Color cyclic homology and Steinberg Lie color algebras

Yongjie WANG1, Shikui SHANG1, Yun GAO1,2
1 School of Mathematical Sciences, University of Science and Technology of China,
Hefei 230026, China
2 Department of Mathematics and Statistics, York University, Toronto, Ontario M3J 1P3,
Canada
c
Higher Education Press and Springer-Verlag Berlin Heidelberg 2015
Abstract The present paper contains two interrelated developments. First,
the basic properties of the construction theory over the Steinberg Lie color
algebras are developed in analogy with Steinberg Lie algebra case. This is
done on the example of the central closed of the Steinberg Lie color algebras.
The second development is that we define the first ε-cyclic homology group
HC1(R, ε) of the Γ-graded associative algebra R (which could be seemed as the
generalization of cyclic homology group and the Z/2Z-graded version of cyclic
homology that was introduced by Kassel) to calculate the universal central
extension of Steinberg Lie color algebras.
Keywords Steinberg Lie color algebra, ε-cyclic homology group, universal
central extension
MSC 17B10, 17B35, 17B70
1 Introduction
Lie color algebras, originally introduced by Rittenberg and Wylerin [10], can
be seen as a direct generalization of Lie (super) algebras. Indeed, the latter are
defined through antisymmetric (commutator) or symmetric (anticommutator)
products, although for the former the product is neither symmetric nor antisymmetric
and is defined by means of a commutation factor. In recent years, Lie
color algebras have become an interesting subject of mathematics and physics.
We assume that k is an algebraically closed field with characteristic zero. Let
Γ be an additive group, and let a skew-symmetric bicharacter of Γ be a map
ε: Γ × Γ → k∗ such that
Received March 7, 2015; accepted March 18, 2015
Corresponding author: Yongjie WANG, E-mail: wyjie@mail.ustc.edu.cn
1180 Yongjie WANG et al.
ε(λ, μ) = ε(μ, λ)−1, ε(λ, μ + ν) = ε(λ, μ)ε(λ, ν), ∀ λ, μ, ν ∈ Γ.
Let R = ⊕γ∈ΓRγ be a Γ-graded k-vector space. A k-bilinear map
[·, ·] : R × R → R
is called an ε-Lie color bracket on R if
[a, b] = −ε(α, β)[b, a] (Skew symmetry),
[a, [b, c]] = [[a, b], c] + ε(α, β)[b, [a, c]] (Jacobi identity)
for all homogeneous elements a ∈ Rα, b ∈ Rβ, and c ∈ R. The algebra structure
(R, [·, ·]) is called a (Γ, ε)-Lie color algebra or simply a Lie color algebra. If Γ = 0
and ε = 1, then a (Γ, ε)-Lie color algebra is a Lie algebra. If Γ = Z/2Z and
ε(i, j) = (−1)ij , ∀ i, j ∈ Z/2Z,
then a (Γ, ε)-Lie color algebra is a Lie superalgebra.
Let R be a Γ-graded associative algebra, and fix a skew-symmetric
bicharacter ε. Then R becomes a (Γ, ε)-Lie color algebra, which we denote
by Rε,−. In fact, the bilinear bracket on R is defined by
[a, b] = ab − ε(α, β)ba
for the homogeneous elements a ∈ Rα and b ∈ Rβ. In particular, let Mn(R) be
the n × n matrix with coefficients in R, and let
deg(Eij(a)) = deg(a)
for any homogeneous element a ∈ R. Under the Lie color bracket,
[Eij(a),Ekl(b)] = δjkEil(ab) − δliε(α, β)Ekj (ba),
where Eij, Ekl are standard matrix units for 1 i, j, k, l n and a ∈ Rα, b ∈
Rβ, becomes a Lie color algebra, denoted by gln(R, ε). The elementary (Γ, ε)-Lie
color algebra sln(R, ε) is the subalgebra of gln(R, ε) generated by the elements
Eij(a), a ∈ R, 1 i = j n.
This article is devoted to the study of the basic properties of Steinberg
Lie color algebras and the universal central extensions of Lie color algebras.
Our present work is motivated by the results and methods in the Lie algebra
case [2–4,12], but central extensions of Steinberg Lie algebra, Steinberg Lie
superalgebra, and matrix Lie superalgebras over Z/2Z-graded algebras could
be seem as special cases of this article. To study the universal central extension
of sln(R, ε) in the category of (Γ, ε)-Lie color algebras, we define Steinberg Lie
color algebra stn(R, ε) by generators corresponding to Eij(a) and those same
canonical relations in analogy with the Lie algebra case. So one has a Lie color
algebra homomorphism φ from stn(R, ε) to sln(R, ε). It is known that φ yields
Color cyclic homology and Steinberg Lie color algebras 1181
a central extension. When n 5, they are the universal central extensions;
moreover, the kernel of φ can be identified with the first ε-cyclic homology
group HC1(R, ε) of the Γ-graded associative algebra R. When n = 3 or n = 4,
stn(R, ε) is not the universal central extension of sln(R, ε). We also construct
the universal central extension of the latter two cases.
2 Basicsonsln(R, ε) and stn(R, ε)
For convenience, we denote all the homogeneous elements of R by S(R). If
a, b ∈ S(R), we simply write
ε(a, b) = ε(α, β), a∈ Rα, b ∈ Rβ.
Let R = ⊕γ∈ΓRγ be a Γ-graded associative algebra over a commutative ring
K with identity 1 ∈ R0. We have the (Γ, ε)-Lie color algebra gln(R, ε) of n × n
matrices with coefficients in R (see Section 1). For n 3, the elementary (Γ, ε)-
Lie color algebra sln(R, ε) is a subalgebra of gln(R, ε), which have the following
three equivalent definitions.
Lemma 1 For (Γ, ε)-Lie color algebra sln(R, ε), the following statements are
equivalent:
(i) sln(R, ε) is the subalgebra of gln(R, ε) generated by all the elements
Eij(a), a ∈ S(R), 1 i = j n;
(ii) sln(R, ε) = [gln(R, ε), gln(R, ε)];
(iii) sln(R, ε) = {X ∈ gln(R, ε) | trX ∈ [R,R]}.
Proof (i) ⇒ (ii). Since 1 ∈ R0 and n 3, we can take distinct 1 i, j, k n
such that
Eij(a) = [Eik(a),Ekj(1)], ∀ a ∈ S(R), 1 i = j n.
Obviously, [gln(R, ε), gln(R, ε)] is a subalgebra of gln(R, ε). Thus, the
subalgebra of gln(R, ε) generated by Eij(a) is contained in [gln(R, ε), gln(R, ε)].
(ii) ⇒ (iii). Recall the bracket product of gln(R, ε), we have the trace of
any element of [gln(R, ε), gln(R, ε)] belongs to [R,R].
(iii) ⇒ (i). Denote by L the subalgebra generated by Eij(a), a ∈ S(R).
First, Eij(R) ⊆ L for 1 i = j n. Since ε is a bicharacter, ε(α, 0) = 1 for
all α ∈ Γ. Thus,
Eii(a) − Ejj(a) = [Eij(a),Eji(1)] ∈ L

, ∀ a ∈ R, 1 i = j n.
Finally, if a, b ∈ S(R), then
E11(ab − ε(a, b)ba) = [E1i(a),Ei1(b)] − ε(a, b)[E1i(ba),Ei1(1)], i = 1.
We have
E11([R,R]) ⊆ L

.
1182 Yongjie WANG et al.
Let
X =

1 i,j n
Eij(aij),
n
i=1
aii ∈ [R,R].
Then
X =
n
i=1
Eii(aii) +

1 i =j n
Eij(aij)
= E11
n
i=1
aii

−
n
i=2
(E11(aii) − Eii(aii)) +

1 i =j n
Eij(aij).
Therefore, X ∈ L . We complete the proof.
Also, for any a, b ∈ R, we have
[Eij(a),Ejk(b)] = Eik(ab), i = j = k = i,
[Eij(a),Ekl(b)] = 0, j = k, i = l.
Clearly, sln(R, ε) has a vector space decomposition
sln(R, ε) = ξ0

1 i =j n
ξij , (1)
where
ξij = {Eij(a) | a ∈ R}, 1 i = j n,
and ξ0 is the subalgebra of diagonal matrix whose trace belongs to [R,R].
Definition 2 For n 3, the Steinberg Lie color algebra stn(R, ε) is defined
to be the (Γ, ε)-Lie color algebra over k generated by the homogeneous
elements Xij(a) with deg(Xij(a)) = deg a for any a ∈ S(R), 1 i = j n,
subject to the Γ-homogeneous relations:
a → Xij(a) is a k-linear map, (2)
[Xij(a),Xjk(b)] = Xik(ab), i, j, k distinct, (3)
[Xij(a),Xkl(b)] = 0, j = k, i = l, (4)
where a, b ∈ S(R), 1 i, j, k, l n.
Both the (Γ, ε)-Lie color algebras sln(R, ε) and stn(R, ε) are perfect. There
exists a surjective (Γ, ε)-Lie color algebra epimorphism
φ: stn(R, ε) → sln(R, ε)
such that
φ(Xij(a)) = Eij(a).
Color cyclic homology and Steinberg Lie color algebras 1183
Set
Tij(a, b) = [Xij(a),Xji(b)], a,b∈
S(R), 1 i = j n.
Then Tij(a, b) is a homogeneous element with degree of deg a+deg b. Moreover,
we have the following lemma.
Lemma 3 For any a, b, c ∈ S(R) and distinct i, j, k, l, we have
Tij(a, b) = −ε(a, b)Tji(b, a), (5)
[Tij(a, b),Xkl(c)] = 0, (6)
[Tij(a, b),Xik(c)] = Xik(abc), [Tij(a, b),Xki(c)] = −ε(a + b, c)Xki(cab), (7)
[Tij(a, b),Xjk(c)] = −ε(a, b)Xjk(bac),
[Tij(a, b),Xkj(c)] = ε(a, b)ε(a + b, c)Xkj (cba),
(8)
[Tij(a, b),Xij (c)] = Xij(abc + ε(a, b)ε(a + b, c)cba), (9)
Tij(ab, c) = Tik(a, bc) − ε(a + b, c)Tjk(ca, b), (10)
Tkj(c, 1) = −Tjk(c, 1) = Tkj(1, c). (11)
Proof By the ε-skew symmetry, one has
Tij(a, b) = −ε(a, b)[Xji(b),Xij (a)] = −ε(a, b)Tji(b, a).
From the ε-Jacobi identity, we have
[x, [y, z]] = [[x, y], z] + ε(x, y)[y, [x, z]]
for the homogeneous elements x, y, z in the (Γ, ε)-Lie color algebra.
Therefore, (6) is obvious, and
[Tij(a, b),Xik(c)] = [[Xij(a),Xji(b)],Xik(c)] = Xik(abc),
[Tij(a, b),Xki(c)] = ε(b, c)[[Xij (a),Xki(c)],Xji(b)]
= − ε(b, c)ε(a, c)[[Xki(c),Xij (a)],Xji(b)]
= − ε(a + b, c)Xki(cab),
which gives (7).
(8) also comes from the ε-Jacobi identity.
For (9), we have
[Tij(a, b),Xij (c)] = [Tij(a, b), [Xik(c),Xkj (1)]]
= [[Tij(a, b),Xik(c)],Xkj (1)] − [[Tij(a, b),Xkj (1)],Xik(c)]
= [Xik(abc),Xkj (1)] − ε(a, b)[Xkj (ba),Xik(c)]
= Xij(abc + ε(a, b)ε(a + b, c)cba).
1184 Yongjie WANG et al.
Using ε-Jacobi identity and (3), we have
Tij(ab, c) = [[Xik(a),Xkj(b)],Xji(c)]
= [Xik(a), [Xkj (b),Xji(c)]] + ε(b, c)[[Xik(a),Xji(c)],Xkj (b)]
= [Xik(a),Xki(bc)] − ε(a + b, c)[Xjk(ca),Xkj (b)]
= Tik(a, bc) − ε(a + b, c)Tjk(ca, b).
Taking a = b = 1 in (10), we get
Tij(1, c) = Tik(1, c) − Tjk(c, 1),
and exchanging j and k, we have
Tik(1, c) = Tij(1, c) − Tkj(c, 1).
Thus,
Tkj(c, 1) + Tjk(c, 1) = 0,
which gives us
Tkj(c, 1) = −Tjk(c, 1) = Tkj(1, c).
By Lemma 3, we have the following result.
Proposition 4 Let
T :=

1 i<j n
[Xij(R),Xji(R)].
Then T is a subalgebra of stn(R, ε) with [T,Xij(R)] ⊆ Xij(R), and
stn(R, ε) = T

1 i =j n
Xij(R). (12)
Moreover, T contains the center Z of stn(R, ε) and ker φ ⊆ Z. Thus, (stn(R, ε),
φ) is a central extension of sln(R, ε).
Proof By (6)–(9), we have [T,Xij(R)] ⊆ Xij(R). Using ε-skew symmetry
and ε-Jacobi identity, T is closed under the bracket, i.e., it is a subalgebra
of stn(R, ε).
From decomposition (1) of sln(R, ε) and the epimorphism φ, we have the
direct sum decomposition (12). Especially, we see that φ is injective restricted
on Xij(R) for 1 i = j n and φ(T) = ξ0. Since the center of sln(R, ε) is
contained in ξ0, we have Z ⊆ T.
Finally, ker φ ⊆ T. For any 1 i < j n, [ker φ,Xij(R)] ⊆ Xij(R). On the
other hand, ker φ is an ideal of stn(R, ε). One has
[ker φ,Xij (R)] ⊆ ker φ ⊆ T.
Thus,
[ker φ,Xij(R)] ⊆ T ∩ Xij(R) = {0}.
Color cyclic homology and Steinberg Lie color algebras 1185
Since stn(R, ε) is generated by Xij(a), a ∈ R, we have
[ker φ, stn(R, ε)] = 0, ker φ ⊆ Z.
Now, setting b = 1 in (10), we have
Tij(a, c) = Tik(a, c) − ε(a, c)Tjk(ca, 1). (13)
For a, b ∈ S(R), we set
t(a, b) = T1j(a, b) − ε(a, b)T1j (ba, 1). (14)
Then, by (10), (11), and (13), we have
t(a, b) = T1k(a, b) − ε(a, b)Tjk(ba, 1) − ε(a, b)T1j (ba, 1)
= T1k(a, b) − ε(a, b)(T1j (ba, 1) − Tkj(ba, 1))
= T1k(a, b) − ε(a, b)T1k(ba, 1),
which shows that t(a, b) does not depend on the choice of j. Then, both Tij(a, b)
and t(a, b) are homogeneous elements with degree deg a+deg b for a, b ∈ S(R).
The following lemma is immediate.
Lemma 5
ker φ =

i
t(ai, bi)


i
[ai, bi] = 0, ai, bi ∈ S(R)

.
Proof First, (10) is equivalent to
Tjk(ca, b) = ε(c, a + b)(Tik(a, bc) − Tij(ab, c)).
Taking c = 1 and i = 1, we have
Tjk(a, b) = T1k(a, b) − T1j(ab, 1) = h(a, b) + ε(a, b)T1k(ba, 1) − T1j(ab, 1).
It follows that every x ∈ T can be written as
x =

i
t(ai, bi) +
n
j=2
T1j(cj , 1), ai, bi ∈ S(R), cj ∈ R.
In Proposition 4, we have seen ker φ ⊆ T. φ(x) = 0 implies

i
E11([ai, bi]) +

2 j n
(E11(cj) − Ejj(cj))
= E11

i
[ai, bi]

+

2 j n
(E11 − Ejj)(cj)
= 0.
1186 Yongjie WANG et al.
It means that
i
[ai, bi] = 0, cj = 0, 2 j n.
Therefore,
x =

i
t(ai, bi)
with
i
[ai, bi] = 0, ai, bi ∈ S(R).
Notice that
[a, b] = ab − ε(a, b)ba, a, b ∈ S(R).
For more information of the structure of ker φ, we have the following
proposition.
Proposition 6 For a, b, c ∈ S(R), the following identities hold:
t(a, 1) = t(1, a) = 0, (15)
t(a, b) + ε(a, b)t(b, a) = 0, (16)
t(ab, c) − t(a, bc) + ε(a + b, c)t(ca, b) = 0. (17)
Proof By (14), we have
t(a, 1) = t(1, a) = 0.
By the definition of t(a, b), using (13), one has
t(ab, c) = T1j(ab, c) − ε(a + b, c)T1j(1, cab)
= T1k(a, bc) + ε(a, b + c)Tkj(b, ca) − ε(a + b, c)T1j (1, cab)
= t(a, bc) + ε(a, b + c)(T1k(1, bca) + Tkj(b, ca)) − ε(a + b, c)T1j(1, cab)
= t(a, bc) + ε(a, b + c)T1j(b, ca) − ε(a + b, c)T1j(1, cab)
= t(a, bc) + ε(a, b + c)(T1j (b, ca) − ε(b, a + c)T1j(1, cab))
= t(a, bc) + ε(a, b + c)t(b, ca). (18)
Since t(a, 1) = 0, setting c = 1 in (18), we have
0 = t(a, b) + ε(a, b)t(b, a),
which gives (16), and it follows that, by rewrite (18),
t(ab, c) = t(a, bc) − ε(a, b + c)ε(b, a + c)t(ca, b)
= t(a, bc) − ε(a + b, c)t(ca, b),
which is (17).
Color cyclic homology and Steinberg Lie color algebras 1187
3 ε-Hochschild homology and ε-cyclic homology
The tensor vector space R⊗(n+1) could be viewed as a Γ-graded vector space by
deg(a0 ⊗ a1 ⊗· · ·⊗an) =
n
i=0
αi,
where ai ∈ S(R) and deg(ai) = αi ∈ Γ for i = 1, 2, . . . , n.
One has the ε-Hochschild boundary dε with respect to ε,
dε
n : R
⊗(n+1) → R
⊗n,
defined by
dε
n(a0 ⊗ a1 ⊗· · ·⊗an) =
n −1
i=0
(−1)ia0 ⊗· · · ⊗aiai+1 ⊗· · ·⊗an
+ (−1)nε
n −1
i=0
ai, an

ana0 ⊗ a1 ⊗· · ·⊗an−1
for homogeneous elements ai, 0 i n. We always assume R⊗0 = k and
dε
0 = 0. Note that dε
n is a grading-preserving linear map of Γ-graded vector
spaces.
Lemma 7
dε
n−1dε
n = 0, ∀ n ∈ N.
Proof Consider dε
n−1dε
n(a0 ⊗ a1 ⊗ · · · ⊗ an) for homogeneous elements ai.
Comparing with the Hochschild homology, we only need notice the coefficients
of ana0a1 ⊗· · ·⊗an−1 and an−1ana0 ⊗· · ·⊗an−2 in the result. One is
(−1)nε
n −1
i=0
ai, an

+ (−1)n−1ε
n −1
i=0
ai, an

= 0,
and the other is
ε
n −2
i=0
ai, an−1 + an

− ε
n −1
i=0
ai, an

ε
n −2
i=0
ai + an, an−1

= ε
n −2
i=0
ai, an−1 + an

− ε
n −2
i=0
ai, an

ε
n −2
i=0
ai, an−1

= ε
n −2
i=0
ai, an−1 + an

− ε
n −2
i=0
ai, an−1 + an

= 0.
1188 Yongjie WANG et al.
Definition 8 The ε-Hochschild homology of the Γ-graded associative algebra
R is H∗(R⊗(∗+1), dε).
Denote by
[R,R] = spank
{ab − ε(a, b)ba | a, b ∈ S(R)}
the linear span of all the ε-commutator. Then we have
H1(R, ε) ∼=
R/[R,R].
Let
ξn+1 =
tn+1 | tn+1
n+1 = 1
be the cyclic group and define an action of ξn+1 on R⊗(n+1) by setting
tn+1(a0 ⊗ a1 ⊗· · ·⊗an) = (−1)nε

a0,
n
i=1
ai

a1 ⊗ a2 ⊗· · ·⊗an ⊗ a0
on the generators of R⊗(n+1). It is then extended to R⊗(n+1) by linearity. One
can see that the action of
tn+1
n+1(a0 ⊗ a1 ⊗· · ·⊗an) = a0 ⊗ a1 ⊗· · ·⊗an
by using the identity
n
i=0
ε

ai,

0 j n, j =i
aj

= 1.
Let Δε
n+1 be the subspace of R⊗(n+1) spanned by elements
a0 ⊗ a1 ⊗· · · ⊗an + (−1)n+1ε

a0,
n
i=1
ai

a1 ⊗ a2 ⊗· · ·⊗an ⊗ a0
for all homogeneous elements ai ∈ S(R), 0 i n.
Lemma 9
dε
n(Δε
n+1) ⊆ Δε
n, ∀ n ∈ N.
Proof For homogeneous elements ai ∈ S(R), by direct computation, we have
dε
n

a0 ⊗ a1 ⊗· · · ⊗an + (−1)n+1ε

a0,
n
i=1
ai

a1 ⊗ a2 ⊗· · ·⊗an ⊗ a0

=
n −1
i=0
(−1)ia0 ⊗· · ·⊗aiai+1 ⊗· · ·⊗an
Color cyclic homology and Steinberg Lie color algebras 1189
+ (−1)n+1ε

a0,
n
i=1
ai
n −1
i=0
(−1)ia1 ⊗· · ·⊗ai+1ai+2 ⊗· · · ⊗an ⊗ a0
+ (−1)nε
n −1
i=0
ai, an

ana0 ⊗ a1 ⊗· · ·⊗an−1
− ε

a0,
n
i=1
ai

ε
n
i=1
ai, a0

a0a1 ⊗ a1 ⊗· · ·⊗an
=
n −1
i=1
(−1)i

a0 ⊗· · ·⊗aiai+1 ⊗· · ·⊗an
+ (−1)nε

a0,
n
i=1
ai

a1 ⊗· · ·⊗aiai+1 ⊗· · ·⊗an ⊗ a0

.
Therefore,
dε
n(Δε
n+1) ⊆ Δε
n.
Denote by
Cn(R, ε) = R
⊗(n+1)/Δε
n+1
the quotient k-space. Then dε
n induces a linear map
dε
n : Cn(R, ε) → Cn−1(R, ε).
The complex (C∗(R, ε), dε
n) gives that the homology of R is H∗(C∗(R, ε), dε),
which we call it ε-cyclic homology and denote it by HC∗(R, ε). For our use, we
rewrite HC1(R, ε) as follows.
Proposition 10 Let

R,R ε = (R ⊗k R)/Iε
be the quotient space, where Iε is the subspace of R ⊗k R spanned by elements
a ⊗ b + ε(a, b)b ⊗ a, ab ⊗ c − a ⊗ bc + ε(a + b, c)ca ⊗ b, ∀ a, b, c ∈ S(R).
Set

a, b ε = a ⊗ b + Iε, ∀ a, b ∈ S(R).
Then
HC1(R, ε) =

i

ai, bi ε


i
(aibi − ε(ai, bi)biai) = 0, ai, bi ∈ S(R)

is a subspace of
R,R ε. Moreover, we have the following exact sequence:
0 −→ HC1(R, ε) −→
R,R ε = C1(R)
Imdε
2
dε
−→1 [R,R] −→ 0.
Remark 11 We would like to emphasize the Γ-grading of
R,R εc
and
1190 Yongjie WANG et al.
HC1(R, ε) here. First, R⊗(n+1) is naturally graded by Γ induced by R. Since Δε
n
is spanned by homogeneous elements of R⊗(n+1), the quotient k-space Cn(R, ε)
is also Γ-graded. We have mentioned that dε
n is a grading-preserving linear
map, which shows that the kernel and image of dε
n are graded subspaces. Thus,
their quotient spaces are also Γ-graded. More directly, the Γ-grading of
R,R ε
is given by
(
R,R ε)c = spank
{
a, b ε ∈
R,R ε | a, b ∈ S(R), deg a+deg b = c}, ∀ c ∈ Γ.
Remark 12 If we take Γ = {0} and ε = 1, then the ε-Hochschild (cyclic)
homology is nothing but the common Hochschild (cyclic) homology of the
associative algebra R (see [7] or [8]). For the super case, we need take Γ = Z2
and
ε(α, β) = (−1)αβ, α, β ∈ Z2.
The Hochschild (cyclic) homology of superalgebras was first introduced by
Kassel (see [6] or [1]).
Now, we state one of the main theorem of this section. In order to prove
this theorem, we need the following definitions.
Definition 13 Let R be a Lie color algebra. An R-module M is a Γ-graded
vector space
M =

c∈Γ
Mc
such that
RaMb ⊆ Ma+b, ∀ a, b ∈ C,
and
[x, y] · m = x(y ·m) − ε(x, y)y · (x ·m), ∀m ∈ M.
Definition 14 Let R be a Lie color algebra, and let C be a Γ-graded free
module over K. A bilinear form ψ on R ×R → C is called a color 2-cocycle on
R if it satisfies
ψ(Ra,Rb) ⊆ Ca+b,
ψ(x, y) = −ε(x, y)ψ(y, x),
ε(z, x)ψ(x, [y, z]) + ε(x, y)ψ(y, [z, x]) + ε(y, z)ψ(z, [x, y]) = 0
for all x, y, z ∈ S(R).
Theorem 15 ker φ HC1(R, ε).
Proof Consider the following two exact sequences:
0 −→ HC1(R, ε) −→
R,R ε dε
−→1 [R,R] −→ 0
↓ η ↓ E11
0 −→ ker φ −→ stn(R, ε) −φ→ sln(R, ε) −→ 0
Color cyclic homology and Steinberg Lie color algebras 1191
Define
η :
R,R ε → stn(R, ε)
by
η(
a, b ε) = t(a, b), a,b∈
S(R).
From Proposition 10, (16), and (17), we know that η is well defined. Moreover,
φ ◦ η(
a, b ε) = φ(t(a, b)) = E11(ab − ε(a, b)ba) = E11(dε
1(
a, b ε)),
which shows that the above diagram commutes and defines, by restriction, a
surjective homomorphism of HC1(R, ε) onto kerφ.
To get the injectivity of η, first we begin by finding a color 2-cocycle
f : gln(R, ε) × gln(R, ε) →
R,R ε.
More precisely, if x = Eij(a), y = Ekl(b) ∈ gln(R, ε), a,b ∈ S(R), 1 i, j, k, l
n, we define
f(x, y) = δjkδli
a, b ε.
Extending f on the whole gln(R, ε) by k-bilinearity, one can check that
f(x, y) + ε(x, y)f(y, x) = 0,
ε(z, x)f(x, [y, z]) + ε(x, y)f(y, [z, x]) + ε(y, z)f(z, [x, y]) = 0
for the homogeneous elements x, y, z ∈ gln(R, ε).
Therefore, f defines a color 2-cocycle on gln(R, ε) with values in
R,R ε.
We can restrict f on sln(R, ε), which is also a color 2-cocycle. Now, we can
form a (Γ, ε)-Lie color algebra
L = sln(R, ε) ⊕
R,R ε
with Lie color bracket
[(x, c), (y, c
)] = ([x, y], f(x, y)), ∀ x, y ∈ sln(R, ε), ∀ c, c
∈
R,R ε.
L is a (Γ, ε)-Lie color algebra as f is a color 2-cocycle. Clearly,
f(Eij(a),Ekl(b)) = 0
only if j = k and l = i.
Thus, the elements (Eij(a), 0) in L satisfy relations (2)–(4). By the universal
property of stn(R, ε), there exists a (unique) Lie algebra homomorphism
ρ: stn(R, ε) → L
such that
ρ(Xij (a)) = Eij(a).
1192 Yongjie WANG et al.
Hence,
ρ(Tij(a, b)) = [(Eij(a), 0), (Eji(b), 0)]
= (Eii(ab) − ε(a, b)Ejj(ba), f(Eij(a),Eji(b)))
= (Eii(ab) − ε(a, b)Ejj(ba),
a, b ε).
Therefore,
ρ(t(a, b)) = ρ(T1j(a, b) − ε(a, b)T1j (1, ba))
= (E11(ab − ε(a, b)ba),
a, b ε − ε(a, b)
1, ba ε)
= (E11(dε
1(
a, b ε)),
a, b ε)
as
1, ba ε = 0 in
R,R ε.
It follows that
ρ(η(
a, b ε)) = (E11(dε
1(
a, b ε)),
a, b ε),
and so
ρ(η(c)) = (E11(dε
1(c)), c), ∀ c ∈
R,R ε.
Thus, η is injective. The proof is complete.
The cohomology theory of the (Γ, ε)-Lie color algebra R is studied
systematically by Scheunert and Zhang [11]. In this paper, we only use the
homology H∗(R, k) of the (Γ, ε)-Lie color algebra R with the trivial coefficient
k. A (Γ, ε)-Lie color algebra R over k is called perfect if [R,R] = R. It is well
known that R is perfect means that the first homology H1(R, k) is trivial. The
pair (
 R, π), where
R is a (Γ, ε)-Lie color algebra and π :
R → R an epimorphism,
is called a central extension of R if [ker(π),
R] = 0. A central extension ( R,ϕ)
of R is called universal if for any central extension (
 R, π) of R, there exists a
unique homomorphism ρ: R →
R such that π◦ρ = ϕ. A (Γ, ε)-Lie color algebra
R is called centrally closed if any central extension of R splits. Similarly, with
the Lie algebras, we have the following result.
Theorem 16 If R is a perfect (Γ, ε)-Lie color algebra, then there exists a
unique universal central extension ( R,ϕ) of R and R is centrally closed. If R
is centrally closed and R is a central extension of R, then R is the universal
central extension of R and the kernel is isomorphic to H2(R, k).
Remark 17 The details for the theory of the universal central extension of
Lie algebras can be found in [5], while Neher [9] studied the universal central
extension of Lie superalgebras.
We have known that
φ: stn(R, ε) → sln(R, ε) (19)
Color cyclic homology and Steinberg Lie color algebras 1193
is a central extension. And both stn(R, ε) and sln(R, ε) are perfect when n 3.
We will see that φ determines a universal central extension except in a few
cases. On one hand, we will prove that
φ: stn(R, ε) → sln(R, ε)
is a universal central extension if n 5. On the other hand, we will use color
2-cocycle to construct the universal central extensions for the rest of two cases
n = 3 and n = 4.
Definition 18 Let ξ be a color 2-cocycle on stn(R, ε) with value in C as
above. Now, we define a Lie algebra stn(R, ε) to be the Lie color algebra
generated by the symbols X
ij(a), a ∈ R, 1 i = j n, and the vector space
C , satisfying the following relations:
a → X
ij(a) is a K-linear map, (20)
[C ,C] = [X
ij(a),C] = 0, (21)
[X
ij(a),X
jk(b)] = X
ik(ab) + ξ(Xij (a),Xjk(b)), i, j, k distinct, (22)
[X
ij(a),X
kl(b)] = ξ(Xij(a),Xkl(b)), j = k, i = l, (23)
where a, b ∈ R, 1 i, j, k, l n.
The following lemma shows that this construction of a central extension of
stn(R, ε) is actually the same as the classical construction if ξ is surjective.
Lemma 19 If
0 −→ C −→ st

n(R, ε) −π→ stn(R, ε) −→ 0
is a central extension of stn(R, ε) constructed from a surjective super 2-cocycle
ξ, then there is an isomorphism
ρ: stn(R, ε) → st

n(R, ε)
with
ρ(X
ij(a)) = Xij(a), ρ(c) = c, a ∈ R, c ∈ C .
Proof The proof is similar to [2,4].
4 Central extension of st4(R, ε)
We define a color 2-cocycle on st4(R, ε) and construct the Lie color algebra

st4(R, ε) as a covering (eventually a universal covering) of st4(R, ε). For any
positive integer m, let Im be the 2-sided Γ-graded ideal of R generated by the
1194 Yongjie WANG et al.
elements mc and ab−ε(a, b)ba for any homogeneous elements a, b ∈ S(R), c ∈ R.
Then, we have the following lemma.
Lemma 20
Im = mR + R[R,R], [R,R]R = R[R,R].
Let Rm := R/Im be the quotient color algebra over K; it is color
commutative. Write
a = a + Im, a∈ R.
If m = 2 and 2 is invertible in R, then R2 = 0. For {i, j, k, l} = {1, 2, 3, 4}, let
ijkl(R2) denote a copy of R2 and identify ijkl(r), ilkj(r), kjil(r), klij(r) for
r ∈ R2. Thus, we have six distinct copies of R2 whose direct sum is denote by
W . Using decomposition (12) of st4(R, ε), we define a K-bilinear map
ψ: st4(R, ε) × st4(R, ε) → W
by
ψ(Xij (a),Xkl(b)) = ijkl(ab), {i, j, k, l} = {1, 2, 3, 4}, a,b ∈ R,
and by ψ(x, y) = 0 for all other pairs of elements from the summand of (12).
Note that, if m = 2, then a = −a in R2.
Lemma 21 The bilinear map ψ is a color 2-cocycle.
Proof Since there is a Lie color algebra homomorphism
a: st4(R, ε) → st4(R2, ε)
with
a(Xij (a)) = Xij(a)
and
ψ(a(x), a(y)) = ψ(x, y)
is well defined for x, y ∈ st4(R, ε), it suffices to verify the lemma for R2. Since
ψ(Xkl(b),Xij (a)) = klij(ba), ab= ±ε(a, b)ba,
we have
ψ(Xij (a),Xkl(b)) = ijkl(ab) = −ε(a, b) klij(ba) = −ε(a, b)ψ(Xkl(b),Xij (a)),
where a, b ∈ S(R). Now, let
J(x, y, z) = ε(z, x)ψ(x, [y, z]) + ε(x, y)ψ(y, [z, x]) + ε(y, z)ψ(z, [x, y])
for homogeneous elements x, y, z ∈ st4(R, ε). We will show J(x, y, z) = 0 by
taking homogeneous elements x, y, z in summands of (12). If a term J(x, y, z)
is not 0, then we can reorder to assume that z = Xkl(d) and 0 = [x, y] ∈ Xij(R)
with {i, j, k, l} = {1, 2, 3, 4}.
Color cyclic homology and Steinberg Lie color algebras 1195
Case 1 x or y is in T.
Without loss of generality, we assume x = Tpq(a, b) and y = Xij(c). Since
R (= R2) is commutative with 2R = 0, [x, y] = 0 forces precisely one of p or q
to be in {i, j} (so the other to be in {k, l} by equations (7)–(9)). Moveover, in
this case,
[x, y] = Xij(abc), [y, z] = 0, [z, x] = Xkl(dab),
and so
J(x, y, z) = ε(a + b, c)ψ(Xij (c), [Xkl(d), Tpq(a, b)]) + ε(c, d)ψ(Xkl(d),Xij (abc))
= ε(a, c)ε(b, c)ε(d, a)ε(d, b)ε(a, b) ijkl (cbad) + ε(c, d) klij (dabc)
= 0.
Case 2 Neither x nor y is in T.
We can assume that x = Xip(a) and y = Xpj(b) with p ∈ {k, l}, and so
[x, y] = Xij(ab).
For p = k, we have
[y, z] = 0, [z, x] = −ε(d, a)Xkl(ad),
and so
J(x, y, z) = ε(a, b)ψ(Xkj (b), [Xkl(d),Xik(a)]) + ε(b, d)ψ(Xkl(d),Xij (ab))
= − ε(a, b)ε(d, a) kjil(bad) + ε(b, d) klij(dab)
= 0.
For p = l, we have
[y, z] = −ε(b, d)Xkj (db), [z, x] = 0,
and so
J(x, y, z) = − ε(d, a)ε(b, d)ψ(Xil (a),Xkj (db)) + ε(b, d)ψ(Xkl(d),Xij (ab))
= − ε(d, a)ε(b, d) ilkj (adb) + 0 + ε(b, d) klij(dab)
= 0.
Therefore, we obtain a central extension of Lie color algebra st4(R, ε):
0 −→ W −→ s
t4(R, ε) −π→ st4(R, ε) −→ 0,
i.e.,

st4(R, ε) = st4(R, ε) ⊕ W ,
with Lie color bracket
[(x, c), (y, c
)] = ([x, y], ψ(x, y)), ∀ x, y ∈ st4(R, ε), ∀ c, c
∈ W .
1196 Yongjie WANG et al.
Here, π is the projection on the first summand:
π : st4(R, ε) ⊕ W → st4(R, ε).
We can now apply Definition 18 with C = W and ξ = ψ to obtain the Lie
color algebra st
4(R, ε). Since 1 ∈ R and st
4(R, ε) is perfect, by Lemma 19, there
exists a unique Lie color algebra isomorphism
ρ: st
4(R, ε) →
st4(R, ε)
such that
ρ(X
ij(a)) = Xij(a), ρ|
W = id.
5 Central extension of st3(R, ε)
In this section, we will handle st3(R, ε). Recall that
I3 = 3R + R[R,R]
and R3 = R/I3 is an associative commutative K-algebra. For {i, j, k} =
{1, 2, 3}, let ijpq(R3) for (p, q) = (i, k) or (k, j) denote a copy of R3 and identify
ijpq(r) with pqij(−r). Thus, we have six distinct copies of R3 whose direct sum
is denoted by U . Using decomposition (12) of st3(R, ε), we define a K-bilinear
map
ψ: st3(R, ε) × st3(R, ε) → U
by
ψ(Xij(r),Xpq(s)) = ijpq(rs)
for (p, q) = (i, k) or (k, j) with {i, j, k} = {1, 2, 3} and r, s ∈ R, and by ψ(x, y) =
0 for all other pairs of elements from the summands of (12).
Lemma 22 The bilinear map ψ is a color 2-cocycle.
Proof As in the proof of Lemma 21, we can assume R = R3; i.e., R is
commutative and 3R = 0. By the definition, we know that ψ is skew-symmetric
and ψ(x, x) = 0 for x in a summand of (12), and hence, for all x ∈ st3(R, ε).
We will show
J(x, y, z) = ε(z, x)ψ(x, [y, z]) + ε(x, y)ψ(y, [z, x]) + ε(y, z)ψ(z, [x, y])
for homogeneous elements x, y, z ∈ st3(R, ε). We will show J(x, y, z) = 0 by
taking homogeneous elements x, y, z in summands of (12). If a term J(x, y, z)
is not 0, then we can reorder to assume that z = Xpq(d) and 0 = [x, y] ∈ Xst(R)
with (p, q) = (s, u) or (u, t) and {i, j, k} = {1, 2, 3}.
Case 1 x or y is in T.
Color cyclic homology and Steinberg Lie color algebras 1197
Without loss of generality, we assume x = Tij(a, b) and y = Xst(c). From
equations (7)–(9) and R = R3, we have
J(x, y, z) = stpq(θabcd) = 0
with θ = 0, 3,−3, and thus, J(x, y, z) = 0.
Case 2 Neither x nor y is in T.
We can assume that x = Xsu(a) and y = Xut(b). If (p, q) = (s, u), then we
have
J(x, y, z) = − ε(d, a)ε(b, d)ψ(Xsu(a),Xst(db)) + ε(b, d)ψ(Xsu(d),Xst(ab))
= − ε(d, a)ε(b, d) sust(adb) + ε(b, d) sust(dab)
= 0.
While, if (p, q) = (u, t), then we have
J(x, y, z) = − ε(a, b)ε(d, a)ψ(Xut (b),Xst(ad)) + ε(b, d)ψ(Xut(d),Xst(ab))
= − ε(a, b)ε(d, a) utst(bad) + ε(b, d) utst(dab)
= 0.
As in the st4(R, ε) case, we have a central extension of st3(R, ε):
0 −→ U −→ s
t3(R, ε) −π→ st3(R, ε) −→ 0,
i.e.,

st3(R, ε) = st3(R, ε) ⊕ U ,
with Lie color bracket
[(x, c), (y, c
)] = ([x, y], ψ(x, y)), ∀ x, y ∈ st3(R, ε), ∀ c, c
∈ W .
Here, π is the projection on the first summand:
π : st3(R, ε) ⊕ W → st3(R, ε).
We can now apply Definition 18 with C = U and ξ = ψ to obtain the Lie color
algebra st
3(R, ε). Since 1 ∈ R and st
3(R, ε) is perfect, by Lemma 19, there
exists a unique Lie color algebra isomorphism
ρ: st
3(R, ε) →
st3(R, ε)
such that
ρ(X
ij (a)) = Xij(a), ρ|
U = id.
Theorem 23 The universal covering of stn(R, ε) is (stn(R, ε), id) if n 5
and (
stn(R, ε), π) if n = 3 or 4.
Proof Suppose that
0 −→ V −→ stn(R, ε) −χ→ stn(R, ε) −→ 0
1198 Yongjie WANG et al.
is a central extension of stn(R, ε). We will show that we can choose a preimage
X
ij(a) of Xij(a) under χ for 1 i = j n, a ∈ R, and a linear map μ: C → V ,
which satisfy relations (20)–(23) for st
n(R, ε) with ξ = 0 and C = 0 if n 5
and with ξ = ψ and C = U ,W for n = 3, 4, respectively. Thus, we will have a
homomorphism
θ : st
n(R, ε) → stn(R, ε)
with θ(X
ij(a)) = Xij(a) so χ ◦ θ = π ◦ ρ as in Lemma 19. Thus, the Lie color
homomorphism
θ ◦ ρ
−1 :
stn(R, ε) → stn(R, ε)
satisfies
χ ◦ (θ ◦ ρ
−1) = π
and (st
n(R, ε), π) is the universal covering. Since
(st

n(R, ε), π) =


(stn(R, ε), id), n 5,
(
stn(R, ε), π), n= 3, 4,
the result will follow. We begin by choosing any preimage Xij(a) for a in a
K-basis of R and extend linearly to all a ∈ R. We observe that
Tij(a, b) := [ Xij(a), Xji(b)]
is independently of the choice of Xij(a). If
[Tpq(a, b),Xij (c)] = Xij(d)
in (6)–(10), then
[ Tpq(a, b), Xij (c)] ∈ Xij(d) + V .
Also, we have
[Tpq(1, 1),Xij (c)] = Xij(mc),
where m may be an integer in {0,±1,±2}. Nevertheless, we always have
[Tpq(1, 1),Xji(c)] = Xji(−mc),
so
[ Tpq(1, 1), Tij (a, b)] ∈ [ Xij(ma) + V , Xji(b)] + [ Xij(a), Xji(−mb) + V ] = {0}.
Fix some k = i, j and replace Xij(a) by [T ik(1, 1),X ij (a)]. Then
[ Tik(1, 1), Xij (a)] = Xij(a).
We see that
[T pq(a, b),X ij (c)] = [T pq(a, b), [T ik(1, 1),X ij (c)]]
= [[T pq(a, b), T ik(1, 1)],X ij (c)] + [T ik(1, 1), [T pq(a, b),X ij (c)]]
= Xij(d)
Color cyclic homology and Steinberg Lie color algebras 1199
if
[Tpq(a, b),Xij (c)] = Xij(d).
In particular, taking (p, q) = (i, l), we can observe that Xij(a) does not depend
on the choice of k = i or j. Applying ad( Tij(1, 1)) to
[ Xij(a), Xjk(b)] ∈ Xik(ab) + V
gives
[ Xij(2a), Xjk(b)] + [ Xij(a), Xjk(−b)] = Xik(ab)
i.e.,
[ Xij(a), Xjk(b)] = Xik(ab).
Thus, Xij(a), V satisfy relations (20)–(22).
For relation (23), we first apply ad( Xik(1)) to [ Xkj(a), Xij (b)] ∈ V giving
[ Xij(a), Xij (b)] + [ Xkj(b), V ] = 0
for i, j, k are distinct, i.e.,
[ Xij (a), Xij (b)] = 0.
If there exist distinct i, j, k, l,m, then applying ad( Tim(1, 1)) to [ Xij(a), Xkl(b)]
∈ V gives
[ Xij(a), Xkl(b)] + 0 = 0,
by [T im(1, 1),X kl(b)] ∈ V . If i, j, k, l are distinct, applying ad(T st(1, 1)) to
[ Xij(a), Xpq(b)] ∈ V , where (p, q) = (i, k) and (s, t) = (l, j) or (p, q) = (k, j)
and (s, t) = (i, l), gives
[ Xij(a), Xpq(b)] = 0.
If n 5, we have shown that Xij(a), V satisfy relations (20)–(23) for ξ = 0
and V = 0. Moreover, for n = 3, 4, all relations are satisfied, except possibly
[ Xij(a), Xpq(b)] = μ( ijpq(ab)), n= 3, (24)
with (p, q) = (i, k) or (k, j);
[ Xij(a), Xkl(b)] = μ( ijpq(ab)), n= 4, (25)
where i, j, k, l are distinct.
For (24), we set
μijpq(c) = [ Xij(c), Xpq(1)]
for (p, q) = (i, k) or (k, j). Applying ad( Tij(a, b)) gives
μijpq(abc + ε(a, b)ε(a + b, c)cba) + ε(a + b, c)[ Xij (c), Xpq(d)] = 0, (26)
1200 Yongjie WANG et al.
where d = ab if (p, q) = (i, k) and d = ba if (p, q) = (k, j). Now, a = b = 1 gives
μijpq(3c) = 0,
where c = 1 gives
μijpq(ab + ε(a, b)ba) + [ Xij(1), Xpq(d)] = 0,
i.e.,
μpqij(d) = μijpq(ab + ε(a, b)ba).
In particular, b = 1 gives
μpqij(a) = −μijpq(a),
so
μijpq(2d) = μpqij(−d) = μijpq(ab + ε(a, b)ba),
which implies
μijpq(ab) = ε(a, b)μijpq(ba),
whenever d = ab or ε(a, b)ba. Letting b = 1 in (26) gives
ε(a, c)[ Xij (c), Xpq(a)] = −μijpq(ac + ε(a, c)ca) = μijpq(ac).
Furthermore, (26) is equivalent to
μijpq(a[b, c]) = μijpq(abc − ε(b, c)acb) = 0.
Thus,
μijpq(I3) = 0.
Since I3 is linearly spanned by elements 3a and a[b, c] for a, b, c ∈ R by Lemma
20, we can define
μ: U → V
by
μ( ijpq(a)) = μijpq(a)
such that (24) holds.
For (25), we set
μijkl(c) = [ Xij(c), Xkl(1)], {i, j, k, l} = {1, 2, 3, 4}.
Applying ad( Tij(a, b)) gives
μijkl(abc + ε(a, b)ε(a, c)ε(b, c)cba) = 0.
Now, b = c = 1 gives
μijkl(2a) = 0,
Color cyclic homology and Steinberg Lie color algebras 1201
and c = 1 gives
μijkl(ab) = −ε(a, b)μijkl(ba) = μijkl(ba).
Moreover,
μijkl(a[b, c]) = μijkl(abc) − ε(b, c)μijkl(acb)
= μijkl(abc) − ε(b, c)μijkl(−ε(a, b + c)cba)
= μijkl(abc + ε(a, b)ε(b, c)ε(a, c)cba)
= 0.
Thus,
μijkl(I2) = 0.
Applying ad( Xkl(−b)) to [ Xik(a), Xkj (1)] gives
[ Xij(a), Xkl(b)] = [ Xil(ab), Xkj (1)] = μilkj(ab).
In particular, b = 1 gives
μilkj(a) = μijkl(a)
and
[ Xij (a), Xkl(b)] = μijkl(ab).
Clearly,
μklij(a) = [ Xkl(1), Xij (a)] = [ Xij(−a), Xkl(1)] = μklij(a)
and
μilkj(a) = μijkl(a) = μkjil(a).
We can now define
μ: W → V
by
μ( ijkl(a)) = μijkl(a),
and so (25) holds.
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