1 Introduction
Derivations and generalized derivations are important in the study of structure of various algebras [Reference Benkovic2–Reference Brešar, Martindale and Miers5, Reference Du and Wang14, Reference Leger and Luks18, Reference Zhang, Feng, Li and Wu28]. In particular, Bre
$\breve{\mathrm{s}}$
ar et al. introduced the notion of biderivations of rings in [Reference Brešar, Martindale and Miers5], and they showed that all biderivations of noncommutative prime rings are inner. In [Reference Wang, Yu and Chen26], biderivations of Lie algebras were introduced, and the authors proved that all skew-symmetric biderivations of finite-dimensional simple Lie algebras over an algebraically closed field of characteristic zero are inner. Furthermore, in [Reference Tang21], it was proved that biderivations (without the skew-symmetry restriction) of finite-dimensional complex simple Lie algebras are inner. In recent years, many scholars gave their attention to the study of biderivations of many other Lie (super) algebras using case by case discussion, see [Reference Dilxat, Gao and Liu12, Reference Liu, Guo and Zhao19, Reference Tang22, Reference Tang and Yang23, Reference Wang and Yu25, Reference Xia, Wang and Han27]. There is no uniform method to determine all biderivations of some classes of Lie (super) algebras.
As is well known, any biderivation can be decomposed into a sum of a skew-symmetric biderivation and a symmetric biderivation. Skew-symmetric biderivations that are connected with linear commuting maps have been deeply studied, such as all skew-symmetric biderivations on any perfect and centerless Lie algebras are inner biderivations [Reference Brešar and Zhao6]. Symmetric biderivations can determine commutative post-Lie algebra structures, which are related to the homology of partition posets and Koszul operads [Reference Vallette24]. However, up to now, there is no efficient tool to determine all symmetric biderivations on Lie algebras. In the present paper, by introducing the concepts of the symmetric biderivation radical and the characteristic subalgebra, we establish a simple approach to determine all symmetric biderivations of finite-dimensional classical simple Lie algebras over a field of characteristic different from
$2, 3$
, and the Witt algebras
$\mathcal {W}^+_n$
over a field of characteristic
$0$
. We summarize our main result in the following theorem:
Theorem 1.1 Let
$\mathfrak {g}$
be a finite-dimensional classical simple Lie algebra with nondegenerate killing form over a field
$\mathbb {F}$
of characteristic not
$2$
or
$3$
, or the Witt algebras
$\mathcal {W}^+_n$
over a field
$\mathbb {F}$
of characteristic
$0$
. Then the following statements hold.
-
(1) Every biderivation
$\delta$
of
$\mathfrak {g}$
is inner, that is, there exists
$\lambda \in \mathbb {F} $
such that
$$ \begin{align*}\delta(x,y) = \lambda [x,y], \quad \forall\, x,y\in\mathfrak{g}.\end{align*} $$
-
(2) Every commutative post-Lie algebra structure on
$\mathfrak {g}$
is trivial.
It is worthwhile to point out that our method is conceptual avoiding a lot of computations and being distinct from those in other existing papers. We believe that our approach may be used to deal with many other Lie algebras.
The paper is organized as follows. In Section 2, we give some fundamental definitions (symmetric biderivation radical and characteristic subalgebra of a Lie algebra L) and establish some related properties. In Section 3, we prove that every symmetric biderivation of a finite-dimensional classical simple Lie algebra over arbitrary fields of characteristic not
$2$
or
$3$
is trivial. In Section 4, we show that every symmetric biderivation of the Witt algebras
$\mathcal {W}^+_n$
over fields of characteristic
$0$
is trivial. Finally, as applications, we determine the commutative post-Lie algebra structure on the aforementioned Lie algebras.
2 Symmetric biderivation radicals and characteristic subalgebras
Throughout the paper, we denote by
$\mathbb {Z},\,\mathbb {Z}_+,\,\mathbb {N}$
the sets of integers, nonnegative integers, and positive integers, respectively. Let L be a Lie algebra over an arbitrary field
$\mathbb {F}$
in the following, unless otherwise stated. Denote by
$\mathrm {Aut\,}(L)$
the automorphism group of L.
Definition 2.1 A bilinear map
$\delta :L\times L\to L$
is called a biderivation if for any
$x,y,z\in L$
,
$$ \begin{align*}\delta([x,y],z)=[x,\delta(y,z)]-[y,\delta(x,z)],\quad \delta(x, [y,z])=[\delta(x,y),z]+[y, \delta(x, z)].\end{align*} $$
If furthermore
$\delta (x,y) = \delta (y,x)$
for all
$x,y\in L$
, we call
$\delta $
a symmetric biderivation. A skew-symmetric biderivation can be similarly defined.
Remark 2.2
-
(1) A bilinear map
$\delta :L\times L\to L$
is a biderivation if and only if both
$\delta (x,\cdot )$
and
$\delta (\cdot ,x)$
are derivations for every
$x\in L$
. -
(2) The set of all biderivations on L is a vector space.
Lemma 2.3 If
$\delta : L\times L\to L$
is a symmetric biderivation, then
$$ \begin{align*}\delta(x,[y,z]) +\delta(y,[z,x])+\delta(z,[x,y])=0,\ \ \ \ \forall x,y,z\in L.\end{align*} $$
Proof A direct computation gives
$$ \begin{align*} &\delta(x,[y,z]) +\delta(y,[z,x])+\delta(z,[x,y])\\ &\quad=[y,\delta(x,z)]-[z,\delta(x,y)]+[z,\delta(y,x)]-[x,\delta(y,z)]\\ &\qquad+[x,\delta(z,y)]-[y,\delta(z,x)]\\ &\quad=0.\\[-30pt] \end{align*} $$
Lemma 2.4 Let L be a Lie algebra admitting a Cartan decomposition
$L=\oplus _{\alpha \in \frak {h}^*}L_{\alpha }$
with respect to a fixed Cartan subalgebra
$\frak {h}$
. Let
$x\in L_{\alpha }, y\in L_{\beta }$
with
$\alpha \neq \beta \in \frak {h}^*$
and
$\delta : L\times L\to L$
be a symmetric biderivation. If
$[x,y]=0$
, then
$\delta (x,y)=0.$
Proof Since
$\alpha \neq \beta $
, there exists
$h\in \frak {h}$
such that
$\alpha (h)\neq \beta (h)$
. By Lemma 2.3, we have
$$ \begin{align*}\delta(x,[y,h])+\delta(y,[h,x])+\delta(h,[x,y])=0,\end{align*} $$
which forces
$\left (\alpha (h)-\beta (h)\right )\delta (x,y)=0$
, that is,
$\delta (x,y)=0$
, as desired.
The following result asserts that the set of symmetric biderivations on L is invariant under the action of
$\mathrm {Aut\,}(L)$
.
Lemma 2.5 Let
$\delta : L\times L\to L$
be a symmetric biderivation. Then for any
$\sigma \in \mathrm {Aut\,}(L)$
,
$\delta _{\sigma }: L\times L\to L$
is a symmetric biderivation, where
$$ \begin{align*}\delta_{\sigma}(x,y)=\sigma(\delta(\sigma^{-1}(x),\sigma^{-1}(y))), \ \ \ \ \forall x,y\in L.\end{align*} $$
Proof The assertion follows from a straightforward computation.
Remark 2.6 It follows from Lemma 2.5 that there is a group action of
$\mathrm {Aut\,}(L)$
on the space of symmetric biderivations.
Now we introduce a very useful tool for the study of symmetric biderivations.
Definition 2.7 Denote
$$ \begin{align*}\mathrm{Rad\,}(L)=\{x\in L\mid \delta(x, L)=0\;\mbox{for every symmetric biderivation}\; \delta: L\times L\to L\},\end{align*} $$
which is called the symmetric biderivation radical of the Lie algebra L.
Next we assemble a few simple properties about
$\mathrm {Rad\,}(L)$
.
Proposition 2.8 The following statements hold.
-
(1)
$\mathrm {Rad\,}(L)$
is a subalgebra of L; -
(2) If L is a finite-dimensional Lie algebra, then
$\mathrm {Rad\,}(L)$
is a closed subset of L with respect to the Zariski topology; -
(3)
$\mathrm {Aut\,}(L)$
stabilizes
$\mathrm {Rad\,}(L)$
.
Proof Statements
$(1)$
and
$(2)$
follow from a direct computation.
Since for every symmetric biderivation
$\delta $
on L, we have
$$ \begin{align*}\delta(\sigma(x), y))=\sigma(\delta_{\sigma^{-1}}(x,\sigma^{-1}(y)))=0,\ \ \ \ \forall x\in \mathrm{Rad\,}(L), y\in L, \sigma\in \mathrm{ Aut\,}(L).\end{align*} $$
Then Statement
$(3)$
follows.
In order to better investigate the last property of symmetric biderivation radical listed above, we give the following definitions in Lie algebra case, analogous to those which arise in group theory.
Definition 2.9
-
(1) A subalgebra K of L is called a characteristic subalgebra if
$\sigma (K)\subseteq K$
for any
$\sigma \in \mathrm {Aut\,}(L)$
; -
(2) An ideal I of L is called a characteristic ideal if
$\sigma (I)\subseteq I$
for any
$\sigma \in \mathrm {Aut\,}(L)$
; -
(3) A Lie algebra L is called characteristically simple if L has no proper characteristic ideal, that is, L has no characteristic ideals other than L and
$0$
.
According to the above definitions and Proposition 2.8, we see that
$\mathrm {Rad\,}(L)$
is a characteristic subalgebra of L and any simple Lie algebra is characteristically simple. In group theory, a finite characteristically simple group is a direct sum of isomorphic simple groups. For the Lie algebra case, we have the following parallel result.
Proposition 2.10 A finite-dimensional Lie algebra L is characteristically simple if and only if L is a direct sum of isomorphic simple Lie algebras.
Proof First, we assume that
$L=S_1 \oplus S_2 \oplus \cdots \oplus S_k$
and
$S_i's$
are isomorphic simple Lie algebras. If
$k=1$
, then L is a simple Lie algebra. It is certainly characteristically simple. In the following, suppose
$k>1$
. For any
$i,j\in \{1,\ldots ,k\}$
with
$i\neq j$
, it is apparent that there exists
$\sigma _{i,j}\in \mathrm {Aut\,}(L)$
such that
$$ \begin{align*}\sigma_{i,j}(S_i)= S_j, \sigma_{i,j}(S_j)= S_i, \sigma_{i,j}(S_l)=S_l, \forall \,\,l\neq i,j.\end{align*} $$
Noticing that L is semisimple and each ideal of L is a sum of certain
$S_i$
’s. Then for any nonzero proper ideal I of L, there exists some
$\sigma _{i,j}\in \mathrm {Aut\,}(L)$
such that
$\sigma _{i,j}(I) \not \subseteq I$
. Consequently, L is characteristically simple.
Conversely, suppose that L is characteristically simple and S is a minimal ideal of L. So
$S\neq 0$
and it is possible that
$S=L$
. Let
$$ \begin{align*}\mathcal{V}=\{N\unlhd L\mid N=S_1 \oplus S_2 \oplus \cdots \oplus S_k, k\in\mathbb{N}, S_i\unlhd L, S_i\cong S, \forall\,1\leq i\leq k\},\end{align*} $$
where
$N\unlhd L, S_i\unlhd L$
mean that N and
$S_i$
are ideals of L. Note that each
$S_i$
is a minimal ideal of L. As
$S\in \mathcal {V}$
,
$\mathcal {V}$
is certainly nonempty. Let
$N=S_1\oplus S_2 \oplus \cdots \oplus S_k \in \mathcal {V}$
be of largest possible dimension. We assert that
$N=L.$
Otherwise, N is not a characteristic ideal since L is characteristically simple. There must exists a
$\sigma \in \mathrm {Aut\,}(L)$
such that
$\sigma (N) \not \subseteq N.$
Namely, there exists i such that
$\sigma (S_i) \not \subseteq N.$
Note that
$\sigma (S_i)\cong S$
is a minimal ideal of L. Since
$N\cap \sigma (S_i)$
is an ideal of L, and
$N\cap \sigma (S_i)$
is properly contained in
$\sigma (S_i)$
, we see that
$N\cap \sigma (S_i)=0$
by minimality of
$\sigma (S_i)$
. So
$$ \begin{align*}N\oplus\sigma (S_i)= S_1 \oplus S_2 \oplus \cdots \oplus S_k\oplus\sigma (S_i),\end{align*} $$
is an ideal of L, which means that
$N\oplus \sigma (S_i)\in \mathcal {V}$
, contradicting the choice of N. Therefore,
$$ \begin{align*}L=N= S_1 \oplus S_2 \oplus \cdots \oplus S_k.\end{align*} $$
It remains to check that S is simple. We may assume that
$S=S_1$
. If I is an ideal of
$S_1$
, then I is an ideal of L. Note that
$S_1$
is a minimal ideal of L, we see either
$I=0$
or
$I=S_1$
. Hence
$S_1$
and also S is simple, completing the proof.
3 Symmetric biderivations of finite-dimensional classical simple Lie algebras
Let
$\mathfrak {g}$
be a finite-dimensional classical simple Lie algebra over a field
$\mathbb {F}$
with nondegenerate Killing form and
$\mathrm {char\,} \mathbb {F}\neq 2, 3$
. If we fix a Cartan subalgebra
$\mathfrak {h}$
, then
$\mathfrak {g}$
has a root space decomposition
$\mathfrak {g}=\mathfrak {h}\oplus (\oplus _{\alpha \in \Phi }\mathfrak {g}_{\alpha })$
, where
$\Phi $
is the root system determined by
$\mathfrak {h}$
, and
$\mathfrak {g}_{\alpha }$
is the root space corresponding to the root
$\alpha \in \Phi $
with
$\mathrm {dim\,}\mathfrak {g}_{\alpha }=1$
. Denote by
$\Phi _{+}$
the set of positive roots,
$\Phi _{-}$
the set of negative roots, and
$\Pi $
a fixed set of simple roots, respectively. Let
$\{e_{\alpha }, f_{\alpha }, h_{\beta }\mid \alpha \in \Phi _+,\beta \in \Pi \}$
be a Chevalley basis of
$\mathfrak {g}$
. Denote by
$\theta $
the highest root in
$\Phi _{+}$
.
For
$\mathrm {Rad\,}(\mathfrak {g})$
, we have the following.
Proposition 3.1 If
$\mathrm {char\,}\mathbb {F}\neq 2, 3$
, then
$\mathrm {Rad\,}(\mathfrak {g})$
is an ideal of
$\mathfrak {g}$
.
Proof Keep in mind that
$\mathrm {Rad\,}(\mathfrak {g})$
is a characteristic subalgebra by Proposition 2.8. It suffices to show that
$\mathrm { Rad\,}(\mathfrak {g})$
is also an idea of
$\mathfrak {g}$
, i.e., we need to show that
$[x,r]\in \mathrm {Rad\,}(\mathfrak {g})$
for any
$r\in \mathrm { Rad\,}(\mathfrak {g})$
and
$x\in \mathfrak {g}$
. Note that
$\mathfrak {g}$
is generated by
$\mathfrak {g}_{\alpha }(\alpha \in \Phi )$
as a Lie algebra. We just need to prove that
$[x_{\alpha },r]\in \mathrm {Rad\,}(\mathfrak {g})$
for any
$\alpha \in \Phi $
. Recall that any root string is of length at most
$4$
. So, for any
$\alpha \in \Phi $
, we have
$(\mathrm {ad\,}x_{\alpha })^4=0$
. It follows from [Reference Benkart, Gregory and Premet1, Lemma 2.8] that
$$ \begin{align} \mathrm{exp\,}(\lambda \mathrm{ad\,}x_{\alpha})=\textbf{id}+\lambda\,\mathrm{ad\,}x_{\alpha}+\lambda^2\frac{(\mathrm{ ad\,}x_{\alpha})^2}{2}+\lambda^3\frac{(\mathrm{ad\,}x_{\alpha})^3}{6}\in \mathrm{Aut\,}(\mathfrak{g}),\ \ \ \ \forall\,\, \lambda \in \mathbb{F}. \end{align} $$
Applying
$\mathrm {exp\,}(\lambda \mathrm {ad\,}x_{\alpha })$
to r and using the fact that
$\mathrm {Rad\,}(\mathfrak {g})$
is a characteristic subalgebra, we get
$$ \begin{align*}r+\lambda[x_{\alpha},r]+\lambda^2\frac{(\mathrm{ad\,}x_{\alpha})^2r}{2}+\lambda^3\frac{(\mathrm{ad\,}x_{\alpha})^3r}{6}\in \mathrm{ Rad\,}(\mathfrak{g}),\ \ \ \ \forall\,\, \lambda \in \mathbb{F},\end{align*} $$
from which we can obtain a linear equation system whose coefficient matrix is exactly the Vandermonde matrix. Thus,
$[x_{\alpha },r]\in \mathrm { Rad\,}(\mathfrak {g})$
and hence the proposition follows.
Remark 3.2 We hope that
$\mathrm {Rad\,}(\mathfrak {g})$
is always an ideal of any Lie algebra
$\mathfrak {g}$
. However, we are not able to do so in this paper, since (3.1) is not always true. In fact, for the p-dimensional simple Witt algebra
$W_1=\text {span}_{\mathbb {F}}\{d_i\mid -1\leq i\leq p-2\}$
over a field
$\mathbb {F}$
of prime characteristic p with the Lie bracket defined as for
$-1\leq i,j\leq p-2$
,
$$ \begin{align*}[d_i, d_j]= \begin{cases} (j-i)d_{i+j}, &\text{if}\,\,-1\leq i+j\leq p-2;\cr 0, & otherwise, \end{cases}\end{align*} $$
we know that
$\mathrm {exp\,}(\mathrm {ad\,}d_{-1})\notin \mathrm {Aut\,}(W_1)$
. More generally, for a Lie algebra
$\mathfrak {g}=X_n$
of Cartan type
$X\in \{W,S,H,K\}$
over a field of prime characteristic, the exponential operator
$\mathrm {exp\,}(\mathrm {ad\,}x)$
does not make sense even for an ad-nilpotent element x in
$\mathfrak {g}$
.
We are now in a position to present the following main result, which together with [Reference Brešar and Zhao6] recovers and generalizes the main results in [Reference Tang21] and [Reference Wang, Yu and Chen26] where finite-dimensional simple Lie algebras over an algebraically closed field
$\mathbb {F}$
of characteristic
$0$
were considered.
Theorem 3.3 Let
$\mathfrak {g}$
be a finite-dimensional classical simple Lie algebra over a field
$\mathbb {F}$
with nondegenerate Killing form and
$\mathrm {char\,} \mathbb {F}\neq 2, 3$
. Then every symmetric biderivation of
$\mathfrak {g}$
is trivial.
Proof Let
$\delta : \mathfrak {g}\times \mathfrak {g}\to \mathfrak {g}$
be a symmetric biderivation. It follows from Lemma 2.3 that
$$ \begin{align}\delta(h_{\alpha}, h)=\delta([e_{\alpha},f_{\alpha}], h)=-2\alpha(h)\delta(e_{\alpha}, f_{\alpha}),\ \ \ \ \forall \alpha\in\Phi_+, h\in\mathfrak{h}. \end{align} $$
For any fixed
$\alpha \in \Phi _+$
, taking
$h=h_{\alpha }$
in the above equation gives
$$ \begin{align} \delta(h_{\alpha}, h_{\alpha})=-4\delta(e_{\alpha}, f_{\alpha}),\ \ \ \ \forall \alpha\in\Phi_+. \end{align} $$
Since the Killing form on
$\mathfrak {g}$
is nondegenerate by the assumption, it follows from Theorem 7 on Page 74 in [Reference Jacobson17] that every derivation on
$\mathfrak {g}$
is inner. Consequently,
$\delta (y,\cdot )=\mathrm {{ad}} u_y$
is an inner derivation for any
$y\in \mathfrak {g}$
, where
$u_y\in \mathfrak {g}$
is determined by y, and
$u_y$
can be written as a sum of elements in the Cartan subalgebra and the root spaces. It follows that
$\delta (h_{\alpha }, h_{\alpha })\in \sum _{\beta \in \Phi }\mathfrak {g}_{\beta }$
for any
$\alpha \in \Phi _+$
and
$$ \begin{align*}\delta(e_{\theta}, f_{\theta})=\delta(f_{\theta}, e_{\theta})\in\big(\mathfrak{h}\oplus(\oplus_{\alpha\in\Phi_{+}}\mathfrak{g}_{\alpha})\big)\cap \big(\mathfrak{h}\oplus(\oplus_{\beta\in\Phi_{-}}\mathfrak{g}_{\beta})\big)=\frak{h},\end{align*} $$
where
$\theta $
is the highest root, and the image of the operator
$\mathrm {{ad}} e_{\theta }$
(resp.
$\mathrm {{ad}}f_{\theta }$
) lies in
$\mathfrak {h}\oplus (\oplus _{\alpha \in \Phi _{+}}\mathfrak {g}_{\alpha })$
(resp.
$\mathfrak {h}\oplus (\oplus _{\beta \in \Phi _{-}}\mathfrak {g}_{\beta })$
. These along with (3.3) yield
$\delta (h_{\theta }, h_{\theta })=\delta (e_{\theta }, f_{\theta })=0$
. From the fact that each long positive root can become a highest root by the action of Weyl group (cf. [Reference Humphreys16, Section 10.4, Lemma C]) and Lemma 2.5, we see that
$\delta (h_{\alpha }, h_{\alpha })=\delta (e_{\alpha }, f_{\alpha })=0$
for any long positive root
$\alpha $
. Then it follows from (3.2) that
$\delta (h_{\alpha }, h)=0$
for any
$h\in \mathfrak {h}$
and long positive root
$\alpha $
. Since
$\Phi $
is spanned by all long positive roots [Reference Carter10, Proposition 8.18], we further obtain
$$ \begin{align*}\delta(\frak{h}, \frak{h})=0,\quad \delta(e_{\alpha}, f_{\alpha})=0,\ \ \ \ \forall\alpha\in \Phi_+.\end{align*} $$
Take
$\beta ,\gamma \in \Phi _+$
such that
$[ce_{\beta }, e_{\gamma }]=e_{\theta }$
for some
$c\in \mathbb {F}$
. By Lemma 2.3, we have
$\delta (e_{\theta },e_{\theta })=\delta ([ce_{\beta }, e_{\gamma }], e_{\theta })=0$
. Then for any long positive root
$\alpha $
,
$\delta (e_{\alpha },e_{\alpha })=0$
, which in turn forces
$$ \begin{align} \delta(h_{\alpha},e_{\alpha})=\delta([e_{\alpha},f_{\alpha}],e_{\alpha})=0, \end{align} $$
by the definition of biderivations. It is well known that for every
$\alpha \in \Phi $
, the set
$\mathrm {Ker\,}\alpha =\{h\in \mathfrak {h}\mid \alpha (h)=0\}$
is a proper subspace of
$\mathfrak {h}$
. Therefore, one can find a vector
$h_0\in \mathfrak {h}\setminus \cup _{\alpha \in \Phi }\mathrm {Ker\,}\alpha $
. For any
$h\in \frak {h}$
, let
$\delta (h, \cdot )=\sum _{\beta }a_{\beta }\mathrm {ad\,}x_{\beta }+\mathrm {ad}h'$
with
$a_{\beta }\in \mathbb {F}, x_{\beta }\in \mathfrak {g}_{\beta }, h'\in \frak {h}$
. Since
$\delta (h, h_0)=0$
, it follows that
$a_{\beta }=0$
for
$\beta \in \Phi $
. That is,
$\delta (h, \cdot )=\mathrm {ad}h'$
. Now for any long positive root
$\alpha $
, from (3.4), we see that
$$ \begin{align*}2\alpha(h')e_{\alpha}=2\delta(h,e_{\alpha})=\delta([h_{\alpha},e_{\alpha}],h)=\delta([h,e_{\alpha}],h_{\alpha})=0,\end{align*} $$
which yields
$h'=0$
by the fact that
$\Phi $
is spanned by the long roots. Consequently,
$\mathfrak {h}\subseteq \mathrm {Rad\,}(\mathfrak {g})$
. Thanks to Proposition 2.8 together with [Reference Collingwood and McGovern11, Corollary 2.1.13] or Theorem 3.1 together with the assumption that
$\mathfrak {g}$
is a simple Lie algebra,
$\mathrm {Rad\,}(\mathfrak {g})=\mathfrak {g}$
, that is,
$\delta $
is trivial. We complete the proof.
We can also derive the assertion
$\mathrm {Rad\,}(\mathfrak {g})=\mathfrak {g}$
from the fact
$\mathfrak {h}\subseteq \mathrm {Rad\,}(\mathfrak {g})$
by [Reference Collingwood and McGovern11, Corollary 2.1.13] and Proposition 2.8 in the following remark.
Remark 3.4 Since any regular semisimple element is conjugate to an element in
$\mathfrak {h}$
under
$\mathrm {Aut\,}(\mathfrak {g})$
by [Reference Collingwood and McGovern11, Corollary 2.1.13] and
$\mathrm {Rad\,}(\mathfrak {g})$
is invariant under the action of
$\mathrm {Aut\,}(\mathfrak {g})$
by Proposition 2.8(3), it follows from the fact
$\mathfrak {h}\subseteq \mathrm {Rad\,}(\mathfrak {g})$
that
$\mathrm {Rad\,}(\mathfrak {g})$
contains all regular semisimple elements in
$\mathfrak {g}$
. Note that the set of all regular semisimple elements in
$\mathfrak {g}$
is dense by [Reference Collingwood and McGovern11, Corollary 2.1.13], and
$\mathrm {Rad\,}(\mathfrak {g})$
is Zariski-closed by Proposition 2.8(2). Therefore,
$\mathrm {Rad\,}(\mathfrak {g})=\mathfrak {g}$
.
4 Biderivations of
$\mathcal {W}^+_n$
In this section, we assume that
$\mathbb {F}$
is a field of characteristic
$0$
. For
$n\in \mathbb {N}$
, let
$A_n=\mathbb {F}[t_1, t_2,\ldots ,t_n]$
be the polynomial algebra and
$\mathcal {W}^+_n=\mathrm {Der\,}(A_n)$
be the Witt algebra. For any
$\alpha =(\alpha _1, \alpha _2,\ldots ,\alpha _n)\in \mathbb {Z}_+^n$
, we denote
$t^{\alpha }=t_{1}^{\alpha _1}\cdots t_{n}^{\alpha _n}$
and
$|\alpha |=\alpha _1+\cdots +\alpha _n$
, respectively. Set
$d_i=\frac {\partial }{\partial t_i}$
and
$\mathcal {D}=\mathrm {span}_{\mathbb {F}}\{d_1,d_2,\ldots ,d_n\}.$
Then
$\mathcal {W}^+_n$
is a free
$A_n$
-module with basis
$d_i, 1\leq i\leq n$
, i.e.,
$\mathcal {W}^+_n=\oplus _{i=1}^nA_nd_i$
with the Lie bracket as follows:
$$ \begin{align*}[fd_i,gd_j]=fd_i(g)d_j-gd_j(f)d_i, \ \ \ f,g\in \mathbb{F}[t_1,\ldots,t_n], i,j=1,\ldots,n.\end{align*} $$
It is known that
$\oplus _{i=1}^n\mathbb {F}t_id_i$
is a Cartan subalgebra of
$\mathcal {W}^+_n$
.
Lemma 4.1 Keep notations as above. Then
$\mathcal {D}\subseteq \mathrm {Rad\,}(\mathcal {W}^+_n).$
Proof To find elements in the symmetric biderivation radical, we choose some commutative elements in
$\mathcal {W}^+_n$
, and constantly apply Lemma 2.4 and Lemma 2.5. Precisely consider first the situation when
$n>1$
. It is sufficient to show that
$d_i\in \mathrm {Rad\,}(\mathcal {W}^+_n)$
for any
$i=1,\ldots ,n$
. Observe that
$\mathrm {exp\,}(\mathrm {ad\,} d_i)\in \mathrm {Aut\,}(\mathcal {W}^+_n)$
for any
$i\in \{1,\ldots ,n\}$
because
$\mathrm {ad\,}d_i$
is locally nilpotent. Take any
$i\neq j\in \{1,\ldots ,n\}$
, then
$[t_i^kd_i,t_i^kt_jd_i]=0$
. From Lemmas 2.4 and 2.5, for any symmetric biderivation
$\delta $
of
$\mathcal {W}^+_n$
and
$k\in \mathbb {Z}_+$
, we have
$\delta (t_i^kd_i,t_i^kt_jd_i)=0$
and
$$ \begin{align*} \delta\big((\mathrm{exp\,}(\mathrm{ad\,} d_j))(t_i^kd_i),(\mathrm{exp\,}(\mathrm{ad\,} d_j))(t_i^kt_jd_i)\big)=\delta(t_i^kd_i,t_i^kt_jd_i+t_i^kd_i)=0, \end{align*} $$
which imply
$$ \begin{align*}\delta(t_i^kd_i,t_i^kd_i)=0,\ \ \ \ \forall\, k\in\mathbb{Z}_+.\end{align*} $$
In particular, we have
$$ \begin{align}\delta(d_i,d_i)=0\ \ \ \ \ \mathrm{and\,}\ \ \ \ \delta(t_id_i,t_id_i)=0.\end{align} $$
Since each derivation of
$\mathcal {W}^+_n$
is inner [Reference Dokovic and Zhao13], we may assume that
$\delta (d_i,\cdot )=\mathrm {ad}(\sum _{p=1}^nf_pd_p)$
with
$f_1,\ldots ,f_n\in A_n$
. Substituting this into the first equality of (4.1), we obtain
$$ \begin{align*}\left[d_i,\sum_{p=1}^nf_pd_p\right]=\sum_{p=1}^nd_i(f_p)d_p=0,\end{align*} $$
which forces
$$ \begin{align}d_i(f_p)=0, \ \ \ \ \ \forall\, p=1,\ldots,n.\end{align} $$
Meanwhile, the second equality of (4.1) together with Lemma 2.5 gives
$$ \begin{align*} \delta\big((\mathrm{exp\,}(\mathrm{ad\,} d_i))(t_id_i),(\mathrm{exp\,}(\mathrm{ad\,} d_i))(t_id_i)\big) =\delta(t_id_i+ d_i,t_id_i+ d_i)=0, \end{align*} $$
yielding that
$\delta (d_i,t_id_i)=0$
. In addition, since
$[d_i,t_qd_i]=0$
provided that
$q\neq i$
, we know that
$\delta (d_i,t_qd_i)=0$
by Lemma 2.4. Putting these together gives
$\delta (d_i,t_sd_i)=0, s=1,\ldots ,n$
. Combining this with the expression of
$\delta (d_i,\cdot )$
and (4.2), we get for any
$s=1,\ldots ,n$
,
$$ \begin{align*}0=\left[\sum_{p=1}^nf_pd_p,t_sd_i\right]=f_sd_i,\end{align*} $$
which in turn forces
$f_p=0, p=1,\ldots ,n$
. As a result,
$\delta (d_i,\cdot )=0$
.
Assume now that
$n=1$
. Denote
$\partial _i=t_1^{i+1}d_1$
for any
$i\geq -1$
with
$\partial _{-2}=0$
. Then we have
$[\partial _i,\partial _{j}]=(j-i)\partial _{i+j}$
for any
$i,j\geq -1$
. Since each derivation of
${\mathcal {W}}_1^+$
is inner [Reference Dokovic and Zhao13], we can write
$$ \begin{align} \delta(\partial_0,\cdot)=\mathrm{ad}\left(\sum_{j\geq -1}a_j\partial_j\right)\quad\mathrm{and\,}\quad \delta(\partial_{-1},\cdot)=\mathrm{ad}\left(\sum_{l\geq -1}b_l\partial_l\right), \end{align} $$
with
$a_j,b_l\in \mathbb {F}$
for
$j,l\geq -1$
. By Lemma 2.3, we have
$$ \begin{align*} (i+1)\delta(\partial_0,\partial_{i-1})&=\delta(\partial_0,[\partial_{-1},\partial_i])\\ &=-\delta(\partial_i,[\partial_{0},\partial_{-1}])-\delta(\partial_{-1},[\partial_{i},\partial_0])\\ &=\delta(\partial_i,\partial_{-1})+i\delta(\partial_{-1},\partial_{i})=(i+1)\delta(\partial_i,\partial_{-1}), \end{align*} $$
forcing
$\delta (\partial _0,\partial _{i-1})=\delta (\partial _i,\partial _{-1})$
for any
$i> -1$
. Inserting (4.3) into this equality, we obtain
$$ \begin{align*}\sum_{j\geq -1}a_j(i-1-j)\partial_{i+j-1}=\sum_{l\geq -1}b_l(i-l)\partial_{l+i},\ \ \ \ \forall i> -1.\end{align*} $$
By comparing the coefficients of
$\partial _{i-2}$
and
$\partial _{2i}$
of both sides, we, respectively, get
$a_{-1}=0$
and
$a_{i+1}=0$
for any
$i> -1$
. That is,
$\delta (\partial _0,\cdot )=a_0\mathrm {ad}\partial _0$
, and then
$\delta (\partial _0,\partial _0)=0$
. Further, by Lemmas 2.5, we have
$$ \begin{align*} &\delta\big((\mathrm{exp\,}(\lambda\mathrm{ad\,}\partial_{-1})) (\partial_0),(\mathrm{exp\,}(\lambda\mathrm{ad\,}\partial_{-1} )) (\partial_0)\big)\\&\quad=\delta(\partial_0+\lambda\partial_{-1},\partial_0+\lambda\partial_{-1})=0,\ \ \ \ \forall \lambda\in\mathbb{F}, \end{align*} $$
which means
$\delta (\partial _0,\partial _{-1})=0$
, i.e.,
$a_0[\partial _0,\partial _{-1}]=-a_0\partial _{-1}=0$
. Hence,
$a_0=0$
and
$\partial _0\in \mathrm { Rad\,}(\mathcal {W}^+_1)$
. Finally, for any
$\lambda \in \mathbb {F}$
and
$ i\geq -1$
, we have
$$ \begin{align*} &\delta\big((\mathrm{exp\,}(\lambda\mathrm{ad\,}\partial_{-1})) (\partial_0),(\mathrm{exp\,}(\lambda\mathrm{ad\,}\partial_{-1})) (\partial_i)\big)\\ &=\delta\left(\partial_0+\lambda\partial_{-1},\partial_i+\lambda(i+1)\partial_{i-1}+\cdots+\lambda^{i+1}\frac{(i+1)\ldots 1}{(i+1)!}\partial_{-1}\right)\\ &=\delta\left(\lambda\partial_{-1},\partial_i+\lambda(i+1)\partial_{i-1}+\cdots+\lambda^{i+1}\frac{(i+1)\ldots 1}{(i+1)!}\partial_{-1}\right)=0. \end{align*} $$
Taking
$\lambda =1,\ldots ,i+2$
, we can obtain a linear equation system whose coefficient matrix is the Vandermonde matrix. So
$\delta (\partial _{-1}, \partial _{i})=0$
and then
$d_1\in \mathrm {Rad}(\mathcal {W}^+_1)$
. We complete the proof.
Lemma 4.2 Let
$x\in \mathcal {W}^+_n$
. If
$[x, y]\in \mathrm {Rad}(\mathcal {W}^+_n)$
for all
$y\in \mathcal {D}$
, then
$x\in \mathrm {Rad}(\mathcal {W}^+_n)$
.
Proof For any
$y\in \mathcal {D}, z\in \mathcal {W}_n^+$
, we have
$$ \begin{align*}0=\delta([x,y],z)=[\delta(x,z),y]+[x,\delta(y,z)]=[\delta(x,z),y],\end{align*} $$
where the last equation holds by Lemma 4.1. This implies that
$\delta (x,z)\in \mathcal {D}$
. There exists
$X\in \mathcal {W}^+_n$
such that
$\delta (x,\cdot )=\mathrm {ad}X$
. Then
$[X,z]\in \mathcal {D}$
for any
$z\in \mathcal {W}^+_n$
. This yields that
$X=0$
. Consequently,
$\delta (x,\cdot )=0$
, as desired.
Now we are in a position to present our main result in this section.
Theorem 4.3 Every symmetric biderivation of
$\mathcal {W}^+_n$
over any field of characteristic
$0$
is trivial.
Proof It is enough to prove that
$t^{\alpha }d_i\in \mathrm {Rad}(\mathcal {W}^+_n)$
for any
$\alpha \in \mathbb {Z}_+^n$
and
$1\leq i\leq n$
. For any fixed i, we proceed by induction on
$|\alpha |$
. The case
$|\alpha |=0$
is given by Lemma 4.1. Assume that
$t^{\beta }d_i\in \mathrm {Rad}(\mathcal {W}^+_n)$
for any
$|\beta |<|\alpha |$
. According to the assumption, for
$1\leq k\leq n$
,
$[d_k,t^{\alpha }d_i]=\alpha _k t^{(\alpha _1,\ldots ,\alpha _k-1,\ldots , \alpha _n)}d_i\in \mathrm {Rad}(\mathcal {W}^+_n)$
if
$\alpha _k>0$
, and
$[d_k,t^{\alpha }d_i]=0\in \mathrm {Rad}(\mathcal {W}^+_n)$
if
$\alpha _k=0$
. Then by Lemma 4.2, we have
$t^{\alpha }d_i\in \mathrm {Rad}(\mathcal {W}^+_n)$
for any
$\alpha $
, completing the proof.
Combining this with the results in [Reference Brešar and Zhao6] asserting that every skew-symmetric biderivation of
$\mathcal {W}^+_n$
is inner, we obtain the following consequence.
Corollary 4.4 Every biderivation
$\delta $
of
$\mathcal {W}^+_n$
over any field
$\mathbb {F}$
of characteristic
$0$
is of the form
$\delta (x,y) = \lambda [x,y]$
,
$x,y\in \mathcal {W}^+_n$
, for some
$\lambda \in \mathbb {F} $
.
We remark that the biderivations of the Witt algebras
$\mathcal {W}_n$
over Laurent polynomials are proved to be inner in [Reference Tang and Yang23].
5 Applications
Post-Lie algebra structure is an important generalization of left-symmetric algebra structure, which arised in many areas of algebra and geometry [Reference Burde7]. Post-Lie algebras, which are related to homology of partition posets and the study of Koszul operads, have been studied by Vallette [Reference Vallette24] and Loday [Reference Loday20]. In addition, post-Lie algebras have been studied in connection with isospectral flows, Yang-Baxter equations, Lie-Butcher Series and Moving Frames [Reference Ebrahimi-Fard, Lundervold, Mencattini and Munthe-Kaas15]. The existence of post-Lie algebra structure on a given pair of Lie algebras turned out to be very meaningful and quite challenging. The authors in [Reference Burde and Dekimpe8] introduced a special class of post-Lie algebra structures, namely commutative post-Lie algebra. Using the Levi decompositions, it was proved that any commutative post-Lie algebra structure on a complex (finite-dimensional) perfect Lie algebra is trivial [Reference Burde and Moens9]. As an application of the previous results, we shall give commutative post-Lie algebra structures on finite-dimensional simple Lie algebras over arbitrary fields of characteristic not
$2$
or
$3$
, and the Witt algebras
$\mathcal {W}^+_n$
over fields of characteristic
$0$
. Let us recall the following definition of a commutative post-Lie algebra.
Definition 5.1 A commutative post-Lie algebra structure on a Lie algebra L over a field
$\mathbb {F}$
is an
$\mathbb {F}$
-bilinear product
$x\cdot y$
on L satisfying the following identities:
$$ \begin{align*} x\cdot y&=y\cdot x,\\ [x,y]\cdot z&=x\cdot (y\cdot z)-y\cdot (x\cdot z),\\ x\cdot [y,z]&=[x\cdot y,z]+[y,x\cdot z],\ \ \ \ \forall x,y,z\in L. \end{align*} $$
We also say that
$(L,[,],\cdot )$
is a commutative post-Lie algebra.
There is always the trivial commutative post-Lie algebra structure on L, given by
$x\cdot y=0$
for all
$x,y\in L$
. However, in general, it is not obvious whether or not a given Lie algebra admits a nontrivial commutative post-Lie algebra structure. The following lemma shows the connection between commutative post-Lie algebra structure and symmetric biderivation of a Lie algebra.
Lemma 5.2 (cf. [Reference Tang22]) Let
$(L,[,],\cdot )$
be a commutative post-Lie algebra. If we define a bilinear map
$\delta :L\times L\to L$
by
$\delta (x,y)=x\cdot y$
for all
$x,y\in L$
, then
$\delta $
is a symmetric biderivation of L.
As a consequence of Theorems 3.3, 4.3 and Lemma 5.2, we now give the main result of this section as follows.
Theorem 5.3 Let L be a finite-dimensional classical simple Lie algebra with nondegenerate Killing form over an arbitrary field of characteristic not
$2$
or
$3$
, or the Witt algebras
$\mathcal {W}^+_n$
over a field of characteristic
$0$
. Then every commutative post-Lie algebra structure on L is trivial.
Acknowledgments
The authors would like to thank professors Shujuan Wang and Hengyun Yang for their helpful discussions, and are very grateful to the referee for carefully reading the manuscript and giving helpful suggestions and comments.
