The Dzhumadildaev brackets: a hidden supersymmetry of commutators and the Amitsur-Levitzki-type identities

TL;DR

This paper explores the Dzhumadildaev brackets, revealing a hidden supersymmetry in commutators and generalizing Amitsur-Levitzki identities across various Lie algebras and superalgebras.

Contribution

It uncovers a hidden supersymmetry in commutators via Dzhumadildaev brackets and discusses potential generalizations of Amitsur-Levitzki identities.

Findings

Discovered hidden supersymmetry in commutators.

Generalized Amitsur-Levitzki identities to Lie superalgebras.

Outlined open problems for future research.

Abstract

The Amitsur-Levitzki identity for matrices was generalized in several directions: by Kostant for simple finite-dimensional Lie algebras, by Kirillov (later joined by Kontsevich, Molev, Ovsienko, and Udalova) for simple vectorial Lie algebras with polynomial coefficients, and by Gie, Pinczon, and Ushirobira for the orthosymplectic Lie superalgebra $\mathfrak{osp}(1|n)$. Dzhumadildaev switched the focus of attention in these results by considering the algebra formed by antisymmetrizors and discovered a hidden supersymmetry of commutators. We overview these results and their possible generalizations (open problems).

Equations

$$n\times n$$

$$a_r(X_1,\dots ,X_r):=\sum _{\sigma \in {\mathfrak{S}}_n}(-1{)}^{\mathrm{sign}\sigma }{X}_{\sigma (1)}\dots X_{\sigma (r)}.$$

$$a_r(X_1,\dots ,X_r)=0\text{ for any r≥2n.}$$

$$A_k(x_1,\dots ,x_k):=\sum _{\sigma \in S_k}(-1{)}^{\mathrm{sign}\sigma }{\mathrm{ad}}_{x_{\sigma (1)}}\dots {\mathrm{ad}}_{x_{\sigma (k)}}.$$

$$a_3({\mathrm{ad}}_{X_1},\dots ,{\mathrm{ad}}_{X_3})(Y)=-2\det \left(\begin{array}{ccc}{x}_1& x_2& x_3\\ x_1^{\prime }& x_2^{\prime }& x_3^{\prime }\\ x_1^{\prime \prime }& x_2^{\prime \prime }& x_3^{\prime \prime }\end{array}\right)\cdot Y,$$

$$A_N(X_1,\dots ,X_N)=0$$

$$\begin{array}[]{l}(a_{k}\ast a_{l})(X_{1},\ldots,X_{k+l-1}):=\\ \mathop{\sum}\limits_{\tiny\begin{array}[]{l}\sigma\in S_{k+l-1}\text{ such that}\\ \sigma(1)<\dots<\sigma(l)\;\text{ and }\\ \sigma(l+1)<\dots<\sigma(k+l-1)\end{array}}\mathop{\mathrm{sign}}\nolimits(\sigma)\,a_{k}(a_{l}(X_{\sigma(1)}\dots X_{\sigma(l)}),X_{\sigma(l+1)}\dots X_{\sigma(k+l-1)}).\end{array}$$

$$a_k\ast a_l=\{\begin{array}{ll}0& \text{if k,l are even},\\ k{a}_{k+l-1}& \text{if l is odd},\\ a_{k+l-1}& \text{if l is even and k is odd.}\end{array}$$

$$X(f{\mathrm{vol}\text{/}}^{\lambda })=(X(f)+f\lambda \mathrm{div}(X)){\mathrm{vol}\text{/}}^{\lambda }\text{ for any f∈F and X∈vect(n)}.$$

$$a_3(X_1,X_2,X_3)=0\text{ is an identity.}$$

$$\begin{array}[]{ll}a_{N}\equiv 0&\text{if $N\geq(n+1)^{2}$ for ${\mathfrak{vect}}(n)$}\\ a_{N}\equiv 0&\text{if $N\geq n(2n+5)$ for ${\mathfrak{h}}(2n)$}\end{array}$$

$$\footnotesize\begin{array}[]{l}A({\mathfrak{sl}}(n))=\mathop{\mathrm{{Span}}}\nolimits\{a_{2k}\mid k=1,2,\dots,n-1\},\text{~{}~{}in particular, $a_{2k+1}\equiv 0$ for any $k$};\\ A({\mathfrak{o}}(2n+1))=\mathop{\mathrm{{Span}}}\nolimits(\{a_{4k+1}\mid k=1,2,\dots,n-1\}\cup\{a_{4k+2}\mid k=0,1,2,\dots,n-1\}),\\ A({\mathfrak{sp}}(2n))=\mathop{\mathrm{{Span}}}\nolimits(\{a_{4k+1}\mid k=1,2,\dots,n-1\}\cup\{a_{4k+2}\mid k=0,1,2,\dots,n-1\}),\\ A({\mathfrak{o}}(2n))=\mathop{\mathrm{{Span}}}\nolimits(\{a_{4k+1}\mid k=1,2,\dots,n-1\}\cup\{a_{4k+2}\mid k=0,1,2,\dots,n-2\}\cup\{a_{4n-2}\}),\\ A({\mathfrak{g}}_{2})=\mathop{\mathrm{{Span}}}\nolimits(\{a_{2};\;\;a_{10}\}).\end{array}$$

$$m_i$$

$$A({\mathfrak{f}}_4)=\mathrm{Span}(\{a_2,a_{10},a_{14},a_{22}\})$$

$$\footnotesize\begin{array}[]{ll}\mathop{\mathrm{Q}}\nolimits(n)=&\{X\in\mathop{\mathrm{Mat}}\nolimits(n|n)\mid[X,J]=0\;\text{ for the odd invertible operator $J$}\}=\\ &\footnotesize{\left\{\begin{pmatrix}A&B\\ B&A\end{pmatrix}\mid A,B\in\mathop{\mathrm{Mat}}\nolimits(n)\right\}.}\end{array}$$

$$\mathrm{qtr}:\left(\begin{array}{cc}A& B\\ B& A\end{array}\right)⟼\mathrm{tr}B.$$

$$BX+(-1{)}^{p(X)p(B)}{X}^{st}B=0,$$

$$st:\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)⟼\left(\begin{array}{cc}{A}^t& -C^t\\ B^t& D^t\end{array}\right).$$

$$a_N(X_1,\dots ,X_N):=\sum _{s\in S_N}\mathrm{sign}(s,P)X_{s(1)}\dots X_{s(N)},$$

$$\mathrm{sign}(s,P):=\prod _{1\le i<j\le k,s(i)>s(j)}(-1{)}^{p_i{p}_j}.$$

$$s_1\circ s_2=(s_1(s_2(1)),\dots ,s_1(s_2(k))).$$

$$\mathrm{sign}(s_1\circ s_2,P)=\mathrm{sign}(s_1,P)\mathrm{sign}(s_2,s_1(P)),\text{ where s1(P)=(ps1(1),…,ps1(k)).}$$

$$\begin{array}[]{l}\mathop{\mathrm{sign}}\nolimits(s^{\prime})\mathop{\mathrm{sign}}\nolimits(s^{\prime},P)\mathop{\mathrm{str}}\nolimits(X_{s^{\prime}(1)}\dots X_{s^{\prime}(2l)})=\\ \mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s_{0})\mathop{\mathrm{sign}}\nolimits(s,P)\mathop{\mathrm{sign}}\nolimits(s_{0},s(P))\mathop{\mathrm{str}}\nolimits(X_{s(2)}\dots X_{s(2l)}X_{s(1)})=\\ -(-1)^{p_{1}(p_{2}+\dots+p_{2l})}\mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s,P)\times\\ (-1)^{p_{s(1)}(p_{s(2)}+\dots+p_{s(2l)})}\mathop{\mathrm{str}}\nolimits((-1)^{p_{s(1)}(p_{s(2)}+\dots+p_{s(2l)})}X_{s(1)}\dots X_{s(2l)})=\\ -\mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s,P)\mathop{\mathrm{str}}\nolimits(X_{s(1)}\dots X_{s(2l)}).\end{array}$$

$$B{a}_k(X_1,\dots ,X_k)+(-1{)}^{p_1+\cdots +p_k}{a}_k(X_1,\dots ,X_k{)}^{st}B=0.$$

$$\begin{array}[]{l}a_{k}(X_{1},\dots,X_{k})=\sum\limits_{s\in S_{k}}\mathop{\mathrm{sign}}\nolimits(s\circ s^{I})\mathop{\mathrm{sign}}\nolimits(s\circ s^{I},P)X_{(s\circ s^{I})(1)}\dots X_{(s\circ s^{I})(k)}=\\ \sum\limits_{s\in S_{k}}\mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s^{I})\mathop{\mathrm{sign}}\nolimits(s,P)\mathop{\mathrm{sign}}\nolimits(s^{I},s(P))X_{s(k)}\dots X_{s(1)}.\end{array}$$

$$B{X}_{s(k)}\dots X_{s(1)}=-(-1{)}^{p_{s(k)}}{X}_{s(k)}^{st}B{X}_{s(k-1)}\dots X_{s(1)}=\cdots =(-1{)}^{k+p_1+\cdots +p_k}{X}_{s(k)}^{st}\dots X_{s(1)}^{st}B,$$

$$\begin{array}[]{l}Ba_{k}(X_{1},\dots,X_{k})=\\ \sum\limits_{s\in S_{k}}\mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s^{I})\mathop{\mathrm{sign}}\nolimits(s,P)\mathop{\mathrm{sign}}\nolimits(s^{I},s(P))(-1)^{k+p_{1}+\dots+p_{k}}X_{s(k)}^{st}\dots X_{s(1)}^{st}B.\end{array}$$

$$(X_{s(1)}\dots X_{s(k)}{)}^{st}=\mathrm{sign}(s^I,s(P))X_{s(k)}^{st}\dots X_{s(1)}^{st},$$

Full text

Dzhumadildaev switched the focus of attention in these results by considering the algebra formed by antisymmetrizors and discovered a hidden supersymmetry of commutators.

We overview these results and their possible generalizations (open problems).

We are thankful to A. Dzhumadildaev for help and inspiring comments.

1. Introduction

Hereafter, the ground field is ${\mathbb{C}}$ although several statements are true over fields ${\mathbb{K}}$ of characteristic $p>2$.

1.1. On an experience of superizing

Consciously superizing various notions and statements since 1971, people observed that there are, usually, several ways and results of superizations: a straightforward one (usually, not a breath-catching one) and one or several other, often quite amazing, ways bringing up totally new notions (examples: the Poisson and anti- brackets, the supertrace and the queer trace on supermatrices, and the “quasi-classical limit” of these traces, and the corresponding superdeterminants, see [LSoS] and [DBS], p. 476).

A difficulty to be able to superize something by at least one method (to say nothing of several) usually indicates that we do not understand, actually, even the allegedly well-understood “nonsuper” situation. A prime example is the integration theory on supermanifolds which is still far from being completely constructed, see [LSoS] and [Lint]. Other examples are two somewhat related topics personified by the following two theorems:

Theorem** (Cayley-Hamilton).**

[TABLE]

Its first superization is due to Yastrebov [Ya]. For various (seemingly completely unrelated) super versions of the Cayley-Hamilton Theorem, see [KT, Del, KV, OP, GPS].

Theorem** (Amitsur–Levitzki).**

Let $C$ be a commutative and associative algebra. For any $X_{1},\ldots,X_{r}\in\mathop{\mathrm{Mat}}\nolimits(n;\,C)$, define antisymmetrizors $a_{r}$ by setting

[TABLE]

Then the Amitsur-Levitzki Identity (ALI) takes place:

[TABLE]

An interesting paper [GPU] was allegedly the final word concerning superization of ALI, but later a no less interesting paper [Sa] appeared. In this note, we also discuss superizations of ALI; for the proof of the classical ALI with the help of a Grassmann superalgebra, see §5.

1.3.1. Amitsur-Levitzki type theorem for vectorial Lie algebras.

A. A. Kirillov formulated the following analog of the Amitsur-Levitzki theorem, for its proof, see Preprints of Keldysh Inst. of Applied Math. in 1980s; for a translation of one such preprint, see [KOU]; the other preprints with related results by Kirillov, Kontsevich and Molev were never translated; Molev reviewed them in [Mo].

Theorem** ([Ki]).**

Let ${\mathfrak{g}}$ be a simple Lie algebra of vector fields over a field of characteristic [math]. Let

[TABLE]

For any $x_{1},\dots,x_{k}\in{\mathfrak{g}}$, the identity $A_{k}(x_{1},\dots,x_{k})\equiv 0$ holds

a) for $k\geq(n+1)^{2}$ if ${\mathfrak{g}}={\mathfrak{vect}}(n)$,

b) for $k\geq n(2n+5)$ if ${\mathfrak{g}}={\mathfrak{h}}(2n)$,

c) for $k\geq 2n^{2}+5n+5$ if ${\mathfrak{g}}={\mathfrak{k}}(2n+1)$.

1.4. Facts that inspired us

Let ${\mathfrak{vect}}(n)$ be the Lie algebra of vector fields (for simplicity, with polynomial coefficients).

[TABLE]

In [D1], Dzhumadildaev revealed a hidden supersymmetry of this well-known Fact(5) and posed a problem natural from this super point of view: quest for “higher” supersymmetries on the good old Lie algebras. Let us recall the less popular definitions and Dzhumadildaev’s construction.

Dzhumadildaev called the antisymmetrizor (2) of vector fields $X_{1},\ldots,X_{N}\in{\mathfrak{vect}}(n)$ an $N$-commutator if $a_{N}(X_{1},\ldots,X_{N})\in{\mathfrak{vect}}(n)$ for any $X_{1},\ldots,X_{N}\in{\mathfrak{vect}}(n)$ and $a_{N}$ does not vanish identically. If $a_{N}(X_{1},\ldots,X_{N})$ is an $N$-commutator, the number $N=N(n)$ is said to be critical.

The $N$-commutator is subcritical if $A_{N}(X_{1},\ldots,X_{N}):=a_{N}(\mathop{\mathrm{ad}}\nolimits_{X_{1}},\ldots,\mathop{\mathrm{ad}}\nolimits_{X_{N}})$ is multiplication by a function for any $X_{1},\ldots,X_{N}\in{\mathfrak{vect}}(n)$. For example, in [KOU], it is shown that for ${\mathfrak{vect}}(1)$, the antisymmetrizor $a_{3}$ acts as an operator of multiplication by a function:

[TABLE]

where $X_{i}=x_{i}(t)\frac{d}{dt}$ for $i=1,2,3$, $f^{\prime}:=\frac{df}{dt}$, and $Y=y(t)\frac{d}{dt}$.

1.4.1. Problems.

1. Is the following analog to the case for $n=1$ true?

[TABLE]

2. The number $N(n)=2$ is always critical for any $n$; we will call it the standard critical number. In [D1], Dzhumadildaev conjectured that the numbers $N(n)=(n+1)^{2}-3$ are also critical for $n>1$, proved the conjecture for $n=2$ and 3, and raised a natural problem: List all critical numbers. The problem is open, except for $n=3$, where Dzhumadildaev established that $N=10$ is also critical, and there are no more critical numbers.

Before we start considering this problem, let us discuss one more of Dzhumadildaev’s results. To present it, we need one more fact. Although we are sure that this fact was known since at least 1960s (for example, to I. Kantor and/or M. Gerstenhaber), the first reference we know is due to Dzhumadildaev [D0]:

[TABLE]

More precisely, we have ([D0])

[TABLE]

Thus, the antisymmetrizors define a ${\mathbb{Z}}$-graded superring $A=\oplus A_{i}$, where $A_{i}=\mathop{\mathrm{{Span}}}\nolimits(a_{i})$, such that $A_{\bar{0}}=\mathop{\oplus}\limits_{i\equiv 1\mod 2}A_{i}$ and the product of any two odd elements of $A_{\bar{1}}=\mathop{\oplus}\limits_{i\equiv 0\mod 2}A_{i}$ is zero. Clearly, $A$ can be considered as a superalgebra over any field by tensoring over ${\mathbb{Z}}$. What is the meaning of the superring or superalgebra $A$?

1.5. Dzhumadildaev’s approach to antisymmetrizors

In a series of papers, Dzhumadildaev changed the emphasis of the interpretation of the result by Amitsur and Levitzki from the search of the identity of the least order to the description of the superalgebra or the superring constructed from the antisymmetrizors in the classical Lie algebras. This approach revealed a hidden relation of the commutators with a certain universal odd superderivation. We overview various possible generalizations of Dzhumadildaev’s result.

Let ${\mathcal{F}}$ be an associative commutative algebra, ${\mathcal{A}}=\mathop{\mathrm{End}}\nolimits{\mathcal{F}}$ the associative algebra of its endomorphisms, and ${\mathcal{A}}_{L}$ the Lie algebra constructed by replacing the associative product by the bracket.

If111In Geometry, ${\mathcal{F}}$ is the algebra of functions on an $n$-dimensional manifold; it is interesting to generalize Dzhumadildaev’s approach to such cases, e.g., to the algebra of Laurent polynomials, i.e., the algebra of functions on the torus. ${\mathcal{F}}={\mathbb{K}}[x_{1},...,x_{n}]$, one can identify the elements of $\mathop{\mathrm{End}}\nolimits{\mathcal{F}}$ with differential operators. If $\mathop{\mathrm{End}}\nolimits{\mathcal{F}}$ is considered as associative algebra, its elements satisfy no identity except associativity. The Lie algebra $L=\mathop{\mathrm{Der}}\nolimits{\mathcal{F}}$ is a Lie subalgebra of $(\mathop{\mathrm{End}}\nolimits{\mathcal{F}})_{L}$ naturally identified with the Lie algebra ${\mathfrak{vect}}(n)$ of vector fields with polynomial coefficients.

Among numerous irreducible representations of $L$ (for their overview, super setting including, see [GLS]), there are two “smallest” ones: in the space of functions (or, more generally, $\lambda$-densities) and the adjoint representation.

Initially, people were interested in polynomial identities in the adjoint representations, see Theorem Theorem. Instead, Dzhumadildaev considered polynomial identities in the “smallest” representation, which for ${\mathfrak{vect}}(n)$ is the representation in the space of functions ${\mathcal{F}}$. It is very interesting to generalize Dzhumadildaev’s approach on the representations in the space of $\lambda$-densities, which is a rank 1 module over the algebra ${\mathcal{F}}$ generated by the $\lambda$-th power of the volume element with the following ${\mathfrak{vect}}(n)$-action (here $\lambda\in{\mathbb{C}}$ is fixed):

[TABLE]

It seems that this approach is more natural than the initial one for the following reasons:

1. If one knows identities in the “natural” representation (of the smallest dimension or — for infinite-dimensional algebras — its analog), then it is easy to construct identities in other representations, in particular in the adjoint representation. For example, $a_{n^{2}+2n-1}=0$ is identity in the space of functions ${\mathcal{F}}$, and since $\mathop{\mathrm{ad}}\nolimits_{X}=r_{X}-l_{X}$, where $r_{X}$ and $l_{X}$ are right and left actions in ${\mathcal{F}}$, it is easy to deduce that $a_{n^{2}+2n+1}=0$ is an identity in the adjoint representation of ${\mathfrak{vect}}(n)$.

2. If $a_{N}=0$ is identity, then one can ask “is $a_{N-1}$ a new operation on ${\mathfrak{vect}}(n)$?”

To consider $a_{N-1}$ as a multi-operation on ${\mathfrak{vect}}(n)$ is meaningless: $a_{N-1}$ maps $\wedge^{N-1}{\mathfrak{vect}}(n)$ to the whole ${\mathcal{A}}=\mathop{\mathrm{End}}\nolimits{\mathcal{F}}$, not just to ${\mathfrak{vect}}(n)$. Dzhumadildaev suggested to consider $a_{N}$ on the space of differential operators making the question “is $a_{N-1}$ a new operation on ${\mathfrak{vect}}(n)$?” meaningful: in some special cases $a_{N-1}$ maps $\wedge^{N-1}{\mathfrak{vect}}(n)$ to ${\mathfrak{vect}}(n)$ once again!

Now consider eq. (6). It means that 3-antisymmetric sum of the adjoint derivations on ${\mathfrak{vect}}(1)$ is a multiplication operator (not the adjoint operator). Certainly, it is an interesting observation, but it is another topic. It has no connection with $N$-commutators: in this setting to speak about $N$-commutator is meaningless. Under the natural action

[TABLE]

Let us retell Dzhumadildaev’s comments on observations due to Kirillov, Molev, Razmuslov, Bergman, and others on identities in ${\mathfrak{vect}}(n)$. The identities

[TABLE]

are not of the smallest degree. Moreover, these are “easy” identities. For example, for the Lie algebra ${\mathfrak{h}}(2)$ of Hamiltonian vector fields in two indeterminates, there are two identities in degree 7. Kirillov’s identity is not minimal and it is a consequence of these two identities. A similar situation with ${\mathfrak{vect}}(n)$. Dzhumadildaev conjectured that the minimal identity for representation of ${\mathfrak{vect}}(n)$ in the space of functions is of degree $(n+1)^{2}-2$ whereas the degree of Kirillov’s identity is $(n+1)^{2}$.

1.6. Antisymmetrizors for simple finite dimensional Lie algebras.

Exponents.

The classical Amitsur-Levitzki theorem states that $a_{2n}=0$ is the minimal identity for ${\mathfrak{gl}}(n)$. For ${\mathfrak{o}}(2n+1)$ and ${\mathfrak{sp}}(2n)$, the minimal identity is $a_{4n}=0$; for ${\mathfrak{o}}(2n)$, the minimal identity is $a_{4n-2}=0$ (see [AL, K1, K2]). Dzhumadildaev formulated the following theorem (known for the serial algebras) and gave explicit formulas for 10-antisymmetrisors in terms of the Chevalley basis for the 7-dimensional representation of ${\mathfrak{g}}_{2}$.

Theorem** ([D2]).**

*Let $A({\mathfrak{g}})$ be the algebra with respect to (8). Then *

[TABLE]

Problem**.**

1)* The indices of the antisymmetrizors are doubled exponents of the respective Lie algebras in the cases ${\mathfrak{sl}}(n)$ and ${\mathfrak{g}}_{2}$, but not for ${\mathfrak{o}}$ or ${\mathfrak{sp}}$:*

[TABLE]

What precisely is the relation between the indices of the nonvanishing identically operations $a_{i}$ and the exponents?

2)* For the matrix realizations in the irreducible module $R(\pi_{1})$ of the least dimension (see the right column in table (11)), is the following conjectural left column in table (11) correct?*

[TABLE]

3)* Clearly, the algebras $A({\mathfrak{g}})$ may depend on the realization of ${\mathfrak{g}}$, i.e., on the representation. And this does happen: the algebras $A({\mathfrak{sl}}(4))$ (corresponding to $R(\pi_{1})$) and $A({\mathfrak{o}}(6))$ (corresponding to $R(\pi_{2})$) are different. Theorem Theorem corresponds to matrix realizations of the Lie algebras ${\mathfrak{g}}$ in the irreducible module of the least (except for ${\mathfrak{o}}(6)$) dimension.*

Conjecture**.**

For the Lie algebras with the natural matrix realization, the above approach is reasonable. However, it seems no less reasonable to consider Lie algebra ${\mathfrak{g}}$ embedded into their universal enveloping algebras and look for $k$-commutators on ${\mathfrak{g}}$ inside $U({\mathfrak{g}})$, not inside a particular representation. For the finite-dimensional simple Lie algebras, only $k=2$ remains.

The proof of Theorem Theorem is based on the particular cases of Lemma **2.2.   **Lemma and [K1, K2].

2. Superizations of Theorem Theorem

First, let us superize the notions involved. For details of superization, see [LSoS]; we only recall here some basics. The supermatrices are considered in the standard format. The associative algebra $\mathop{\mathrm{Mat}}\nolimits(n)$ of $n\times n$ matrices has two super analogs: $\mathop{\mathrm{Mat}}\nolimits(n|m)$ and

[TABLE]

Accordingly, the general linear Lie algebra ${\mathfrak{gl}}(n)$ has two superanalogs: the Lie superalgebras ${\mathfrak{gl}}(n|m)$ and ${\mathfrak{q}}(n)$ obtained from the associative superalgebras $\mathop{\mathrm{Mat}}\nolimits(n|m)$ and $\mathop{\mathrm{Q}}\nolimits(n)$, respectively, by replacing the dot product by the super-bracket.

On the queer Lie superalgebra ${\mathfrak{q}}(n)$, the queer trace is defined:

[TABLE]

The Lie superalgebra ${\mathfrak{sq}}(n)$ is the Lie subsuperalgebra of ${\mathfrak{q}}(n)$ consisting of queertraceless supermatrices.

The supermatrix $X$ is said to preserve the bilinear form $B$ if

[TABLE]

where the supertransposition $st$ describing the matrix of the dual operator, see [LSoS], is defined as follows (in the standard format):

[TABLE]

Thanks to linearity, it suffices to consider only homogeneous with respect to parity elements.

The Lie superalgebras ${\mathfrak{osp}}(n|2m)$ and ${\mathfrak{pe}}(n)$ consist of elements preserving the nondegenerate symmetric bilinear form (even and odd, respectively) the normal forms of their Gram matrices being $\mathop{\mathrm{diag}}\nolimits(1_{n},J_{2m})$, where $J_{2m}=\mathop{\mathrm{antidiag}}\nolimits(1_{m},-1_{m})$, and $J_{n|n}=\mathop{\mathrm{antidiag}}\nolimits(1_{n},-1_{n})$, respectively (i.e., $J_{n|n}$ coincides with $J_{2n}$ but is odd). The same Lie superalgebras preserve antisymmetric nondegenerate bilinear forms.

For the composition $X_{1}\dots X_{k}$ of any $k$ operators $X_{1},\dots,X_{k}$ (supermatrices or vector fields, or whatever) of parities $P=(p_{1},\dots,p_{k})\in({\mathbb{Z}}/2)^{k}$, define its antisymmetrizor to be

[TABLE]

where $\mathop{\mathrm{sign}}\nolimits(s,P)=\mathop{\mathrm{sign}}\nolimits(s)\mathop{\mathrm{sign}}\nolimits(s^{\prime})$ and $s^{\prime}$ is the permutation induced by $s$ on the ordered subset of odd elements among $X_{1},\dots,X_{k}$. In other words, if $x_{1},\dots,x_{k}$ are elements of a supercommutative superalgebra whose respective parities are $p_{1}+{\bar{1}},\dots,p_{k}+{\bar{1}}$, then $x_{s(1)}\dots x_{s(k)}=\mathop{\mathrm{sign}}\nolimits(s,P)x_{1}\dots x_{k}$. One can express $\mathop{\mathrm{sign}}\nolimits(s,P)$ in another form, more convenient for computations. We define

[TABLE]

Define the composition of permutations by setting

[TABLE]

The function $\mathop{\mathrm{sign}}\nolimits(s,P)$ is a 1-cocycle on $S_{k}$ ([L]):

[TABLE]

Lemma**.**

The Lie superalgebra ${\mathfrak{sl}}(m|n)$ is closed under the $a_{2l}$ for any $m,n\geq 0$ and $l>0$. Moreover, $a_{2l}(X_{1},\dots,X_{2l})\in{\mathfrak{sl}}(m|n)$ for any $X_{1},\dots,X_{2l}\in\mathop{\mathrm{Mat}}\nolimits(m|n)$. For $mn=0$, the nonvanishing identically operations $a_{k}$ are listed in Theorem Theorem.

Proof.

We need to prove that $\mathop{\mathrm{str}}\nolimits\,a_{2l}(X_{1},\dots,X_{2l})=0$. Let $P=(p_{1},\dots,p_{2l})$ be the vector of parities of $X_{1},\dots,X_{2l}$.

For any $s\in S_{2l}$, set $s^{\prime}=(s(2),\dots,s(2l),s(1))$ (i.e., $s^{\prime}=s\circ s_{0}$, where $s_{0}=(2,\dots,2l,1)$. Then the terms in the sum (14) corresponding to $s$ and $s^{\prime}$ have opposite supertraces:

[TABLE]

Since $2l$ — the order of $s_{0}$ — is even, $S_{2l}$ can be represented as the disjoint union of two sets of equal cardinalities; and the set of elements of $S_{2l}$ can be divided in pairs of the form $(s,s\circ s_{0})$. Thus, the total supertrace of the sum (14) is equal to [math].∎

Lemma**.**

The Lie superalgebras ${\mathfrak{osp}}(m|2n)$ and ${\mathfrak{pe}}(n)$ are closed under $a_{k}$ for $k=4l+1$ and $4l+2$ for any $m,n,l\geq 0$.

For ${\mathfrak{osp}}(m|2n)$ and $mn=0$, the nonvanishing identically operations $a_{k}$ are listed in Theorem Theorem; for ${\mathfrak{osp}}(1|2n)$, we have $a_{4n}=0$ ([GPU]). For ${\mathfrak{osp}}(m|2n)$ and $mn\neq 0$ but not ${\mathfrak{osp}}(1|2n)$, and for ${\mathfrak{pe}}(n)$, the $a_{k}$ for $k=4l+1$ and $4l+2$ never vanish identically.

Proof.

Let $B$ be the Gram matrix of the bilinear form. Let $X_{1},\dots,X_{k}\in{\mathfrak{aut}}(B)$ be of parities $p_{1},\dots,p_{k}$. Then $BX_{i}+(-1)^{p_{i}}X_{i}^{st}B=0$, and we need to show that

[TABLE]

Set $s^{I}=(k,k-1,\dots,1)$. Then we can rewrite (14) as

[TABLE]

Since

[TABLE]

we have

[TABLE]

On the other hand,

[TABLE]

so

[TABLE]

The two sums are opposite if $\mathop{\mathrm{sign}}\nolimits(s^{I})(-1)^{k}=-1$, and then

[TABLE]

Since $\mathop{\mathrm{sign}}\nolimits(s^{I})=(-1)^{[k/2]}$, this is true for $k=4l+1,4l+2$.∎

Problem**.**

What is the analog of Lemma **2.2.   **Lemma for ${\mathfrak{spe}}(n)$?

Lemma**.**

The Lie superalgebra ${\mathfrak{q}}(n)$ is closed under $a_{k}$ and ${\mathfrak{sq}}(n)$ is closed under $a_{2k}$ for any $n$ and $k$.

Proof.

The associative algebra $\mathop{\mathrm{Q}}\nolimits(n)$ is closed with respect to the dot product; hence the result about ${\mathfrak{q}}$.

Since $\mathop{\mathrm{qtr}}\nolimits(XY)=(-1)^{p(X)p(Y)}\mathop{\mathrm{qtr}}\nolimits(YX)$ ($=\mathop{\mathrm{qtr}}\nolimits(YX)$, since $X$ and $Y$ should be of different parities in order to have $\mathop{\mathrm{qtr}}\nolimits(XY)\neq 0$) and so the same arguments as for ${\mathfrak{sl}}$ are applicable. ∎

2.4. Questions

What is the super analog of eq. (8) for the super-antisymmetrizor (14)?

3. Vectorial Lie algebras

3.1. ${\mathfrak{vect}}(n)$

In [FF], Feigin and Fuchs proved, among other things, that for $n=1$, the only critical pair is the standard one: $(1,2)$.

In [D1], Dzhumadildaev showed that for $n=2$, the complete list of critical pairs consists only of the standard pairs $(2,2)$ and $(2,6)$. Dzhumadildaev gave the following explicit expression of the 6-commutator: the 6-tuple $(X_{1},X_{2},X_{3},X_{4},X_{5},X_{6})$, where $X_{i}=u_{i,1}\partial_{1}+u_{i,2}\partial_{2}$ for $i=1,\ldots,6$, goes to

[TABLE]

where the coefficient of $\partial_{2}$ is obtained from that of $\partial_{1}$ by interchanging the subscripts $1$ and $2$ if there is only one subscript (as in $\partial_{1}u_{5,2}$), only second subscripts $1$ and $2$ should be interchanged when dealing with $u_{ij}$.

3.2. How to write the $k$-commutator for any $n$?

Let $X_{1},...,X_{k}\in{\mathfrak{vect}}(n)$ with coefficients $u_{i,j}$ (i.e., $X_{i}=\mathop{\sum}\limits_{1\leq j\leq n}u_{i,j}\partial_{j}$). Let $a=(a_{1},\dots,a_{k})$, where $a_{i}\in\{1,2,\dots,n\}$; let $(b_{ij})$ be a $k\times n$ matrix with elements in ${\mathbb{Z}}_{\geq 0}$; let $D((a_{i}),(b_{ij})$ be the determinant of the $k\times k$ matrix whose $(i,j)$-th slot is occupied by $\partial_{1}^{b_{i1}}...\partial_{n}^{b_{in}}u_{j,a_{i}}$. Considering the $k$-commutator of the fields $X_{1},...,X_{k}$ as a differential operator, its 1-st degree component is equal to

[TABLE]

where the $E^{i,j}$ are matrix units.

Accordingly, if the $k$-commutator is a first order operator, then (17) is its expression. Unfortunately, this expression is not user-friendly: first, it is longish ($n^{k}\times(k-1)!$ summands) which even for $n=2,k=6$ is $>7000$), second, it is very redundant: some of the summands vanish, some are equal to each other, some are equal in absolute value but are of different signs (so there are just 14 distinct types of summands for $n=2$, not $>7000$).

In [D3], Dzhumadildaev showed that for $n=3$, in addition to the standard pairs $(3,2)$ and $(3,13)$, there is exactly one more critical pair, $(3,10)$.

3.3. The other series of simple vectorial Lie algebras with

polynomial coefficients.

It is equally natural to list all critical pairs for the other types of simple vectorial Lie algebras. For these Lie algebras, only the pairs $(n,2)$ will be called standard.

For the Lie algebras ${\mathfrak{svect}}(n)$ of divergence-free vector fields, Dzhumadildaev proved [D1, D3] that the only nonstandard critical pairs are $(2,5)$ and $(3,10)$ (for $n=2$ and 3, respectively). Since ${\mathfrak{svect}}(2)\simeq{\mathfrak{h}}(2)$ the result for this Lie algebra might be pertaining to the Hamiltonian series, rather than to the divergence-free one.

For the Lie algebras ${\mathfrak{h}}(2n)$ of Hamiltonian vector fields, Dzhumadildaev proved [D1] that the only nonstandard critical pair for $n=1$ is $(2,5)$. In terms of generating functions in $p$ and $q$, the 5-commutator is proportional to the following beautiful map

[TABLE]

Problem**.**

What are the $N$-commutators for the Lie algebra of contact vector fields ${\mathfrak{k}}(2n+1)$?

Dzhumadildaev’s guiding idea is very simple and brought to the title of [D3]: it is a certain odd derivation of a certain superalgebra associated with the problem, which is in the heart of this matter, see the next Section.

4. The universal odd derivation and $N$-commutators (after [D3])

Let $L$ be a Lie (super)algebra, $U(L)$ its enveloping algebra, $\Pi$ the change of parity functor. Take the associative supercommutative superalgebra $K=S^{\mathbin{\hbox{\raise 1.20552pt\hbox{\bf.}}}}(\Pi(L))$; in particular, if $L$ is purely even, then $K$ is a Grassmann superalgebra. In $L$, select an arbitrary basis $B$ and set

[TABLE]

Lemma**.**

The $N$-commutator on $L$ yields an element of $L$, if and only if $D^{N}\in K\otimes L$. The $N$-commutator does not vanish identically if and only if $D^{N}\neq 0$.

In particular, for $L\subset{\mathfrak{vect}}(n)={\mathfrak{der}}({\mathbb{K}}[x])$, we clearly have

[TABLE]

The $N$-commutator on $L$ yields an element of $L$ if and only if $D^{N}\in{\mathfrak{vect}}(n|n)={\mathfrak{der}}({\mathbb{K}}[x,\Pi(x)])$.

Comment. For superspaces, the following modification of Fact (5) takes place:

[TABLE]

Fact (5) is, therefore, a corollary of Fact (19) for $N=2$. This is the hidden supersymmetry of the anticommutator mentioned in the title of this paper.

Conjecture**.**

We only considered Lie algebras of vector fields with polynomial coefficients. We conjecture that the answer will be same for any type of coefficients (at least, if polynomials are dense in the space of coefficients).

4.2. Discussion and setting of the problem.

As noted in Introduction, attempts to superize a problem or a notion usually reveal two roads: a straightforward one (not of much interest) and a totally unexpected one. The problem Dzhumadildaev posed (describe all critical pairs for (simple) Lie algebras of vector fields) is the one which we do not know how to superize. In particular, what is the answer for any of the simple Lie superalgebras of vector fields (with polynomial coefficients to begin with)?

Recall steps of Dzhumadildaev’s proof. Let $l$ be the length function on $\mathop{\mathrm{Diff}}\nolimits_{n}$ defined by Dzhumadildaev, namely:

[TABLE]

Let us extend $l$ to a grading (Dzhumadildaev’s definition is slightly different but equivalent). Note that the possibility of such extension is a little less evident than in the case of $\mathcal{L}_{n}$ because the elements $\eta_{i_{j},\alpha_{j}}\partial^{\beta}_{j}$ do not supercommute.

Let $X_{1},\dots,X_{k}$ be some abstract vector fields (considered as variables here) of $n$ indeterminates. Define the following map $F$ from $\mathop{\mathrm{Diff}}\nolimits_{n}^{[k]}$ to the algebra of differential operators (of arbitrary degree) in $n$ indeterminates:

[TABLE]

Here $(\partial^{\alpha_{j}}X_{s(j),\;i_{j}})$ is a function, $(\partial^{\alpha_{j}}X_{s(j),\;i_{j}})\partial^{\beta_{j}}$ is a differential operator (possibly of zero degree, if $\beta_{j}=0$), and the whole term is the composition of differential operators.

Statement**.**

The map $F$ is faithful.

The idea of a proof: the map preserves commutation relations. Note that

a) $F(D^{k})$ is just the $k$-commutator of the $X_{j}$ (considered as a differential operator of arbitrary degree);

b) the map $F$ preserves the degree of the differential operator (for generic $X_{i}$).

So the $k$-commutator is of degree $1$ for any $X_{j}$ if and only if $\deg D^{k}=1$.

5. Appendix: A proof of the classical Amitsur–Levitzki identity

Let $A$ be a supercommutative superalgebra and $X\in\mathop{\mathrm{Mat}}\nolimits(n|0;\,A)_{\bar{1}}$. It is clear that $X^{r}=0$ for any $r>n^{2}$. It turns out that $r$ can be considerably diminished.

Proposition**.**

$X^{2n}=0$* for any $X\in\mathop{\mathrm{Mat}}\nolimits(n|0;\,A)_{\bar{1}}$.*

First of all, let us discuss what does this identity mean from the “ordinary”, i.e., nonsuper, algebra point of view. Let $C$ be commutative algebra and $X_{1},\ldots,X_{r}\in\mathop{\mathrm{Mat}}\nolimits(n;\,C)$. Set $A=C[\xi_{1},\ldots,\xi_{r}]$, where the $\xi_{i}$ are odd and let $X:=\sum\xi_{i}X_{i}\in\mathop{\mathrm{Mat}}\nolimits(n|0;\,A)_{\bar{1}}$. Clearly,

[TABLE]

Hence, Proposition **5.1.   **Proposition implies the Amitsur–Levitzki identity (3).

Exercise**.**

The Amitsur–Levitzki identity implies Proposition **5.1.   **Proposition.

Proof

of Proposition **5.1.   **Proposition.222This proof is due to J. Bernstein, 1975. At about the same time V. Drinfeld also noticed the equivalence proved here. Set $Y=X^{2}$. The elements of $Y$ belong to the commutative algebra $A_{{\bar{0}}}$, and therefore, we may consider the characteristic polynomial $P(\lambda)=\det(\lambda 1_{n}-Y)$ with coefficients in $A_{\bar{0}}$. Let us prove that $P(\lambda)=\lambda^{n}$. Since the Cayley–Hamilton theorem implies $P(Y)=0$, we have $Y^{n}=0$, i.e., $X^{2n}=0$. We will prove that $P(\lambda)=\lambda^{n}$ by three different methods.

1. If $\text{char }k=0$, then the coefficients of $P(\lambda)$ can be expressed in terms of $\mathop{\mathrm{tr}}\nolimits Y^{r}$ for $r=1,2,\ldots$. Therefore, it suffices to verify that $\mathop{\mathrm{tr}}\nolimits Y^{r}=0$. Indeed,

[TABLE]

Hence, $\mathop{\mathrm{tr}}\nolimits Y^{r}=0$ for $r=1,2,\ldots$.

2. Let us show that $P(\lambda)^{2}=\lambda^{2n}$. If $2$ is invertible in $A$, we see that $P(\lambda)=\lambda^{n}$. We have to show that $\det^{2}(1_{n}-\lambda X^{2})=1$. This follows from a more general statement.

Lemma**.**

Let $U\in\mathop{\mathrm{Mat}}\nolimits(p\times q;\,A)$ and $V\in\mathop{\mathrm{Mat}}\nolimits(q\times p;\,A)$ be matrices whose entries are odd elements of $A$. Then

[TABLE]

Proof.

Let Z=\text{\footnotesize\begin{pmatrix}1_{p}&U\cr V&1_{q}\end{pmatrix}}\in\mathop{\mathrm{{GL}}}\nolimits(p|q;\,A) and \text{\footnotesize\begin{pmatrix}A&B\ C&D\end{pmatrix}}^{\Pi}:=\text{\footnotesize\begin{pmatrix}D&C\ B&A\end{pmatrix}}. From [LSoS] we know that $\mathop{\mathrm{Ber}}\nolimits Z^{\Pi}=(\mathop{\mathrm{Ber}}\nolimits Z)^{-1}$, so $\mathop{\mathrm{Ber}}\nolimits Z^{\Pi}=\det(1_{q}-VU)$ because $\mathop{\mathrm{Ber}}\nolimits Z=\det(1_{p}-UV)$. ∎

3. Let Z=\text{\footnotesize\begin{pmatrix}1_{n}&\lambda X\cr\lambda X&1_{n}\end{pmatrix}}\subset\mathop{\mathrm{{GQ}}}\nolimits(n;\,A[\lambda]). From [LSoS] we know that $\mathop{\mathrm{Ber}}\nolimits Z=1$. But $\mathop{\mathrm{Ber}}\nolimits Z=\det(1_{n}-\lambda^{2}X^{2})$, hence, $\det(1_{n}-\lambda^{2}Y)=1$, and we have $\det(1_{n}-\lambda Y)=1$. Thus, $\det(\lambda 1_{n}-Y)=\lambda^{n}$. ∎

5.2. How to superize the Cayley–Hamilton theorem?

The degree of the polynomial equation a given $n\times n$ matrix satisfies given by the Amitsur–Levitzki identity can be diminished even more (Cayley–Hamilton theorem, see (1)).

Conjecture** ( [GPS]).**

The analog of the Cayley–Hamilton theorem for supermatrices was unknown, except for small values of $n$ (equal to $2$ or $1|1$), until recently. Now we have a conjectural formula suggested by the study of quantum algebras and passage to the appropriate “super” limit.