THE ADJOINT REPRESENTATION OF GROUP ALGEBRAS AND ENVELOPING ALGEBRAS

THE ADJOINT REPRESENTATION OF GROUP ALGEBRAS AND ENVELOPING ALGEBRAS

that Z(G) 9~0 .As we will see, the converse of this statement is definitely not true.Nevertheless, be begin by proving that Z(G) is controlled by the f.c .center of G; namely 0=0(G)={xEG1 iG :CG(x)j<oo} .
Theorem 1 .1 .If Z(G) is the kernel of the adjoint representation of K[G], then In particular, if A(G) = 1, then Z(G) = 0 .
To proceed further we require a better understanding of Z(G) n K[A] .
As we see below, there is an easy description of this ideal, and of Z(G) for that matter, based on the fact that the adjoint representation is a permutation representation .Suppose G permutes a set 9 and consider the K-vector space V = KQ having 52 as a basis .Then V is naturally a left K[G]-module, the permutation module associated with (G, S2) .Furthermore, if x E 2, then the annihilator of x in K[G] is clearly equal to K[G]W(Gx ) where Gx is the stabilizer of x in G and where w(G) denotes the augmentation ideal of K [Gx] .In particular, since the adjoint representation of K[G] corresponds to the conjugation permutation module with 52 = G and with Gx = CG(x), the preceding considerations applied to G and to 0 yield EEG It is a simple consequence of this result that if F is a field extension of K, then Z(F[G]) = F ® Z(K [G]).Thus the vanishing of Z(G) depends only upon G and the characteristic of the field .Solee interesting examples are as follows .
Lemma 1 .3 .i. Suppose G has a nonidentity no7mal torsion-free abelian subgroup A with I G/Al < oo and Cc (A) = A .Then I(G) = 0 if and only if some element of G/A acts without fieed points on A .
ii .Let p and cl be primes with p 1 q -1 and form the group G = Q A P. Here P is an elementary abelian p-group of order p2 , Q is an elementary abelian q-group of order qn+r and each subgroitp of P of order p is the kernel of the action of P on a cyclic facto7-of Q .Then 7(G) = 1, but I(G) :~0.Proof.. (i) Since A is selfcentralizing, CIA acts faithfully on A. Flrrthermore, Since CIA is finite and A is torsion free, it follows that A = ,~,(G) and that K[A] is a commutative domain .Now the preceding lernma irnplies that gEGIA In particular, if some CA(g) = 1; then I(G) n K[0] = 0 and I(G) = 0 by Theorem 1 .1 .On the other hand, if al] CA(g) =A 1, then each of the finitely many ideals K[A]w(CA(g)) is nonzero .Thus, Since K[A] is a dornain, it follows that I(G) n K[0] 7~0 and hence that I(G) 0 0.
(ii) Write Q = Qo x Qr x . . .x Qn where each Qi is cyclic of order q.Since p 1 q -1 and P has precisely p + 1 subgroups of order p, say Po, Pr , . . ., P. we can certainly define the action of P on Q so that the cyclic group P/Pz acts faithfully on Qi .It then follows easily that for each g E G = Q m P, there exists a subscript i with CC(g) _? Q2 .Thus the preceding lernma irnplies that z(G) 2 11Z' o w(Qj) 5 E 0. Since 7(G) = 1, the result follows.
There are of course numerous variants of the construction in (ii) .Lernma 1.2 also yields a second characterization of Z(G) .Proof.For any x E G we have I(G) C_ K[G]w(CG(x)) and hence 7r x (Z(G)) C w(CG(x)) .Conversely, suppose that I is a two-sided ideal of K[G] with irx (I) C_ w(CG(x)) for all x E G. Since I is a left ideal, it follows from [P,Lemma 1.1 .3]that I C_ K[G]w(Cc(x)) for all such x and therefore that I C_ I(G) by Lemma 1 .2.Finally, this description of I(G) is right-left symmetric, and thus we conclude that I(G) is also the kernel of the right adjoint representation .Now Theorem 1 .1 reduces the study ofI(G) to K[0] and we know that 0(G) contains all the finite normal subgroups of G. Thus an appropriate first step in studying K [,] is to consider the case where G is finite.
We assume throughout the remainder of this section that G is a finite group.If H is a subgroup of G, define Notice that gHg-i = gHg -1 and that if Hl C H2, then Hl divides H2.Furthermore, let A(G) denote the two-sided ideal of K[G] generated by the elements Cc(x) for all x E G. Finally, since A(G) is sylnmetrically defined, we can use the right-left symmetry of I(G) given by the previous lernma to reverse the roles of the right and left annihilators in the above formulas.Thus and similarly as required .Alternately, we can use the fact that K[G] is a symmetric algebra and that in such algebras right and left annihilators of two-sided ideals are equal.
For conveniente, we say that a subset X of G "spans 1" if the element 1 is contained in the two-sided ideal of K[G] generated by the elements CC(x) for all x E X. Obviously, if some such subset spans 1, then 1 E A(G) and I(G) = 0. Our next examples are Frobenius groups .Lemma 1 .6 .i. Let K be a field of characteristic p > 2 and let G be the diliedral grovp of order 2p.Then I(G) 7~0 even though 7L(G) = 1 .
ii.Let K be a field of characteristic 0 and let G = NC be a Frobenius group with kernel N and complernent C. If 1 :7~x E N and 1 0 y E C, then { x, y } sparcs 1 and hence I(G) = 0.
Proof.(i) Let .P be the subgroup of G of order p and let Cl , C2, . . ., Cr be the subgroups of G of order 2 .Then these subgroups and G itself are the centralizers of elements of G. Now let A: K[G] -> K be the K-hornomorphism defined by A(g) = 1 if g E P and A(g) _ -1 if g E G \ P .Since K has characteristic p, it follows that P, Ci and G are all contained in the kernel of A. Thus A(G) q¿ K[G] and I(G) yÉ 0 .Indeed, 0 7~G -275 E I(G) since this element is central and annihilates all generators of A(G) .
(ii) Basic properties of G imply that _ _ CC(x) C N and CC(y) C C. Therefore N, G, and gCg -1 are all contained in the ideal I of K[G] generated by OG(x) and OC(y) .Finally, if ¡NI = n, then we know that G is the "disjoint" union of N and of the n conjugates C,, C2, . ., C,, of C. Thus and, since K has characteristic 0, we conclude that 1. = 1 E I.
Since the dihedral group of order 2p with p > 2 is a Frobenius group, the preceding lemma shows that the vanishing of 7(G) does indeed depend on the characteristic of the field .For the remainder of this section we will asume that K has characteristic 0. For the most part, we will be concerned with the symmetric groups Sym, .To start with, we have Lernma 1.7.Let K be a field of characteristic 0, let G = Sym,, and let p be an odd prime.
i. Suppose n >_ p, let P be the cyclic subgroup of G generated by a p-cycle and let Q be the czlclic .subgroup of G generated by a ii .If n = p, then the set consisting of any p-cycle and any (p -1)cycle spans 1 .iii.If n = 4; then the set consisting of any nonidentity 2-element and any non-identity 0-element spans 1.
Proof.(i)(ii) Assume that n >_ p.Let F = GF(p) and consider the usual group H of linear functions on F described by ( F--+ a( + b for all a; b E F with a 7É 0 .Then H = NC is a Frobenius group with kernel N of order p and complernent C of order p-1 .Since H faithfully permutes the elements of F, we can view H as being embedded in Sym1, C Sym,, = G.In this embedding, N is generated by a p-cycle x and C is generated by a (p-1)-cycle y.Hence, since Gii(x) = N and GH(Y) = C, Lernma 1 .6(ii)implies that and (i) is proved .Finally, if n = p, then P is the centralizar in G of a p-cycle and Q is the centralizar in G of a (p -1)-cycle .Thus (ii) follows .
(iii) In G = Sym4, let A = Alto and let N denote the Klein four subgroup.Furthermore let Dl ; D2 and D3 be the Sylow 2-subgroups of G and let Cl, C2, C3 and C4 be the Sylow 3-subgroups .Notica that Di n DI = N and Ci n c,= 1 for all appropriate i -7~j .Thus since all elements of G are either 2-elements or 3-elements, we have Di C;=G+2Ñ+41 Similarly, by considering the alternating group A, we obtain =A+41.
Thus, adding twice the second equation to the first and canceling the N terms yields ~I-i+3~U, =G+2A+1.21 .i Finally, if x is a nonidentity 2-elernent of G and if y is a nonidentity 3-element, then CG(x) C_ Di and CC, (y) = Cj for some i, j.Thus the ideal I of K[G] generated by _ Cc (x) and Cc(y) contains Di and all its conjugatesas well as C; and all its conjugates.Furthermore, since Cj C A C G, we sea that I contains A and G .Thus 121 E I and, since K has characteristic 0, we conclude that 1 = 1 E I.
It is obvious that I(Sym1) = 0 and that I(Sym2) = w(Sym2) :,Z 0. Our goal is te show that I(Symj = 0 for all n > 3 and tire proof will proceed by inductivn en n.The following lemma shows how the inductiva llypothesis is usad .Part (i) is an unpleasant teclmical formulation which is needed te handle a few small cases .
Lemma 1 .8 .Let G = Syrn, let C be the cyclic subgroup of G gen- erated by a k-cycle c and let H = Symn_k .
i. Suppose that X is the set of elements of H which do not Nave a k-cycle in their cycle decomposition and assume that X spans 1 Proof. .(i) Consider the usual embedding of Symk x Symn _k in G = Sym,, and suppose that c E Symk and that H = Sym _k .If x E X C H, then CG(cx) = C x CII(x) since c is a k-cycle and since, by hypothesis, x has no k-cycle in its decomposition .Thus, since cx E Y, we sea that CCII Furthermore, since C commutes with K[H], it follows that I contains by assumption, and therefore C E I C A(G) as required .
Furthermore, since k > n/2, we have k > n -k and therefore X consists of the elements of H which do not have a k-cycle in their cycle decomposition .The result now follows from (i) .
Next we consider two special cases with small values of n.
i .If n = 6 and if X is the set of all elements of G which are not 6-cycles, then X spans 1 .
Proof.: We start with a general observation .Let a: G + { fl } be the natural group homomorphism determined by the parity map o, and extend this to a K-algebra homomorphism u: K To this end, let I denote the left-hand ideal and set W = {g E G 1 go ,(g) E I}-Then W is easily seen to be a normal subgroup of G and; since t E W, we have W = G and hence clearly I = Ker(o,) .Define T to be the subgroup of G of order 2 generated by t.
(i) Let n = 6 and continue with the above notation.Furthermore, let Y be the set of elements of G which contain precisely one 2-cycle in their decomposition .If H = Sym4 ; then we know from Len-una 1.7(iii) that the set X. C_ H consisting of a 4-cycle and a 3-cycle spans 1 .Thus since t is a 2-cycle, Lemma 1 .8(i)implies that and, by the observation of the first paragraph, the preceding ideal cont ains Ker(o,) and has codimensi on at most 1 .But if y = (12)(3 4 5 6) E Y, then CG(y) C Alts and hence CC(y) 1 Ker(o,) .With this, we conclude that Y spans 1 and, since Y contains no 6-cycle, this part is proved .
(ii) Similarly, let n = 9 and set H = Sym7.Then, by Lemma 1.7(ii), the set X C_ H consisting of a 7-cycle and a 6-cycle spans 1.Again, Lemma 1.8(i) implies that 1 + t = T E A(G) and therefore A(G) 2_ Ker(Q) .But if y E G is a 9-cycle, then CC(y) C Alt9 and therefore CG(y) 0 Ker(a) .As before, it follows that A(G) = K [G] .
We now use Bertrand's Postulate to handle the remaining symmetric groups .
Theorem 1 .10 .If K has character°istic 0, then Z(Symj = 0 for all n>3 . Proof.Write G = Syrn,, and proceed by induction on n >_ 3 .In view of Lernma 1.5, our goal is to show that A(G) = K[G] .If n is a prime, then this is certainly the case by Lernma 1 .7(ii)and, in particular, this starts the induction wheu 7i = 3. VVe can now assume that n is not a prime and that n > 4.
Suppose that there exists a prime number p with n/2 < p < n and with p 7~n -1, n -2, (rc + 1)/2 or (n + 2)/2 .In this case, since n = 'Z p by assumption, we sea that if k. = p or p -1, then n > nk >_ 3 and furthermore k > n/2 .By induction we know that 1. E A(Sym,_ti)and therefore Lermna 1..8(ii) iniplies that A(G) contains _ and _ where P is generated by a p-cycle and Q is generated by a (p -1)-cycle .Since p > n/2 is odd, we conclude from Lernma 1.7(i) that A(G) = K[G] and trence that Z(G) = 0. I:t remains to show that such a prime p exists except for a few small values of n .For any positive real nurriber (, let r«) denote the number of primes less than or edual to ~.Then the set of primes p with n/2 < p < n has size 7r(n) -rr(n/2) .With this, it is simple to check that 7r(n) -rr(n/2) > 3 for all n > 17.
In particular, if n >_ 17 then there are at least three primes p with n/2 < p < n.On the other hand, we note that n, -1 asid n -2 cannot both be prime and that (n+1)/2 and (n+2)/2 cannot both be integers .Thus the restriction p :,P'~n-1, n-2, (-n+1)/2 or ('n+2)/2 eliminates at most two primes from this segment and we conclude that an appropriate p does indeed exist for n >_ 17 .Therefore the induction stop carries through in this range .R.ecall that n > 4 and that n is not a prime.Rrrthermore, note that we do not really need f (n) -7r(n/2) to be > 3. Rather, we just need the existente of an appropriate prime and clearly p = 11 works for n = 16,15,14 and p = 7 works for n = 10.This leaves only the special cases n = 4, 6, 8, 9, 12 to be considerad and we already have affirmative answers for n = 4 from Lenuna 1 .7(iii)and for n = 6,9 from Lemma 1.9.Thus there are only two integers left .
If n = 12, let p = 7.Then it follows from Lemma 1 .8(ii)that P E A(G) where P is the subgroup of G generated by a 7-cycle .Furthermore, by Lemma 1.9(i), the set of elements of Symb which are not 6-cycles spans 1 .Thus Lemma 1.8(i) with k = p -1 = 6 and n -k = 6 implies that Q E A(G) where Q is generated by a 6-cycle .Lemma 1 .7(i)therefore yields the result in this case.Similarly, if n = 8 then we let p = 5 and use the fact, given in Lemma 1 .7(iii),that a 2-cycle and a 3-cycle span 1 in K _ [Sy_m4] .This again guarantees that A(G) contains an appropriate P and Q and the theorem is proved .a If K is an algebraically closed field of characteristic 0, then the representation theory of K[G] is equivalent, in some sense, to the character theory of the finite group G. Thus it is reasonable to translate the problem studied here into character-theoretic language.Since this translation is well known, we will just mention a few basic facts.We follow the notation of [I] .where the sum is over representatives x of the conjvgacy classes of G. ii .
where ~(g) = O(g -I ) for all g E G.
Proof: (i) Since 0 is the character of the conjugation permutation representation, it is clear that O(g) = ICG(g)I, the number of elements x E G which are fixed by g.The second formula is also clear Since is the sum of the characters which correspond to the action of G on individual conjugacy classes.
(ii) The expression 0 = L.'OElrr(G) 00 can be deduced from the char- acter orthogonality relations [I,Theorem 2.18] and the fact that O(g) = ICG(9)1 .A more natural proof starts by writing K[G]=ZiPElrr(G) M a direct sum of full matrix rings over K, with M(O) corresponding to the character 0. Then each M(O) is a submodule of K[G] under the adjoint action and, as is well known, the character of this submodule is ." It is obvious that the adjoint representation of K[G] is faithful if and only if every irreducible character X E Irr(G) is a constituent of 0. Thus the above and Frobenius reciprocity yield Lemma 1 .12.If K is algebraically closed, ten I(G) = 0 if and only if,, for each X E Irr(G), we have i .X is a constituent of (le~i,,;i)G for Borne x E G, or ii .le, :(, ;) is a constituent of X restricted to CC(x) for some x, E G, or iii.X is a constituent of zpo for soirr,e ,0 E Irr(G) .
There is obviously a good deal yet to be done on this topic .

. Enveloping Algebras
Now let L be a Lie algebra over the field K and let U(L) denote its universal enveloping algebra.Then 7-t = U(L) is a Hopf algebra with comultiplication and antipode determinad by e H P ® 1 + 1 ® e and S(e) = -P., respectively, for all e E L. Of course, 1 H 1 ® 1 and S(1) = 1 .It follows that the Hopf left adjoint action of U(L) on itself is determinad by e -x = ex -xe for all e E L and x E U(L) .In other words, e -x = [e, x] = (ad e)x .Again we are concerned with the kernel of the adjoint representation, namely It is clear that I(L) n L = 7L(L) and therefore, as with groups, 7L(L) :7~0 implies that I(L) =,b 0.
One might expect the study of the adjoint actions of U(L) and of K[G] to be somewhat similar because of the common Hopf algebra root .However, this is apparently not the case and indeed Lie results seem to be considerably more elusivo.Nevertheless, since delta methods are now available for U(L), there does exist an appropriate Lie analog of Theorem 1 .1.For this, let 0 = {e E L 1 dirnK [e, L] < co } and let AL denote the subspace of L generated by all finita-dimensional Lie ideals of L. Then AL C ~i are characteristic Lie ideals of L and we have Theorem 2.1.If L is a Lie algebra over a field K of characteristic 0, then In particular, if AL = 0 then Z(L) = 0 .
Proof-We follow the notation of [BP2] .Let a E Z(L), choose a com- plementary basis X for AL in L and use it to write cx = El~pa, based on AL .Thus each aN, E U(AL) and each p is a distinct straightened monomial in X .Our goal is to show that %, E Z(L) for all p .
It is convenient to use the Hopf comultiplication notation.For example, since p is a monomial in X, we can write the comultiplication of p as E( ,,) p1 ®p2 where the sum contains all ordered pairs { p1, p2 } which are complementary partial products of p. Since p is a straightened monomial in X, so also are p1 and p2 .Furthermore, since %, E U(OL), the comultiplication of % is contained in U(AL) ®U(AL) .Therefore we can assume that the sum E(«j (aN )1 ® (au)2 involves only terms from U(OL) .We use both these assumptions when we write the comultiplication of par, as Now a E Z(L), so for all r E U(L) we have and hence 0 = a -r = (Pa,) -r = 1: (Pa, .)1r S((pa, .)2)Y" mi (nj1 r S(p2(a~)2) 0 = P1(%)1 r S((%)2)S(p2) for all r E U(L) .
Notice that the latter formula is a linear identity in U(L).Furthermore, each (a,)1 is contained in U(AL) and each p1 is a straightened monomial in X.Thus, for any monomial u, [BP2,Theorem 4.5(i)] implies that 0= ~l E(a,)1rS((aM)2)S(p2) wllere indicates the partial sum of those terms with MI = a .In particular, if we take o, = 1, then pq = 1 implies that /-I2 = ¡~and the preceding displayed equation becomes 0= E(a,,)Ir5((%)2)S(h) IL (ai.) Again, the latter formula is a linear identity in U(L) and this time we observe that S((arJ2) E U(InkL) and that S(p) = fp is plus or minus a straightened monomial in X .Thus, for any monomial T, [BP2,Theorem 4.5(ii)] yields 0 = ~(a,), r S((a,)2) = aT -r for all r E U(L) .
In other words, aT E I(L) n U(AL) and, since this holds for all such T, we have T for all r E U(L) .
Thus, since I(L) n U(AL) is an L-stable ideal Of U(AL), we conclude that and the result follows .Furthermore, if AL = 0, then clearly we have I(L) n U(OL) = I(L) n K = 0 and therefore I(L) = 0 as required.
The corresponding result in characteristic p > 0 is decided1y falso .For example, let L = A x B wllere A = (al, a2. . . . ) is an infinito-dimensional abelian ideal, B = Kb is a one-dimensional complemerit and [a¡ , b] = al+I for all i .Then it follows easily that AL = 0 and that ar E 71 (U (L» for all a E A .Thus 0 = ad ar = (ad a)'', so ar' E I(L) and therefore Z(L) 0 0.
As is well known, such examples occur because, in characteristic p > 0, ordinary enveloping algebras do not adequately reflect their Hopf algebra structure .Indeed, in this context, one knows that it is more appropriate to consider restricted Lie algebras .Furthermore, if L is any Lie algebra, then there exists a restricted Lie algebra L with U(L) = u(L) .Thus, in positive characteristics, the restricted case is really the general case.
If L is a restricted Lie algebra over a field K of.characteristic p > 0, then we let u(L) denote its restricted enveloping algebra .The formulas for the comultiplication, antipode and left adjoint action are of course the lame as in the ordinary case and we write for the kernel of the adjoint representation .Again AL denotes the subalgebra of L generated by al] finite-dimensional (not necessarily restricted) Lie ideals and we let A be as before .Now delta methods exist for u(L), but unfortunately their statements in the literatura are not quite what we require .Nevertheless, the necessary results are true, follow fairly easy from what is known and will be published at soma later time.For now we just make two quick observations .First, it is an easy exercise to show that AL = 0 in the restricted case and, in particular, we sea that AL is a characteristic restricted ideal of -L.Second, the argument of [BP2, Proposition 4.2] applies equally well to restricted enveloping algebas and yields the necessary sharpening of [BP1,Theorem 5 .1] .With this, the exact restricted arralog of [BP2; Theorern 4.5] holds and therefore a direct application of the preceding proof yields Theorem 2.2.If L is a restricted Lie algebra over a field K of characteristic p > 0, then The similarity of the preceding two theorems suggests that we should modify our notation somewhat .Thus, for the remainder of this section, we will observe the following conventions .
(1) If K has characteristic 0, then L is an ordinary Lie algebra and U(L) is, as usual, its ordinary enveloping algebra.(2) If K has characteristic p > 0, then L will be a restricted Lie algebra and we changa notation to let U(L) denote its restricted enveloping algebra .
As will be apparent, this chango allows us to avoid unnecessary repetitions.
Next, we also modify our direction somewhat .We know that L is a submodule of U(L) under the adjoint action and, for many reasons, this is a more interesting module structure to study.Thus we will be concerned with faithfulness in this context and we let ,7(L) = { a E U(L) 1 a -L = 0 } denote the ideal of U(L) which is the kernel of the action of U(L) on L. Obviously ,7(L) D Z(L) and ,7(L) :A 0 when L is abelian .
If L is a finite-dimensional Lie algebra, then it is well known that ,7(L) :A 0. Indeed, in characteristic 0, every 0 :,A 2 E L is transcendental as an element of U(L), but of course ad Q is an algebraic linear transformation en the finito-dimensional space L. On the other hand, in characteristic p > 0, the dimension of U(L) is just too largo.To be precise, if dimK L = n >_ 2, then we have dimK U(L) = pl > n2 = dimK EndK(L) .FIrthermore, equality can only occur when p = n = 2 and this case is then easily checked by hand.As we will see in the next two lemmas, ,7(L) =, ,É 0 seems te be the rulo even in infinito-dimensional situations .Nevertheless, in the final result of this section we will apply delta methods to show that there is an abundant supply of Lie algebras L with ,7(L) = 0. Lemma 2 .3.Let A be an associative algebra over the field K and let L be a K-subspace of A closed under the Lie bracket [x, y] = xy -yx .Furthermore, if K has characteristic p > 0, assume that L is closed under taking pth powers .Then ,7(L) :y~0 if either i.K has characteristic 0 and L contains a nonzero algebraic element of A, or ii .K has characteristic p > 5 and L contains a nonzero idempotent of A or a nonzero element of A of square 0.
Proof. .Notice that L acts on A via P -a = (ad Q)a = [e, a] for all P E L, a, E A and hence the adjoint action of U(L) on L extends to an action on A. Dirthermore, we have ad £ = Ge -Re where Ge : A -> A denotes left multiplication by 2 and where Re : A -~A denotes right multiplication .Of course Ge and Re comrnute as operators .Now if 0 q¿ ~E L is an algebraic element of A, then clearly Ge and Re are algebraic operators on A and, since they comrnute, so is ad e = ,Ce -Re.Thus the result follows in characteristic 0 since 2 is transcendental as an element of U(L) .
Finally, if P is an idempotent element of A, then Ge and Re are commuting idempotent operators and it follows easily that (ad P)3 = ad B. Similarly, if P is an element of A of square 0, then ,e2 = R2 = 0 and hence (ad 2)3 = 0.In either case, ad B is algebraic of degree <_ 3. On the other hand, if K has characteristic p > 0 and if 2 ~0, then 1, P, P2, . . ., QP-1 are K-linearly independent when viewed as elements of U(L) .Thus, if p > 5, we conclude that U(L) is not faithfu1 on L .
In other words, the usual constructions for infinito-dimensional Lie algebras lead to situations which are not faithful .One such construction of particular interest is as follows .If W is a subspace of the vector space V, we let End K (V ;W) = {t E EndK(V) 1 t(V) C w} .
Then End(V ; W) is a subalgebra (without 1) of Énd(V) and, in particular, it gives rise to a (restricted) Lie algebra which we denote by gl(V ; W) .Lemma 2.4.Let L = gl(V; W) with W an infinite-dimensional proper subspace of V. Then 0 = 0, but ,7(L) 7É 0 .
Proof. .We can easily see that 0 = 0 by looking at matrices, but a simple direct proof is as follows .Let t be a nonzero element of £ = End(V ; W) .Since t q¿ 0 and V q¿ W, it follows that t does not vanish on V \ W. Thus we can choose v E V \ W with t(v) = w 0 0 and we note that { v, w } is linearly independent since w E W. Now let S be the subalgebra (without 1) of £ consisting of all s E £ with .s(v)= 0. Then the independence of { v, w } easily implies that the evaluation map 1: S -> W given by .sH-> .s(w) is a K-linear epimorphism and therefore the kernel of ~has infinite codimension in S. Furthermore ; if s E Es(t), then s(w) = st(v) = ts(v) = 0 and thus CS(t) C Ker(~) .Putting this all together, we see that Cs(t) has infinite codimension in S, so Q(t) has infinite codimension in £ = End(V, W) and this, in turn, implies that 0=0 .
For the second part, we note that ,7(L) 0 0 follows irrimediately from the previous lemma except when K Iras characteristic 2 or 3 .However, we can offer a simple direct proof which applies to all characteristics .Te start with, since V 7~W and dinIK W >_ 2, it follows that there exist two linearly independent transformations x, y E End(V; W) with x(W) = y(W) = 0.The latter conditions then imply that xEnd(V ; W) = yEnd(V ; W) = 0 and hence; for all t E End(V; W), me have [x, t] = -tx and [y, t] = -ty .Thus (ad x) (ad y)t, = tyx = 0 and we conclude that xy E ,7(L) .On the other hand, { x, y } is linearly independent ; so xy =,A 0 in U(L) and therefore ,7(L) :,A 0 as required .
Thus we see that 0 = 0 does not imply that ,7(L) vanishes and therefore the natural analog of Theorerns 2 .1 and 2.2 cannot hold in this context .
For any Lie algebra L, let w (L) denote the augmentation ideal of U (L), namely the ideal of U(L) generated by L. Of course, w(L) is also the kernel of the algebra epimorphism U(L) --> K determined by L -> 0 and hence U(L) = K + w (L) .Lemma 2.5.'Let L be a Lie algebra over the field K. Proof (i) We can asume that L is an infinite-dimensional Lie algebra and therefore that dinIK U(L) = oo .Now let V = U(L) and set W = w(L).Then, via the left regular representation of U(L), we see that w(L) embeds in EndK(V ; W) .Tlrus L C_ gl(V; W) = 111 and the preceding lernma yields tire result.
(ii) We are given ~L(L) = 0 and we can clearly asume that L :7~0 so that dimK L = oo .If N = w(L) then, via the usual Lie bracket [x, y] = xy -yx, it follows that N is a (restricted) Lie algebra containing L. The goal is to show that 0(N) = 0 and that ,7(N) n U(L) = 0.
To start with, if x E 0(N), then certainly x is an elernent of U(L) satisfying dinIK L/CL(x) < oo .[BP1,Lernrria 7 .1]therefore implies that x E U(0) = K and hence x E K n w(L) = 0 as required .
Next ; consider the adjoint action o of U(N) on N and notice that L aets on N via, e " a = [~, a] = Pa -aB for all B E L and a E N = w(L) .Since the action of U(L) on N is the unique algebra extension of this, it follows that the o action of U(L) is identical to the adjoint action -of U(L) on w(L) .Therefore, we have ,7(N) n U(L) = {p E U(L) 1 0 -w(L) = 0 } and, for conveniente, we denote the latter ideal by Z'(L) .Now it is clear that {,3 E U(L) 1 3 -K = 0 } = w(L) and therefore U(L) = K + w(L) implies that w(L) n Z'(L) = Z(L) .Furthermore, 0(L) = 0 so Z(L) = 0 by Theoorms 2.1 and 2 .2.Thus T(L) C 1.ann w (L) = 0, since L is infinite dimensional, and the result follows .
It is now a simple matter to prove Theorem 2.6 .Tf L is a Lie algeb-ra in characteristic 0 or a restricted Lie algebra in characteristic p > 0, lhen there exists a (restricted) Lie algebra L D L such that 0(L) = 0 and ,7(L) = 0 .
Proof-According to Lemma, 2.5(i), we can embed L in a Lie algebra L r with A(L1) = 0. Furthermore, starting with Lr, we can then use Lemma 2.5(ii) to inductively construct a chain Lr C L2 C --of Lie algebras satisfying A(Lj = 0 and ,7(L ,+r) n U(L,z ) = 0 for all n > 1 .Finally, we set L = U' L. so that L is a (restricted) Lie algebra containing L. Tire goal is to show that 0(L) = 0 and that ,7(L) = 0.Both of these are quite simple .
First, suppose that x E 0(L) .Then x E L, for sorne n _> 1 and therefore x E A(L) n L,, C 0(Lj = 0 as required .Next, suppose that a E ,7(L).Since U(L) is clearly equal to U1°U(Ln), it follows that a E U(Lj for sorne n > 1. Furthermore ; note that the ad action of Lemma 1 .4.For each x E G, let 7 rx : K[G] -> K[CG(x)] denote the natural projection map.Then I(G) is the largest two-sided ideal I of K[G] with zrx (I) C w(CG(x)) for all x E G.In particular, I(G) is also the kerizel of the right adjoint representation of K[G] .
Lemma 1.11 .Assume that K is algebraically closed and let 0: G -> K denote the character of the adjoint representation of K[G] .Then i .O(g) = ¡CG(g)j for all g E G and G 0 = E(1C,,(-)) i.There exists a Lie algebra M :,b 0 containing L with 0(M) = 0. ii.If 0(L) = 0, then there exists a Lie algebra N containing L with 0(N) = 0 and ,7(N) n U(L) = 0.