SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE

D. S. Passman

University of Wisconsin-Madison

Abstract. Let K be a field and let ε: Γ × Γ → K• be a bicharacter defined on

the multiplicative group Γ. We suppose that A is a Γ-graded, associative K-algebra

which is color commutative with respect to ε. Furthermore, let ∆ be a nonzero Γ-

graded, K-vector space of color derivations of A and suppose that ∆ is also color

commutative with respect to the bicharacter ε. Then, with a rather natural definition,

A ⊗K ∆ = A∆ becomes a Lie color algebra, and we obtain necessary and sufficient

conditions here for this Lie color algebra to be simple. With two minor exceptions

when dimK ∆ = 1, simplicity occurs if and only if A is graded ∆-simple and A∆⊗∆ =

A∆∆ acts faithfully as color derivations on A.

§1. Construction of the Lie Algebra

In the recent paper [Pa], we took a ring theoretic approach to the constructionof simple Lie algebras of Witt-type. Here, we use similar methods to show thatanalogous results hold in the context of Lie color algebras. Since the color aspectsof this construction may be new, we will include all definitions and carefully verifyall required identities.

To start with, let K be a field and let Γ be a multiplicative abelian group. Recallthat a bicharacter ε: Γ × Γ → K• is a map which satisfies

E1. ε(x, y)ε(y, x) = 1 for all x, y ∈ ΓE2. ε(x, yz) = ε(x, y)ε(x, z) for all x, y, z ∈ Γ.

Furthermore, it is clear that these imply

E3. ε(xy, z) = ε(x, z)ε(y, z) for all x, y, z ∈ Γ.

We will fix K, Γ and ε throughout this work.Now let A be a Γ-graded associative K-algebra. Thus A = ⊕

∑

x∈Γ Ax is a directsum of the K-subspaces Ax, indexed by the elements x ∈ Γ, and these summands

2000 Mathematics Subject Classification. 16W25, 16W55, 17B60, 17B70.

The author’s research was supported in part by NSF Grant DMS-9622566.

Typeset by AMS-TEX

1

2 D. S. PASSMAN

satisfy AxAy ⊆ Axy for all x, y ∈ Γ. Of course, 1 ∈ A1. As usual, the homogeneouselements H(A) of A are precisely the nonzero elements in the various componentsAx. Thus each homogeneous element a ∈ H(A) determines a unique group elementa# ∈ Γ by a ∈ Aa# . Fortunately, we can almost always drop the symbol # here,since confusion rarely occurs. In particular, if a is homogeneous, then we havea ∈ Aa and also aAy ⊆ Aay for all y ∈ Γ.

As is well known (see [B, page 15]), the K-linear map [ , ]:A×A → A defined by[a, b] = ab − ε(a, b)ba for all a, b ∈ H(A), gives rise to a Lie color algebra structureon A. Here, of course, we have abbreviated ε(a#, b#) by ε(a, b). We will actuallyrequire very little about this color algebra, since the appropriate graded analog ofthe commutativity of A in [Pa] is that A be color commutative. By this we meanthat [A,A] = 0 or equivalently that

C1. ab = ε(a, b)ba for all a, b ∈ H(A).

Lemma 1.1. If A is a color commutative, Γ-graded algebra, and if V is a gradedsubset of A, then AV = V A. In particular,

i. If V is a right or left ideal of A, then V is a two-sided ideal of A.ii. If V is nilpotent, then V A is a nilpotent ideal of A.iii. If H(V ) consists of nilpotent elements, then V A is a nil ideal of A.

Proof. If x, y ∈ Γ, then VxAy = ε(x, y)AyVx = AyVx. Thus, since V =∑

x∈Γ Vx

and A =∑

y∈Γ Ay, we conclude that V A = AV . Parts (i) and (ii) are now obvious,

and (iii) follows since V A =∑

v∈H(V ) vA and each vA is a nilpotent ideal of A. �

Next, a nonzero K-linear map ∂:A → A is said to be a homogeneous colorderivation of grade ∂# ∈ Γ if

D1. ∂(Ax) ⊆ A∂x for all x ∈ ΓD2. ∂(ab) = ∂(a)b + ε(∂, a)a∂(b) for all a, b ∈ H(A).

As usual, we dropped the # symbol in the above formulas, so that A∂x is reallyA∂#x and similarly ε(∂, a) = ε(∂#, a#). Of course, when a = b = 1, (D2) impliesthat ∂(1) = 0, and hence

D3. ∂(K) = 0.

For convenience, we denote the K-linear span of all such homogeneous deriva-tions by Derε(A), so that Derε(A) = ⊕

∑

x∈Γ Derεx(A), where Derε

x(A) is the K-subspace consisting of all such derivations of grade x ∈ Γ along with the derivation0. Note that the directness of the sum follows immediately from (D1). Now, ifα, β ∈ H(Derε(A)), then [α, β] = αβ − ε(α, β)βα is easily seen to be either 0 or ahomogeneous color derivation of grade α#β#. In particular, the K-linear extensionof this bracket yields a map [ , ]: Derε(A) × Derε(A) → Derε(A), and in this way,Derε(A) becomes a Lie color algebra.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 3

Lemma 1.2. Assume that A is a color commutative, Γ-graded algebra. Let a, b ∈H(A) and let α, β ∈ H(Derε(A)).

i. The map aα:A → A given by aα: r 7→ aα(r) for all r ∈ A is either 0 or ahomogeneous color derivation of A of grade a#α#.

ii. If [α, β] = 0, then [aα, bβ] = aα(b)β − ε(aα, bβ)bβ(a)α as operators on Aand hence as elements of Derε(A).

Proof. (i) We can suppose that aα 6= 0. If r, s ∈ H(A), then the color commutativ-ity of A yields ar = ε(a, r)ra and hence

(aα)(rs) = aα(r)s + aε(α, r)rα(s)

= aα(r)s + ε(a, r)ε(α, r)raα(s)

= (aα)(r)s + ε(aα, r)r(aα)(s)

for all r, s ∈ H(A). Thus, since (aα)Ax ⊆ aAαx ⊆ Aaαx for all x ∈ Γ, aα is indeeda homogeneous color derivation of A of grade a#α#.

(ii) Again, we can assume that aα and bβ are nonzero. If r ∈ H(A), then

(aα)(bβ)(r) = aα(b)β(r) + abε(α, b)(αβ)(r)

(bβ)(aα)(r) = bβ(a)α(r) + baε(β, a)(βα)(r).

Now A is color commutative, so ab = ε(a, b)ba and [α, β] = 0 implies that αβ =ε(α, β)βα. Thus abε(α, b)(αβ)(r) = baε(β, a)(βα)(r)·k where

k = ε(a, b)ε(α, b)ε(α, β)ε(a, β) = ε(aα, b)ε(aα, β) = ε(aα, bβ).

With this, it follows that [aα, bβ] = (aα)(bβ) − ε(aα, bβ)(bβ)(aα) maps r to theelement aα(b)β(r)−ε(aα, bβ)bβ(a)α(r). Thus, as operators on A, we have [aα, bβ] =aα(b)β − ε(aα, bβ)bβ(a)α, as required. �

For the remainder of this paper, we assume that A is a color commutative,Γ-graded K-algebra so that condition (C1) holds. Furthermore, we let ∆ be acolor commutative Lie color subalgebra of DerεA. Thus ∆ = ⊕

∑

x∈Γ ∆x where∆x = ∆ ∩ Derε

x(A), and [∆,∆] = 0. In particular, the latter is equivalent to

C2. αβ = ε(α, β)βα for all α, β ∈ H(∆).

Note that the tensor product A ⊗K ∆ = A∆ is Γ-graded by defining

L1. (A∆)x =∑

yz=x Ay∆z for all x ∈ Γ.

In view of Lemma 1.2(i), we now have a graded action of A∆ on A, where aα ∈ A∆acts via its corresponding operator in Derε(A). Moreover, Lemma 1.2(ii) motivatesus to define the binary operation [ , ] on A∆ as the K-linear extension of

L2. [aα, bβ] = aα(b)β − ε(aα, bβ)bβ(a)α for all a, b ∈ H(A), α, β ∈ H(∆).

With all of this, we obtain

4 D. S. PASSMAN

Proposition 1.3. Let A and ∆ be as above. Then A⊗∆ = A∆ is a Lie color alge-bra with Γ-grading given by (L1) and Lie color bracket given by (L2). Furthermore,the action of A∆ on A yields a Lie color homomorphism θ:A∆ → Derε(A).

Proof. It is clear that [ , ] is K-linear in each variable. Thus it remains to verifythe color versions of skew symmetry and the Jacobi identity as given in [S, (3.3)and (3.4)]. To this end, let a, b, c ∈ H(A) and let α, β, γ ∈ H(∆). Then

[aα, bβ] = −ε(aα, bβ)[bβ, aα]

follows immediately from (L2) and (E1). For the Jacobi identity, we note that

[aα, [bβ, cγ]] = [aα, bβ(c)γ − ε(bβ, cγ)cγ(b)β].

Hence (D2) yields

ε(cγ, aα)[aα, [bβ, cγ]] = J1 + J2 + J3 − J4 − J5 − J6

where

J1 = ε(cγ, aα)aα(b)β(c)γ

J2 = −ε(cγ, aα)ε(bβ, cγ)ε(α, c)ac(αγ)(b)β

J3 = ε(cγ, aα)ε(aα, bβcγ)ε(bβ, cγ)cγ(b)β(a)α

= ε(aα, bβ)ε(bβ, cγ)cγ(b)β(a)α

J4 = ε(cγ, aα)ε(aα, bβcγ)bβ(c)γ(a)α

= ε(aα, bβ)bβ(c)γ(a)α

J5 = −ε(cγ, aα)ε(α, b)ab(αβ)(c)γ

J6 = ε(cγ, aα)ε(bβ, cγ)aα(c)γ(b)β.

For convenience, set f(aα, bβ, cγ) = J1 + J2 + J3 and note that f(bβ, cγ, aα) isobtained from f(aα, bβ, cγ) by a cyclic shift of the variables. Now it is easy to seethat the shifted version of J1 is equal to J4 and that the shifted version of J3 is J6.Furthermore, it follows from (C1) and (C2) that the shifted version of J2 is

−ε(aα, bβ)ε(cγ, aα)ε(β, a)·ba·(βα)(c)γ

= −ε(aα, bβ)ε(cγ, aα)ε(β, a)·ε(b, a)ab·ε(β, α)(αβ)(c)γ

= −ε(cγ, aα)ε(α, b)ab(αβ)(c)γ = J5.

In other words,

ε(cγ, aα)[aα, [bβ, cγ]] = f(aα, bβ, cγ) − f(bβ, cγ, aα)

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 5

and the color Jacobi identity follows by adding the three shifted versions of thisexpression.

Of course, it is clear from Lemma 1.2 and (L2) that θ is indeed a Lie colorhomomorphism. �

As we observed earlier, if a = 1, then β(a) = β(1) = 0. Thus, in this specialcase, equation (L2) reduces to [1α, bβ] = α(b)β. Furthermore, since the bicharacterdoes not appear here, it follows from linearity that

L3. [1α, bβ] = α(b)β for all b ∈ A, α, β ∈ ∆.

§2. Necessary Conditions for Simplicity

The goal of this paper is to determine when A ⊗ ∆ = A∆ is a simple Lie coloralgebra. Recall that L is a Lie color ideal of A∆ if

I1. L is a Γ-graded K-subspace of A∆.I2. [A∆, L] = [L,A∆] ⊆ L.

Furthermore, if A∆ has only the trivial Lie color ideals, namely L = 0 and L = A∆,then A∆ is said to be simple. Note that if the bicharacter satisfies ε(Γ,Γ) = 1,then A∆ is really an ordinary Lie algebra. But, in view of (I1), we see that A∆ isa simple Lie color algebra if and only if it is a graded simple Lie algebra.

We continue with the notation of the preceding section. As in [Pa], there aretwo obvious necessary conditions for simplicity. To start with, A must be graded∆-simple. This means that A has no nontrivial graded ∆-stable ideals. Indeed, if Iis a graded ∆-stable ideal of A, then I ⊗∆ = I∆ is clearly a Lie color ideal of A∆.Hence, if A∆ is simple, then I must 0 or A. Note that the graded ∆-simplicity ofA also forces the ring of constants A∆ to have a rather nice structure.

Lemma 2.1. If A is graded ∆-simple, then all elements of H(A∆) are invertible

in this subalgebra. In particular, A∆ is a crossed product B∗Γ̃ where B = (A∆)1 is

the identity component of the subalgebra and where Γ̃ is some subgroup of Γ.

Proof. We know that A∆ is a graded subalgebra of A. Furthermore, if a ∈ H(A∆),then aA = Aa is a nonzero graded ∆-stable ideal of A, by Lemma 1.1(i), and henceaA = Aa = A. Thus a is invertible in A and let b = A−1 ∈ H(A). Finally, for any∂ ∈ H(∆), equation (D2) implies that 0 = ∂(1) = ∂(ba) = ∂(b)a, so ∂(b) = 0 andb ∈ A∆, as required. �

The graded ∆-simplicity of A is also somewhat related to the parity functionon Γ. Specifically, note that ε(x, x)2 = 1 by (E1) and therefore ε(x, x) = ±1.Furthermore, (E2) implies that the map Γ → {±1 } given by x 7→ ε(x, x) is a grouphomomorphism. Thus the kernel of this map, Γ+ = {x ∈ Γ | ε(x, x) = 1 }, is asubgroup of Γ of index ≤ 2, and if this subgroup has index 2 then its second coset

6 D. S. PASSMAN

is given by Γ− = {x ∈ Γ | ε(x, x) = −1 }. Of course, if charK = 2, then we alwayshave Γ = Γ+. For convenience, if V is a Γ-graded, K-vector space, then we useH+(V ) to denote the set of homogeneous elements of V with grade in Γ+, and welet V+ be the K-linear span of H+(V ). Similarly, we let H−(V ) denote the set ofhomogeneous elements with grade in Γ−, and V− is its span. Clearly, V = V+⊕V−.

Lemma 2.2. AA− = A−A is a graded nil ideal of A. In particular, if A is graded∆-simple and if ∆ = ∆+, then A = A+.

Proof. If a ∈ H−(A), then (C1) implies that a2 = aa = ε(a, a)aa = −a2. Thusa2 = 0 since char K 6= 2 in this case. It now follows from Lemma 1.1(iii) that A−Ais a graded nil ideal and hence A−A 6= A. Finally, if ∆ = ∆+, then it is clearthat A−A is ∆-stable. In particular, if A is graded ∆-simple, then A−A = 0 andtherefore A = A+. �

A second necessary condition for the simplicity of A∆ is that this Lie coloralgebra act faithfully on A. Indeed, if A∆ is simple, then the kernel of the Liecolor homomorphism θ of Proposition 1.3 is either 0 or A∆. But, in the latter case,1∆ = ∆ acts trivially on A, contradicting its definition as a nonzero subspace ofDerε(A). Thus A∆ must act faithfully and hence so must its subspace A∆ ⊗ ∆ =A∆∆. As we see below, all that is required is this weaker assumption. Specifically,we will prove

Theorem 2.3. Let ε: Γ × Γ → K• be a bicharacter for the multiplicative group Γ.Suppose A is a color commutative Γ-graded, K-algebra and let ∆ be a nonzero colorcommutative Γ-graded, K-vector space of color derivations of A. If dimK ∆ = 1,assume in addition that char K 6= 2 and that ∆ = ∆+. Then A ⊗ ∆ = A∆ is asimple Lie color algebra if and only if A is graded ∆-simple and A∆ ⊗ ∆ = A∆∆acts faithfully on A.

As is apparent, there is two missing cases here which occur when dimK ∆ = 1.In this situation, we have

Lemma 2.4. If A is graded ∆-simple and dimK ∆ = 1, then A∆ ⊗∆ = A∆∆ actsfaithfully on A.

Proof. Say ∆ = K∂ and consider the restriction of θ to A∆ ⊗∆ = A∆∆. Since θ isa graded homomorphism, the kernel of θ is generated by its homogeneous elements.But 0 6= ∂ ∈ H(∆) and all elements of H(A∆) are units. Thus ker θ contains nononzero homogeneous elements and A∆∆ acts faithfully, as required. �

Thus, when ∆ is 1-dimensional, the assumption that A∆∆ acts faithfully can bedropped. However, now another condition comes into play.

Theorem 2.5. Suppose A is a color commutative Γ-graded, K-algebra and let ∆be a nonzero color commutative Γ-graded, K-vector space of color derivations of A.Let dimK ∆ = 1 and assume that either char K = 2 or char K 6= 2 and ∆ = ∆−.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 7

Then A⊗∆ = A∆ is a simple Lie color algebra if and only if A is graded ∆-simpleand ∆(A) = A.

At this point, an example would be helpful. Here we use the usual bicharacterassociated with superalgebras.

Example 2.6. Let Γ = { 1, g } have order 2 and let ε be defined so that ε(g, g) =−1. Suppose F is a field extension of K and let δ be an ordinary K-derivation ofF . If A = F [t]/(t2), then A is a Γ-graded, color commutative algebra with A1 = Fand Ag = F t̄, where t̄ is the image in A of the variable t. Now let ∂:A → A begiven by ∂(r + st̄) = s + δ(r)t̄ for all r, s ∈ F . Then ∂ is a color derivation of A ofgrade ∂# = g. Furthermore, if ∆ = K∂, then A is graded ∆-simple, and A∆ is asimple Lie color algebra if and only if δ is onto.

Proof. Since ε(x, y) = 1 if one of the group elements x or y is 1, it follows easilythat A is Γ-graded and color commutative. Furthermore, it is easy to verify that∂ is a color derivation of A of grade ∂# = g. Now the only proper ideal of A isF t̄, and this ideal is not ∆-stable. Consequently, A is graded, ∆-simple. Finally,since ∂(A) = A if and only if δ(F ) = F , the preceding theorem implies that A∆ issimple if and only if δ is onto. �

If δ(F ) = 0 in the above, then it is easy to see that F∂ is a Lie color ideal of A∆.Indeed, this follows since [F∂, F∂] = 0 and [F∂, F t̄∂] = F∂. On the other hand, aswas shown in [KN], numerous purely transcendental field extensions F ⊇ K exist,in any characteristic, with δ(F ) = F . Note that Example 2.6 also applies whenchar K = 2. Here Γ = Γ+ and ∂ is an ordinary derivation of A.

We close this section by isolating some specific computations which will be re-quired later. They, of course, apply to an arbitrary A∆.

Lemma 2.7. Let a, b ∈ H(A) and let ∂ ∈ H(∆).

i. ε(∂, a)[a∂, b∂] =[

∂(ab) − 2∂(a)b]

∂ if ε(∂, ∂) = 1.ii. ε(∂, a)[a∂, b∂] = ∂(ab)∂ when ε(∂, ∂) = −1.iii. ∂(bn) = n∂(b)bn−1 if n ≥ 1 and ε(b, b) = 1.

Proof. Write ε(∂, a)[a∂, b∂] = c∂ for a suitable c ∈ A. Then

c = ε(∂, a)a∂(b) − ε(∂, a)ε(a∂, b∂)b∂(a)

= ε(∂, a)a∂(b) − ε(∂, ∂)ε(∂(a), b)b∂(a)

= ε(∂, a)a∂(b) − ε(∂, ∂)∂(a)b

since (C1) implies that ∂(a)b = ε(∂(a), b)b∂(a). Thus

c = ∂(ab) − ∂(a)b − ε(∂, ∂)∂(a)b,

by (D2), so parts (i) and (ii) follow.

8 D. S. PASSMAN

For (iii), we proceed by induction on n ≥ 1, and we suppose that the result holdsfor n. Then

∂(bn+1) = ∂(bbn) = ∂(b)bn + ε(∂, b)b∂(bn)

= ∂(b)bn + nε(∂, b)b∂(b)bn−1 = (n + 1)∂(b)bn

since ε(b, b) = 1 and (C1) imply that ε(∂, b)b∂(b) = ε(b∂, b)b∂(b) = ∂(b)b. �

Of course, if b ∈ H−(A), then we know that bn = 0 for n ≥ 2. We will also need

Lemma 2.8. Let a, r ∈ H(A) and let ∂1, ∂2 ∈ H(∆). Then

ε(a, ∂1)[r∂1, a∂2] − [ra∂1, 1∂2] = ε(a∂1, ∂2)r∂2(a)∂1 + ε(a, ∂1)r∂1(a)∂2.

Proof. To start with, we have

ε(a, ∂1)[r∂1, a∂2] = ε(a, ∂1)r∂1(a)∂2 − ε(a, ∂1)ε(r∂1, a∂2)a∂2(r)∂1

= ε(a, ∂1)r∂1(a)∂2 − ε(ra∂1, ∂2)ε(r∂2, a)a∂2(r)∂1.

Furthermore,

−[ra∂1, 1∂2] = ε(ra∂1, ∂2)∂2(ra)∂1

= ε(ra∂1, ∂2)ε(∂2, r)r∂2(a)∂1 + ε(ra∂1, ∂2)∂2(r)a∂1

= ε(a∂1, ∂2)r∂2(a)∂1 + ε(ra∂1, ∂2)ε(r∂2, a)a∂2(r)∂1,

since ∂2(r)a = ε(r∂2, a)a∂2(r) by (C1). Thus, by adding the latter two displayedequations, we obtain

ε(a, ∂1)[r∂1, a∂2] − [ra∂1, 1∂2] = ε(a∂1, ∂2)r∂2(a)∂1 + ε(a, ∂1)r∂1(a)∂2,

as required. �

§3. Sufficient Conditions for Simplicity

In this section, we prove the main theorems. Thus, we assume throughout thatΓ is a multiplicative abelian group and that ε: Γ × Γ → K• is a fixed bicharacterto the field K. Furthermore, A is a color commutative, Γ-graded, associative K-algebra and ∆ is a nonzero color commutative Lie color subalgebra of Derε(A). Ofcourse, we suppose that

A1. A is graded ∆-simpleA2. A∆∆ acts faithfully on A.

Finally, we let L be a nonzero Lie color ideal of A ⊗ ∆ = A∆.If V is a K-subspace of A, then the sum of all left (or right) ideals of A contained

in V is the unique largest left (or right) ideal of A contained in V . In particular,the ideals I mentioned in part (i) of the following lemma always exist.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 9

Lemma 3.1. Let V be a ∆-stable, Γ-graded K-subspace of A.

i. If I is the largest left (or right) ideal of A contained in V , then I is a ∆-stable, graded two-sided ideal. In particular, if V contains a nonzero left(or right) ideal, then V = A.

ii. AV and V A are ∆-stable, graded two-sided ideals of A. Thus V 6= 0 impliesthat AV = V A = A and hence that AV+ = V+A = A.

Proof. (i) Assume that I is the unique largest left ideal of A contained in V .For each x ∈ Γ, let πx:A → Ax denote the natural projection, and set J =⊕

∑

x∈Γ πx(I). Since Axπy(I) ⊆ πxy(AxI) ⊆ πxy(I) for all x, y ∈ Γ, it followsthat AxJ ⊆ J and hence that J is a left ideal of A. But I ⊆ J ⊆ V , since V isgraded, and hence I = J is a graded left ideal.

Now let ∂ ∈ H(∆). Then I + ∂(I) ⊆ V and I + ∂(I) is a left ideal of A since,for all a ∈ H(A) and r ∈ H(I), we have ε(∂, a)a∂(r) = −∂(a)r + ∂(ar) ∈ I + ∂(I).Thus I + ∂(I) ⊆ I by the maximal nature of I, and hence I is ∆-stable. Note thatI / A by Lemma 1.1(i). Finally, if V contains a nonzero left (or right) ideal, then Iis a nonzero graded, ∆-stable two-sided ideal of A. But A is graded ∆-simple, soI = A and therefore V = A.

(ii) Certainly AV is a graded left ideal of A and hence AV = V A / A byLemma 1.1(i). Furthermore, AV is ∆-stable since both A and V are ∆-stableand graded. In particular, if V 6= 0, then the graded ∆-simplicity of A impliesthat AV = A. Finally, note that A = AV = AV+ + AV−. Furthermore, byLemma 1.1(iii), AV− is a nil ideal of A and hence it is contained in the Jacobson rad-ical of the algebra. With this, the previous sum implies that A = AV+ = V+A. �

Such subspaces V are of interest because of

Lemma 3.2. Let ∂1, ∂2, . . . , ∂n be K-linearly independent elements of H(∆) andlet V be the set of all a ∈ A such that there exists α = a1∂1 +a2∂2 + · · ·+an∂n ∈ Lwith a = an. Then V is a graded ∆-stable, K-subspace of A.

Proof. Since L is a K-subspace of A∆, it is clear that V is a K-subspace of A.Furthermore, if x ∈ Γ, then since L is a graded subspace of A∆, it follows that∑n

i=1(ai)x∂−1i

∂i = αx ∈ L. Thus Vx∂−1n

⊆ V and V is a graded subspace of A

since x ∈ Γ is arbitrary. Finally, if ∂ ∈ ∆, then 1∂ ∈ A∆ and equation (L3)implies that ∂(a1)∂1 + ∂(a2)∂2 + · · · + ∂(an)∂n = [1∂, α] ∈ L since L / A∆. Thus∂(a) = ∂(an) ∈ V and V is ∆-stable. �

It is now a simple matter to prove

Proposition 3.3. A∆ acts faithfully on A.

Proof. Let L / A∆ be the kernel of the action of A∆ on A and assume, by wayof contradiction, that L 6= 0. Then we can let n ≥ 1 be the minimal homoge-neous support size of a nonzero element of L. In other words, there exist linearly

10 D. S. PASSMAN

independent homogeneous elements ∂1, ∂2, . . . , ∂n in ∆ with L ∩∑n

i=1 A∂i 6= 0and such that all nonzero elements in this intersection have all their A-coefficientsnonzero. Let V be defined as in the previous lemma. Then V is a nonzero graded∆-stable, K-subspace of A. Furthermore, the nature of the action of A∆ on Aimplies that AL ⊆ L and hence AV ⊆ V . Thus V / A by Lemma 1.1(i) and, sinceA is graded ∆-simple, we have V = A. In particular, 1 ∈ V and we can findα = a1∂1 + a2∂2 + · · · + an∂n ∈ L with an = 1. If ∂ ∈ ∆ is arbitrary, then equa-tion (L3) implies that ∂(a1)∂1 + ∂(a2)∂2 + · · · + ∂(an−1)∂n−1 = [1∂, α] ∈ L since∂(an) = ∂(1) = 0. Thus the minimality of n implies that ∂(ai) = 0 for all i and all∂ ∈ ∆. In other words, ai ∈ A∆ for all i, and α ∈ A∆∆. But, by assumption, A∆∆acts faithfully on A, so we have the required contradiction. We conclude, therefore,that L = 0 and hence that A∆ acts faithfully on A. �

Recall that L is a fixed nonzero Lie color ideal of A∆. As a consequence of theabove, we obtain

Lemma 3.4. If ∂ ∈ H(∆), then L ∩ A∂ 6= 0.

Proof. Since L 6= 0, we can choose n ≥ 1 minimal so that there exist linearly inde-pendent homogeneous elements ∂1, ∂2, . . . , ∂n ∈ ∆ with ∂ = ∂1 and L∩

∑ni=1 A∂i 6=

0. The goal is to show that n = 1. Thus suppose that n ≥ 2 and let V be defined asin Lemma 3.2. Then V is a graded ∆-stable, K-subspace of A and the minimalityof n implies that V 6= 0.

Now let α =∑n

i=1 ai∂i be a nonzero homogeneous element contained in thenonzero graded vector space L∩

∑ni=1 A∂i. If r ∈ H(A), then [α, r∂1] ∈ L and it is

clear from equation (L2) that this element has support in { ∂1, ∂2, . . . , ∂n }. Thuswe can write [α, r∂1] =

∑ni=1 ci(r)∂i. The most complicated coefficient here is that

of ∂1. Specifically, c1(r) = −ε(a1∂1, r∂1)r∂1(a1) +∑n

i=1 ai∂i(r) and consequentlyc1(H(A)) 6= 0. Indeed, if c1(H(A)) = 0, then r = 1 implies that ∂1(a1) = 0 andhence 0 = c1(r) =

∑ni=1 ai∂i(r) = α(r) for all r ∈ H(A). In other words, α acts

trivially on H(A), so linearity implies that it acts trivially on all of A. Thus α = 0by the previous proposition, a contradiction.

We have therefore shown that [α, r∂1] is not identically 0, so the minimal natureof n ≥ 2 implies that cn(H(A)) 6= 0. Here cn(r) = −ε(an∂n, r∂1)r∂1(an), so∂1(an) 6= 0 and V contains H(A)∂1(an). Thus V contains the nonzero left idealA∂1(an), and it follows from Lemma 3.1(i) that V = A. Thus there exists β ∈L ∩

∑ni=1 A∂i with β =

∑ni=1 bi∂i and bn = 1. Furthermore, by taking α to be the

∂#n component of β, we see that we could have chosen α with an = 1. But then

∂1(an) = 0, a contradiction. Thus n = 1 and 0 6= L∩A∂1 = L∩A∂, as required. �

Next, we see that only a small part of ∆ has to be studied in detail. Indeed, weprove

Lemma 3.5. If L ⊇ A∂ for some ∂ ∈ H(∆), then L = A∆.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 11

Proof. Fix b ∈ H(A) with ∂(b) 6= 0, let ∂ ′ ∈ H(∆) be arbitrary and define V ′ ={ a ∈ A | a∂′ ∈ L }. As usual, V ′ is a Γ-graded, ∆-stable K-subspace of A. SinceL ⊇ A∂, it follows that

a∂(b)∂′ = [a∂, b∂′] + ε(a∂, b∂′)b∂′(a)∂ ∈ L

for all a ∈ H(A). Hence V ′ contains the nonzero left ideal A∂(b), and Lemma 3.1(i)implies that V ′ = A. In other words, L contains A∂ ′ for all ∂′ ∈ H(∆), so L = A∆as required. �

Lemma 3.6. Let ∂1 and ∂2 be K-linearly independent elements of H(∆).

i. There exist a, b ∈ H(A) with

ε(∂2, a)∂1(a)∂2(b) + ε(∂1, a∂2)∂2(a)∂1(b) 6= 0.

ii. The composition ∂1∂2 is not zero.

Proof. (i) Choose a ∈ H(A) with ∂1(a) 6= 0. Then

α = ε(∂2, a)∂1(a)∂2 + ε(∂1, a∂2)∂2(a)∂1

is a nonzero element of A∆. Thus, since A∆ acts faithfully on A by Proposition 3.3,there exists b ∈ H(A) with ε(∂2, a)∂1(a)∂2(b) + ε(∂1, a∂2)∂2(a)∂1(b) = α(b) 6= 0.

(ii) Let a, b ∈ H(A) be given by part (i) above. Then

∂1∂2(ab) = ∂1

(

∂2(a)b + ε(∂2, a)a∂2(b))

= ∂1∂2(a)b + ε(∂1∂2, a)a∂1∂2(b) + c

where c = ε(∂2, a)∂1(a)∂2(b)+ ε(∂1, a∂2)∂2(a)∂1(b) 6= 0. It follows that ∂1∂2(r) 6= 0for r = a or b or ab, and hence ∂1∂2 6= 0. �

We can now obtain the following crucial fact.

Lemma 3.7. Let charK 6= 2. If ∂ ∈ H+(∆), then L ⊇ A∂.

Proof. Let V = { a ∈ A | a∂ ∈ L }. Then, by Lemma 3.2, V is a graded ∆-stable,K-subspace of A. Of course, Lemma 3.4 implies that V 6= 0. If v ∈ H(V ) andr ∈ H(A), then [v∂, r∂] ∈ L and, since ∂ ∈ H+(∆), Lemma 2.7(i) yields

(1) ∂(vr) − 2∂(v)r ∈ V for all v ∈ H(V ), r ∈ H(A).

Suppose in addition that v ∈ H+(V ). Then ∂(v2) = 2∂(v)v by Lemma 2.7(iii), soreplacing r by vr in (1) implies that the expression

∂(v2r) − 2∂(v)vr = ∂(v2r) − ∂(v2)r = ε(∂, v2)v2∂(r)

12 D. S. PASSMAN

is contained in V . Consequently, we have

(2) v2∂(r) ∈ V for all v ∈ H+(V ), r ∈ H(A).

Fix a ∈ H+(V ) and b ∈ H+(A). Then, by equation (2), we have a2∂(b) ∈ V .Also ∂(b2/2) = ∂(b)b by Lemma 2.7(iii), so a2∂(b)b ∈ V . Using v = a2∂(b) andr = bs in (1), we get

(3) ∂(a2∂(b)bs) − 2∂(a2∂(b))bs ∈ V for all s ∈ H(A).

Similarly, if we use v = a2∂(b)b and r = s in (1), we get

(4) ∂(a2∂(b)bs) − 2∂(a2∂(b)b)s ∈ V for all s ∈ H(A).

Thus, by subtracting expression (4) from expression (3), we obtain

2ε(∂, a2∂(b))a2∂(b)2s = 2[

∂(a2∂(b)b) − ∂(a2∂(b))b]

s ∈ V

for all s ∈ H(A), and therefore V ⊇ a2∂(b)2A.It remains to show that we can choose suitable a and b with a2∂(b)2 6= 0. To

this end, note that V 6= 0 so V+A = A by Lemma 3.1(ii). Similarly, ∂(A) 6= 0and this is a graded ∆-stable, K-subspace of A by the color commutativity rule(C2). Thus, since ∂# ∈ Γ+, we have ∂(A)+ = ∂(A+) and hence ∂(A+)A = A byLemma 3.1(ii) again. Consequently, A = V+A = V+∂(A+)A, and it follows fromLemma 1.1(iii) that not all homogeneous generators of this ideal can be nilpotent.In particular, there exist a ∈ H+(V ) and b ∈ H+(A) with a∂(b) not nilpotent.Hence, by (C1), a2∂(b)2 6= 0, and therefore V contains the nonzero right ideala2∂(b)2A. We conclude from Lemma 3.1(i) that V = A, as required. �

The rest of the argument is somewhat more technical. To start with, we need

Lemma 3.8. Let ∂ ∈ H(∆) and assume that either charK = 2 or ∂ ∈ H−(∆).Then L∩A∂ ⊇ ∂(A)∂ = [A∂,A∂]. In particular, if dimK ∆ = 1, then A∆ is simpleif and only if ∆(A) = A.

Proof. Say L ∩ A∂ = V ∂ so that, as usual, V is a nonzero graded ∆-stable,K-subspace of A with V A = A by Lemma 3.1(ii). Now if a, b ∈ H(A), thenLemma 2.7(i)(ii) implies that [a∂, b∂] = ∂(ab)∂ under either hypothesis, and con-sequently [A∂,A∂] = ∂(A)∂. Furthermore, if a ∈ H(V ) and b ∈ H(A), then[a∂, b∂] ∈ L, so V ⊇ ∂(V A) = ∂(A) and hence L ⊇ ∂(A)∂ = [A∂,A∂].

Finally, suppose ∆ = K∂ is 1-dimensional so that ∆(A) = ∂(A). If ∂(A) = A,then the above implies that L ⊇ A∂ = A∆, and hence A∆ is simple. On the otherhand, if ∂(A) is properly smaller than A, then [A∂,A∂] is a nonzero Lie color idealof A∂ properly smaller than A∂, and therefore A∂ = A∆ is not simple. �

Finally, we use some bicharacter computations to prove

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 13

Lemma 3.9. Let ∂1, ∂2 ∈ H(∆) be K-linearly independent, and assume that eitherchar K = 2 or ∂1, ∂2 ∈ H−(∆). Then L ⊇ A∂1 + A∂2.

Proof. Write M = A∂1 + A∂2 so that M is a Lie color subalgebra of A∆. Weproceed in a series of two steps.

Step 1. L ⊇ [M,M ].

Proof. Let V2 be the set of ∂2-coefficients in L ∩ M = L ∩ (A∂1 + A∂2). ByLemma 2.7(ii), there exists an element a ∈ H(A) with ∂1∂2(a) 6= 0 and we setb = ∂2(a). Then b∂2 ∈ L∩A∂2 by the previous lemma, so for all r ∈ H(A) we have

r∂1(b)∂2 − ε(r∂1, b∂2)b∂2(r)∂1 = [r∂1, b∂2] ∈ L.

In other words, V2 contains the left ideal A∂1(b) and note that ∂1(b) = ∂1∂2(a) 6= 0.Lemmas 3.2 and 3.1(i) now imply that V2 = A, and hence A∂1 + L ⊇ A∂2. Butthen A∂1 +(M ∩L) = M , so M/(M ∩L) ∼= A∂1/(A∂1 ∩L) and the latter Lie coloralgebra is abelian by Lemma 3.8. �

Step 2. L ⊇ M .

Proof. Let a, b, r ∈ H(A). Since L ⊇ [M,M ], we conclude from Lemma 2.8 that

(5) ε(a∂1, ∂2)r∂2(a)∂1 + ε(a, ∂1)r∂1(a)∂2 ∈ L for all a, r ∈ H(A).

Furthermore, replacing r by r∂1(b) in (5) yields

(6) ε(a∂1, ∂2)r∂1(b)∂2(a)∂1 + ε(a, ∂1)r∂1(b)∂1(a)∂2 ∈ L for all a, b, r ∈ H(A)

and, if we interchange the order of the products ∂1(b)∂2(a) and ∂1(b)∂1(a) in theabove, using (C1), we obtain

(7) λ1r∂2(a)∂1(b)∂1 + λ2r∂1(a)∂1(b)∂2 ∈ L for all a, b, r ∈ H(A)

where

λ1 = ε(a∂1, ∂2)ε(b∂1, a∂2) and λ2 = ε(a, ∂1)ε(b∂1, a∂1).

Of course, if we interchange a and b in (6), we get

(8) λ3r∂1(a)∂2(b)∂1 + λ4r∂1(a)∂1(b)∂2 ∈ L for all a, b, r ∈ H(A)

whereλ3 = ε(b∂1, ∂2) and λ4 = ε(b, ∂1).

14 D. S. PASSMAN

Finally, set µ = ε(a∂1∂2, b)ε(∂2, a∂1), multiply expression (7) by µλ4 and sub-tract µλ2 times expression (8). This yields

r[

η1∂2(a)∂1(b) − η2∂1(a)∂2(b)]

∂1 ∈ L

where

η1 = µλ1λ4

= ε(a∂1∂2, b)ε(∂2, a∂1)ε(a∂1, ∂2)ε(b∂1, a∂2)ε(b, ∂1)

= ε(∂1, a∂2)

and

η2 = µλ2λ3

= ε(a∂1∂2, b)ε(∂2, a∂1)ε(a, ∂1)ε(b∂1, a∂1)ε(b∂1, ∂2)

= ε(∂2, a)ε(∂1, ∂1) = −ε(∂2, a),

since ε(∂1, ∂1) = −1 when either char K = 2 or ∂1 ∈ H−(∆).We have therefore shown that L ∩ A∂1 ⊇ I∂1 where I is the principal left ideal

Ac with c = ε(∂2, a)∂1(a)∂2(b)+ε(∂1, a∂2)∂2(a)∂1(b). But Lemma 3.6(i) guaranteesthe existence of homogeneous elements a and b with c 6= 0. Thus we conclude fromLemmas 3.2 and 3.1(i) that L ⊇ A∂1 and, by symmetry, it follows that L ⊇ A∂2. �

There is presumably some natural reason why the element c above is so intimatelyrelated, as in Lemma 3.6(ii), to ∂1∂2(ab). However, we will not pursue this here. Itis now a simple matter to offer the

Proof of the Main Theorems. We know that if A∆ is color simple, then A is graded∆-simple and A∆∆ acts faithfully on A. Furthermore, if dimK ∆ = 1 and if eitherchar K = 2 or charK 6= 2 and ∆ = ∆−, then it follows from Lemma 3.8 that∆(A) = A. With this, the necessity aspects of Theorems 2.3 and 2.5 are proved.

For the sufficiency part of these two theorems, assume that A is graded ∆-simpleand that A∆∆ acts faithfully on A if dimK ∆ ≥ 2. Then, by Lemma 2.4, A∆∆acts faithfully on A in all situations. Now let L be a nonzero Lie color ideal ofA∆. If charK 6= 2 and ∆+ 6= 0, then Lemma 3.7 implies that L ⊇ A∂ for some∂ ∈ H(∆). On the other hand, if either char K = 2 and dimK ∆ ≥ 2 or charK 6= 2and dimK ∆− ≥ 2, then Lemma 3.9 implies that L ⊇ A∂ for some ∂ ∈ H(∆).Thus, in either case, we conclude from Lemma 3.5 that L = A∆ and hence thatA∆ is simple. Finally, suppose that dimK ∆ = 1 and that either char K = 2 orchar K 6= 2 and ∆ = ∆−. Then Theorem 2.5 follows from Lemma 3.8 and theadditional hypothesis that ∆(A) = A. �

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 15

§4. Examples and Discoloration

We begin with a canonical example, intimately related to the ordinary Witt Liealgebra. As usual, we let ε: Γ×Γ → K• be a fixed bicharacter. Furthermore, in thefollowing, we use U(g) to denote the universal enveloping algebra of the Lie coloralgebra g when char K = 0 or the restricted enveloping algebra of the restrictedLie color algebra g when char K = p > 0. See [B] for the basic properties of theseassociative algebras.

Example 4.1. Let V =∑

x∈Γ Vx be a nonzero Γ-graded K-vector space, and viewV as a Lie color algebra by defining [V, V ] = 0. Furthermore, when char K = p > 0,we suppose that V is restricted with trivial pth power map. If A = U(V ), thenA is a Γ-graded, color commutative K-algebra. Now let ∆ =

∑

x∈Γ ∆x, where∆x−1 = HomK(Vx,K), and view ∆ as a Lie color algebra by defining [∆,∆] = 0.Then ∆ acts naturally and faithfully as color derivations on A, and with respectto this action, A ⊗ ∆ = A∆ is a simple Lie color algebra unless dimK V = 1 andeither char K = 2 or charK 6= 2 and V = V−.

Proof. Let g = Z ⊕ V ⊕∆, where 0 6= Z = Kz and where z has grade 1 ∈ Γ. Theng is a Γ-graded K-vector space, and we define [ , ] : g × g → g in such a way thatthe only possible nonzero terms, involving the homogeneous generators of g, aregiven by

[λ, v] = λ(v)z, [v, λ] = ε(x, x)λ(v)z for all x ∈ Γ, v ∈ Vx, λ ∈ ∆x−1 .

Since [g, [g, g]] = 0, it is clear that, with this definition, g is now a Lie color algebra.

Furthermore, if charK = p > 0, we let g be restricted by defining g[p]+ = 0.

Next, via the adjoint map, we know that g acts as color derivations on U(g), andin particular, ∆ ⊆ g acts as color derivations on B = U(Z ⊕ V ). Furthermore, z iscentral in B and there is a natural graded algebra epimorphism φ:B = U(Z⊕V ) →A = U(V ) determined by φ(z) = 1. Thus, since ∆(z) = 0, φ gives rise to aderivation action of ∆ on A. Indeed, this action is faithful, since ∆ is alreadyfaithful on φ(Z ⊕ V ) = K ⊕ V ⊆ A. It now makes sense to form the Lie coloralgebra A ⊗ ∆ = A∆, and we will use the main theorems here to determine whenA∆ is simple.

To start with, any element a ∈ A can be written as

a =∑

α

kαtα1

1 tα2

2 · · · tαn

n ,

a PBW polynomial in the variables t1, t2, . . . , tn. Here { t1, t2, . . . , tn } is a linearlyindependent subset of H(V ) and

αi ∈

{ 0, 1 }, if ti ∈ H−(V )

{ 0, 1, 2, . . . }, if ti ∈ H+(V ) and charK = 0

{ 0, 1, . . . , p − 1 }, if ti ∈ H+(V ) and charK = p > 0.

16 D. S. PASSMAN

Of course, this polynomial expression for a is unique once t1, t2, . . . , tn is chosen,and deg α =

∑ni=1 αi is well-defined, independent of this choice. Note that, if

t1, t2, . . . , tn are given, then the definition of ∆ implies that the “partial derivatives”

∂1, ∂2, . . . , ∂n ∈ H(∆) exist with ∂#i = (t#i )−1 and with ∂i(tj) = δi,j , the Kronecker

delta.Now suppose that the element a above is contained in A∆. Then for all i,

0 = ∂i(a) =∑

α

kαε(∂i, tα1

1 · · · tαi−1

i−1 )αitα1

1 · · · tαi−1i · · · tαn

n

by (D2) and Lemma 2.7(iii). In particular, if kα 6= 0, then the uniqueness aspect ofthis expression implies that αi = 0, and consequently a = k0 ∈ K. Thus A∆ = Kand A∆ ⊗ ∆ = K ⊗ ∆ = ∆ acts faithfully on A. Furthermore, let I be a nonzerograded ∆-stable ideal of A and let 0 6= a ∈ I be an element of minimal degree.Since deg ∂(a) < deg a for all ∂ ∈ H(∆), it follows that ∂(a) = 0 and hence thata ∈ A∆ = K. But this implies that I = A, so A is graded ∆-simple and the resultfollows from Theorems 2.3 and 2.5. �

The above construction, when A∆ is not simple, yields a familiar example. Tostart with, we have dimK V = 1, so say V = Kt. Thus, since either char K = 2 orchar K 6= 2 and t ∈ H−(V ), it follows that A = K ⊕ Kt with t2 = 0. Furthermore,by definition of the action, we have ∆ = K∂ with ∂(r + st) = s for all r, s ∈ K. Inparticular, this is precisely Example 2.6 with F = K and δ = 0.

Of course, starting with an arbitrary A and ∆, we can obtain additional Liestructures by considering group rings and formal exponentials as in [Pa, §1]. Sincethe arguments for this are fairly routine, and since these new examples may not beparticularly interesting, we prefer to take a somewhat different tack. Specifically,we consider what the discoloration functor has to say about Witt-type Lie coloralgebras.

Let Γ and K be given, and recall that a 2-cocycle is a map τ : Γ × Γ → K •

satisfying

H1. τ(x, yz)τ(y, z) = τ(x, y)τ(xy, z) for all x, y, z ∈ Γ.

Furthermore, using τ , we can construct the twisted group algebra K τ [Γ] of Γ overK. The latter is, of course, the K-algebra with basis Γ = {x | x ∈ Γ } and withmultiplication defined distributively by

H2. x y = τ(x, y)xy for all x, y ∈ Γ.

Note that (H1) is precisely equivalent to the associativity of multiplication, andhence Kτ [Γ] is an associative K-algebra which is Γ-graded if we set K τ [Γ]x = Kxfor all x ∈ Γ. Of course, if τ is trivial, that is if τ(Γ,Γ) = 1, then K τ [Γ] = K[Γ] isthe ordinary group algebra of Γ.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 17

Lemma 4.2. Let Kτ [Γ] be given as above and define µτ (x, y) = τ(x, y)/τ(y, x) forall x, y ∈ Γ. Then

i. x y = µτ (x, y)y x for all x, y ∈ Γ.ii. µτ : Γ × Γ → K• is a bicharacter with µτ (x, x) = 1 for all x ∈ Γ.

Proof. (i) This follows from (H2) since

x y = τ(x, y)xy = τ(x, y)yx = τ(x, y)/τ(y, x) y x.

(ii) Certainly µτ (x, y)µτ (y, x) = 1 and µτ (x, x) = 1. Moreover, if x, y, z ∈ Γ,then yz = ky z for some k ∈ K• and hence

µτ (x, yz)yz x = x yz = kx y z = kµτ (x, y)y x z

= kµτ (x, y)µτ (x, z)y z x = µτ (x, y)µτ (x, z)yz x.

Thus µτ (x, yz) = µτ (x, y)µτ (x, z), as required. �

If V = ⊕∑

x∈Γ Vx and W = ⊕∑

x∈Γ Wx are Γ-graded K-vector spaces, then wedefine the diagonal product V � W to be the diagonal subspace of V ⊗W given byV �W = ⊕

∑

x∈Γ Vx ⊗Wx. Of course V �W is graded with (V �W )x = Vx ⊗Wx.Furthermore, if A and B are Γ-graded K-algebras, then A � B is a subalgebra ofA ⊗ B, and it is Γ-graded with (A � B)x = Ax ⊗ Bx. Since the tensor productis associative and since (U � V ) � W is the diagonal subspace of (U ⊗ V ) ⊗ W ,it follows that the diagonal product is also associative. Note that A � K[Γ] ∼= Aas Γ-graded algebras. Furthermore, if τ is a 2-cocycle, then so is τ−1 = 1/τ , and

clearly we have µτ−1 = µ−1τ and Kτ [Γ] � Kτ−1

[Γ] ∼= K[Γ].In the following, we will be concerned with more that one bicharacter of Γ. Thus,

to be precise, we will say, for example, that an algebra is ε-commutative if it is colorcommutative with respect to the bicharacter ε.

Lemma 4.3. Let ε be a bicharacter of Γ and let τ be a 2-cycle. Then ε′ = εµτ

is also a bicharacter of Γ and the operator T on the set of Γ-graded associativeK-algebras given by T (A) = A � Kτ [Γ] yields a one-to-one correspondence fromthe isomorphism classes of ε-commutative algebras to the isomorphism classes of

ε′-commutative algebras. Here T−1 is given by T−1(A′) = A′� Kτ−1

[Γ].

Proof. If S is defined by S(A′) = A′� Kτ−1

[Γ], then

S(T (A)) = (A � Kτ [Γ]) � Kτ−1

[Γ]

= A � (Kτ [Γ] � Kτ−1

[Γ]) ∼= A � K[Γ] ∼= A.

Thus ST is the identity on isomorphism classes of algebras, and similarly so isTS. In other words, S = T−1 and T is a one-to-one correspondence. Finally, the

18 D. S. PASSMAN

comment on ε- and ε′-commutative algebras follows immediately from ε′ = εµτ andLemma 4.2(i). �

Next, we turn to derivations. Let A be a Γ-graded algebra and let ∂ be ahomogeneous ε-derivation of A of grade ∂# = z. Then ∂ ⊗ z acts on A � Kτ [Γ] by

H3. (∂ ⊗ z): a ⊗ x 7→ ∂(a) ⊗ z x for all x ∈ Γ, a ∈ Ax,

and it is easy to see that ∂ ⊗ z is an ε′-derivation of A�Kτ [Γ]. Indeed, let x, y ∈ Γand let a ∈ Ax, b ∈ Ay. Since

ε(z, x)z x = ε(z, x)µτ (z, x)z x = ε′(z, x)x z,

by Lemma 4.2(i), it follows that

(∂ ⊗ z)[

(a ⊗ x)(b ⊗ y)]

= (∂ ⊗ z)(ab ⊗ x y) = ∂(ab) ⊗ z x y

=[

∂(a)b + ε(z, x)a∂(b)]

⊗ z x y

= ∂(a)b ⊗ z x y + ε(z, x)a∂(b) ⊗ z x y

= ∂(a)b ⊗ z x y + ε′(z, x)a∂(b) ⊗ x z y

=[

(∂ ⊗ z)(a ⊗ x)]

(b ⊗ y) + ε′(z, x)(a ⊗ x)[

(∂ ⊗ z)(b ⊗ y)]

,

as required. More to the point, we have

Lemma 4.4. Let A be a Γ-graded associative K-algebra. Then the map T :∆ 7→∆ � Kτ [Γ] yields a one-to-one correspondence from the family of graded vectorspaces of ε-commutative, ε-derivations of A to the family of graded vector spaces ofε′-commutative, ε′-derivations of T (A) = A�Kτ [Γ]. Here T−1 is essentially given

by T−1:∆′ 7→ ∆′� Kτ−1

[Γ].

Proof. By the above, we know that ∆′ = ∆ � Kτ [Γ] acts as ε′-derivations on thealgebra A′ = A � Kτ [Γ], and this action is clearly faithful. Furthermore, if ∆ isε-commutative, then Lemma 4.2(i) implies that ∆′ is ε′-commutative. Finally, if

we let S(∆′) = ∆′� Kτ−1

[Γ] and S(A′) = A′� Kτ−1

[Γ], then

S(T (∆)) = (∆ � Kτ [Γ]) � Kτ−1

[Γ] ∼= ∆ � K[Γ],

S(T (A)) = (A � Kτ [Γ]) � Kτ−1

[Γ] ∼= A � K[Γ] ∼= A,

and ∆� K[Γ] acts on A � K[Γ] in precisely the same way that ∆ acts on A. Thus,S is essentially equal to T−1. �

Now it is clear from the preceding two lemmas that the maps � K τ [Γ] yielda one-to-one correspondence from the isomorphism classes of ε-Lie color algebrasof Witt type to the isomorphism classes of ε′-Lie color algebras of Witt type. Tomake this precise, we first need the following variant of [S, Theorem 2].

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 19

Lemma 4.5. The map T : g 7→ g � Kτ [Γ] yields a one-to-one correspondence fromthe isomorphism classes of Γ-graded ε-Lie color algebras to the isomorphism classesof Γ-graded ε′-Lie color algebras. Here the color brackets are related by

[a ⊗ x, b ⊗ y]′ = [a, b] ⊗ x y for all x, y ∈ Γ, a ∈ gx, b ∈ gy,

and T−1 is given by T−1: g′ 7→ g′ � Kτ−1

[Γ]. Furthermore, g is simple if and onlyif T (g) is simple.

Proof. Let g be an ε-Lie color algebra and let U(g) denote its universal envelopingalgebra. Then g ⊆ U(g) and in fact, g is an ε-Lie color subalgebra of Lε(U(g)),the ε-Lie color algebra of U(g). Now observe that g � K τ [Γ] is contained in theassociative algebra U(g) � Kτ [Γ] and we show that it is an ε′-Lie color subalgebraof Lε′(U(g) � Kτ [Γ]). Indeed, if x, y ∈ Γ, a ∈ gx, b ∈ gy, and if [ , ]′ denotes theε′-Lie bracket, then Lemma 4.2(i) yields

[a ⊗ x, b ⊗ y]′ = (a ⊗ x)(b ⊗ y) − ε′(a, b)(b ⊗ y)(a ⊗ x)

= (ab ⊗ x y) − ε(a, b)µτ (x, y)(ba ⊗ y x)

= (ab ⊗ x y) − ε(a, b)(ba ⊗ x y)

=(

ab − ε(a, b)ba)

⊗ x y = [a, b] ⊗ x y.

In otherwords, we have shown that g′ = g � Kτ [Γ] is an ε′-Lie color algebra withε′-Lie bracket as above. Furthermore, if H is a color ideal of g, then since H mustbe Γ-graded, it is clear that H′ = H � Kτ [Γ] is a color ideal of g′. Finally, since

Kτ [Γ] � Kτ−1

[Γ] ∼= K[Γ], it is easy to see that g′� Kτ−1

[Γ], with the appropriateLie bracket, is naturally isomorphic to g. �

Putting this all together, we quickly obtain

Proposition 4.6. The map T : g 7→ g � Kτ [Γ] yields a one-to-one correspondencefrom the isomorphism classes of ε-Lie color algebras of Witt type to the isomorphismclasses of ε′-Lie color algebras of Witt type. Specifically, if g = A∆, then g′ =g � Kτ [Γ] ∼= A′∆′ where A′ = A � Kτ [Γ] and ∆′ = ∆ � Kτ [Γ]. Furthermore, g issimple if and only if g′ is simple.

Proof. Let A be an ε-commutative, graded associative K-algebra and let ∆ be anε-commutative, graded K-vector space of ε-derivations of A. If g = A∆, then itsuffices to show that g′ = g � Kτ [Γ] is naturally isomorphic to A′∆′ where A′ and∆′ are given by Lemmas 4.3 and 4.4. For this, it is best to view the constructionof A∆ as formal multiplication.

Let x, y, z, w ∈ Γ, let a ∈ Ax, α ∈ ∆y, b ∈ Az and β ∈ ∆w. Then a′ = a ⊗ xand b′ = b ⊗ z are elements of A′, while α′ = α ⊗ y and β′ = β ⊗ w are in ∆′. For

20 D. S. PASSMAN

convenience, write xy = sx y and zw = tz w where s = τ(x, y)−1 and t = τ(z, w)−1.Then, we have

(aα)′ = aα ⊗ xy = s(a ⊗ x)(α ⊗ y) = sa′α′

(bβ)′ = bβ ⊗ zw = t(b ⊗ z)(β ⊗ w) = tb′β′.

In particular, if [[ , ]]′ denotes the ε′-Lie bracket of A′∆′, then

[[(aα)′, (bβ)′]]′ = st[[a′α′, b′β′]]′ = st(

a′α′(b′)β′ − ε′(a′α′, b′β′)b′β′(a′)α′)

.

Next, observe that

(st)a′α′(b′)β′ = (st)(a ⊗ x)(α(b) ⊗ y z)(β ⊗ w)

= aα(b)β ⊗ (sx y)(tz w) = aα(b)β ⊗ xy zw,

and

(st)ε′(a′α′,b′β′)b′β′(a′)α′

= ε(aα, bβ)µτ (xy, zw)(st)(b ⊗ z)(β(a) ⊗ w x)(α ⊗ y)

= ε(aα, bβ)µτ (xy, zw)(

bβ(a)α ⊗ (tz w)(sx y))

= ε(aα, bβ)(

bβ(a)α ⊗ µτ (xy, zw)zw xy)

= ε(aα, bβ)(bβ(a)α ⊗ xy zw).

Thus, by the above computations and the previous lemma,

[[(aα)′, (bβ)′]]′ =(

aα(b)β − ε(aα, bβ)bβ(a)α)

⊗ xy zw

= [aα, bβ] ⊗ xy zw = [(aα)′, (bβ)′]′,

so g′ ∼= A′∆′ as ε′-Lie color algebras. Lemma 4.5 now yields the result. �

Of course, the payoff here comes from the discoloration functor. If ε: Γ×Γ → K •

is a bicharacter, let Γε+ denote the kernel of the parity function x 7→ ε(x, x). Then

we have

Theorem 4.7. Let ε, ε′: Γ × Γ → K• be two bicharacters with Γε+ = Γε′

+. Thenthere is a one-to-one correspondence between the isomorphism classes of (simple)ε-Lie color algebras of Witt type and the isomorphism classes of (simple) ε′-Liecolor algebras of Witt type.

Proof. If Γε+ = Γε′

+, then it follows from [S, Lemma 2] and [Po, Theorem] that thereexists a 2-cocycle τ : Γ × Γ → K• with ε′ = εµτ (see Corollary 5.2 of this paper).Now apply Proposition 4.6. �

Of course, one can view the preceding result in two ways. In the positive direc-tion, it offers a method of constructing simple Lie color algebras of Witt type fromthe corresponding super algebras. In the negative direction, it indicates that therewill be no new surprises.

SIMPLE LIE COLOR ALGEBRAS OF WITT-TYPE 21

§5. Appendix

At present, the result of [Po] quoted in the proof of Theorem 4.7 seems to requirethat the field K be algebraically closed. Fortunately, we do not need as strong aconclusion as is contained in [Po, Theorem], where it is shown that a 2-cocycleexists which is multiplicative in each factor. Indeed, we just require the existenceof a 2-cocycle, and this can be proved without any assumptions on the field. For thesake of completeness, we offer an elementary demonstration of this fact, changingnotation somewhat and using a group theoretic argument.

Proposition 5.1. Let A and Z be multiplicative abelian groups and let ε:A×A →Z be a map. The following are equivalent.

i. ε(ab, c) = ε(a, c)ε(b, c), ε(b, a) = ε(a, b)−1 and ε(a, a) = 1 for all a, b, c ∈ A.ii. There exists a group G with Z ⊆ Z(G), G/Z = A and such that the com-

mutator map ( , ):G/Z × G/Z → Z is equal to ε.iii. There exists a 2-cocyle τ :A × A → Z with ε(a, b) = τ(a, b)/τ(b, a) for all

a, b ∈ A.

Proof. We will show that (ii) is equivalent to (i) and to (iii). Most of this is routine.The only nontrivial argument occurs in the proof of implication (i) ⇒ (ii).

Suppose first that (ii) holds. Then G is a class 2 group and consequently thecommutator map G×G → Z given by x×y 7→ (x, y) = x−1y−1xy satisfies (xy, z) =(x, z)(y, z), (y, x) = (x, y)−1 and (x, x) = 1 for all x, y, z ∈ G. Furthermore, (x, y)is constant on the cosets of Z. Thus the commutator gives rise to a map fromA×A = G/Z ×G/Z to Z and if this map is ε, then ε satisfies the conditions of (i).

Next, for each a ∈ A, choose a ∈ G with Za = a ∈ G/Z = A. Then, for alla, c ∈ A, we have a c = τ(a, c)ac with τ :A×A → Z a 2-cocycle. Furthermore, sincec a = τ(c, a)ca and ca = ac, we obtain

(a, c) = a−1c−1a c = (c a)−1a c

= τ(a, c)τ(c, a)−1ca−1ac = τ(a, c)τ(c, a)−1 .

Thus ε(a, c) = (a, c) has the appropriate form and (iii) is proved.Next, we show that (iii) ⇒ (ii). To this end, let τ :A × A → Z be a 2-cocycle.

Then τ defines a group G with central subgroup Z and with G/Z = A. Indeed, foreach a ∈ A, there exists a ∈ G with Za = a and with a c = τ(a, c)ac. As in theabove displayed equation, we have (a, c) = τ(a, c)τ(c, a)−1 = ε(a, c), by assumption,and thus the commutator map ( , ):G/Z × G/Z → Z is equal to ε, as required.

It remains to prove that (i) ⇒ (ii). To this end, we first embed Z in a divisibleabelian group D. Next, we let X be the group generated by D and the symbols{xa | a ∈ A } with D central and with additional relations given by (xa, xb) =ε(a, b) ∈ D for all a, b ∈ A. Since ε(b, a) = ε(a, b)−1 and ε(a, a) = 1, it is clear thatX is just the central product of the free class 2 group on the symbols {xa | a ∈ A }

22 D. S. PASSMAN

with the group D, identifying (xa, xb) with ε(a, b). In particular, X/D is isomorphicto the free abelian group on the elements Dxa, and consequently there exists anepimorphism θ:X → A given by θ(D) = 1 and θ(xa) = a for all a ∈ A.

Let N = ker θ, so that X/N ∼= A. If x = d∏

i xei

ai∈ N with d ∈ D, then

1 = θ(x) =∏

i aei

i . Hence, for all b ∈ A, we have

(x, xb) = (d∏

i xei

ai, xb) =

∏

i(xai, xb)

ei

=∏

i ε(ai, b)ei = ε(

∏

i aei

i , b) = ε(1, b) = 1,

since ε is multiplicative in its first variable. It follows that N is central in X and, inparticular, that N is abelian. But N ⊇ D and D is divisible, so N = D×M for somesubgroup M . Furthermore, since M is central in X, it is normal and we can setH = X/M . Note that D = N/M /X/M = H and H/D ∼= X/N = A. Indeed, since(xa, xb) = ε(a, b) ∈ D, it follows that the commutator map ( , ):H/D ×H/D → Dis equal to ε.

Finally, note that H ′ = (H,H) ⊆ Z and that Z is central in H. Thus H/Z is anabelian group containing the divisible group D/Z as a subgroup. It therefore followsthat H/Z = G/Z × D/Z for some subgroup G of H. Of course, G/Z ∼= H/D ∼= Aand it is clear that ( , ):G/Z × G/Z → Z is equal to ε. Thus (ii) is proved. �

Reverting to the notation of the previous section, we have

Corollary 5.2. Let ε, ε′: Γ × Γ → K• be two bicharacters with Γε+ = Γε′

+. Thenthere exists a 2-cocycle τ : Γ × Γ → K• with

ε′(x, y) = ε(x, y)µτ (x, y) = ε(x, y)τ(x, y)/τ(y, x) for all x, y ∈ Γ.

Proof. It is clear that ε′ε−1: Γ × Γ → K• is a bicharacter with ε′ε−1(x, x) = 1 forall x ∈ Γ. Thus, by (i) ⇒ (iii) of the previous proposition, with A = Γ and Z = K •,there exists a 2-cocycle τ : Γ × Γ → K• with ε′(x, y)ε(x, y)−1 = τ(x, y)τ(y, x)−1 =µτ (x, y) for all x, y ∈ Γ, and the result follows. �

References

[B] Yu. A. Bahturin, A. A. Mikhalev, V. M. Petrogradsky and M. V. Zaicev, Infinite Dimen-

sional Lie Superalgebras, de Gruyter, Berlin, 1992.

[KN] K. Kishimoto and A. Nowicki, On the image of derivations, Comm. Algebra 23 (1995),

4557–4562.

[Pa] D. S. Passman, Simple Lie algebras of Witt-type, J. Algebra (to appear).

[Po] Horia C. Pop, A generalization of Scheunert’s theorem on cocycle twisting of color Lie

algebras (to appear).

[S] M. Scheunert, Generalized Lie algebras, J. Math. Phys. 20 (1979), 712–720.

Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706

E-mail address: passman@math.wisc.edu