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Biderivations of the higher rank Witt algebra without anti-symmetric condition

Xiaomin Tang
• Corresponding author
• Department of Mathematics, Heilongjiang University, Harbin 150080, China
• Heilongjiang Provincial Key Laboratory of the Theory and Computation of Complex Systems, Heilongjiang University, Harbin 150080, China
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/ Yu Yang
Published Online: 2018-04-30 | DOI: https://doi.org/10.1515/math-2018-0042

Abstract

The Witt algebra 𝔚d of rank d(≥ 1) is the derivation algebra of Laurent polynomial algebras in d commuting variables. In this paper, all biderivations of 𝔚d without anti-symmetric condition are determined. As an applications, commutative post-Lie algebra structures on 𝔚d are obtained. Our conclusions recover and generalize results in the related papers on low rank or anti-symmetric cases.

MSC 2010: 17B05; 17B40

1 Introduction

Let 𝔽 be a field of characteristic zero. We denote by ℤ the sets of all integers. We fix a positive integer d ≥ 1 and denote by 𝔚d the derivation Lie algebra of the Laurent polynomial algebra $\begin{array}{}A=\mathbb{F}\left[{z}_{1}^{±1},\cdots ,{z}_{d}^{±1}\right]\end{array}$ in d commuting variables z1, …, zd over 𝔽. It is well known that the infinite-dimensional Lie algebra 𝔚d is called the Witt algebra of rank d. Its representations have attracted a lot of attention from many mathematicians [1,2,3,4,5]. According to their notations, the Witt algebra can be described as follows.

For i ∈ {1, 2, …, d}, set $\begin{array}{}{\mathrm{\partial }}_{i}={z}_{i}\frac{\mathrm{\partial }}{{z}_{i}}\end{array}$ . For any n ∈ ℤd (considered as row vectors, i.e., n = (n1, …, nd) where ni ∈ ℤ), set zn = $\begin{array}{}{z}_{1}^{{n}_{1}}{z}_{2}^{{n}_{2}}\cdots {z}_{d}^{{n}_{d}}.\end{array}$ We fix the vector space 𝔽d of 1 × d matrices. Denote the standard basis by e1, e2, …, ed, which are the row vectors of the identity matrix Id. Let (·, ·) be the standard symmetric bilinear form such that (u, v) = uvT ∈ 𝔽, where vT is the matrix transpose. For u ∈ 𝔽d and r ∈ ℤd, we denote D(u, r) = $\begin{array}{}{z}^{r}\sum _{i=1}^{d}{u}_{i}{\mathrm{\partial }}_{i}.\end{array}$ Then we have

$Du,r,Dv,s=Dw,r+s,u,v∈Fd,r,s∈Zd,$(1)

where w = (u, s)v − (v, r)u. Therefore, the Witt algebra of rank d is the 𝔽-linear space

$Wd=spanF{D(u,r)|u∈Fd,r∈Zd}$

with brackets determined by (1). Note that D(u, r) is linear only with respect to the first component u. It is clear that 𝔚d has a basis as

${D(e1,r),D(e2,r),⋯,D(ed,r)|r∈Zd}.$

Recall that

$h=Du,0|u∈Fd=spanF{D(e1,0),D(e2,0),⋯,D(ed,0)}$

is the Cartan subalgebra of 𝔚d.

It is well-known that derivations and generalized derivations are very important subjects in the research of both algebras and their generalizations. In recent years, biderivations have interested a great number of authors, see [6,7,8,9,10,11,12,13,14,15,16], , Brešar et al. showed that all biderivations on commutative prime rings are inner biderivations, and determined the biderivations of semiprime rings. The notion of biderivations of Lie algebras was introduced in . Since then, biderivations of Lie algebras have been studied by many authors. It may be useful and interesting for computing the biderivations of some important Lie algebras. In particular, the authors in  determined anti-symmetric biderivations for all 𝔚d. All biderivations of 𝔚1 without anti-symmetric condition were later obtained in . In the present paper, we shall use the methods of  to determine all biderivations of 𝔚d for all d ≥ 1.

Next, let us introduce the definition of biderivation. For an arbitrary Lie algebra L, a bilinear map f : L × LL is called a biderivation of L if it is a derivation with respect to both components. Namely, for each xL, both linear maps ϕx and ψx form L into itself given by ϕx = f(x, ·) and ψx = f(·, x) are derivations of L, i.e.,

$f([x,y],z)=[x,f(y,z)]+[f(x,z),y],f(x,[y,z])=[f(x,y),z]+[y,f(x,z)]$(2)

for all x, y, zL. Denote by B(L) the set of all biderivations of L. For λ ∈ ℂ, it is easy to verify that the bilinear map f : L × LL given by f(x, y) = λ [x, y] for all yL is a biderivation of L. Such biderivation is said to be inner. Recall that f is anti-symmetric if f(x, y) = -f(y, x) for all x, yL.

In this paper, we will prove that every biderivation of 𝔚d without anti-symmetric condition is inner. As an application, we characterize the commutative post-Lie algebra structures on 𝔚d.

2 Biderivations of the Witt algebras

We first give some lemmas which will be useful for our proof.

Lemma 2.1

(). Every derivation of 𝔚d is inner.

Lemma 2.2

Suppose that fB(𝔚d). Then there are linear maps ϕ and ψ from 𝔚d into itself such that

$f(x,y)=[ϕ(x),y]=[x,ψ(y)]$

for all x, y ∈ 𝔚d.

Proof

Since f is a biderivation of 𝔚d, then for a fixed element x ∈ 𝔚d the map ϕx : LL given by ϕx(y) = f(x, y) is a derivation of 𝔚d by (2). Therefore, from Lemma 2.1 we know that ϕx is an inner derivation of 𝔚d. Therefore, there is a map ϕ : 𝔚d → 𝔚d such that ϕx = adϕ(x), i.e., f(x, y) = [ϕ(x), y]. Since f is bilinear, it is easy to verify that ϕ is linear. Similarly, if we define a map ψz from 𝔚d into itself given by ψz(y) = f(y, z) for all y ∈ 𝔚d, then one can obtain a linear map ψ from 𝔚d into itself such that f(x, y) = ad(−ψ (y))(x) = [x, ψ(y)]. The proof is completed. □

Lemma 2.3

Let fB(𝔚d) and ϕ, ψ be determined by Lemma 2.2. For any i, j ∈ {1, …, d} and r, s ∈ ℤd, we assume that

$ϕ(D(ei,r))=∑k=1d∑n∈Zdak,n(i,r)D(ek,n),$(3)

$ψ(D(ej,s))=∑k=1d∑n∈Zdbk,n(j,s)D(ek,n)$(4)

where $\begin{array}{}{a}_{k,n}^{\left(i,r\right)},{b}_{k,n}^{\left(j,s\right)}\end{array}$ ∈ 𝔽. Then

$∑k=1d∑n∈Zdak,n(i,r)skD(ej,n+s)−∑k=1d∑n∈Zdak,n(i,r)njD(ek,n+s)=∑k=1d∑n∈Zdbk,n(j,s)niD(ek,n+r)−∑k=1d∑n∈Zdbk,n(j,s)rkD(ei,n+r).$(5)

Proof

Lemma 2.2 tells us that

$f(D(ei,r),D(ej,s))=[ϕ(D(ei,r)),D(ej,s)]=[D(ei,r),ψ(D(ej,s))]$

for all i, j ∈ {1, …, d} and r, s ∈ ℤd. From (3) and (4), the conclusion follows by direct computations. □

Lemma 2.4

Let fB(𝔚d) and ϕ, ψ be determined by Lemma 2.2. Then the Cartan subalgebra 𝔥 is an invariant subspace of both maps ϕ and ψ.

Proof

Note that D(ei, 0), i = 1, … d span the Cartan subalgebra 𝔥, so it is enough to prove that ϕ(D(ei, 0)), ψ(D(ei, 0)) ∈ 𝔥 for each i ∈ {1,2, …, d}. For any fixed i ∈ ℤ, applying (3) for r = 0 we have that

$ϕ(D(ei,0))=∑k=1d∑n∈Zdak,n(i,0)D(ek,n).$(6)

We will prove that $\begin{array}{}{a}_{k,n}^{\left(i,0\right)}\end{array}$ = 0 in (6) for all n ∈ ℤd ∖ {0}, and so that ϕ(D(ei, 0)) = $\begin{array}{}\sum _{k=1}^{d}{a}_{k,0}^{\left(i,0\right)}D\left({e}_{k},0\right)\in \mathfrak{h}.\end{array}$ The proof of ψ(D(ei, 0)) ∈ 𝔥 is similar.

Now for an arbitrary s ∈ ℤd ∖ {0}, we assume that sj ≠ 0 for some j ∈ {1,2, …, d}. It follows by letting r = 0 in (5) that

$∑k=1d∑n∈Zdak,n(i,0)skD(ej,n+s)−∑k=1d∑n∈Zdak,n(i,0)njD(ek,n+s)=∑k=1d∑n∈Zdbk,n(j,s)niD(ek,n).$(7)

It is clear that the right-hand side of (7) does not contain any non-zero elements in 𝔥, thereby the left-hand side is so. From this, one has that

$∑k=1dak,−s(i,0)skD(ej,0)+∑k=1dak,−s(i,0)sjD(ek,0)=0,$

which implies that

$(2aj,−s(i,0)sj+∑k=1k≠jdak,−s(i,0)sk)D(ej,0)+∑k=1k≠jdak,−s(i,0)sjD(ek,0)=0.$(8)

Thanks to sj ≠ 0, we have by (8) that $\begin{array}{}{a}_{k,-s}^{\left(i,0\right)}\end{array}$ = 0 for every kj. Once again applying (8), we see that $\begin{array}{}2{a}_{j,-s}^{\left(i,0\right)}{s}_{j}\end{array}$ = 0, i.e., $\begin{array}{}{a}_{j,-s}^{\left(i,0\right)}\end{array}$ = 0. In other words, $\begin{array}{}{a}_{k,-s}^{\left(i,0\right)}\end{array}$ = 0 for all k = 1, …, d. Notice the arbitrariness of s, the proof is completed. □

Lemma 2.5

Let fB(𝔚d) and ϕ, ψ be determined by Lemma 2.2. Then we have

$ϕ(D(ei,r))≡ai,r(i,r)D(ei,r)modh,$(9)

$ψ(D(ej,s))≡bj,s(j,s)D(ej,s)modh$(10)

for all i, j ∈ {1, …, d} and r, s ∈ ℤd ∖{0}.

Proof

We will only prove (10), the proof for (9) is similar. Continuing the use of the assumptions (3) and (4), we also have that (7) holds. This, together with Lemma 2.4 meaning $\begin{array}{}{a}_{k,n}^{\left(i,0\right)}\end{array}$ = 0 for all n ∈ ℤd ∖{0}, yields that

$(∑k=1dak,0(i,0)sk)D(ej,s)=∑k=1d∑n∈Zdn≠0bk,n(j,s)niD(ek,n).$(11)

Therefore, from (11) we see that $\begin{array}{}{b}_{k,n}^{\left(j,s\right)}{n}_{i}\end{array}$ = 0 for all i = 1, … d and (k, n) ≠ (j, s) with n ≠ 0. This implies that $\begin{array}{}{b}_{k,n}^{\left(j,s\right)}\end{array}$ = 0 for all (k, n) ≠ (j, s) since n ≠ 0. It has been obtained that

$ψ(D(ej,s))=∑k=1dbk,0(j,s)D(ek,0)+bj,s(j,s)D(ej,s),$

which proves (10). □

Lemma 2.6

Let fB(𝔚d) and ϕ, ψ be determined by Lemma 2.2. Then there is λ ∈ 𝔽 such that

$ϕ(D(ei,r))=λD(ei,r),ψ(D(ej,s))=λD(ej,s)$

for all i, j ∈ {1, …, d} and r, s ∈ ℤd ∖ {0}.

Proof

We use the assumptions (3) and (4). With Lemmas 2.4 and 2.5, Equation (5) becomes

$(∑k=1dak,0(i,r)sk)D(ej,s)+ai,r(i,r)siD(ej,r+s)−ai,r(i,r)rjD(ei,r+s)=−(∑k=1dbk,0(j,s)rk)D(ei,r)+bj,s(j,s)siD(ej,r+s)−bj,s(j,s)rjD(ei,r+s).$

It follows that

$∑k=1dak,0(i,r)sk=∑k=1dbk,0(j,s)rk=0,∀s,r∈Zd∖{0},withs≠r,$(12)

$ai,r(i,r)si=bj,s(j,s)si,ai,r(i,r)rj=bj,s(j,s)rj,∀s,r∈Zd∖{0},i≠j.$(13)

Although r ≠ 0, but we still can find a subset {(1), …, (d)} of ℤd such that (1), …, (d) are 𝔽-linearly independent with (t)r, t = 1, …, d. Let s run over the vectors (1), …, (d) in (12), then we see that

$s~(1)⋮s~(d)a1,0(i,r)⋮ad,0(i,r)=0,$

which implies that

$a1,0(i,r)=⋯=ad,0(i,r)=0.$

Similarly, we have

$b1,0(j,s)=⋯=bd,0(j,s)=0.$

Next, by taking s = (1, …, 1) ≐ e in (13), we have $\begin{array}{}{a}_{i,r}^{\left(i,r\right)}={b}_{j,e}^{\left(j,e\right)}\end{array}$ for all ij. This tells us that $\begin{array}{}\begin{array}{}{a}_{i,r}^{\left(i,r\right)}={b}_{1,e}^{\left(1,e\right)}\end{array}\end{array}$ for any i ≠ 1 and $\begin{array}{}\begin{array}{}{a}_{i,r}^{\left(i,r\right)}={b}_{2,e}^{\left(2,e\right)}\end{array}\end{array}$ for any i ≠ 2. It follows that $\begin{array}{}{a}_{i,r}^{\left(i,r\right)}\end{array}$ is a constant denoted by λ for all i = 1, …, d and r ∈ ℤd ∖ {0}. Similarly, we obtain that $\begin{array}{}{b}_{j,s}^{\left(j,s\right)}\end{array}$ is a constant denoted by μ for all j = 1, …, d and s ∈ ℤd ∖ {0}. Finally, by $\begin{array}{}{a}_{2,e}^{\left(2,e\right)}={b}_{1,e}^{\left(1,e\right)}\end{array}$ we have λ = μ, which completes the proof. □

Lemma 2.7

Let fB(𝔚d), and ϕ, ψ be determined by Lemma 2.2, λ ∈ 𝔽 be given by Lemma 2.7. Then

$ϕ(D(ei,0))=λD(ei,0),ψ(D(ej,0))=λD(ej,0)$

for all i, j ∈ {1, …, d}.

Proof

We use the assumptions (3) and (4). By Lemma 2.4, we have

$ϕ(D(ei,0))=∑k=1dak,0(i,0)D(ek,0),ψ(D(ej,0))=∑k=1dbk,0(j,0)D(ek,0)$

for all i, j ∈ ℤ. Namely, it follows that, in (3) and (4), $\begin{array}{}{a}_{k,n}^{\left(i,0\right)}={b}_{k,n}^{\left(j,0\right)}=0\end{array}$ for all n ∈ ℤd ∖ {0}. Note that Lemma 2.6 tells us that, in (3) and (4), $\begin{array}{}{a}_{k,n}^{\left(i,r\right)}={\delta }_{i,k}{\delta }_{n,r}\lambda \end{array}$ and $\begin{array}{}{b}_{k,n}^{\left(j,s\right)}={\delta }_{j,k}{\delta }_{n,s}\lambda \end{array}$ for any i, j ∈ ℤ and r, s ∈ ℤd ∖ {0}. All these together with letting r = 0 in (5), deduce that

$∑k=1dak,0(i,0)skD(ej,s)=λsiD(ej,s).$

Then we have

$s1a1,0(i,0)+⋯+si(ai,0(i,0)−λ)+⋯+sdad,0(i,0)=0$

for all s ∈ ℤd ∖ {0}. Let s run over the vectors e1, e2, …, ed, we have $\begin{array}{}{a}_{i,0}^{\left(i,0\right)}=\lambda \end{array}$ and $\begin{array}{}{a}_{k,0}^{\left(i,0\right)}=0\end{array}$ for every ki. This proves that ϕ(D(ei, 0)) = λD(ei, 0). Similarly, we can obtain that ψ(D(ej, 0)) = λD(ej, 0). The proof is completed. □

Our main result is the following.

Theorem 2.8

Every biderivation of 𝔚d without anti-symmetric condition is inner.

Proof

Suppose that f is a biderivation of 𝔚d. Let ϕ be determined by Lemma 2.2, λ ∈ 𝔽 be given by Lemma 2.7. Note that 𝔚d is spanned by D(u, 0), D(u, r) for all u ∈ 𝔽d and r ∈ ℤd ∖ {0}. Then by Lemmas 2.6 and 2.7, we see that ϕ(x) = λx for all x ∈ 𝔚d. Now, it follows by Lemma 2.2 that

$f(x,y)=[ϕ(x),y]=[λx,y]=λ[x,y]$

for all x, y ∈ 𝔚d, as desired. □

3 An application

The anti-symmetric biderivation can be applied to linear commuting maps, commuting automorphisms and derivations, see . Another application of biderivation without the anti-symmetric condition is the characterization of post-Lie algebra structures. Post-Lie algebras have been introduced by Valette in connection with the homology of partition posets and the study of Koszul operads . As  point out, post-Lie algebras are natural common generalization of pre-Lie algebras and LR-algebras in the geometric context of nil-affine actions of Lie groups. Recently, many authors have studied some post-Lie algebras and post-Lie algebra structures [19, 20, 21, 22, 23]. In particular, the authors of  study the commutative post-Lie algebra structure on Lie algebra. Let us recall the following definition of a commutative post-Lie algebra.

Definition 3.1

Let (L,[,]) be a Lie algebra over 𝔽. A commutative post-Lie algebra structure on L is a 𝔽-bilinear product xy on L and satisfies the following identities:

$x∘y=y∘x,[x,y]∘z=x∘(y∘z)−y∘(x∘z),x∘[y,z]=[x∘y,z]+[y,x∘z].$

for all x, y, zL. It is also said that (L,[,],∘) is a commutative post-Lie algebra.

Lemma 3.2

(). Let (L,[,],∘) be a commutative post-Lie algebra. If we define a bilinear map f : L × LL given by f(x, y) = xy for all x, yL, then f is a biderivation of L.

Theorem 3.3

Any commutative post-Lie algebra structure on the generalized Witt algebra 𝔚d is trivial. Namely, xy = 0 for all x, y ∈ 𝔚d.

Proof

Suppose that (𝔚d, [, ], ∘) is a commutative post-Lie algebra. By Lemma 3.2 and Theorem 2.8, we know that there is λ ∈ 𝔽 such that xy = λ [x, y] for all x, y ∈ 𝔚d. Since the post-Lie algebra is commutative, so we have λ[x, y] = λ[y, x]. It implies that λ = 0. The proof is completed. □

Acknowledgement

We would like to thank the referee for invaluable comments and suggestions. This work was supported in part by the NNSFC [grant number 11771069], the NSF of Heilongjiang Province [grant number A2015007] and the Funds of the Heilongjiang Education Committee [grant numbers 12531483 and HDJCCX-2016211).

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Accepted: 2018-02-26

Published Online: 2018-04-30

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 447–452, ISSN (Online) 2391-5455,

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