 51 7 0 unknown unknown -73+62 unknown unknown 1 -1

Heptagonal Lie Multiplication Table

The above calculations allow us to fill in the first column of the multi­ plication table for £ . Notice that some of the entries in Table 4.3 are unknown, that is, they may not be determined from the relations in £ (R, II7). We shall see later that this is because £ (R, II7) is actually infinite dimensional. Completing the multiplication table would mean that the algebra is finite dimensional. Thus we have to add more relations in order to complete the table. The motivation for the extra relations we are adding to the Heptagonal Lie algebra is not entirely clear (even to those who added the extra relations). We will attempt to describe the process the authors went through. We first tried to copy the Hexagonal algebra for entries such as 141 = 1 + 1 and x 111 = 0. It was hoped that a product such as 2 5 1, 5 1 1 , and 4 71 could be expressed as linear combinations of vertex elements. By trial and error it was discovered that the multiplication table could be filled in as in Table 4.3. After completing the table it is,of course, necessary to show that this table actually defines a Lie algebra of dimension 14 that satisfies the heptagonal Lie algebra relations. To do this we will construct the adjoint representation of this algebra which consists of 14 x 14 matrices. This will be done in a later chapter where we will be able to check that the multiplication table is correct, that the Lie algebra defined is 14 dimensional, and that the associated Lie

Linkage Lie Algebras and Groups

118

[,] 1 2 3 4 5 6 7 14 25 36 47 51 62 73 Table 4.3

1 0 0 -2 -14 51 7 0 -1-1 -1-2 -73+62 1+7 1+1 1 -1

Heptagonal Lie Multiplication Table

group satisfies the heptagonal group linkage relations. These facts will all be checked using the computer language Mathematica. We will also demonstrate that all of this works for any commutative ring R. At this time we will only give the adjoint representation of the action of vertex 1. It is I 0 0 0 0 0 0 0 -2 -1 0 1 2 1 V- l

0 0 -1 0 0 0 0 0 -1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0 0

0 0 \ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 o }

By the symmetry in the table we can obtain the matrices corresponding to the

4-3 The Heptagonal Algebra

119

other vertices by cyclically permuting the first seven rows and columns and the last seven rows and columns of the adjoint representation of vertex 1. We can do this by conjugating the above matrix by the matrix // 00 11 00 00 0 0 00 00 00 00 00 0 0 0 00 00 o\ \ 00 0 11 00 00 00 00 00 00 00 00 00 00 0 0 00 0 0 0 11 00 00 00 00 00 00 00 00 00 0 0 00 0 0 0 0 00 11 00 00 00 00 00 00 00 00 0 0 00 0 0 0 00 00 11 00 00 00 00 00 00 00 0 0 00 0 0 0 00 00 00 11 00 00 00 00 00 00 0 0 „ 1 10 00 0 0 0 00 00 00 00 00 00 00 00 00 0 0 B = 00 0 0 0 00 00 00 00 00 11 00 00 00 00 0 0 00 0 0 0 00 00 00 00 00 00 11 00 00 00 0 0 0 0 0 0 0 0 00 00 00 00 00 00 00 11 00 00 0 0 0 0 0 0 0 00 00 00 00 00 00 00 00 11 00 0 0 0 0 0 0 0 00 00 00 00 00 00 00 00 00 1 1 0 0 0 0 0 0 0 0 00 00 00 00 00 00 00 00 00 0 0 1 1 V o 0o 0o 0o 0o 0o o0 i1 o0 o0 o0 o 0o 0o 0/ )

\o

Let us denote the 14 dimensional Lie algebra defined above as £7. We would like to know what Lie algebra it is according to the classical description of Lie algebras. One could check that it is a simple algebra and from its dimensions and the classical classification one would quickly arrive at the conclusion that it must be G^. However, we want to construct an explicit isomorphism between £7 and G 2 . This isomorphism will be constructed in the next section and by its nature will reveal that these two Lie algebras are not isomorphic in any way that would be considered natural. Now consider the Lie group obtained from £7, which we will denote by QrfirQ7 = (exp(r ad(i) \| r € G R, i G V{Il V{U77)) ) This group may be defined for any commutative ring R and satisfies the group linkage relations of the heptagon II7. Hence, Qj is an image of L(R, H7). This is another proof that the group L(R, U7) actually has the ring R embedded at each vertex. This proof does not depend on the combinatorics of the normal form described in Chapter 3. It also has the advantage that the elements of the group can be described explicitly as matrices, and thus amenable to computer calculations.

120

Linkage Lie Algebras and Groups

1

GU 1 2 3 4 5 6 7 Table 4.4

0 0 -2 -1+2+3-4+5-6+7 1-2+3-4+5-6-7 7 | 0

Characteristic 3 Lie Multiplication Table

We mention in passing that there is a 7 dimensional representation of the heptagonal algebra over a commutative ring in which 3 = 0. For example, this representation would exist over any field of characteristic 3. In this representation the basis for the algebra consists of the vertex element 1,2,3,4,5,6,7. The first column for the multiplication table is given in Table 4.4. The other columns in the table can be computed by symmetry. We also mention that the Pentagonal algebra C(R, II5) is equal to the zero algebra as a simple calculation shows. This is analagous to the collapsing of the Pentagonal group L(R, II5) as described in Lemma 2.4.1. 4.4

A Description of G2

To demonstrate the isomorphism between £7 and G2 we will first give a description of G2 in terms of generators and relations. We will use the relations given in Section 4.2 except that we will now us right normed commutators in keeping the notation in [12]. ;j. Since oince the me Cartan t a r t a n matrix matr for G2 is

(V,1)

it follows that G2 is generated by xi,x2,yi,y2,hi,h2 with the relations 1. [hi,hj] = 0 2. [xi,yi] = ^

4-4 A Description of G2

121

3. [x2, 3/2] = h2 4. [xuy2]

=0

5- [X2, Vi] = 0

6. [fti,Xi] = 2xi 7. [/»i,x2] = —3x2 8. [/i2,a;i] = —x\ 9. \h
10- [/n,2/i] = -2t/i 11- [^1,2/2] = 3y 2 12. [A2,«/i] = «/i 13. [A2,2/2] = - 2 y 2

14. [*i,asi,si,ari,*3] = 0 15. [a?2>a%Si] = 0 16- [2/1,2/1,2/1,2/1,2/2] = 0

17- [2/2,2/2,2/1] = 0. To simplify notation we will denote terms such as [*i,afi,*i, asj] by s l l l 2 and [2/1,2/1,2/1,2/2] by 2/1H2, etc. We need a partial multiplication table for G2 which we give in Table 4.5. We will now give the calculations needed to construct this table. We will make extensive use of the Jacobi identity in the form a (be) = b(ac) + (ab)c. Remember that these calculations are taking place over the Complex Numbers. These calculations are tedious but necessary for the sake of completeness. First note that a simple induction implies that / v n w •pJU g

■ ■ xt =(a(a +sj asjH+ ■ rj rj +a

hi atj)x Q>tjJX Xs ' ' ■x ' Xf T X$ T

t

and hjVrVs■ ■■■y■t hjUrVs 1. X = 00 a^a'ixia^ — 2XiX1X2

+ asj+ +asj ■-\■ ■=Vt—(a = rj~{arj

\- atj)yTy■•Vi■ + atjhrVs s ■■■yt.

122

Linkage Lie Algebras and Groups

[,] xl x2 xl2 xll2 xlll2 X21112 yi y2 yl2 yll2 ylll2 y21112

xl x2 0 xl2 -xl2 0 -xll2 0 -xlll2 0 0 -x21112 0 0 -hi 0 0 -h2 -3y2 yi -4yl2 0 -3 yll2 0 0 -ylll2 Table 4.5

xl2 xll2 0 0 x21112 0 0 3x2 -xl

xll2

xlll2 0 -x21112 0 0 0 4x12 0 4x1 hl+3 h2 -8 hi - 12 h2 -4yl 0 12 yl -3 yll2 -12 yl2

xlll2 0 x21112 0 0 0 0 3x112 0 0 -12x1 36 hi +36 h2 -36 y2

Partial Multiplication Table for G2

Proof. x2112 = - x l 2 2 1 = = -xl(x221) = -a:l(0) = 0. 2. (xix 2 )(xixix 2 ) = — x2XiXiXiX2

Proof. (xl2)(xll2) = xl ((xl2)(xl2) + ((xl2)xl)(xl2) = = 0 - (xll2)(xl2) = = - ( x l ( x l l 2 ) + ((xll2)xl)x2) = = xl(x2112,) - ((xll2)xl)x2 = o = (xlll2)x2 = -x21112. 3. (x 1 x 2 )(xix 1 x 1 x 2 ) = 0

X21112 0 0 0 0 0 0 0 -3 x l l l 2 3x112 12x12 36x2 -36 hi -72 h2

4-4 A Description

of G2

Proof. (xl2) ( x l l l 2 ) = xl ( x l 2 x l l 2 ) + (xl2a:l)(a:112) = = xl(-x21112) - arll2xll2 = 0

= -xl21112 = = -x2

xllll2-xl2xlll2 = 0

= -xl2xlll2. Hence, 2x12x112 = 0. Thus, x l 2 x l l l 2 = 0.

4. (xix 2 )(x 2 XiXiXiX 2 ) = 0

Proof.

xl2 x21112 = x2 1x12x1112

+ (xl2x2)xlll2 =

= 0-(x2xl2)xlll2 = = (x2x21)xlll2 = x221xlll2 = 0.

5. (xiXiX2)(xiXia;iX2) = 0

123

124

Linkage Lie Algebras and Groups

Proof. xll2xlll2 = xl(xll2xll2) + (xll2xl)xll2 o = -xlll2xll2 = = -xl(xlll2xl2)-(xlll2xl)xl2 = o -£11112, xYl = 0. o 6. x i ( x 2 X i X i X i x 2 ) = 0

Proof. xlx21112 = x2 .xini2, + ,xl2xlll2=0. 0

0

7. X2 (x 2 XiX 1 X 1 X2) = 0

Proof. x2x21112 = x2(xl x2112 +x21 xll2) = o = x2 (x21xll2) = = x21 x2112 +(x221 x2)xlll2 = 0. o

8. ( X i X 1 X 2 ) ( x 2 X i X 1 X i X 2 ) = 0

Proof. xll2x21112 = = x2(xll2xlll2,) + (xll2x2)xlll2 = 0. o

9. (xiXiXix2)(x2a;iXiXiX2) = 0

4-4 A Description

of G2

Proof. xlll2x21112 = = xlll2(xl x2112 + x21 xll2) := = == xlll2(x21xll2) xllU(z2lxll2) = -xlll2(xll2x21) = = = xlll2(xll2xl2) = = xll2(xlll2xl2) xll2(xlll2a:12) + (xJJ12xll2)xl2 (xlll2xll2):cl2 = 0

= 0. 10. J/l(x!X 2 ) = 3X2

Proof. = xl xlt/2x2 ylxl2 = y2x2 + ylxl2 = — = 0 --hlx2 / t l x 2 = 3x2.

1 1 . y2(xix2)

=

-X]

Proof. j/2 xl2 xlj/2 x2 + (y2i (t/2 xl)x2 y2: rl2 = xlj/2, cl)x2 == == xl(-h2) xl(--h2)--= = h2xl h2xl =- xxll.. 12. J / i ( x ! X i X 2 ) = 4 X ! X 2

125

126

Linkage Lie Algebras and Groups

Proof.

j/la :112 = x l j/1 xl2 + (j/1 x2)xl2 = ylxll2 = xl j/1 xl2 + (j/1 x2)xl2 =

= = xl (3 x2)-hlxl2 = xl (3 x2) - hi xl2 = 3xl2-(2-3)xl2 = = 3xl2-(2-3)xl2 = = 4x12.

= 4x12. 13. y2(xix1x2) 13. y2(xix1x2)

= 0 = 0

Proof.

2/2:Cll2: = x l j/2:nl2+(j/2: cl)xl2 = 2/2x112 = x l y 2 x l 2 + (j/2xl)xl2 = = x l ( --xl) = 0. = x l ( - x l ) = 0.

14. t/i(a;iXiXiX2) = 3

X\Xix2

14. t/i(xiXiXiX2) = 3 X i X i x 2

Proof.

yl: r l l l 2 = xl 2/I x l l 2 + (yl x l ) x l l 2 = j/1 x l l l 2 = xl 2/I x l l 2 + (2/I x l ) x l l 2 = = xl ( 4 x l 2 ) - M x l l 2 = = xl ( 4 x l 2 ) - M x l l 2 = = 4.x l l 2 - ( 2 + 2 - 3 ) x l l 2 = = 4 x 1 1 2 - (2 + 2 - 3 ) x l l 2 = = 3x112. = 3x112. 15. ^ ( a ^ i ^ i ^ ) = 0 15. 2/2(siXix1x2) = 0 Proof. 2/2: c l l l 2 ===xxl 2/2x1112 l 2/2x112+(i/2xl)xll2 1/2x112+(i/2xl)xll2 = 00

= 0. 16. j/i(x 2 / i ( x 22xx11Xxixix i X i X 22)) = == 00

00

4-4 A Description of Gi

Proof. 2/13:21112 = x2 j/1 x l l l 2 + (t/1 x2) x l l l 2 =

= xl (3x1112) = 3x11112 = 0. 17. y2{x2XiX-iX\X2)

= —Z X\X\XxX2

Proof. 2/2x21112 = x2 (t/2 x l l l 2 ) + (y2x2) x l l l 2 = o = -^2x1112 = = - ( 2 + 2 + 2-3)a;1112 = = -3x1112. 18- (2/12/2)3:1 =- - 3 2/2

Proof. 2/12x1 = - x l j / 1 2 = -2/1 xl j/2 - (xl y\ (t/2) = 0 - / d y 2 = - 3 j/2. 19. (2/12/2)3:2 = 2/1

Proof. 2/12 x2 = - x 2 2/12 = -j/1 x2 j/2 - (x2 j/1 )j/2 = = - 2 / U 2 - ( x 2 2/1)2/2 = = -j/1fe2- 0 = h.2 2/1 =2/1. 20. (2/12/2X3:1X2) = Ai + 3 / J 2

\11

128

Linkage Lie Algebras and Groups

Proof. 2/12x12 = xl(2/12x2) + {y\2xl)x2

=

= xl 3/1 + ( - 3 j/2) x2 = hi + 3 ft2. 21- (2/iS/2)(a:iZiX2) = 4 x i

Proof. 2/12x112 = xl(2/12 xl2) + (j/12 xl)xl2 = = xl(fcl +3fe2) +(-3)2/2x12 =

= -hlxl-3h2xl

-3(-xl) =

= -2x1 - 3 ( - l ) x l + 3 x l = = (-2 + 3 + 3)xl = 4 x 1 . 22. (yiy2)(xix1x1x2)

= 0

2/12x1112 = xl (2/12x112) + (j/12 x l ) x l l 2 = = xl (4x1) + (-3)2/2x112 = 0 + 0 = 0. 23. {yiy2)(x2XiXiXiX2) = 3 xixix 2

Proof. 2/12x21112 = x2 2/12x1112 + (2/12x2)xlll2 = = x2(0)+ylxlll2 = = 3x112. 24. (2/12/12/2)2:1 = -42/i2/2

4-4 J4A DescripU Description G% 4-<4 ion of of G% 2/112x1 = --x ■xl l 1/112 2/112 = = —2/1 -2/1 xl x l 2/12 2/12 - - ((xl x l 2/1)2/12 yl)yl2 =

= 2/1(2/12x1) = yl(yl2xl)-hl(yl2) -hl(yl2)

=

= 2y /ll((--33yy 2 ) ++ ( 22--33))2y/ 1l 2 = = --44 j/12. 2/12.

25. (2/12/12/2)3:2 === 0

Proof. 22/112x2 / 1 1 2 x 2== = xx2 2 2/112 =

--ylx2 2/12 -- ( (a;2 x 2 2/1)2/12 2/1)2/12 = = - 2/1x2 2/120

= 2/12/12x2 = ylyl2x2 == J/12/1 ylyl == 0.

26. (j/i2/i2/2)(xix 26. (J/I2/I2/2)(XIX22)) ==

--44 j2/1 /l

Proof. P roof. 2/112x12 +(2/112xl)x2 x l 2/112x2 + ( 2 / H 2 : cl)x2 = 2/112:c l 2 = xl 0

(-4j/12)x2 === - 4 2/1. == (-Ayl2)x2

27. (2/ij/i2/2)(a;iXiX2) (2/i3/i2/2)(a;ia;iX2) = --8h8 / 1 - 1-I2h 2 /2i 2

129 129

130

Linkage Lie Algebras and Groups

Proof. 2/112x112 = xl

J/112X12

+ (j/112xl)xl2 =

= x l ( y l l 2 x l 2 ) + (i/112xl)xl2 = = x l ( - 4 j / l ) + (-4)2/12x12 = = -4 hi -4(hl =

+ 3h2) =

-8hl-12h2.

28. (2/i2/i2/2)(xiXiXix2) = - 1 2 X!

Proof. 2/112x1112 = xj/112 x l l 2 + ( y l l 2 x l ) x l l 2 = = xl ( - 8 hi - 12 h2) + ( - 4 j/12)xll2 = = 8 M x l + 12h2xl-4(4xl) = = 16x1 - 12x1 - 16x1 = - 1 2 x 1 . 29. (2/i2/i2/2)(x2Xix1x1x2) = 12xix 2

Proof. 2/112x21112 = = x22/112xlll2 + (2/112x2)xlll2 = = x 2 ( - 1 2 x l ) + 0 = 12x1x2. 30. (j/i2/i2/i»/2)xi = -32/i2/i2/2

4-4 A Description

of G2

Proof. 2/1112x1 = - x l 2/1112 = 1/1112x1 = - x l J/1112 =

= -j/1 xl j/112 - (xl j/l)2/112 = = -y\ xl y l l 2 - (xl yl)yll2 = = 2/12/112x1 = Mj/112 = = ylyll2*l =fclyll2 = = 2j //ll(( - 4 y l 2 )) + (2 + 2 - 3 ) y l l 2 = = -32/112. -3yll2. 31. (yiviym)®i

= 0

Proof. 2/112x2 2/112, x2 = -x2|/1112 -x22/1112 = -!/12s2j/112-(x2j/l)j/112 ?/l 1/112x2 = = -2/12: c2 2/112 - ( x 2 2/1)2/112 = 2/13/112x2 = 2/1 (0) = 0. 32. (viyipiyi)(xixz)

= 0

Proof. 2/1112x12 = = xl 1/1112x2 + (2/1112 xl)x2 = = xl(0) + ( - 3 2/112)x2 = = -3(0) = 0. 33. (yiyiyiy2)(xix1x2)

= I2yi

131

132

Linkage Lie Algebras and Groups

Proof.

2/1112x112 = 2/1112x112 = = xl 2/1112x12 + (2/1112xl)xl2 = = xl 2/1112x12 + (j/1112xl)xl2 = = xl(0) + ( - 3 y l l 2 ) x l 2 = = xl(0) + (-3»/112)xl2 = = - 3 ( - 4 j / l ) = 122/1. = - 3 ( - 4 y l ) = 122/1. 34. (yiyiyiy2)(xix x x ) = 36 1 1 2 34. (yiyiyiy2)(xix1x1x2) - 36 hi hx + + 36 36 hh22 Proof.

2/1112x1112 = ylll2xlll2 = = xl 1/1112 x l l 2 + (t/1112xl)xll2 = = xl2/1112xll2 + (i/1112xl)xll2 = = xl(12 2/l) + (-32/112)xll2 = = xl(12 t/1) + (-3j/112)xll2 = = 12 hi - 3 ( - 8 f c l - 1 2 f c 2 ) = = 12/il-3(-8ftl-12/i2) = = 36 hi + 36 h2. = 36 hi + 36 h2. 35. 3 6 x 22 35. (3/13/12/12/2X3:2X1X1X1X2) (j/iyij/iy2)(a;2XiXiXiX2) = = 36x P rnof Proof.

3/1112x21112 = 3/1112x21112 = = x22/1112x1112 + (2/1112 x2)xlll2 = = x21/1112x1112 + (2/1112 x2)xlll2 = = x2(36 hi + 36 h2) + (0)xlll2 = = x2(36 hi + 36 h2) + (0)xlll2 = = -36 Alx2-36^2x2 = = -36 A l x 2 - 3 6 ^ 2 x 2 = = - 3 6 ( - 3 + 2)x2 = = - 3 6 ( - 3 + 2)x2 = = 36x2. = 36x2.

36. fayiymyzjx! = 36. (2/2«/ii/i?/i2/2)zi = 00

4-4 A Description

of G2

Proof. 2/21112x1 = - x l i/21112 = y21112xl = - x l 2/21112 = = - y 2 x l y l l l 2 - x l 2 y l l l 2 =. = -y2xly\l\2 - x l 2 y l l l 2 =. = y2ylll2xl = y2(-3yll2) = = y2ylll2xl = y2(-3yll2) = = -3y2112 = 0. = - 3
37. (y2Viyiym)x2 = -ymym Proof. y21112x2 = --x2y21112 = 2/21112x2 = -x2y21112 = = - y 2 x 2 y l l l 2 -- x 2 y 2 2/1112 = = -3/2x22/1112 - x2j/22/1112 = = y2ylll2x2 -fe2ylll2 = = 2/22/1112x2-^22/1112 = )ylll2 = = 0 + ( (- -1 l- -1l --■ 1 l + 22)2/1112 = -t/1112. -ylll2. 3 8 . (2/2^1 J/i2/i«/2)(xiX 2 ) = - 3 2/i2/i2/2

Proof. y21112xl2 = 2/21112x12 = = xly21112x2 + (y21112a l ) x 2 = = xly21112x2 + (2/21112x1) x2 = = x l ( - y l l l 2 ) =: y l l l 2 x l = - 3 y l l 2 . = xl (-2/1112) = 2/1112x1 = -32/112. 3 9 . (y2j/i2/i2/i2>2)(ziZiZ2) (2/2j/i2/i2/i2>2)(zia;iX2) == -12yiy2 -12yiy2

133

134

Linkage Lie Algebras and Groups

Proof.

2/21112x112 = 2/21112x112 = = xl 2/21112x12 + (2/21112:cl)xl2 = = xl j/21112xl2 + (2/21112xl)xl2 = = x l ( - 3 2/112) = = x l ( - 3 j/112) = = 32/112x1 =3(-42/12) =. 122/12. = 32/112x1 = 3(-4j/12) = 121/12.

40. {y2yiyiyiy2)(xix1xix2) = -361/2 4 0 . («/22/iyi!/i2/2)(ziZiZiZ2) = - 3 6 J / 2

Proof.

2/21112x1112 = 2/21112x1112 = = xl 2/21112x112 + (2/21112xl)xll2 = = xl 2/21112x112 + (2/21112xl)xll2 = = xl(-12j/12)+0 = = xl ( - 1 2 r / 1 2 ) + 0 = = 12 2/12x1 = 12(—3y2) = - 3 6 2/2. = 12 2/12x1 = 12(—3y2) = - 3 6 2/2. 41. {y2yiy\yiy2){x2x1x1x1x2) = - 3 6 hi -- 7 2 h2 41- (j^j/ij/ij/i^Xa^iziZi^) = - 3 6 / ^ - 72 /i2 Proof.

2/21112x21112 = 3/21112x21112 = = x22/21112xlll2 + (j/21112 x2)xlll2 = = x22/21112xlll2 + (2/21112 x2)xlll2 = = x2(-36j/2) + (-2/1112) x l l l 2 = = x2(-36j/2) +(-2/1112) x l l l 2 = = - 3 6 h2 - (36 hi + 36 h2) = = - 3 6 h2 - (36 hi + 36 h2) = = - 3 6 hi -72h2. = - 3 6 hi -72h2. The above calculations give us enough information to compute all the entries of Table 4.5. We want to construct the adjoint representation for this algebra. To do this we need the matrices which represent a
4-4 A Description of G2

135

representation will make it much easier to check that the isomorphism we are constructing between £7 and G2 is actually correct. Note that the matrix representation corresponds to the ordered basis x Xl, 3>1, X 1i 2 , XiXiX\X X2X\XiX\X22, X!X XlXiX XiXiXjX 2, 2 , ^\^\ 2, 2 . , X 2 XiXiXiX x22,,X\X

2/1,, 2/2 j/2,, 3/12/2, 3/12/2 , 2/1

2/12/12/12/2, 2/22/12/^12/2, j/2yi2/iyij/2, yiym, ymy-i, ymym,

hi, h2 We do need some more tedious calculations to complete the construction of our adjoint representations. 1. (x1x2)yi - - 3 x 2 Proof.

xl2yl = --2/1x12 = xl2yl

= -2/1x12 =

= — xl yl x2 -(ylxl)x2 -

=

= — xl yl x2 — (yl xl)x2 =

= hlx2 = -3x2. = hlx2 = - 3 x 2 . 2. (xiXix2)yi Proof.

= —4xix2

x l l 2 j / l = -2/1x112 = xll2j/l = - j / l x l l 2 = = - x l 2/1 xl2 - (yl xl)xl2 = = - x l J/1 xl2 - (yl xl)xl2 = = x l ( - 3 x 2 ) + felxl2 = = x l ( - 3 x 2 ) + felxl2 = = -3xl2 + (2-3)xl2 = = -3xl2 + (2-3)xl2 = = -4x12. = -4x12.

3. (xixixix 2 )2/i = - 3 x i X i x 2

136

Linkage Lie Algebras and Groups

Proof.

4. {x2x1x1x1x2)yi

4.

xlll2yl = -ylxlll2 = xlll2yl = -ylxlll2 = = - x l yl x l l 2 - (yl x l ) x l l 2 = = -xl yl x l l 2 - (yl x l ) x l l 2 = = x l x l l 2 y l + fclxll2 = = x l x l l 2 y l + fclxll2 = = -3x112. = -3x112. --

(X2X1XIXIX2)J/I =

Proof.

5.

(y2yiyiyiy2)yi

Proof.

6. ( x j x 2 ) y 2 = X! 6. (x2x2)y2 = X! Proof.

=

0

x21112yl = -ylx21112 = x21112yl = -j/lx21112 = = - y l (xl a:2112 + (yl x2)xlll2) = = - y l (xl a:2112 + (j/1 x2)xlll2) = = - y l (0 + 0x1112) = 0 . = - j / l (0 + 0x1112) = 0 .

j/21112j/l = -yly21112 = y21112yl = - y l j/21112 = = - y l ( y l y 2 1 1 2 + (yly2)yll2) = = - y l ( y l y 2 1 1 2 + (yly2)yll2) = = -yl(0 + yl2yll2) = = -yl(0 + yl2yll2) = = - y l l 2 y l l 2 = 0. = - y l l 2 y l l 2 = 0.

xl2y2 = - y 2 x l 2 = == -—xly2x2 x l y 2 x 2 --((yy22xxll)) x 2 = = -h2xl -h2:cl == x l .

7. (x1x1x2)y2 = 0

4-4 A Description

of G 2

Proof. xll2j/2 = -2/2x112 = xll2j/2 = - j / 2 x l l 2 = = - x l j/2 xl2 - (j/2 xl)xl2 = = - x l «/2 xl2 - (j/2 xl)xl2 = = xl(0) = 0. = xl(0) = 0. 8. 8.

(xiXiXiX22)2/2 )j/2 = 0 (xiX!XiX

Proof. x l l l 2 y 2 = - j / 2 x : 1112 = xlll2j/2 = -2/2x1112 = = - x l 2/12x112 - ( » 1 x l ) x l l 2 = = -xlj/12a;112-(ylxl)a;112 = = 0. = 0. _ (x 2 XiXiXiX 2 )j/2 = —a;iXia;iX2 9. {x,2X\X\X\X%)]ji = —X1X1X1X2 9.

Proof. x21112j/2 = --2/2x21112 = a;21112y2 = -j/2x21112 = = - x 2 2/22/1112--(2/2x2)xlll2 = = - x 2 2/2j/1112 - (2/2x2)xlll2 = = A2xlll2 = -a:1112. = /i2xlll2 = - x l l l 2 . 0. (2/13/12/2)2/2 10. (2/12/12/2)2/2 = = 0o Proof. 2 / 2 2/112 = y2/112y2 l l 2 y 2 ==- -2/2 = -2/ly2112-(2/2yl)2/12 -2/12/2112 -(j/22/l)2/12 = = 0. 1. (j/22/i«/i2/i)j/2 11(j/22/i2/i2/i)2/2 == 0

137

138

Linkage Lie Algebras and Groups

[ ,] xl x2 yl y2 xl 0 xl2 hi 0 x2 -xl2 0 0 h2 xl2 -xll2 0 -3 x2 xl xll2 -xlll2 0 -4x12 0 xlll2 0 -x21112 -3x112 0 x21112 0 0 0 -xlll2 yl -hi 0 0 yl2 y2 0 -h2 -yl2 0 yl2 -3y2 yl -yl2 0 yll2 -4yl2 0 -ylll2 0 ylll2 -3yll2 0 0 -y21112 y21112 0 -ylll2 0 0 hi 2x1 -3x2 -2yl 3 y2 h2 | -xl 2x2 yl -2 y2 Table 4.6 Adjoint Table for G2 Proof.

j/21112t/2 = -j/2j/21112 = j/21112j/2 = -3/21/21112 = = -2/2(2/12/22/112 + (y2»l)yll2) = = - y 2 (yl j/2j/112 + (s/22/l)j/112) = = - 2 / 2 ( 0 - y l 2 2/112) = = -2/2(0 -2/12 2/112) = = 2/2 2/122/112 = = 2/2 2/122/112 = = j/122/2112 + (y21/12)2/112 = = 2/122/2112+ (2/22/12)2/112 = = 0. = 0. With these calculations we can fill in enough of the multiplication table With these calculations we can fill in This enough of the multiplication table for G2 to compute all the adjoints we need. information is given in Table for G to compute all the adjoints we need. This information is given in Table 4.6. 2 4.6. We now have enough information to construct the adjoint representaWe now have enough information to construct the adjoint representation of G 2 . In the chapter on computer calculations we will implement this reption of G 2 . In the chapter on computer calculations we will implement this representation in a Mathematica notebook. We will be able to use this notebook resentation in a Mathematica notebook. We will be able to use this notebook to check all the computations needed demonstrate the Isomorphism between to check all the computations needed demonstrate the Isomorphism between £7 and G 2 . £7 and G 2 .

4-5 The Isomorphism

4.5

between £7 and G2

139

The Isomorphism Between £7 and G2

We need to perform numerous calculations in order to demonstrate the iso­ morphism between £7 and G%. These calculations will be done by computer in a later chapter using a matrix representation for G 2 . In this section we will describe what the calculations are. It can be shown that there is a hexagonal algebra such as that in Figure 2.7 in the algebra G 2 . Its vertices are

-2/2

1 -ggy2«/i2/i2/i3/2

1 — yiyiym OD

X2X\X\X\X2 X\X\X\X2-

140

Linkage Lie Algebras and Groups

There is also a heptagonal algebra in G2 whose vertices are -2/2 -2/2

1 ~36

- ^ g 2 / 22/2j/iS/iyi2/2 j/iJ/iyiJ/2 11 myivi¥2 -^ 36 ymvm

xX22 Z\ *1

zz22 xX11Xx11Xx11Xx2 2

— J/2 y2 —

where zi ixx11xix x1x22 + ByiByxZ\ = x22 + A xxxx22 + x221

z2 = x2X\X\X\X2

+ 3 5 X\X\X2

11 — t/ij/ij/ij/2 yxyiyiy2 3D do

— hi — 2h2

x1xxxix22 - - BByyiy —y2 2 yiyiymyiyiym-A x% xi ++i ! xix xy2 2 - - —y Computer calculations reveal that the above vertices form a heptagonal algebra in G2 except possibly for the relation [21,22] = 0. For this final relation one needs specific values of the constants A and B. The appropriate values are

A ==-#6 -^6 and

"k as will be checked by computer in Chapter 6. When all of these relations are checked it can be seen that we have a Lie algebra homomorphism $$ :C : £ 77 -> -> G G22

4-5 The Isomorphism

between £7 and G 2

141

given by 1 n. -j/2

2

I—>•

1 -jjg!/2j/iS/iSfi2/2

, 1 3 •-» gg ymym 4 i-> aj 2

5 !-»• 2l 6 h-> 2 2

7 h-» xia;isia;2 - j/ 2 By taking linear combinations and Lie brackets of the images of the vertices of £7 it can be shown that the image of £7 under $ contains Zi,a;2,3/1,3/2- Since these generate all of G2 it follows that $ is surjective. Since £7 and G2 are both 14 dimensional it follows that $ is an isomorphism. The isomorphism described above was constructed under the assump­ tion that the underlying field was the complex numbers. Note that the multi­ plication table for G2 expresses each Lie product as an integer linear combina­ tion of the basis elements. Hence, the multiplication table can be defined for any field (or any commutative ring). The same is true for all the classical Lie algebras defined by the Cartan matrices given above and is also true for £7. If R is a commutative ring then we denote by G2 (R) and C^R) the algebras corresponding to G2 and £7 with scalars coming from R. The isomorphism we described can be defined as long as R contains — , \/6 and —7=. This is true 36 V6 for example is R is a an algebraically closed field of characteristic other than 2 or 3. For another description of this isomorphism see [11].

Chapter 5 COMBINATORICS IN LINKAGE LIE ALGEBRAS 5.1

The Reduction Process

In this chapter we will deal with linkage graphs which give rise to linkage Lie algebras whose combinatorial structure is not too difficult to study. From these Lie algebras we can define the associated linkage Lie groups. While groups are the main interest of the authors, the Lie algebras constructed on the path to constructing these groups may be of interest to others. These Lie algebras are generally infinite dimensional. We now say that two vertices i and j of a linkage graph T react with each other if there is an arrow or edge joining them in Y or if i = j . If i and j react with each other then we denote this by i ~ j . Also recall that a linkage graph T must satisfy the following rules: 1. If (w, v, w) is a linkage of T, then there is no edge joining u and w. 2. If (u,v,w) is a linkage of T, then there is an edge joining u and v, and an edge joining v and w. 3. If (u, v, w) is a linkage, and there is an edge joining u and z, and an edge joining w and z in T, then v = z.

Definition 1. A linkage graph T is special if 1. For any set of 3 vertices that all react with each other, either for each of the 3 vertices there is an edge joining it to each other vertex in the set, or all of the vertices lie in a single linkage, and if

143

144

Combinatorics

in Linkage Lie Algebras

2. For any vertices i,j,k,l ifi ~ j , j ~ Jfc, k ~ I, I ~ i, and j ~ /, then each of these vertices is joined to each other vertex by an edge, or else all 4 vertices lie in a single linkage. The associated linkage group L(R, T) and the Lie algebra C(R, T) will be referred to as special linkage groups and special linkage algebras. Notice that in a special linkage Lie algebra L= L(R, T) if a and b are vertices of T connected by an edge and if 6 and c are connected by an edge, then if a and c react with each other, then either all three vertices are connected by edges or else they lie in the same linkage. If they are distinct and lie in the same linkage then there is an arrow joining a and c. Hence, [a, c] — ±6. We will construct £ = C(R, T) for a special linkage graph T by finding a representation of £ as an algebra of transformations of a certain free module over the commutative ring R. Since £ will have a concrete representation it will be possible to find a normal form for it elements. Then we will consider the question of adding local nilpotence conditions on the vertices of £ in order to construct the adjoint map. This will allow us to construct the associated Lie group which will be an image of L(R, T). Consider Ad — Ai(R, T) the free module over R which has as its basis all n-tuples of vertices of T for n > 1. We will denote the n-tuple (t>i, t>2, • * • > vn) by v1v2---vn. Definition 2. The following are called transformation moves on elements of M.{R,T) :

1. Replacing uwviv2 ■ • ■ vn with vv^v? ■ ■ ■ vn when (u, v, w) is a linkage ofT 2. Replacing uvviv? ■ ■ ■ vn with 0 when u and v are joined by an edge in V. 3. Replacing uvv\V2 • ■ • vn with —vuviv? ■ ■ ■ vn for any vertices u and v. 4. Replacing vlv2--- vnuwvn+1 ■ ■ ■ vm+n with vxv2 ■ ■ ■ vnwuvn+2 ■ ■ ■ vm+n if u and w are connected by an edge in T, for n > 1 and m > 0.

5.1 The Reduction

Process

145

5. Replacing V\V2 ■ ■ ■ VnUWVn+i

■ ■ ■ Vm+n

with VtV2 ■ ■ ■ VnWUVn+i

where (u,v,w)

■ ■ ■ Vm+n +

V}V2 ■ ■ ■ VnV Vn+1 ■ ■ ■ Vm+n

is a linkage ofT,for n > 1 and rn > 0.

6. Replacing VxV2 • • • VnUWVn+i

■ ■ ■ Vm+n

with viv2 • ■ ■ vnwuvn+i • ■ ■ vm+n - vtv2--- vnv vn+1 ■ ■ ■ vm+n where (w,v,u)

is a linkage ofT, for n > 1 and m > 0.

In transformation moves numbers 5 and 6, we refer to Viv2 ■ ■ ■ vnwuvn+i ■ ■ ■ vm as the main part of the result, and ±viv2 ■ ■ ■ vnv vn+i ■ ■ ■ vm+n as the com mutator part of the result. The n-tuple ViV2 ■ ■ ■ vn is meant to correspond to the left-normed com­ mutator [• • • [vi, «a], • • • , vn]- Consider the case in which (u, v, w) is a linkage of T. In £ we have by the Jacobi Identity that VlV2 ■ • ■ VnUWVn+1

■ ■ ■ Vm+n =

= ((v1v2 ■ ■ ■ vnu)w) vn+1 ■ ■ ■ vm+n = - ((U w) V-iV2 ••■Vn)

= -(vviv2

Vn+1 ■ ■ ■ Vm+n - (W (viV2

■ ■ ■ Vn) u) Vn+i

■ • • Vm+n =

■ ■ ■ vn)vn+1 ■ ■ ■ vm+n - (w (vtv2 ■■■vn) u) vn+1 ■ ■ ■ vm+n =

= V1V2 ■ ■ ■ VnWUVn+1

■ ■ ■ Vm+n +

VlV2 • • • VnV Vn+1 ■ • ■ Vm+n.

Thus transformation move 5 corresponds to an equality in C. The same can be said about all of the other transformation moves. Now on the vertices of T we define a relation -< such that if (it, v, w) is a linkage of V, then u -< v u -< w v -< w.

146

Combinatorics

in Linkage Lie Algebras

We then extend -< arbitrarily so that for any two vertices u and v of T, either u -< v or v -< u. As with the relation -< defined in Section 3.2, the relation introduced here is not meant to be transitive or anti-symmetric. However, under this ordering any two vertices can be compared. We let u ■< v stand for "u -< v or u — v." We can now lexicographically order n-tuples for a particulars. We will also write u -4 v if u is an n-tuple and v is an m-tuple and n < m. Thus, under this ordering any n-tuple can be compared to any m-tuple for any ra and n. Definition 3. Let Viv2 ■ ■ -vnbe an n-tuple in M. We say that Vi moves in front of Vj if i > j and Vj ^ v,-; and if Vj reacts with Vk for k = i + \,i + 2,---,i-i- " The definition of moves in front of is similar to the same definition for groups given in Chapter 3. With algebras, however, there is another way for a vertex to move to the front of an n-tuple. Consider for example T = II7 and the 3-tuple 14 7. This can be transformed into —41 7, which can be transformed in to —4 71, which can be transformed into 7 4 1 . Notice that vertex 7 moved to the first spot in the 3-tuple but that it does not react at all with 4. This prompts our next definition.

Definition 4. Let w = V\V2 ■ ■ ■ vn be an n-tuple in M. If j > 2, we say that Vj moves to the first spot in w if Vj reacts with v2, v3, ■ ■ ■, w,-_i, or if Vj reacts with V\ and V3, • ■ ■, U7--1. We will now define a Reduction Process that is similar to the Collection Process of Chapter 3. We use different names in order to distinguish them since they are different.

R E D U C T I O N PROCESS FOR ELEMENTS OF M Step 1 If w = 0 (the 0-tuple) then w is reduced. Step 2 If w = vxv2 ■ ■ ■ vn then find the reordering of the vertices of w which is least in the ordering -< and which can be obtained from w by trans­ formation moves. (This includes transformation moves which shorten the

5.1 The Reduction

Process

147

main part of w.) If w is not in such a least form, then use transformation moves to put the main part in such a form. The main part of w is then reduced. Step 3 Now apply the above steps to each commutator part produced by transformation moves in Step 2. (If an rc-tuple u and its negative — u are produced they, of course, combine to form 0.) When all the commutator parts have been reduced then we say w is reduced and the final result is the reduced form of w.

If w = Yli ri Bi where ri £ R and Bi is an n;-tuple, we reduce w by applying the Reduction Process to Bi and multiplying the result by n for each i. The final result will be denoted by w. Notice that the Reduction Process applied to w will terminate in a finite number of steps since there are only finitely many ways to reorder an n-tuple and one of these ways is in the proper order. This is very different from the Collection Process in Chapter 3 where it was not even clear that the process terminated. This is to be expected because the Lie algebras we are working with here are "linear" versions of the corresponding Lie groups. We do however have to show that the reduced form of w is the same no matter what the order is in which the reduction moves are made. We now give some examples of the Reduction Process in M. = M(R, LTm) where m > 6. The linkage graph here is, of course, special. We will let w —► u mean that u is obtained from w by finitely many transformation moves. The relation -< is 1 ^2 1 -<3 2^3 2 -<4 n -i 1 n A2 If i and j are not in the above list, then -< is just the natural ordering < . For example, 1 -i 4, and

Combinatorics

148

in Linkage Lie Algebras

(i) 3 -< 7. (ii) 431 -> 0 (iii) 413 -»• 431 + 42 ->• 0 + 42 -+ 42 -> - 2 4 -► - 3 (iv) 241 ->• 31 ->■ - 2 1. (m > 7)

361 -> -631 -» - 6 1 3 + 62 -» 163 + 62 -> 163 - 26

Among the Transformation Moves the last four are reversible, that is, they have inverses. The first two do not have inverses. Definition 5. Two elements of M(R,T) are equivalent if one can be obtained from the other by a finite series of transformation moves of types 3 through 6. Since transformation moves 3 through 6 have inverses, Definition 5 actu­ ally defines an equivalence relation. We must prove several results concerning the Reduction Process before we can get to the main point. L e m m a 5.1.1. Suppose that w = v\ ■ ■ ■ vn is an n-tuple in M(R, T) where T is a special linkage graph. Suppose that a number of transformation moves of types 3 through 6 are applied to w in such a way that the main part of w is changed to w', and that some other transformation moves of types 3 through 6 are applied to w in such a way that the main part is changed to w'. Then the commutator parts arising from each set of transformations are equivalent. Proof.

Let w — uabv

where a and b are vertices and u and v are (possibly empty) subwords of w. We consider the result of the various transformation moves applied to w. Upper case C's represent words arising from the commutator parts that come from the transformation moves. The symbol [a b] will represent the commutator of a and 6, for example, if (a, c, b) is a linkage, then [a b] = c. If a and b are joined by an edge, then [ab] will denote 0.

5.1 The Reduction Process

149

We will begin with the simplest case which is a word of the form w = uab-cdv where u and v are (possibly empty) subwords. We will show that it does not matter if we switch a and b first, or if we switch c and d first. The result from switching a and b first is w = uab- ■ -cdv —► —► uba ■ ■ ■ cdv + u[ab] ■ ■ -cdv —> —> uba • • ■ dcv + u[ab] ■ ■ -cdv + uba- ■ ■ [cd]v. The the result from switching c and d first is w = uab- ■ ■ cdv —> —> uab- • -dcv + uab- ■ ■ [cd]v —» —» uba- ■■ dcv + u[a6] •••dcu + uafc--- [ccfjt;. The commutator parts are equivalent by the following steps: u[ab] ■ ■ -cdv + uba- • ■ [cd]v —> —> w[a&] ■ -dcv + uab- ■ ■ [cd]v + u[ab] ■ ■ ■ [cd\v + u[ba] ■ ■ ■ [cd]v

—* u[ab] - - -dcv -\-uab- ■ ■ [cd]v + u[ab] ■ ■ ■ [cd]v — u[ab] • ■ ■ [cd]v —* u[ab] ■ ■ -dcv + uab- ■ ■ [cd]v.

uabcv

The other situation that we must deal with is a word of the form w = where a,b,and c are distinct: w = uabcv —► —> ubacv —> ubcav

+ u[ab]cv —>

+ ub[ac]v + u[ab]cv —»

—► ucbav + u[bc]av + ub[ac]v + u[ab]cv

150

Combinatorics

in Linkage Lie Algebras

and w = uabcv —» —► u a c bv + u a[bc]v —» -^ ucabv + u[ac]bv + ua[bc]v —* —> ucbav

+ uc[ab]v -f u[ac]few + ua[bc]v.

In order for the above transformation moves to take place a, 6, and c must all react with each other. Hence they are all joined by edges in which case there are no commutators, or else they all belong to the same linkage Note that if x, y, z all belong to the same linkage then [xy] = 0 or else [xy] and z are joined by an edge. Thus, for example, if [ab] ^ 0, then [ab]c — c[ab], [ac] = 0 and [be] = 0. This makes all the commutator parts equivalent. In a similar fashion, the commutator parts produced above are equivalent if [ac] ^ 0 or if [6c] ^ 0. Notice that we have used part of the definition of special linkage graph in a crucial way here. We now proceed by induction on the total number of transformation moves applied to w. We have two sequences of transformation moves being applied to w. Suppose that w = ■ ■ ■ abc ■ ■ ■

and that the first set of moves first transforms w into • • ■ cba 1- C\ and that the second set of moves transforms w into • • • cba • - • + C% where C\ and C% a commutators. Then C\ and Ci are equivalent by the above paragraph. From this point on there are fewer total transformation moves applied and thus all of the other commutator parts that will be produced are equivalent. The proof can now be completed for other cases in a similar way. □ We must next deal with the situation in which a transformation move of type 1 or 2 is used in reducing w. Notice first that if we are dealing with vertices that belong to a single linkage we do not have any problems because we are using only relations that hold in the Lie algebra represented by the following matrices:

5.1 The Reduction Process

A A==

151

/I 0 1 0 \\ /' 0 0 0 \ 0 0 1 L, aand 0 0 0 ,B= nd ^^ 0 0 0 / - ( \ ,o0 0o 0o)/

cC = =

/ 00 0 1 \ 0 0 0 V o o :o)/

loo

where the Lie bracket is given by [x,y] = xy — yx. This algebra has the single linkage (A, C, B). An n-tuple in this algebra is a left-normed compound commutator of weight n. Note that such a left-normed commutator of length more than 1 is the zero matrix. Thus any n-tuple is of the form: / 00 r 00 \\ / / 00 0 0 s s\ \ / 0 0 0 \ 0 0 0 , , 00 00 00 , or 0 0 t . \V oo oo oo // \ \ o0 o 0o 0/ j \Voo o o / o o/ Any sum of such commutators can be written as rA + sB rA sB + tC tC for appropriate r, s, and t.Thus single linkages offer no problems in the reduction process. Proposition 5.1.2. Elements of Ai = Ai(R,T), where V is special, have a unique reduced form, that is, not matter what reduction moves and not matter what the order in which the moves are applied to an element of M, there is only one reduced from of the element that will ever be produced. Proof. By the previous result we need only worry about the situation in which reduction moves of type 1 or 2 are applied at some stage in the Reduction Process. Suppose w be an rc-tuple and that at some stage vertices a and b are moved to the first two positions in the main part and then combined in a transformation move of type 1 or 2. Suppose that there are other vertices c and d that can also be moved to the first two positions of the main part and then combined in a transformation move of type 1 or 2. We first assume that w = abcdu where u is a (possibly empty) subword. We assume that a and 6

152

Combinatorics in Linkage Lie Algebras

are joined by an edge or an arrow in T. We also assume that c and d are joined by and edge or an arrow in T. In order for c and d to move into the first two spots either all four of the vertices a, b, c, d must react with each other or they all react with each other except possibly for 6 and c (in which case » —bead —> abed —» —bacd — a6cd —* -■» c6ad cbad —* —►cdab, cdab,

ignoring commutators). In either case there is no a problem since the four vertices are either all connected by an edge or are in a single linkage. If they are connected by an edge then all commutator parts are 0 and the main part also reduces to 0. If they all lie in the same linkage, then abed can be put into standard from according to the comments just previous to this proposition. Note that we have just made use of the second part of the definition of special linkage g r a p h . Suppose now that w — abcu and that a ~ b, and that c can be moved into the first position in w and then reacts with the second position in w which is either a or b. If a, 6, c all react with each other there is again no problem since Y is special. There is essentially one case to consider: a ~ 6, c ~ a. w= to = aabcu 6cu — —»> [ab]c [ab]cu u and

w = abcu —> —bacu —> w = abcu —> —bacu —► —► —beau —► —beau

— 6[ac]u —► — 6[OC]M —►

—► c 6 a u — 6[ac]u —> —► c 6 a u — &[ac]u —> —» c a 6M + c[6a]M — 6[ac]M —» ca bu + c[6a]u — 6[ac]u —> —> [ca]6M + c[6a]M — 6[ac]w -» —> [ca]6u + c[ba]u — b[ac]u —> —► [ca]6u + C[6O]M + [OC]6M

—► [ca]6w + c[ba]u + [ac]bu - —> —> » [ca]6u + c[6a]u C[6O]M — [ca]6u [ca]6u — -*

c[6a]« —> — [6a]CM — ► —> c[6a]M [6a]cu — ► —» [a6]cM.

153

5.1 The Reduction Process

We have now shown that when we apply two sets of transformation moves to w, then if we arrive at the same main part then the commutator parts are equivalent. If we reduce the length of the main part by a move of type 1 or 2, then the commutator paxts axe again equivalent. Hence, if we apply transformation moves to w and get the main part as short as possible, then the remaining parts are equivalent. It follows that every w has only one reduced form. This completes the proof. □ For any w <E A4 = M(R,T) we denote the unique reduced form of w by to. For any vertex we now define the linear transformation Ti given by T, : M ->_M w— i > wi We let A = A(R, T) denote the algebra of transformations generated by T; for each vertex i of T, with the bracket operation [A, B] = AB — BA.

1. T h e o r e m 5.1.3. Suppose that T is a special linkage graph. Then the Lie algebra A(R, T) satisfies the linkage relations ofT as a linkage Lie Algebra. Proof. Suppose that i and j are vertices of T that are either joined by an edge or an arrow. Let w be any element of M.. Then we have w[Ti,Tj] = wTiTj-wTjTi = wij — wji =

=

= wji + w[i, j] — wj i = w[itj]~wT[iji.

This completes the proof. □ Suppose now that we have a special linkage graph T in which for every vertex there is another vertex which does not react with it. For a fixed vertex i consider the mapping $i : R -» A(R,T) r i-» rT{

154

Combinatorics

in Linkage Lie Algebras

Clearly, $,- is a homomorphism from the additive group of R into the additive group of A(R, T). Let j be a vertex of T that does not react with i. Then

(j)rTi = rji. Since i and j do not react, r ji is not 0. Hence, $; is injective. As with linkage groups we say that R is embedded at each vertex.

Remark: We thus may use special linkage graphs to construct Lie algebras. Suppose that i and j are vertices that do not react with each other. Let Un =TiTj ■ ■ ■ TiTj . It is clear that the set s * ' n times

{Un | n a positive integer} is a linearly independent set in A(R, T). Theorem 5.1.3. A(R, T) is an infinite dimensional Lie algebra for any special linkage graph Y that has at least two vertices that do not react with each other. If for ever vertex there is another vertex that does not react with it, then R is embedded at each vertex. Recall that in Section 4.3 we were dealing with the Heptagonal Algebra, which began as the algebra C(R, II7). We added some extra relations to get a 14-dimensional algebra. We can now see that these extra relations were really needed since £(R, II7) maps onto A(R, II7) and this latter algebra is infinite dimensional. Hence, C(R, Ti.7) is also infinite dimensional.

Chapter 6 COMPUTER CALCULATIONS 6.1

Introduction

This chapter is a bit different from the previous chapters in this book. It consists of computer output that verifies some of the claims made in previous chapters. The computer languages used in this chapter are Mathematica and CAYLEY. We will present actual output from these computer languages.

IBS

156

6.2

Computer

Calculations

The Adjoint Representation of £ 7

This is a Mathematica notebook constructing a 14 dimensional matrix repre­ sentation of the Heptagonal algebra described in Section 4.3. We verify all the claims of that section. In particular, we verify that the multiplication table for this algebra given in Section 4.3 is correct. This verification was not done in Section 4.3 because of the large amount of computation required.

6.2 The Adjoint Representation

of £7

157

The Adjoint Repesentation of the Heptagonal Algebra This section contains a Mathematica Notebook which constructs the adjoint representation of the Heptagonal Lie algebra of Secton 4.3 and checks to make sure that various relations hold. This section and several others were created on Mathematica as Notebooks. The commands given here can be run in Mathematica. Anyone interested in obtaining the notebooks in this chapter on disk should contact the first author at [email protected].

Since this is a Mathematica Notebook all the appropriate Mathematica commands are given. The output should be self-explanatory even for those readers who are not familiar with Mathematica. The matrix A given below represents the adjoint operation of vertex 1 in the Hepatagonal Lie algebra of Section 4.3 with respect to the ordered 14-dimensional basis given in that section. The adjoint operator acts on the right, as do all operators in this book. A={{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , { 0 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , {0, - 1 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , { 0 , 0, 0, 0, 0, 0, 0, - 1 , 0, 0, 0, 0, 0, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 ) , {0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) , { - 2 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , { - 1 , - 1 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) , { 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1, - 1 } , { 1 , 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 } , { 2 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) , { 1 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , { - 1 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } } ; The matrix B cyclically permutes rows and columns in groups of seven. Thus, conjugating A by powers of B gives the adjoint representation of all of the vertices of the algebra.

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Calculations

B={{0, 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 } , {0, 0 , 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , {0, 0 , 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0 } , {0, 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 } , {0, 0 , 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 } , {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) , {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 } , {0, 0 , 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 } , {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 } , {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 } } ; In order to make the matrices look better in a printed form we set the following function which will be applied to all output. $PrePrint=MatrixForm[#,TableSpacing->{0,2}]& The matrices A and B can now be printed as follows: A 0 0 0 0 0 0 0 -2 -1 0 1 2 1 -1

0 0 -1 0 0 0 0 0 -1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 1 0 0 0

0 0 0 -1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 -1 0 0 0 0

6.2 The Adjoint Representation of £7

0 0 0 0 0 0 1 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1 0 0

159

0 0 0 0 0 0 0 0 0 0 0 0 1 0

We can now construct a list M of all adjoint representations of all the vertices. We construct the conjugates of A by B to the power i for i= 1 to 9 rather than 1 to 7. This repeats the first two but makes it easier to cycle through the list to check the relations. M=Table[ MatrixPower[Inverse[B],i].A.MatrixPower[B,i], {i,l,9}]; 1 We now check to see whether this adjoint representation acutally gives us a Lie algebra which satisfies the heptagon relations. The first thing we check is that [1,3]=2, [2,4]=4, etc. The vertex i is represented here by M[[i]].

Do[Print[ M[[i]].M[[i+2]]-M[[i+2]].M[[il]==M[[i+l]]], {i,l,7}] True True True True True True True

Next we check to see that [1,2]=0,[2,3]=0, etc.

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Do[Print[M[ [i] ] .M[ [i+1] ] = M [ [i+1] ] .M[ [i] ] ], {i,l,7>] True True True True True True True

That works so we now check to see that ad(i)"2 has only 2's and O's as entries. MatrixPower[M[[1]],2] 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 2 2 0 0 0 0 2 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

Since all other matrices in M are merely M[[l]] with rows and columns switched we do not have to check each matrix in M. We only have to check M[[l]]. Similarly we check that ad(i)*3 is the zero matrix. We only have to check M[[l]].

6.2 The Adjoint Representation

of £7

161

MatrixPower[M [ [ 1 ] ] , 3 ] 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

We will now check the multiplication table for the heptagon algebra. Because of symmetry we only have to check the first column of Table 4.3. We must define the zero matrix and we also define a ring commutator opertor: zero=Table[0,[i,l,14},{j,l,14}]; comm[i_,j_]:=M[[i]].M[[j]]-M[[j]].M[[i]] We can now use the function comm[i,j] to check the commutators of M[[i]] with Mffi]]. comm [1,1] = z e r o True comm [2,1] = z e r o True comm[3,l]=-M[[2]] True comm [4,1] =-comm [1,4] True comm [5,1] = c o m m [5,1] True

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Calculations

comm[6,l]=M[[7]] True comm [ 7 , 1 ] = z e r o True We will now define a more general commutator function to deal with products in the rest of the multiplication table which are products of three vertices. comm[i_, j _ , k _ ] : =conim[i, j ] .M[ [k] ] -M[ [k] ] . c o m m [ i , j ] comm[l,4,l]=-2 M[[l]] True comm[2,5,l]=-M[[l]]-M[[2]] True comm [ 3 , 6 , 1 ] = - c o m m [ 7 , 3 ] +comm [ 6 , 2 ] True comm[4,7,l]=M[[l]]+M[[7]] True comm[5,l,l]=2 M[[l]] True comm[6,2,l]==M[[l]] True comm[7,3,l]=-M[ [1] ] True The above calculations show that the Table 4.3 defines a Lie algebra that satisfies the hepatagonal relations as described in Section 4.3.

6.2 The Adjoint Representation

of £7

163

■ The exponential map. We now define exp(r ad(i)) for each i = l , . . . , 7 and for r in any commutative ring. We will let X[i,r] denote I + r ad(i) + r"2 (1/2) adtf)^. Notice that we are defining the exponential map for a general element r. The symbolic manipulaltion facilities of Mathematica allow us to do this. This shows that the computations are valid for elements r from any commutative ring R. X[i_,r_]:=IdentityMatrix[14]+ r M[[i]]+ r A 2 ((1/2) M[[i]].M[[i]])

We will now check to see that the matrices X[i,r] satisfy the heptagon relations for a group. We deal first with the relation X[i,r].X[i,s]=X[i,r+s]. In order to check that these are true we need to apply the Simplify operator to both sides of the equation. As above we only need to check this for i = 1. However,it does not really take any longer to check for all i.

Do[ Print[X[i,r] .X[i,s]=X[i,r+s]//SimplifY] , {i,l,7}] True True True True True True True

Next we will check that X[i,r].X[i+l,s]=X[i+l,s].X[i,r].

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Calculations

Do[ Print[X[i,r] .X[i+l,s]=X[i+l,s] .X[i,r]//Simplify] , {i,l,7} ] True True True True True True True

Finally we check that [X[i,r],X[i+2,s]]=X[i+l,rs]. We define first the group commutator function gpcomm[-,-].

gpcontm [x ,y ]:=Inverse[x].Inversely].x.y Do[ Print[ gpcomm[X[i,r] , X [ i + 2 , s ] ] = X t i + l , r s ] / / S i m p l i f y ] , {i,l,7}] True True True True True True True

The above calculations show that the adjoint representation of the Heptagonal algebra actually satisfies the Heptagonal Lie relations. Furthermore, the associated Lie group obtained by the exponential map satisfies the Heptagonal group relations. Neither the group nor the algebra collapse. Note that the ring involved is not necessarily a field. The matrix representations for both the algebra and the group involve only integer coefficients. The computations are valid over any commutative ring.

6.3 The Adjoint Representation

6.3

of G2

165

The Adjoint Representation of G2

This section is a Mathematica notebook constructing a 14 dimensional matrix representation of the Lie algebra G2. After the construction is complete it is checked that the algebra actually satisfies the defining relations for G'2 as described in Section 4.4.

166

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Calculations

Adjoint Representation for G2 This section is again a Mathematica Notebook. The comments at the beginning of the previous section apply here as well. Note that this notebook is separate from the previous notebook. Both have defined a commutator function comm. However, this function is diferent in each notebook. Thus one should not run these two notebooks in the same Mathematica session without clearing the definition of comm. We will now construct the adjoint representation for the Lie algebra G2. We will do this by constructing the matrices representing ad(xl), ad(xl), ad(yl), and ad(y2). From these we can construct the whole of the algebra. In order to facility typing elements such as ylyl y2 are written as y l l 2 , etc. Since the matrices we are constructing have mostly O's as entries, we set each of the desired matrices equal to the zero matrix and then change the non-zero entries. xl=Table[0,{i,1,14},{j,1,14}]; xl[[2,3]]=-l; xl[[3,4]]=-l; xl[[4,5]]=-l; xl[[7,13]]=-l; xl[[9,8]]=-3; x l [ [ 1 0 , 9 3 3—4 ; xl[ [11,10]]=-3; xl[[13,l]]=2; xl[[14,l]]=-l; As in the previous section we will again set the default printing style for matrices: $PrePrint=MatrixForm[#,TableSpacing->{0,2}]&

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168

Computer

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We will now construct the commutator function and some of the other basis elements for this algebra. We will also turn off the general spelling checker. We will be naming many elements with similar names and this will cause a lot of warning messages if we do not turn the spelling error message off. Off[General::spell] comm[A_,B_] :=A.B-B.A xl2=comm[xl, x2 ] ; xll2=comm[xl,xl2]; xlll2=cornm[xl,xll2]; x21112=comm[x2,xlll2]; We now construct the zero matrix. zero=Table[0,{14},{14}]; yl=zero; yl[[l,13]]=l; yl[[3,2]]=-3; Yl[[4,3]]=-4; yl[[5,4]]=-3; yl[[8,9]]=-l; yl[[9,10]]=-l; yl[[10,ll]]=-l; yl[[13,7]]=-2; yl[[14,7]]=l;

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170

Computer

Calculations

yll2=coinm[yl,yl2] ; ylll2=comm[yl/yll2]; y21112=comm[y2,ylll2]; hl=connn[xl , y l ] ; h2=comm[x2,y2]; At this we have matrix representations for all the basis elements. It is now neccesary to check that the relations for G2 are satisfied. comm [hi, h2 ] = z e r o comm [ xl, y 1 ] = h l comm [x2, y2 ] = h 2 True comm [ xl, y2 ] = z e r o True comm [x2 ,yl ] = z e r o True comm [hi, xl] = 2 xl True comm [hi, x2] = - 3 x2 True comm [h2, xl ] = - x l True comm[h2,x2]=2 x2 True comm[hl,yl]=-2yl True comm [hi,y2] = 3 y2 True comm [ h2, y 1 ] = y 1 True comm[h2,y2]=-2 y2 True

6.3 The Adjoint Representation

of G2

171

comm [ x l , x l 1 1 2 ] = z e r o True comm [x2, comm [x2, xl ] ] =zero True comm [yl, yl 112 ] =zero True comm [y2, comm [y2, yl ] ] =zero True Since all of therelationsabove are satisfied it follows that we have a matrix representation of the Lie algebra G2. The Matrices are 14 x 14 and the bracket operation is the ring commutator [A,B]=A B - B A. We will use thisrepresentationin the next section to check the isomorphism between L7 and G2.

172

6.4

Computer

Calculations

The Isomorphism Between £7 and G2

This section is a Mathematica notebook which verifies the details concerning the isomorphism between £7 and Gi which was described in Section 4.5. It is also shown that there is a hexagonal subalgebra inside of G2-

6.4 The Isomorphism between £7 and G2

The Isomorphism Between L7 and 6 2 We can now us the adjoint representation of G2 in the previous section to verify the isomorphism between L7 and G2. The Notebook from section 6.3 must be run before this Notebook can be run. Note that to run this notebook we need to run first the notebook of the previous section in order to define the funciton comm. We begin by showing that there is a hexagon algebra inside of G2. comm[-y2,1/36 ylll2]=-l/36 y21112 True commt-y2,-1/36 y21112]=zero True comm[-l/36 y21112,l/36 ylll2]=zero True comm[-l/36 y21112,x2]=l/36 ylll2 True comm[l/36 ylll2,x2]=zero True comm[l/36 ylll2, x21112]=x2 True comm[l/36 ylll2,x21112]=x2 True comm[x21112 , x l l l 2 ] = z e r o True comm[x2,xlll2]==x21112 True comm[x21112,-y2]=xlll2 True comm [ x l 112, -y2 ] = z e r o True

173

174

Computer

Calculations

The above calculations show there is a hexagon in G2. While this is not entirely necessary to the task of demonstrating the isomorphism it is of some interest. We now demonstrate that the first few vertices of the hexagon demonatrated above are part of a heptagon whose vertices are: -y2,-l/36y21112,l/36ylll2, x2, zl, z2, xlll2-y2 where zl and z2 are somewhat more complicated than the other vertices.

comm[-y2,xl12-y2]==zero True comm[xlll2-y2,-l/36 y21112]=-y2 True zl=x2+A xl2+x21112+B yl -1/36 ylll2; z2=x21112 +3 B xll2-hl-2h2-A xl + X1112-B yl2 -1/36 y21112; comm [x2, zl ] = z e r o True comm[x2,zl]==zero True comm[z1,xl112-y2]==z2 True comm [ z2, -y2 ] ==xl 112 -y2 True comm [x2 , z2 ] = z l True comm[z2 ,xlll2-y2]=zero True

At this point we have only one other relation to demonstrate, namely, |zl,z21 = 0. We have not need up to now the actual values of A and B. We now need these values and will substitute them into [zl, z2], simplfiy the result and see if it is 0. values={A->-6 A (l/3),B->1/A} 1/3 1 {A -> -6 , B -> -} A (comm[zl, z2] // . values//Simplify)=zero True

6.4 The Isomorphism

between £7 and G2

The above calculations show that G2 contains a subalgebra that satisfies the relations of L7. Thus G2 contains a subalgebra that is a homomoiphic image of L7. However, both L7 and G2 are 14 dimensional. Hence, they are isomorphic.

175

176

6.5

Computer

Calculations

Embedding Z2 in A5

In this section we will present a verbal embedding of Z t into the alternating group A$. The embedding is as follows: Z2-^A5 0i->e

1->(1,2)(3,4) The embedding word is [xa,yh] where a = (3,5)(4,6) and b = (2,4,3). It is easy to check that this actually is a verbal embedding. The constant group for the embedding is (a ,1) = As ., (a,6)SM t It is not quite so easy to determine ermine the linkage graph that corresponds to the embedding. We do this by running a program in CAYLEY, the group theoretic programming language. This program constructs all the vertices and numbers them from 1 to 30. Then it prints out all of the linkages. Then the vertices are printed out with their corresponding numbering. The edges of the linkages lie on the edges of a truncated dodecahedron. Note that the group of orientation preserving symmetries of a truncated dodecahedron is precisely A 5 . In addition, the vertices that are diagonally opposed on the truncated dodecahedron are identified in the linkage graph. Thus, even though the truncated icosahedron has 60 vertices, the linkage graph has 30 vertices. This section contains the CAYLEY program "library linkages" which constructs the linkages of this verbal embedding and it contains the output of this program. Even if one is not familiar with CAYLEY the commands correspond more or less to standard group theoretic concepts. Also in this section we have included a sketch of the linkage graph drawn in a plane. This is accomplished by identifying the vertices on the outside edge of the graph with vertices that are diametrically opposed. We have also included a sketch of a truncated icosahedron. Notice that the linkage graph is made up of subgraphs which we identified as "basic pieces" in Chapter 3.

6.5 Embedding Zi in At

library linkages; g=alternating(6); x=(l,2)(3,4) of g; a=(3,5)(4,6) of g; b=(2,4,3) of g; cons=; o=order(cons); print g,cons; print 'We have defined G to be Alt(6)'; print 'and CONS to be Alt(5).'; print 'We now check the commutator relations'; print 'among x, xAa, and xAb.'; print (xAAa,xAb),x; print (x a,x),(xAb,x); print 'We now print out all the linkages'; cen = centralizer(cons,x); t = transversal(cons,cen); num = empty; for each c in t do num = append(num,xAc); end; for each c in t do u = position(num,xAA(a*c)); v = position(num,xAc); w = position(num,x (b*c)); print u,v,w; print ' '; end; print '((((((((((((((((((((((((((((('; print ')))))))))))))))))))))))))))))'; print 'We now print out the correspondance between'; print 'the vertices as numbered 1,2, ... and the '; print 'group elements.'; for i = 1 to length (num) do print i,num[i]; end; finish;

177

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V3.8.3

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(ULTRIX/UNIX)

GROUP G OF ORDER 360 = 2A3 * 3 A 2 * 5 IS A SUBGROUP GENERATORS : (1,2)0,4,5,6) (1,2,3) GROUP CONS OF ORDER 60 = 2*2 * 3 * 5 IS A SUBGROUP GENERATORS : (3,5)(4,6) (2,4,3) We have defined G to be Alt(6) and CONS to be Alt(5). We now check the commutator relations among x, x A a, and x A b. (1,2)(3,4) (1,2)(3,4) IDENTITY IDENTITY We now print out all the linkages 2 1 3 1

2

17

5

3

20

17 3

4 5

10 18

18

6

11

26

7

28

20

8

16

27

9

29

22

10

4

23

11

6

21

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30

13

6

28

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24

6.5 Embedding Z2 in A$ 29

15

4

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4

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26

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18

27

25 8

19 20

22 3

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Calculations

((((((((((((((((((((((((((((( ))))))))))))))))))))))))))))) We now print out the correspondance between the vertices as numbered 1,2, ... and the group elements. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

(1,2)( 3,4) (1,2)( 5,6) (1,4)( 2,3) U,6)( 3,4) U,4)( 5,6) U,6)( 2,3) (1,5)( 3,4) (1,3)( 5,6) (1,5)( 2,3) (1,4) (3,6) (1,3) (2,6) (1,2) (4,6) (1,2) (3,6) (1,4) (2,6) (1,3) (4,6) (1,6) (3,5) (1,6) (2,5) (1,6) (4,5) (1,5) (2,4) (1,3) (2,4) (1,2) (3,5) (1,4) (2,5) (1,3) (4,5) (1,6) (2,4) (1,5) (3,6) (1,5) (2,6) (1,5) (4,6) (1,4) (3,5) (1,3) (2,5) (1,2) (4,5)

END OF RUN. 0.863 SECONDS The above calculations show that G2 contains a subalgebra that satisfies the relations of L7. Thus G2 contains a subalgebra that is a homomorphic image of L7. However, both L7 and G2 are 14 dimensional. Hence, they are isomorphic.

6.5 Embedding Z2 in A5

Figure 1

Linkage Graph for A5

181

1"

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Calculations

Figure 2 Truncated Dodecahedron

6.6 Embedding of Z$ into Sg

6.6

183

E m b e d d i n g of Z 3 into Sg

In this section we will show that there is a verbal embedding of Z3 into Sg. The corresponding linkage graph is II7. The verbal embedding is given by Z3 —> Sg Oi->e 1^(1,5,4)(2,3,8)(6,7,9) The embedding word is [a;a~ ,ya] where a = (1,2,5,7,3,6,4). Now <(l,5,4)(2,3,8)(6,7,9),a) = S 9 . However, ((1,5,4)(2,3,8)(6,7,9))W = ^ 9 . Thus the alternating group Ag is the image of L = i(Z 3 ,Il7). From Chapter 3 we know that L is an infinite group. As in the last section the computations here were done on CAYLEY. The CAYLEY program that computes the result in this section is called "li­ brary batch".

184

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VAX/UNIX CAYLEY

Calculations

V3.5-1

> > GROUP G RELATIONS : XA3 AA7 X A -1 A A -1 X A -1 A X A A -1 X A X A -1 A A -2 X A -1 A A 2 X A A -2 X A X A -1 A WE NOW COMPUTE; s=low index subgroups(g,,9;pr=2,limit=l) CLASS

CL.l

LENGTH 1

INDEX 1

SUBGROUP GENERATORS :< X >; PERMUTATION REPRESENTATION A -> IDENTITY X -> IDENTITY CLASS

CL.2

LENGTH 9

INDEX 9

SUBGROUP GENERATORS :< X A -1*A*X A -1*A >; PERMUTATION REPRESENTATION :A-> (1,2,5,7,3,6,4) X -> (1,3,2)(4,5,8)(6,7,9)

*** TERMINATED AFTER THE FIRST CLOSED TABLE TOTAL TIME = 3.000 SECONDS The cosact image of G on s[2] is: GROUP A9 OF ORDER 181440 = 2 A 6 * 3 A 4 * 5 * 7 GENERATORS : (1,4,3,7,5,6,2) (1,5,4)(2,3,8)(6,7,9)

***

6.6 Embedding of Z$ into Sg

I s A9 t h e a l t e r n a t i n g g r o u p of d e g r e e 9? We Al A2 A3 A4 A5 A6 A7

185

TRUE

h a v e l o c a t e d v e r t i c e s A 1 , . . . , A 7 i n t h e g r o u p A9: = (1,5,4)(2,3,8)(6,7,9) = (1,7,8)(2,5,9)(3,4,6) = (1,6,9)(2,7,3)(4,5,8) = (1,5,7)(2,9,4)(3,6,8) = (1,9,3)(2,8,7)(4,6,5) = (1,8,5)(2,6,3)(4,9,7) = (l,2/7)(3,9,5)(4,8/6)

WE NOW CHECK THAT THE ELEMENTS Al,A2,...,A7, SATISFY LINKAGE RELATIONS AND COMPUTE ORDERS OF VERTICES. TRUE 3 TRUE TRUE TRUE 3 TRUE TRUE TRUE TRUE

3 TRUE

TRUE TRUE

3 TRUE

TRUE TRUE

3 TRUE

TRUE TRUE

3 TRUE

TRUE TRUE

3 TRUE

WE NOW CHECK FOR AN INTERNAL HEXAGON. 3 TRUE TRUE TRUE FALSE FALSE

15 TRUE

TRUE FALSE

3 FALSE

FALSE TRUE

3 FALSE

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Calculations

WE NOW COMPUTE THE ORDERS OF 6 6 6 6 6 6 6

A(i)*A(i+3)

FOR i = 1,...7.

HENCE WE SEE THAT THE ALTERNATING GROUP OF DEGREE 9 IS AN IMAGE OF THE HEPTAGON GROUP OVER Z3 AND THAT THE ORDER THE PRODUCT OF DIAGONALLY OPPOSED ELEMENTS IS 6. HOWEVER,THERE IS NOT AN INTERNAL HEXAGON IN THE GROUP. > > END OF RUN. 21.366 SECONDS, 199986 WORDS USED.

library batch; n=7; p=3; g:free(a,x); g.relations:xAp=aAn=(x,xAa)=l,(x,xA(aA2))=xAa; print g; print 'WE NOW COMPUTE;'; print ' s=low index subgroups(g,,9;pr=2,limit=l)'; s=low index subgroups(g,,9;pr=2,liniit=l); a9=cosact image(g,s[2]); o=order(a9); print 'The cosact image of G on s[2] is:'; print a9; print 'Is A9 the alternating group'; print 'of degree 9?'; print alternating(a9); print ''; al=a9.2; b=a9.1A ; a2=al b; a3=a2AAb; a4=a3 b; a5=a4AAb; a6=a5Ab; a7=a6 b;

6.6 Embedding of Z3 into Sg

print 'We have located vertices print 'group A9:'; print 'Al =',al; print 'A2 =',a2; print 'A3 =',a3; print 'A4 =',a4; print 'A5 =',a5; print 'A6 =',a6; print 'A7 =',a7; c=a7*al*(-l); d=(a4,c);

187

hi,. ,.,A7 in the';

print 'WE NOW CHECK THAT THE ELEMENTS, print 'WE NOW CHECK THAT THE ELEMENTS Al,A2,...,A7'; print 'SATISFY'; print 'LINKAGE RELATIONS AND COMPUTE ORDERS OF VERTICES.'; c(al,a2,a3); c(a2,a3,a4); c(a3,a4,a5); c(a4,a5,a6); c(a5,a6,a7); c(a6,a7,al); c(a7,al,a2); print 'WE NOW CHECK FOR AN INTERNAL HEXAGON.'; c(c,al,a2); c(d,c,al); c(a4,d,c); c(a3,a4,d); print 'WE NOW COMPUTE THE ORDERS OF print ' For i = 1,... , 7.' ; o(al,a4) ; o(a2,a5); o(a3,a6); o(a4,a7) ; o(a5,al); o(a6,a2) ; o(a7,a3);

A(i)*A(i+3)

188

print print print print print print

Computer

Calculations

'===================================================='; 'HENCE WE SEE THAT THE ALTERNATING GROUP OF DEGREE 9'; 'IS AN IMAGE OF THE HEPTAGON GROUP OVER Z3 AND THAT'; 'THE ORDER THE PRODUCT OF DIAGONALLY OPPOSED ELEMENTS'; 'IS 6. HOWEVER,THERE IS NOT AN INTERNAL HEXAGON IN'; 'THE GROUP.';

procedure o(x,y); print order(x*y); end; procedure c(x,y,z); print (x,z) eq y,order(x); print (x,y) eq id,(y,z) eq id; print ' '; end; finish;

6.7 Verbal Embedding of Z2 into U(3, 3)

6.7

189

Verbal Embedding of Z2 into 1/(3,3)

In this section we give a CAYLEY program that verifies the claims of Example 3 in Chapter 1. The program that verifies the results is called "library u33". While the program given here verifies the claim it should be pointed out that the result was discovered in a trial and error manner. There are two ways to find verbal embeddings using CAYLEY. The first is to begin with some known group and perform some sort of search within this group. Verbal em­ beddings of Z2 are the easiest to find. First one looks for a Sylow-2-subgroup. It must have a subgroup isomorphic to the group of upper unitriangular 3 x 3 matrices over Z2. Then one searches for a and /? such that [a;", y13] serves as an embedding word. It is easy to write such a program although such a program can run for a very long time in a large group. The other method for finding a verbal embedding is to begin with gen­ erators and relations for a group in which there is a verbal embedding if the group does not collapse. For example, consider the group G=(a,x\

an,x2,[xa~\xa]

=

x,[xa,x]}.

This group (if it does not collapse) embeds Z2 using the constant group (a) = Cn. The embedding is

Z2^G 0i-> e 1 ->x One then runs the CAYLEY command "low index subgroups" which searches for subgroups of relatively small index in G. If H is such a subgroup then CAYLEY can construct the permutation group K acts on the cosets of H in G in the same way that G does. Then if is a finite permutation group which is an image of G. Thus, K will be a concrete group in which Z2 is embedded. It is then necessary to determine exactly which group K is. This is usually fairly easy since CAYLEY has a built-in function to determine the composition factors of a finite permutation group. Note that although we have not included the computations here, it is quite easy to find verbal embeddings of Z 2 in the Mathieu group M n , and of Z 2 and Z3 in the Mathieu group Mi 2 .

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CAYLEY

V3.8.3

Calculations

(ULTRIX/UNIX)

GROUP U 3 3 OF ORDER 6 0 4 8 = 2 A 5 GENERATORS : WA3 WA2 2 WA2 WA7 WA2 WA6 2 WA5

* 3*3

* 7 I S A SUBGROUP OF G

1 WA6 WA3 W 2 1 WA1 W A 1 WA3 0 A

I s U33 simple?

TRUE

We now check t h a t t h e r e l a t i o n s of t h e heptagon h o l d i n U33. The order of X i s : 2 The order of A i s : 7 A =

WA3 WA2 2 WA2 WA7 WA2 WA6 2 WA5

X ==

1 WA6 WA3 WA2 1 WA1 W A 1 WA3 0

(XA(AA-1),XAA)

(XAA,X)

=

1 WA6 WA3 W 2 1 WA1 W A 1 WA3 0

=

A

1

0 0 A

A

(X (A -1),X)

=

0

0

0

1 1

0 0 END OF RUN. 1 5 . 2 2 9 SECONDS

0 1 0

0

0

1

0 1

6.7 Verbal Embedding of Z2 into U(3, 3)

library u33; q = 3; k = field(q*q,w); v = vector space(k,3); g = special linear(v); a = mat(wA3,wA2,wA4:wA2,wA7,wA2:wA6,wA4,wA5) of g; x = raat(l,wA6,wA3:wA2,l,w:w,wA3,0) of g; u33 = ; o = order(u33); print u33; print ''; print 'Is U33 simple?',simple(u33); print ''; print 'We now check that the relations of the'; print 'heptagon hold in U33.'; print 'The order of X is:',order(x); print 'The order of A is:',order(a); print 'A = ',a; print 'X = ',x; print '(XA(AA-1),XAA) = ',(x A (a A -l),x A a); print '(XAA,X) = ',(x A a,x); print '(XA(AA-1),X) = ',(x A (a A -l),x); finish;

191

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6.8

Computer

Calculations

The Relations in G 2 (2)'

We have seen from our work that there is a surjective homomorphism i(Z2,n7)^G2(2)'. In this section we will determine the kernel of this homomorphism. If the generators of this kernel are combined with the linkage relations of £(Z 2 , II7) they will, of course, give a presentation for G 2 (2)'. We give a CAYLEY program which constructs a chain of subgroups I(Z 2 , II7) > L\ > L2 > L3 > L4 > L5 > L6 > L7 with indices 28,2,2,2,3,3,3. It is then shown that L7 is normal in L(Z2,1I7) and that the factor group is G 2 (2)' = (7(3,3), the simple group of order 6048. The program is lengthy and makes use of subroutine called sim which simplifies presentations by performing various Tietze transformations on the presentations. The presentation for Li must be simplified for each i since otherwise the presentations become quite unworkable. The final presentation for L?, which is the kernel we are seeking has 37 generators and 144 relators. The total length of all the relators is 14,668. If the kernel is abelianized it is a free abelian group of rank 34. This suggests that the kernel could be generated by 34 generators rather than 37. However, at this point the computer's memory ran out (as did the authors' patience). Note that this gives another proof that i ( Z 2 , n 7 ) is infinite. The CAYLEY program that produced the output here is called "library kerf72".

6.8 The Relations in G 2 (2)'

VAX/UNIX CAYLEY

193

V3.5-1

> > August 26, 1989 Revised July 9,1991 THIS PROGRAM FINDS THE KERNEL OF THE MAP FROM G ONTO U(3,3) = 2A(2,3) = G(2,2)', THE SIMPLE GROUP OF ORDER 6048. IT CONSTRUCTS A CHAIN OF SUBGROUPS G > LI > L2 > L3 > L4 > L5 > L6 > L7 OF INDICES 28,2,2,2,3,3,3. THE GENERATORS OF L7 ARE REWRITTEN IN TERMS OF THE GENERATORS OF G AND THE RESULTING SUBGROUP OF G IS CALLED KER. THEN G/KER IS COMPUTED AND SHOWN TO BE THE SIMPLE GROUP OF ORDER 6048. Thomas Fournelle Department of Mathematics University of Wisconsin-Parkside Kenosha, WI 53141 U.S.A. fournellivacs.uwp.edu

GROUP G RELATIONS : XA2 AA7 X A -1 A A -1 X A -1 A X A A -1 X A X A -1 A A -2 X A -1 A A 2 X A A -2 X A X A -1 A XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX A subgroup of the above group of index 28 is now found. The group has 17 generators. The group has 52 relators. The total length of the relators is The The The The The

group group total group group

has 10 generators. has 25 relators. length of the relators is has 9 generators. has 22 relators.

199

100

194

Computer

Calculations

The total length of the relators is 94 1 The group The group The total 94 2 substring The group The group The total

has 8 generators. has 21 relators. length of the relators is

95

search has 8 generators. has 13 relators. length of the relators is

49

The group has 7 generators. The group has 12 relators. The total length of the relators is 49 3 The group The group The total 47 4 substring The group The group The total

52

has 6 generators. has 10 relators. length of the relators is

47

search has 6 generators. has 9 relators. length of the relators is

40

The group has 5 generators. The group has 8 relators. The total length of the relators is 40 5 SUBSTRING The group The group The total

SEARCH has 5 has 7 length of

number 1 generators. relators. the relators is

The The The The The The 32

has 5 has 7 length of has 4 has 6 length of

generators. relators. the relators is generators. relators. the relators is

group group total group group total 1

94

The group has

3

generators

46

32

32 36

6.8 The Relations in G 2 (2)'

The group The total 32 2 substring The group The group The total

195

has 4 relators. length of the relators is

39

search has 3 generators. has 4 relators. length of the relators is

35

The group has 3 generators. The group has 4 relators. The total length of the relators is 32 3 SUBSTRING The group The group The total

SEARCH has 3 has 4 length of

number 2 generators. relators. the relators is

The The The The The The 34

group group total group group total 1

has 3 has 4 length of has 3 has 4 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 3 has 4 length of

number 3 generators. relators. the relators is

35

34

34 34

35

ORIGINAL GROUP: The group has 17 generators. The group has 52 relators. The total length of the relators is

199

SIMPLIFIED GROUP The group has 3 generators. The group has 4 relators. The total length of the relators is

35

TIME:

94

Computer

196

Calculations

The subgroup that is being considered is GROUP H IS GENERATORS A A -3 A A -1 A A -2 A X

A SUBGROUP AS WORDS IN SUPERGROUP : X AA3 X A A -1 X A A -2 X A A -1 X X A A -2 X A A -1 X A A -2 X A A -2 X A A -2 X

The simplified presentation for this group has 3 generators and 4 relators. The total length of the relators is 35 The abelian quotient of this group is SEQ( 4, 0 ) The number of O's here is 1 The simplified version of this group is called LI XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX A subgroup of the above group of index 2 is now found. The group has 5 generators. The group has 8 relators. The total length of the relators is The The The The The The 54

group group total group group total 1

The group The group The total 54 2 substring The group The group The total

has 5 has 8 length of has 4 has 6 length of

generators. relators. the relators is generators. relators. the relators is

54

54 74

has 4 generators. has 6 relators. length of the relators is

74

search has 4 generators. has 6 relators. length of the relators is

64

6.8 The Relations in G 2 (2)'

197

SUBSTRING The group The group The total

SEARCH has 4 has 6 length of

number 1 generators. relators. the relators is

The The The The The The 62

group group total group group total 1

has 4 has 6 length of has 4 has 6 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 4 has 6 length of

number 2 generators. relators. the relators is

The The The The The The 58

group group total group group total 1

has 4 has 6 length of has 4 has 6 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 4 has 6 length of

number 3 generators. relators. the relators is

54

ORIGINAL GROUP: The group has 5 generators. The group has 8 relators. The total length of the relators is

54

SIMPLIFIED GROUP The group has 4 generators. The group has 6 relators. The total length of the relators is

54

TIME:

64

62

62 62

58

58 58

198

Computer

Calculations

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .2A-1 .1 .3A-1 .1A2 .1A-1 .3 The simplified presentation for this group has 4 generators and 6 relators. The total length of the relators is 54 The abelian quotient of this group is SEQ( 2, 0, 0 ) The number of 0's here is 2 The simplified version of this group is called L2

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx A subgroup of the above group of index The group has 7 generators. The group has 12 relators. The total length of the relators is The group The group The total The group The group The total 96 1 substring The group The group The total

has 7 has 12 length of has 6 has 11 length of

generators. relators. the relators is generators. relators. the relators is

2

96

96 156

search has 6 generators. has 10 relators. length of the relators is

116

The group has 5 generators. The group has 9 relators. The total length of the relators is

142

is now found.

6.8 The Relations in G 2 (2)'

96 2 substring The group The group The total

search has 5 generators. has 9 relators. length of the relators is

118

The group The group The total 96 3 substring The group The group The total

has 4 generators. has 8 relators. length of the relators is

192

search has 4 generators. has 8 relators. length of the relators is

142

The group The group The total 96 4 substring The group The group The total

has 3 generators. has 7 relators. length of the relators is

256

search has 3 generators. has 7 relators. length of the relators is

244

SUBSTRING The group The group The total

SEARCH has 3 has 7 length of

number 1 generators. relators. the relators is

206

The The The The The The 206

group group total group group total 1

has 3 has 7 length of has 3 has 7 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 3 has 7 length of

number 2 generators. relators. the relators is

186

generators. The group has 3 relators. The group has 7 The total length of the relators is

186

206 206

199

200

Computer

Calculations

The group has 3 generators. The group has 7 relators. The total length of the relators is 186 1 SUBSTRING The group The group The total

SEARCH has 3 has 7 length of

number 3 generators. relators. the relators is

186

174

ORIGINAL GROUP: The group has 7 generators. The group has 12 relators. The total length of the relators is

96

SIMPLIFIED GROUP The group has 3 generators. The group has 7 relators. The total length of the relators is

174

TIME:

99

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .1 .2A-1 .4A-1 .3A2 •3A-1 .2A-1 .3 The simplified presentation for this group has 3 generators and 7 relators. The total length of the relators is 174 The abelian quotient of this group is SEQ( 2, 0, 0 ) The number of 0's here is 2 The simplified version of this group is called

201

6.8 The Relations in G 2 (2)' L3

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx A subgroup of the above group of index The group has 4 generators. The group has 12 relators. The total length of the relators is The The The The The The 148

group group total group group total 1

has 4 has 7 length of has 4 has 7 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 4 has 7 length of

number 1 generators. relators. the relators is

The The The The The The 131

group group total group group total 1

has 4 has 7 length of has 4 has 7 length of

generators. relators. the relators is generators. relators. the relators is

SUBSTRING The group The group The total

SEARCH has 4 has 7 length of

number 2 generators. relators. the relators is

The The The The The The 130

has 4 has 7 length of has 4 has 7 length of

generators. relators. the relators is generators. relators. the relators is

group group total group group total 1

SUBSTRING SEARCH The group has 4

number 3 generators.

2

236

148 148

131

131 131

130

130 130

is now found.

202

Computer

Calculations

The group has 7 relators. The total length of the relators is

130

ORIGINAL GROUP: The group has 4 generators. The group has 12 relators. The total length of the relators is

236

SIMPLIFIED GROUP The group has 4 generators. The group has 7 relators. The total length of the relators is

130

TIME:

66

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .2A-1 .1 .3A-1 .1 .3 .1 .2A-1 .1 The simplified presentation for this group has 4 generators and 7 relators. The total length of the relators is 130 The abelian quotient of this group is SEQ( 3, 3, 0, 0 ) The number of 0's here is 2 The simplified version of this group is called L4

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx A subgroup of the above group of index The group has 10 generators. The group has 21 relators. The total length of the relators is

3

304

is now found.

6.8 The Relations in G 2 (2)'

The The The The The The 304

203

group group total group group total 1

has 10 generators. has 21 relators. length of the relators is has 9 generators. has 20 relators. length of the relators is

328

The group The group The total 304 2 substring The group The group The total

has 8 generators. has 19 relators. length of the relators is

602

search has 8 generators. has 19 relators. length of the relators is

508

The group The group The total 304 3 substring The group The group The total

has 7 generators. has 18 relators. length of the relators is

925

search has 7 generators. has 18 relators. length of the relators is

645

The group The group The total 304 4 substring The group The group The total

has 7 generators. has 18 relators. length of the relators is

645

search has 7 generators. has 18 relators. length of the relators is

645

The group The group The total 304 5 substring The group The group The total

has 7 generators. has 18 relators. length of the relators is

645

search has 7 generators. has 18 relators. length of the relators is

645

304

204

Computer

Calculations

SUBSTRING The group The group The total

SEARCH number 1 has 7 generators. has 18 relators. length of the relators is

The The The The The The 596

has 7 has 18 length of has 7 has 18 length of

group group total group group total 1

596

generators. relators. the relators is generators. relators. the relators is

596

SUBSTRING The group The group The total

number 2 SEARCH generators. has 7 relators. has 18 length of the relators is

523

The group The group The total The group The group The total 523 1 substring The group The group The total

has 7 has 18 length of has 6 has 17 length of

search has 6 generators. has 17 relators. length of the relators is

851

The group The group The total 523 2 substring The group The group The total

has 6 generators. has 17 relators. length of the relators is

851

search has 6 generators. has 17 relators. length of the relators is

851

The group The group The total 523 3 substring The group

has 6 generators. has 17 relators. length of the relators is

851

search has 6

generators. relators. the relators is generators. relators. the relators is

generators.

596

523 1069

205

6.8 The Relations in G 2 (2)'

The group has 17 relators. The total length of the relators is

851

SUBSTRING The group The group The total

SEARCH number 3 has 6 generators. has 17 relators. length of the relators is

758

ORIGINAL GROUP: The group has 10 generators. The group has 21 relators. The total length of the relators is

304

SIMPLIFIED GROUP The group has 6 generators. The group has 17 relators. The total length of the relators is

758

TIME:

274

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .1A-1 .4A-1 .3 .2 .2 .3 .3A-1 .1A-1 .3 .2A-2 .3 .3A-1 .4A-1 .3 The simplified presentation for this group has 6 generators and 17 relators. The total length of the relators is 758 The abelian quotient of this group is SEQ( 3, 3, 0, 0, 0, 0 ) The number of 0's here is 4 The simplified version of this group is called

206

Computer

Calculations

L5

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx A subgroup of the above group of index The group has 16 generators. The group has 51 relators. The total length of the relators is

3

2059

The group The group The total The group The group The total 2059 1 substring The group The group The total

has 16 generators. has 51 relators. length of the relators is has 15 generators. has 50 relators. length of the relators is search has 15 generators. has 50 relators. length of the relators is

2375

The group The group The total 2059 2 substring The group The group The total

has 14 generators. has 49 relators. length of the relators is

3454

search has 14 generators. has 49 relators. length of the relators is

2743

The group The group The total 2059 3 substring The group The group The total

has 13 generators. has 48 relators. length of the relators is

10235

search has 13 generators. has 48 relators. length of the relators is

7315

has 13 generators. has 48 relators. length of the relators is

7315

The group The group The total 2059 4 substring The group

search has 13

generators.

2059 3121

is now found.

6.8 The Relations in G 2 (2)'

The group has 48 relators. The total length of the relators is

207

7315

The group The group The total 2059 5 substring The group The group The total

has 13 generators. has 48 relators. length of the relators is

7315

search has 13 generators. has 48 relators. length of the relators is

7315

SUBSTRING The group The group The total

SEARCH number 1 has 13 generators. has 48 relators. length of the relators is

6852

The group The group The total The group The group The total 6852 1

has 13 generators. has 48 relators. length of the relators is has 13 generators. has 48 relators. length of the relators is

SUBSTRING The group The group The total

SEARCH number 2 has 13 generators. has 48 relators. length of the relators is

The group The group The total The group The group The total 6370 1

has 13 generators. has 48 relators. length of the relators is has 13 generators. has 48 relators. length of the relators is

SUBSTRING The group The group The total

SEARCH number 3 has 13 generators. has 48 relators. length of the relators is

6852 6852

6370

6370 6370

5360

208

Computer

Calculations

ORIGINAL GROUP: The group has 16 generators. The group has 51 relators. The total length of the relators is

2059

SIMPLIFIED GROUP The group has 13 generators. The group has 48 relators. The total length of the relators is

5360

TIME:

2954

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .1 .6 .3 .6A-1 .1A-1 .6A-1 • 2A-1 .3A-2 .6 .6A-1 .4A-1 .6A-1 .5A-1 .3A-1 .6A2 .3A-1 .1A-1 .3A-1 .2A-1 .1A3 .1 .4 .1 .5 .1 . 2 '-1 .3 .6 . 1 A - 1

.6 .6 .6 .6 .3 .3

The simplified presentation for this group has 13 generators and 48 relators. The total length of the relators is 5360 The abelian quotient of this group is SEQ( 3, 3, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0 , 0 The number of 0's here is 10

)

The simplified version of this group is called L6

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

6.8 The Relations in G2(2)'

A subgroup of the above group of index

209

3

The subgroup that is being considered is GROUP H IS A SUBGROUP GENERATORS AS WORDS IN SUPERGROUP : .1A-1 .6A-1 .1A-1 .8A-1 .1A-1 .9A-1 .1A-1 .11A-1 .1A-1 .12A-1 .1A-1 .13A-1 .1A-1 .2 .1A-1 .3 .1A-1 .7 .1A-1 .10 .1A-1 .4A-1 .1 .1A-1 .5A-1 .1 •6A-2 .1 .6A-1 .8A-1 .1 -6A-1 .9A-1 .1 .6A-1 .11A-1 .1 .6A-1 .12A-1 .1 .6A-1 .13A-1 .1 .6A-1 .2 .1 .6A-1 .3 .1 .6A-1 .7 .1 .6A-1 .10 .1 .1A3 .1 .2A-1 .1 .3A-1 .4 .5 .1 .6 .1 .7A-1 .1 .8 .1 .9 .1 .10A-1 .1 .11 .1 .12 .1 .13 The simplified presentation for this group

is now found.

Computer

210

Calculations

has 37 generators and 144 relators. The total length of the relators is

14668

The abelian quotient of this group is SEQ( 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ) The number of 0's here is 34 The simplified version of this group is called L7 The group L7 is now translated into G and called KER. THE ORDER OF G/KER IS 6048 The group IM2 is a faithful image of G/KER but has lower degree and so is easier to work with. The order of IM2 is 6048 COMPOSITION FACTORS OF GROUP IM2 2A(2, 3)

APPEARS ONCE

TOTAL TIME = 5248.850 CPU SECONDS. > > END OF RUN. 5249.033 SECONDS, 199996 WORDS USED.

6.8 The Relations in G 2 (2)'

211

library kerf72; print 'THIS PROGRAM FINDS THE KERNEL OF THE MAP FROM G'; PRINT 'ONTO U(3,3) = 2A(2,3) = G(2,2)$', THE SIMPLE GROUP'; PRINT 'OF ORDER 6048. IT CONSTRUCTS A CHAIN OF SUBGROUPS'; PRINT ' G > LI > L2 > L3 > L4 > L5 > L6 > L7' ; PRINT 'OF INDICES 28,2,2,2,3,3,3. THE GENERATORS OF L7 ARE '; PRINT 'REWRITTEN IN TERMS OF THE GENERATORS OF G AND THE '; PRINT 'RESULTING SUBGROUP OF G IS CALLED KER. THEN G/KER IS'; PRINT 'COMPUTED AND SHOWN TO BE THE SIMPLE GROUP OF ORDER 6048.'; PRINT ''; PRINT 'Thomas Fournelle'; print 'Department of Mathematics'; print 'University of Wisconsin-Parkside'; print 'Kenosha, WI 53141'; print 'U.S.A.'; print ''; print 'fournel1gvacs.uwp.edu'; print '' ; print ''; n=7; p=2; g:free(a,x); g.relations:xAp=aAn=(x,xAa)=l,(x,xA(aA2))=xAa; print g; s=low index subgroups(g,,28;limit=l); f,u33,ker=cosact homomorphism(g,s[2]); h=s[2]; k=rewrite(g,h;simplify=false); sim(k,false,50,2;11) ; info(s[2],ll); print 'LI'; s=low index subgroups(11,0,seq(2);pr=l); hl=s[4]; k=rewrite(ll,hl;simplify=false); sim(k,false,50,2;12) ; info(s[4],12); print 'L2'; s=low index subgroups(12,0,seq(2);pr=l); h2=s[2]; k=rewrite(12,h2;simplify=false); sim(k,false,50,2;13); info(s[2],13);

212

Computer

Calculations

print 'L3'; s=low index subgroups(13,0,seq(2);pr=l); h3=s[6]; k=rewrite(13,h3;simplify=false); sim(k,false,50,2;14); info(s[6],14); print 'L4'; s=low index subgroups(14,0,seq(3);pr=l,limit=14); h4=s[13]; k=rewrite(14,h4;simplify=false); sim(k,false,50,2;15); info(s[13],15); print 'L5'; gen=[15.1A3,15.1*15.4,15.1*15.5,15.1*15.2^-1,15.3*15.6*15.1A-1]; s=low index subgroups(15,,seq(3);pr=l,limit=2); h5=s[2]; k=rewrite(15,h5;simplify=false); sira(k,false,50,2;16); info(s[2],16); print 'L6'; gen=[16.1A3,16.1*16.2^-1,16.1*16.3^-1,16.4,16.5,16.1*16.6, 16.1*16.7A-l,16.1*16.8,16.1*16.9,16.1*16.10A-l, 16.1*16.11,16.1*16.12,16.1*16.13]; s=low index

subgroups(16,,seq(3);pr=l,limit=l);

h6=s[2]; 17=rewrite(16,h6;simplify=false); info(s[2],17); print 'L7'; goback(15,16,17;gp); goback(14,15,gp;gp); goback(13,14,gp;gp); goback(12,13,gp;gp); goback(11,12,gp;gp); goback(g,ll,gp;ker); line; line; print 'The group L7 is now translated into G and called KER.'; f 2 , im,kkk=cosact homomorphisiti(g,ker) ; print 'THE ORDER OF G/KER IS',order(im);

6.8 The Relations in G 2 (2)'

213

gen=null; for each xx in generators(h) do gen=gen join [f2(xx)]; end; im2=cosact image(im,); print ''; print 'The group IM2 is a faithful image of G/KER but has lower'; print 'degree and so is easier to work with.'; print 'The order of IM2 is',order(im2); print compos factors(im2); print ''; print ''; show time; "THE FOLLOWING PROCEDURES ARE USED IN THIS PROGRAM."; procedure sim(g,bool,k,kk;h); t0= cputime(O); inf(g;m,n,lll); if bool eq false then h=g; goto l;end; s=low index subgroups(g,,l); h=rewrite(g,s[l];simplify=false); l:for j=l to 3 do print ' '; h=tietze(h;eliminate); inf(h;mm,nn,ll); i=l; while i le mm do h=tietze(h;eliminate,gen); inf(h;m,n,l); lll=min(111,1); print 111,1; i f 1 g t min(2*11,2*111) or 1 g t m i n ( l l l + k , l l + k ) or ( i mod kk) eq 0

p r i n t 'substring search'; h=tietze(h;search); inf(h;m,n,1111); lll=min(ll,min(lll,llll)); end; lll=min(l,min(lll,ll));

if m eq mm and n eq nn and 1 eq 11 then i = mm+l;end; i=i+l; mm=m;nn=n;11=1; print ' '; end; print 'SUBSTRING SEARCH number',j;

214

Computer

Calculations

h=tietze(h;search); h=tietze(h;newgen); inf(h;m,n,l); 111=1; end; hl=tietze(h;search,equal); hl=tietze(hi;search); hl=tietze(hl;newgen); hl=tietze(hi;search); hl=tietze(h;auto); procedure line; print '================== end; print ''; print " ; line; print 'ORIGINAL GROUP:'; inf(g;a,b,c); line; h=hl ; print 'SIMPLIFIED GROUP'; inf(h;a,b,c); line; print 'TIME:',(cputiroe(0)-t0)/1000; end; procedure inf(h;m,n,l); print 'The group has',ngens(h),'generators.'; print 'The group has' ,card(relations(h)), 'relators.' ; 1=0; for each r in relations(h) do l=l+length(r); end; print 'The total length of the relators is',1; m=ngens(h); n=card(relations(h)); end; procedure info(h,l);

6.8 The Relations in G 2 (2)'

line; print 'The subgroup that is being considered is'; print h; print 'The simplified presentation for this group'; print 'has',ngens(1),'generators'; print 'and',card(relations(l)),'relators.'; 11=0; for each r in relations(l) do ll=ll+length(r); end; print 'The total length of the relators is',11; print ''; print 'The abelian quotient of this group is'; print aqinvariants (1); print ''; print 'The simplified version of this group is called'; end;

procedure goback(g,h,1;gp); hgens=generating words(g,h); lgens=generating words(h,1); define f:h to g by images setseq(hgens); gens=null; for each x in lgens do gens=gens join [f(x)]; end; gp=; end; finish;

215

Computer

216

CAYLEY

V3.8.3

Calculations

(ULTRIX/UNIX)

The group N is: GROUP N OF ORDER 32 = 2 A 5 RELATORS : AA4 BA4 CM DA2 B A 2 A A -2 C A 2 A A -2 DA-1 C D C B A A -1 BA-1 A A -1 A A -1 CA-1 A C A A -1 DA-1 A D BA-1 CA-1 B C BA-1 DA-1 B D The automorphism group of N has order 1920 We will now search for automorphisms of order 5 which satisfy the conditions of our counter example: The order of the automorphism which we found is: 5 We now check that it acts in the proper fashion in N: TRUE TRUE TRUE TRUE

6.8 The Relations in Gi{2)'

We now look for automorphisms which send c to a d. We print the orders of some of these since they give orders of constant groups for verbal embeddings in to N. 5 4 3 4 3 3 2 4 5 4 2 5 2 2 4 END OF RUN. 46.168 SECONDS

217

218

Computer

Calculations

library counter; print 'The group N is:'; n=free(a,b,c,d); A A n.relations:a 4,b 4,cA4,dA2, A A A aA 2=bA 2=c 2, c d=c (-l), a*b=b*aA(-l), (a,c),(a,d),(b,c),(b,d); o=order(n); print n; aut=automorphism group(n); print 'The automorphism group of N '; print 'has order',order(aut); print 'We will now search for automorphisms o f ; print 'order 5 which satisfy the conditions'; print 'of our counter example:'; for each f in if f(a) eg f(b) eg f(c) eg f(d) eg break; end; end;

aut do a*b*b*c*d and c and a and c*b then

print print print print print print print print

'The order of the automorphism'; 'which we found is:',order(f); 'We now check that it acts in'; 'the proper fashion in N:'; f(a) eg a*b*b*c*d; f(b) eg c; f(c) eg a; f(d)eg c*b;

print print print print print print

'We now look for automorphisms'; 'which send c to a d. We '; 'print the orders of some of'; 'these since they give orders of '; 'constant groups for verbal'; 'embeddings in to N.';

6.8 The Relations in G 2 (2)'

for each f in aut do if f(c) eq a*d and (order(f) eq 2 or order(f) eq 3 or order(f) eq 4 or order(f) eq 5) then print order(f); end; end; finish;

219

Chapter 7 QUESTIONS This chapter is a list of questions that arise from the results in this book. The questions are not stated in any particular order. 1. (The Isomorphism Problem) Suppose that Ti and T2 are two linkage graphs and that L(R,Ti) and L(R, r 2 ) are isomorphic for some rings R. Then are Ti and r 2 isomorphic as linkage graphs? We have already seen that in some linkage graphs some of the arrows can have their orientations changed and the linkage groups will remain the same (for commutative rings). If we replace " for some rings i?" by "for all rings i?" then are r x and r 2 isomorphic? Linkage groups are generalized matrix groups and the linkage graph is rather like the dimension of the group. For example, SL(n,R) and SL(m, R) are not expected to be isomorphic unless m = n. This is perhaps a hard problem. Can one say that if L(R, Ti) and L(R, r 2 ) are isomorphic for some rings R, then Ti and r 2 at least have the same number of vertices? It seems that something like this ought to be true. 2. In describing Standard Linkage graphs in Section 3.6 we took II n , n > 7 as a basic graph. We did not let n = 7 because for n — 7 we do not know if torsion elements of L(R, II7) are conjugate to elements in a basic linkage. What can be said about torsion in L(R, II7)? 3. What do the solvable subgroups of L(R, T) look like when T is a standard linkage graph? (See Section 3.6) The torsion in L(R, V) comes from basic linkages, that is, any torsion element is conjugate to an element in a

221

222

Questions

basic linkage. Are solvable subgroups either infinite cyclic or conjugate to subgroups of basic linkages? We can also ask this same question about L(R, II7) (See the previous question.) There is an infinite non-cyclic solvable subgroup of £(Z 2 , II7) but this group arises only because of the especially simple nature of the ring Z 2 . The questions being asked here are really assuming that R has more than 2 elements. 4. (Tits' Alternative) If T is a standard linkage graph, in particular, if V has no hexagon subgraphs, then can it be said that every subgroup of L(R, T) is solvable-by-finite or else contains a free subgroup rank 2? Recall that in Chapter 3 we saw that subgroups of L(R, T) of finite index contain free subgroups of rank 2. 5. From the theory of verbal embeddings we know that many finite simple groups are images of linkage groups. For a particular such group what does the linkage graph look like? By Theorem 2.2.1 and Theorem 2.2.4 if there is a verbal embedding with constant group C into a simple group G, then for every H such that C < H < G there is a linkage group L{R,YH) that maps onto G. Changing H changes the linkage graph TH. In describing the group L(R,TH) is it better to have Has large as possible, namely equal to G, or as small as possible? The mapping of £(Z 2 , II7) onto the group U3 (3) is an illustrative example. The linkage graph is quite simple in structure but the kernel of the mapping is very complex. Would the kernel be easier to describe if II7 were replace by some larger linkage graph? If the kernels are described then this gives a presentation for the image group. Are any of these presentations obtainable in a nice way? Are any of them useful? For example, the Heptagonal description of the groups G 2 (F) yields immediately the existence of an automorphism of order 7. 6. Most of the finite simple described in the book have verbal embeddings of the ring Z 2 . Some have embeddings of some larger rings. How much do the linkage graphs corresponding to the larger rings differ? Are the kernels described in the previous sections easier or harder to describe when the embedded rings are larger?

Questions

223

7. The description of the Lie algebra G2 and the Lie groups G2(F) imply the existence of automorphisms of order 7 in both the group and the algebra. What has 7 got to do with G-p. When experts on Lie algebras are asked this question they, mention a 7 dimensional subalgebra of the Octonians which is naturally related to Gi. This however does not appear to be the answer to the question. 8. Linkage Lie algebras can be constructed by generators and relations and are mostly infinite dimensional. Are these algebras related to Kac-Moody or other infinite dimensional Lie algebras, or are they a new class of infinite dimensional Lie algebras? Even if they are the same as some other algebras they may not be the same in any nice way. For example the Heptagonal Lie algebra L(R, 1T7) (with appropriate extra relations) is isomorphic to G2{R) but the isomorphism is not in any way natural. The information you get from one description is almost completely different from the information you get from the other description. In any event, is the information you get from describing a Lie algebra as a linkage Lie algebra useful in describing new algebras or in redescribing old ones? 9. Linkage groups have many homomorphic images among finite simple groups. What other interesting images do they have? What are the finite simple images of the standard linkage groups of Section 3.6? Are there non-trivial finite normal subgroups of these groups (something that cannot happen for r = II n for n > 7)? What can be said about the infinite simple images of these groups? 10. Suppose that a verbal embedding has a cyclic constant groups C and that C acts regularly on the associated linkage graph. Then is the encoding word of the form

[x\ y a -1]

0 lU -* + this i\,',C n that where a is a generator of C?' Recall need not be the case if the constant group does not act regularly on the linkage graph.

11. In describing the linkage Lie algebras of Chapter 5 we did not define an exponential map which is needed to construct the corresponding linkage Lie group. Can this exponential map be defined?

224

Questions

To define the exponential map one would need to have nilpotence or local nilpotence of ad(i) for every vertex i. This requires adding more relations to the linkage Lie algebra. If these relations are added it is then necessary to show that the algebra (and the group) do no collapse. Suppose that it is possible to define the exponential map for a linkage Lie algebra C over a special linkage graph T for a ring R. If Q is the Lie group corresponding C then Q is the homomorphic image of L(R, T). What is the kernel of this homomorphism? Are these groups isomorphic? The answer here would of course depend on the nature of the nilpotence conditions added to C. 12. In [23] Sidki computed matrix representations of many groups over GL(2, K) and PGL(2, K) for various fields K. The work involved taking matrices with unknowns for entries (one matrix for each generator of the group) and setting up matrix equations corresponding to the relations of the group. If the entries of the matrices can be determined in such a way as a to satisfy the equations then one obtains a matrix representation for the group. This is, of course, not a new idea. A new feature in Sidki's work was the use of a computer algebra system (in his case MAPLE) to deal with the difficult algebraic manipulations. It would probably be useful to investigate more representations (using MAPLE or Mathematica, for example). These computer algebra systems could possibly yield information about 3 x 3 representations which are very complicated to study by hand. The 3 x 3 representations would probably be more fruitful than 2 x 2 representations since verbal embeddings arise more naturally in the 3 x 3 case.

References 1. G. BERGMAN , Embedding arbitrary algebras into groups, Algebra Universalis, 25 ( 1988), 107-120. 2. R . W . C A R T E R , "Simple Groups of Lie Type," John Wiley and Sons, London, New York, 1972. 3. L. P . COMERFORD, J R . , M. F . NEWMAN, Verbal definitions of operation on groups, Algebra Universalis, 28 (1991), 495-499. 4. J. H. CONWAY et al, "Atlas of Finite Groups," Oxford Univ. Press (Claren­ don), London, New York, 1985. 5. W . F E I T , J . G. THOMPSON, Solvability of groups of odd order, Pacific J. Math., 13 (1963), 775-1029. 6. J . FAULKNER, "Groups with Steinberg Relations and Coordinizations of Polygonal Geometries," Memoirs A.M.S. 10, Number 85, 1977. 7. T . A. FOURNELLE, K. W . W E S T O N , Verbal embeddings and a geometric approach to some group presentations, J. Algebra 124 (1989), 300-316. 8. T . A. FOURNELLE, K. W . W E S T O N , A generalization of Malcev's corre­ spondence between rings and groups and a class of Lie groups, Contributions to Contemporary Mathematics (A.M.S.) 131 (1992), 113-122. 9. T . A. FOURNELLE, K. W . W E S T O N , A geometric approach to some group presentations, Combinatorial Group Theory, Conference at the University of Maryland, April 1988, Contributions to Contemporary Mathematics, (A.M.S.) 109 (1990), 25-33. 10. T . A. FOURNELLE, S. SIDKI, K. W . W E S T O N , On algebraic embeddings of

rings into groups, Arch. Math. (Basel), 51 (1988), 425-433. 11. T . A. FOURNELLE, S. SIDKI, K . W . W E S T O N , Cyclically symmetric presen­

tations for SL3 and G2, Arch. Math (Basel), 61 (1993), 409-420. 12. J . H. HUMPHREYS, "Introducton to Lie Algebras and Representation The­ ory," Springer-Verlag, New York, 1972. 13. D GORENSTEIN, "Finite Simple Groups," Plenum Press, New York, 1982. 14. I. GROSSMAN, W . M A G N U S , "Groups and Their Graphs," New Mathematical Library, Number 14, Mathematical Association of America, 1964. 15. H. L A U S C H , W . NOBAUER, "Algebra of Polynomials," North Holland Pub­ lishing Co., Amsterdam-New York-Oxford, 1973. 16. W . M A G N U S , A. K A R A S S , D. SOLITAR, "Combinatorial Group Theory,"

Dover Publications, Nff©dj&jrf?giJ#g6i 225

226

References

17. A. I. MALCEV, A correspondence between rings and groups, Amer. Soc. Transl. (2) 45 (1960), 221-231.

Math.

18. W. D. MAURER, J. L. RHODES, A property of finite non-abelian groups, Proc. Amer. Math. Soc, 16 (1965), 552-554. 19.

J. MILNOR, "Introduction to Algebraic K-Theory," Annals of Mathematics Studies, Number 72, Princeton University Press, Princeton, N.J., 1971.

20. D. J. S. ROBINSON, "Generalized Soluble Groups," Volumes 1 and 2, Springer-Verlag, New York 1972. 21. D. J. S. ROBINSON, "A Course In The Theory of Groups," Graduate Texts in Mathematics, Number 80, Springer-Verlag, New York, 1980. 22. W. R. S C O T T , "Group Theory," Prentice-Hall, Englewood Cliffs, N. J., 1964. 23. S. SiDKi, Solving certain group equations in PGL(2,k) - a computational approach, Matemdtica Contemporanea, (Sociedade Brasileira de Matemdtica) 7 (1994), 59-70. 24. R. STEINBERG, Lectures on Chevalley Groups, Yale University Notes, 1967. 25. R. STEINBERG, Generateurs, relations et revetements de groupes algebriques, Colloq. Theorie des groupes algebriques, Bruxelles 1962, 113-127. 26. S. T H O M A S , Classification Theory of Simple Locally Finite Groups, Doctoral Thesis, Bedford College, University of London, 1983. 27. K. W . W E S T O N , An equivalence between non-associative ring theory and the theory of a special class of groups, Proceedings Amer. Math. Soc., 6 (1968), 11356-1362. 28. K. W . W E S T O N , On nilpotent class 2 groups and the Steinberg groups ST(3,R),Arch. Math. 45 (1985), 207-210.

Index A

E

Adjoint mapping, 98 Alternating group general verbal embedding in, 16 of degree five, 11, 16

Embedded at each vertex, 23, 154 Encoding polynomial of a general verbal embedding, 13 Encoding word, 1 minimal, 18 Exponential map, 98 can be defined, 98

B Basic linkages, 21 Bergman, George, 15 Bracket operator, 96

F

c Campbell-Baker-Hausdorff formula, 99 Cartan matrices, 109 CAYLEY, 6, 11, 42, 155, 176, 183, 192 Center of a linkage, 21 Chevalley group, 115 Chevalley groups twisted, 115 Collection algorithm, 72 Collection process, 72 Comerford, Leo, 16, 64-65 Commutator part, 145 Constant group, 1 cyclic 21 as, 43 cyclic 7 as, 6, 32 free group as, 32, 37 S3 as, 32

227

Finite simple groups polynomial completeness of, 14 First factor, 74 Fixed-point-free action, 44 Fournelle, Thomas, 1, 6, 158 Free products of linkage graphs, 70 Free products with amalgamation, 4, 64 Freely reduced, 71

G G2 verbal embedding in, 7 Gates, 74 General verbal embedding definition of, 13 encoding polynomial, 13 Graph groups, 25 Group commutator, 1 Group conjugate, 1

228

H Hall-Witt identity, 35 Heptagonal Lie algebra, 117, 154 Hexagonal Lie algebra, 98, 108 Hexagonal linkage graph, 23

I Infinite dimensional Lie algebra, 154 Isomorphism problem, 38, 221

J Jacobi identity, 95 Janko's first group, 11 special verbal embedding in, 17

L Lagrange polynomial, 17 Last factor, 75 Lewin, Jacques, 25 Lichtman, Alexander, 4 Lie algebra, 95 Lie bracket, 95 Lie group associated to a Lie algebra, 101 Lie ring, 95 Linkage algebra special, 144 Linkage graph, 9 anti-automorphism, 25 anti-isomorphim, 25 antihomomorphism, 25 automorphism, 25 definition of, 21 homomorphism, 25 isomorphism, 25 n-chain, 39 polygonal, 39 special, 144

Index

standard, 85 Linkage group, 11 Linkage group and simple groups, 33 and Steinberg groups, 33 and verbal embeddings, 29 center of, 83 constructing, 63 definition of, 22 free subgroups in, 82 special, 144 standard, 85 torsion elements in, 83 Linkage Lie algebra, 95 Linkage Lie group, 101 Linkage relations, 96 Linkage subgraph, 39 Linkages, 21 Locally nilpotent transformation, 98

M Main part, 145 Malcev, 1 Mathematica, 155-156, 165, 172 Mathieu groups verbal embeddings in, 189 Minimal encoding word special role of, 18 Moves in front of, 72, 146 Moves to the first spot, 146

N n-chain graph, 39 n-fold commutator, 78 Newman, Michael, 16, 42 Nielsen-Schreier Theorem, 82 Nilpotent transformation, 98 Normal Form Theorem, 80

Index

Not freely reduced, 71

o Odd order paper, 15

P Pentagonal algebra, 120 Pentagonal group, 120 Perfect groups and verbal embeddings, 7 Polygonal linkage graph, 39 Polynomial over a group, 14 Polynomially complete, 14, 17 PSL(3,F) verbal embedding in, 33

R React with each other, 71, 143 Reduced, 72 Reduced word, 72 Reduction process, 146 Regular action, 44 Replacement moves, 72 Residually solvable group, 14 and general verbal embeddings, 14 Roots, 115

s Serre's relations, 111 Serre's Theorem, 112 Sidki, Said, 1, 35, 224 Simple groups and linkage groups, 33 and verbal embeddings, 33 sporadic, 47 Solvable groups and verbal embeddings, 8

229

Special linear group verbal embedding in, 2 Special linkage algbera, 144 Special linkage graph, 144 Special linkage group, 144 Special verbal embedding in A5, 16 definition of, 13 in J l , 17 Sporadic simple groups, 46 Standard epimorphism, 70 Standard isomorphism, 70 Standard linkage graph, 85 Standard linkage group, 85 Steinberg group, 8, 21, 23, 45 Steinberg group generalization, 3 action of symmetric group, 3 and linkage groups, 33 definition of, 2 linkage graph of, 34 verbal embedding in, 2 Symmetric group acting on Steinberg group, 3

T Thomas, Simon, 18 Tits' alternative, 222 Transformation moves, 144 Twisted Chevalley groups, 115

u Unitary group verbal embedding in, 6, 32 Universal algebra, 13 Upper unitriangular matrices, 9, 12, 24

230

V Verbal embedding, 1 Verbal embedding and perfect groups, 6 and linkage groups, 29 and simple groups, 33 general, 13 in G2, 7 in SL(3,F), 2 in sporadic groups, 47 in Steinberg groups, 2 in Unitary group, 6 into PSL(3,F), 33 special, 13 Z in GL(2,C), 18 Z in PSL(2,C), 18 Vertex subgroup, 23, 63 Vertices reacting with each other, 71

w Weston, Kenneth, 1, 35

Index