Supertraces on queerified algebras

TL;DR

This paper characterizes supertraces on various queerified algebras, including matrix, Calogero-Moser, higher spin, and pseudo-differential operator algebras, revealing integrability properties of related Euler equations.

Contribution

It provides a comprehensive description of supertraces on queerified algebras, linking algebraic structures to integrability of associated dynamical systems.

Findings

Supertraces are explicitly described for queerified matrix and Calogero-Moser algebras.

Supertraces on pseudo-differential operator algebras demonstrate integrability of Euler equations.

Results connect algebraic supertrace structures to dynamical system integrability.

Abstract

We describe supertraces on ``queerifications'' (see arxiv:2203.06917) of the algebras of matrices of ``complex size'', algebras of observables of Calogero-Moser model, Vasiliev higher spin algebras, and (super)algebras of pseudo-differential operators. In the latter case, the supertraces establish complete integrability of the analogs of Euler equations to be written.

Equations

$$\Pi (x)\Pi (y):=xy\text{ for any }x,y\in A.$$

$$[x,y]:=xy-yx;[x,\Pi (y)]:=\Pi (xy-yx);[\Pi (x),\Pi (y)]:=xy+yx\text{ for any }x,y\in A.$$

$$(X,Y)⟷\left(\begin{array}{cc}X& Y\\ Y& X\end{array}\right)\in \mathfrak{gl}(n|n),\text{ where X,Y∈Mat(n)}.$$

$$\begin{array}[]{l}{}[(X_{1},0),(X_{2},0)]:=([X_{1},X_{2}],0),\quad[(X,0),(0,Y)]:=(0,[X,Y]),\\ {}[(0,Y_{1}),(0,Y_{2})]:=(Y_{1}Y_{2}+Y_{2}Y_{1},0).\end{array}$$

$${\mathfrak{u}}_{\bar{0}}=[\mathfrak{g},\mathfrak{g}]+[\Pi (\mathfrak{g}),\Pi (\mathfrak{g})].$$

$$\text{qtr}:\left(\begin{array}{cc}X& Y\\ Y& X\end{array}\right)\mapsto \text{tr}Y.$$

$${\mathfrak{g}}^{\prime }=({\mathfrak{g}}^{\prime }{)}_{\bar{0}}\oplus ({\mathfrak{g}}^{\prime }{)}_{\bar{1}}$$

$$(\mathfrak{qg}{)}_{\bar{0},\bar{0}}={\mathfrak{g}}_{\bar{0}},(\mathfrak{qg}{)}_{\bar{0},\bar{1}}={\mathfrak{g}}_{\bar{1}},(\mathfrak{qg}{)}_{\bar{1},\bar{0}}=\Pi ({\mathfrak{g}}_{\bar{1}}),(\mathfrak{qg}{)}_{\bar{1},\bar{1}}=\Pi ({\mathfrak{g}}_{\bar{0}}).$$

$$({\mathfrak{g}}^{\prime }{)}_{\bar{0}}=\text{Span}([x,y]:=xy-yx,[a,b]:=ab+ba),({\mathfrak{g}}^{\prime }{)}_{\bar{1}}=\text{Span}([x,a]:=xa-ax).$$

$${\mathfrak{u}}_{\bar{0},\bar{0}}$$

$${\mathfrak{u}}_{\bar{0},\bar{0}}=(\mathfrak{qg}{)}_{\bar{0},\bar{0}}={\mathfrak{g}}_{\bar{0}}$$

$$u^2\in Z$$

$$D^{n}f:=\sum _{k\ge 0}\left(\genfrac{}{}{0ex}{}{n}{k}\right)D^k(f)D^{n-k},\text{ where (kn):=k!n(n−1)…(n−k+1), for any f∈F}.$$

$$\text{tr}(\sum _{k\le n}{f}_k{D}^k)=\int f_{-1}dx,\text{ where fk∈F}.$$

$$b(X,Y):=\text{tr}(XY)\text{ for any X,Y∈A}.$$

$$\dot{L}=[L,P],\text{ where L,P∈A and dot signifies derivative with respect to time t},$$

$$({\partial }_{\tau }+\tau {\partial }_t)L=[L,H],\text{ where L,H∈A},$$

$$\{f,g\}(X):=X([df(X),dg(X)])\text{ for any X∈g∗}.$$

$$\text{tr}(a_1\otimes a_2):=({\text{tr}}_1{a}_1)({\text{tr}}_2{a}_2)$$

$$T(P):=\{\begin{array}{ll}(P(\frac{1}{x+1})){|}_{x=1}& \text{ if wht(P)=0,}\\ 0& \text{ if wht(P)=0}.\end{array}$$

$$T(PQ)=(-1{)}^{p(P)p(Q)}T(QP).$$

$$\begin{array}[]{ll}T(PQ)&=(x^{n+1}(-1)^{n+1}\frac{(n+1)!}{(x+1)^{n+2}})|_{x=1}=-(-\frac{1}{2})^{n+2}(n+1)!\ ,\\ T(QP)&=(d(x^{n+1}(-1)^{n}\frac{n!}{(x+1)^{n+1}})|_{x=1}\\ &=((-1)^{n}\frac{(n+1)!\ x^{n}}{(x+1)^{n+1}}-(-1)^{n}\frac{(n+1)!\ x^{n+1}}{(x+1)^{n+2}})|_{x=1}=(-\frac{1}{2})^{n+2}(n+1)!\end{array}$$

$$\begin{array}[]{l}T(QP)=(xP(\frac{1}{x+1}))|_{x=1}=(P(\frac{1}{x+1}))|_{x=1},\\ T(PQ)=(P(\frac{x}{x+1}))|_{x=1}=(P(1-\frac{1}{x+1}))|_{x=1}=-(P(\frac{1}{x+1}))|_{x=1},\end{array}$$

$$\sum _{n=0}^{\infty }(-1{)}^n\text{(the coefficient of xn in Pxn)}.$$

$$\sum _n(-1{)}^{n}P{x}^n{|}_{x=1}=P(1-x+x^2-x^3+...){|}_{x=1}=\left(P\left(\frac{1}{1+x}\right)\right){|}_{x=1}.$$

Full text

The traces on Lie algebras, and even (not odd) supertraces on Lie superalgebras, are known to be very useful, for example, in representation theory, see, e.g., [DSB] and references therein. The odd supertraces are less known, hence less popular. In Section 4, we show one of their usages not previously explored: application to the study of integrability of certain hamiltonian dynamical systems.

Inspired by [BLLS], where Lie algebras are queerified over an algebraically closed ground field of characteristic $p=2$ in order to produce the complete list of simple finite-dimensional Lie superalgebras in characteristic $p=2$, this new method — queerification — producing many new simple Lie superalgebras from associative algebras and superalgebras is applied in [L] over $\mathbb{C}$ to several infinite-dimensional algebras of interest in theoretical physics. A  number of papers were devoted to description of traces on these algebras, and supertraces on these algebras considered as superalgebras, see [KS, KT1, KT2, KT3]. In this note, we describe the supertraces on the Lie queerifications of these algebras and superalgebras, having added one more type of examples.

1. Preliminaries

1.1. From associative to Lie

Let $\mathbb{K}$ be an algebraically closed ground field of characteristic $p\neq 2$; unless otherwise stated, we consider $\mathbb{K}=\mathbb{C}$.

Let $A$ be any associative algebra, let $A^{L}$ be the Lie algebra whose space is $A$ but multiplication is given by the commutator $[a,b]:=ab-ba$ for any $a,b\in A$.

Let $A$ be a $\mathbb{Z}/2$-graded algebra; let $p$ denote the parity function. If $A$ a $\mathbb{Z}/2$-graded associative algebra, let $A^{S}$ be the Lie superalgebra whose space is $A$ but the multiplication is given by the supercommutator, which by the modern habitual abuse of notation is also denoted $[a,b]$ although defined to be $[a,b]:=ab-(-1)^{p(a)p(b)}ba$. Let $\mathfrak{g}^{\prime}:=[\mathfrak{g},\mathfrak{g}]$ be the first derived Lie algebra (resp. Lie superalgebra), a.k.a. commutant (resp. supercommutant), of the Lie algebra (resp. Lie superalgebra) $\mathfrak{g}$.

Not every $\mathbb{Z}/2$-graded algebra $\mathcal{A}$ is called superalgebra: only if multiplication in $\mathcal{A}$ or (if $\mathcal{A}$ is associative) in $\mathcal{A}^{S}$ depends on the parity.

Recall, that a trace (called supertrace in the super setting, for emphasis) on a given Liealgebra (resp. Lie superalgebra) $\mathfrak{g}$ is a linear function that vanishes on its commutant (resp. supercommutant), so there are $\dim(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])$ linearly independent traces (resp. supertraces) on  $\mathfrak{g}$, some of the supertraces can even and some of them odd.

1.2. Queerifications in characteristic $p\neq 2$ (from [BLLS])

Let $A$ be an associative algebra. The space of the associative algebra $\text{Q}(A)$ — the associative queerification of $A$ — is $A\oplus\Pi(A)$, where $\Pi$ is the change of parity functor, with the same multiplication in $A$ and the adjoint action of $A$ on $\Pi(A)$; let

[TABLE]

Set $\text{Q}(n):=\text{Q}(\text{Mat}(n))$.

We will be mostly interested in the following Lie queerifications $\mathfrak{q}(A)$:

1. for $A$ an associative algebra;

2. for $A$ a $\mathbb{Z}/2$-graded associative algebra (“Liefication” yields $\mathbb{Z}/2$-graded Lie algebra $A^{L}$);

3. for $A$ an associative superalgebra (this case differs from case 2) because passing to $A^{S}$ the supercommutators instead of commutators are considered; “super Liefication” yields Lie superalgebra $A^{S}$ which is $\mathbb{Z}/2$-graded by parity.

In case 1), as spaces, $\mathfrak{q}(A):={A^{L}\oplus\Pi(A)}$, so $\mathfrak{q}(A)_{\bar{0}}=A^{L}$ and $\mathfrak{q}(A)_{\bar{1}}=\Pi(A)$, with the bracket given by the following expressions and super anti-symmetry, i.e., anti-symmetry amended by the Sign Rule:

[TABLE]

The term “queer”, now conventional, is taken after the Lie superalgebra $\mathfrak{q}(n):=\mathfrak{q}(\text{Mat}(n))$. (The associative superalgebra $\text{Q}(n)$ is an analog of $\text{Mat}(n)$; likewise, the Lie superalgebra $\mathfrak{q}(n)$ is an analog of $\mathfrak{gl}(n)$ for several reasons, for example, due the role of these analogs in Schur’s Lemma and in the classification of central simple superalgebras, see [Lsos, Ch.7].) We express the elements of the Lie superalgebra $\mathfrak{q}(n)$ by means of a pair of matrices

[TABLE]

For any associative $A$, we will similarly denote the elements of $\mathfrak{q}(A)$ by pairs $(X,Y)$, where $X,Y\in A$. The brackets between these elements are as follows:

[TABLE]

We define Lie queerifications in cases 2) and 3) in the next section.

2. Traces and supertraces: generalities

Theorem**.**

Let $A$ be an associative algebra. Then, the estimate of the number of traces in the $3$ cases of its Lie queerification are described below.

2.1.1. Case $1$

Let $A$ be an associative algebra, then $Q(A):=A\oplus\Pi(A)$ is the associative queerification of $A$; let $\mathfrak{g}:=A^{L}$ be the corresponding Lie algebra and let the Lie superalgebra $\mathfrak{q}\mathfrak{g}:=(Q(A))^{S}=\mathfrak{q}(A)$ be the Lie queerification of $A$. Clearly, $\mathfrak{q}\mathfrak{g}=\mathfrak{g}\oplus\Pi(\mathfrak{g})$ as spaces.

Denote, for brevity, $\mathfrak{u}:=(\mathfrak{q}\mathfrak{g})^{\prime}$. By definition the supercommutant $\mathfrak{u}:=\mathfrak{u}_{\bar{0}}\oplus\mathfrak{u}_{\bar{1}}$ is spanned by elements $[a,b]$ for any $a,b\in\mathfrak{q}\mathfrak{g}$. In particular, $\mathfrak{u}_{\bar{1}}$ is spanned by elements of the form $[x,\Pi(y)]=\Pi([x,y])$ for any $x,y\in(\mathfrak{q}\mathfrak{g})_{\bar{0}}$, and hence $\mathfrak{u}_{\bar{1}}=\Pi([\mathfrak{g},\mathfrak{g}])$. Therefore, there are as many odd supertraces on $\mathfrak{q}\mathfrak{g}$ as there are traces on $\mathfrak{g}$.

Clearly, $\mathfrak{u}_{\bar{0}}$ is the sum of two ideals of $\mathfrak{g}$:

[TABLE]

Observe that the second summand does not have to be contained in the first one. Therefore, there are fewer even supertraces on $\mathfrak{q}\mathfrak{g}$ than there are traces on $\mathfrak{g}$.

In particular, if $A$ has unit ${\mathbbm{1}}$, then $\mathfrak{u}_{\bar{0}}=\mathfrak{g}$, because $[\Pi({\mathbbm{1}}),\Pi(x)]=2x$ for any $x\in\mathfrak{g}_{\bar{0}}$.

For example, if $A=\text{Mat}(n)$, then on $Q(A)$ there is an odd trace, nowadays called queertrace; it was first defined in [BL] by the formula

[TABLE]

2.1.2. Case $2$

Let $A$ be a $\mathbb{Z}/2$-graded associative algebra, $\mathfrak{g}:=A^{L}$. Clearly, $\mathbb{Z}/2$-grading of  $A$ makes $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$ a  $\mathbb{Z}/2$-graded Lie algebra.

Now, consider $A$ as an associative superalegbra; let $\mathfrak{s}=\mathfrak{s}_{\bar{0}}\oplus\mathfrak{s}_{\bar{1}}:=A^{S}$. Clearly, $\mathfrak{s}_{\bar{0}}=\mathfrak{g}_{\bar{0}}$ and $\mathfrak{s}_{\bar{1}}=\Pi(\mathfrak{g}_{\bar{1}})$, as spaces. The first equality is, moreover, true as an isomorphism of Lie algebras, whereas the second one as an isomorphism of $\mathfrak{s}_{\bar{0}}=\mathfrak{g}_{\bar{0}}$-modules.

The supercommutants are as follows:

[TABLE]

Therefore, $\mathfrak{s}^{\prime}_{\bar{1}}=\Pi((\mathfrak{g}^{\prime})_{\bar{1}})$, but a priori we can not say anything about relations between $(\mathfrak{g}^{\prime})_{\bar{0}}$ and $(\mathfrak{s}^{\prime})_{\bar{0}}$, except for both contain $(\mathfrak{g}_{\bar{0}})^{\prime}=(\mathfrak{s}_{\bar{0}})^{\prime}$.

2.1.3. Case $3$

Now, let $A=A_{\bar{0}}\oplus A_{\bar{1}}$ be an associative superalgebra, $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}:=A^{S}$.

Let $Q(A):=A\oplus\Pi(A)$ and $\mathfrak{q}\mathfrak{g}:=(Q(A))^{S}$; clearly, $\mathfrak{q}\mathfrak{g}=\mathfrak{g}\oplus\Pi(\mathfrak{g})$ as superspaces. It is also clear that the natural $(\mathbb{Z}/2\times\mathbb{Z}/2)$-grading on $Q(A)$ induces the same grading on $\mathfrak{q}\mathfrak{g}$, where

[TABLE]

Let us now compare the supercommutants of $\mathfrak{g}$ and $\mathfrak{q}\mathfrak{g}$. We will denote the elements of $\mathfrak{g}_{\bar{0}}$ by letters $x,y,\dots$, the elements of $\mathfrak{g}_{\bar{1}}$ by letters $a,b,c,\dots$. We have $\mathfrak{g}^{\prime}=(\mathfrak{g}^{\prime})_{\bar{0}}\oplus(\mathfrak{g}^{\prime})_{\bar{1}}$, where

[TABLE]

The elements that span homogeneous components of $\mathfrak{q}\mathfrak{g}$ are as follows:

[TABLE]

Therefore, $\mathfrak{u}_{\bar{0},\bar{0}}=(A_{\bar{0}})^{2}+(A_{\bar{1}})^{2}$ and $\mathfrak{u}_{\bar{1},\bar{0}}=\Pi(A_{\bar{0}}\cdot A_{\bar{1}})$, as spaces. In particular, if $A$ has unit, then

[TABLE]

In general, a priori we can only say that $\mathfrak{u}_{\bar{0},\bar{1}}=(\mathfrak{g}^{\prime})_{\bar{1}}$.

3. Examples of supertraces on queerified algebras and superalgebras

3.1. Clifford–Weyl algebras and superalgebras

Vasiliev was, most probably, the first to publish that on the Weyl algebra $W_{n}$ of polynomial differential operators in $n$ even indeterminates $x_{i}$ considered as superalgebra with parity given by $p(x_{i})=p(\partial_{i})=\bar{1}$ for all $i$, there is an even supertrace, see [Va1]. For a generalization to the Calogero-Moser case, see [Va2]; the detailed version [Va3] contains an elementary proof of uniqueness of the (super)trace; the construction yields algebras of “matrices of complex size” which first appeared as associative algebras in the book [Di] and as Lie algebras in [Fei]. For a  generalization to symplectic reflection algebras in [KT1, Th.7.1.1]. Alexey Lebedev suggested a  beautiful elementary proof of the existence of the trace on $W_{n}$, see §5.

Obviously unaware of Vasiliev’s works, his results were rediscovered by mathematicians, see [Mon, Proposition 4.3] and [Mus]. Montgomery found out the following sufficient condition to the super version of Herstein’s theorem and formulated it in the infinite-dimensional situation (in the finite-dimensional case, it is also true).

Theorem**.**

([Mon, Th.3.8])* Let $A$ be a $\mathbb{Z}/2$-graded simple associative algebra of characteristic $p\neq 2$ with supercenter $Z$ whose elements supercommute with any $a\in A$. Let the condition*

[TABLE]

hold. Then, $SL(A):=(A^{S})^{(1)}/((A^{S})^{(1)}\cap Z)$ is a simple Lie super algebra.

Observe that the superalgebra $A$ of differential operators in any finite number of indigenously odd indeterminates (a.k.a. the Clifford algebra on $2n$ generators considered as a $\mathbb{Z}/2$-graded associative superalgebra) is isomorphic to the matrix superalgebra $\text{Mat}(2^{n-1}|2^{n-1})$ on which there is an even supertrace, whereas on $\mathfrak{q}(2^{n}):=\mathfrak{q}(\mathfrak{gl}(2^{n-1}|2^{n-1}))$ there is the (well-known today) odd queer trace. Therefore, by arguments in Subsection 2.1.2, there are no even supertraces on $\mathfrak{q}A$, but there is one odd supertrace.

3.2. (Super)algebras of “matrices of complex size”

Absolutely the same arguments as in Example 3.1 are applicable to the queerifications of algebras and superalgebras $A$ of “matrices of complex size”, see [L, Subsection 2.2] with the same answer. (Although we do not need it here, note — just for its beauty — that since $\mathfrak{gl}(\lambda)=\mathfrak{gl}(\lambda)^{\prime}\oplus\mathbb{C}\,1$, one can define the trace on the Lie algebra $\mathfrak{gl}(\lambda)$ by any value on  $1$. J. Bernstein defined the trace on $\mathfrak{gl}(\lambda)$ for $\lambda\in\mathbb{C}\setminus\mathbb{Z}$, see [KhM], so that $\text{tr}(1)=\lambda$; Bernstein’s trace naturally generalizes the trace on $\text{Mat}(|n|)$ for any $n\in\mathbb{Z}\setminus\{0\}$ such that $\text{tr}(1_{|n|})=|n|$.)

3.3. Symplectic reflection algebras and superalgebras

For the classification of traces (resp. supertraces) on these algebras and superalgebras $A$, see [KS, Tables on pp.5,6]. Considering them as algebras (resp. superalgebras) we get the exact number of supertraces on their Lie qeerifications, according to general results established in Case 2.1.1 (resp. Case 2.1.2).

For a description of ideals in these algebras and superalgebras, see [KT1, KT2, KT3]. It is interesting to determine when these ideals are themselves simple algebras and superalgebras, and describe (super)traces on them and on their queerifications.

3.4. (Super)trace on the (super)algebra of pseudo-differential operators

3.4.1. $N=0$

Recall that the associative algebra $\Psi$ of pseudo-differential operators of integer order is $\mathcal{F}((D^{-1}))$, where $D:=\frac{d}{dx}$ and $\mathcal{F}$ is the algebra of functions in $x$, with multiplication given for any integer $n$ by the Leibniz rule

[TABLE]

Adler defined a  trace on the algebra $\Psi$ of pseudo-differential operators, see [A] and a very reader-friendly reviews [R, Ma], as the composition of the residue and the indefinite integral

[TABLE]

This trace (it vanishes on the commutators even before the integral is taken, just residue suffices, see [R]) takes values in $\mathcal{F}$. According to Subsection 2.1.2, there are no even supertraces on $\mathfrak{q}\Psi$, but there are $\geq 1$ odd supertraces; we conjecture there is just one odd supertrace.

3.4.2. $N=1$

On the superalgebra $\Psi_{1}:=\mathcal{F}((D^{-1}))$ of $N=1$-extended pseudo-differential operators, where $D:=\frac{\partial}{\partial\xi}+\xi\frac{\partial}{\partial x}$ and $\mathcal{F}$ is the algebra of functions in the even $x$ and odd  $\xi$, Manin and Radul defined an even supertrace, see [MR]. According to Subsection 2.1.2, there are no even supertraces on $\mathfrak{q}\Psi_{1}$, but there are $\geq 1$ odd supertraces; we conjecture there is just one odd supertrace.

4. An application of traces: integrals in involution

Let $A$ be an associative (super)algebra, tr a (super)trace on  $A$, and $b$ the corresponding invariant symmetric bilinear form (briefly: IS form)

[TABLE]

Let, moreover, $b$ be non-degenerate, briefly: NIS.

Let tr, and hence $b$, be even. Then, for the equation ($L$ and $P$ are in honor of Peter Lax)

[TABLE]

and for its more subtle version, more adequate in the super setting with $1|1$-dimensional Time with even coordinate $t$ and odd one $\tau$, see [Sha]:

[TABLE]

the functions $L\mapsto\text{tr}(L^{k})$ are integrals in involution with respect to the Poisson bracket defined on the space of functions on $\mathfrak{g}^{*}$ as follows, see, e.g., [A]. We identify $\mathfrak{g}$ with the linear functions on $\mathfrak{g}^{*}$ and for any functions $f,g\in\mathcal{F}(\mathfrak{g}^{*})$ the Poisson bracket is defined to be

[TABLE]

It seems, nobody considered yet the Euler equations or Lax pairs (5) related with superalgebras $A$ on which there is an odd trace qtr, and hence an odd $b$. If $b$ is odd, then $\mathfrak{g}\simeq\Pi\mathfrak{g}^{*}$, and an antibracket, rather than a Poisson bracket, is defined on the space $\Pi\mathcal{F}(\mathfrak{g}^{*})$ of functions on $\mathfrak{g}^{*}$. The functions $L\mapsto\text{qtr}(L^{k})$ are integrals in involution, i.e., commuting with the Hamiltonian and each other (themselves including in the case of the anti-bracket) with respect to the antibracket.

On $2n|k$-dimensional superspace on which the Poisson bracket is defined or $n|n$-dimensional superspace which the anti-bracket bracket is defined, let there be $n$ first integrals of the dynamical system  (5) in involution. Then, a theorem of Shander guarantees complete integrability of the system whatever $k<\infty$ is, see [Sh].

The traces on (super)algebras $A$ considered in Subsection 3.4 determine what researchers conceded to call “complete integrability” in the case of infinite-dimensional Hamiltonian system (since there are infinitely many of these traces — “this infinity is a half of the infinite dimension”). Examples: the KdV equations (resp. $(N=1)$-extended KdV, see Subsection 3.4.2) for the case where $L$ is the Schrödinger operator (resp. its $(N=1)$-extension).

Open problem: For any simple associative algebra $A$, give examples of integrable systems related with $Q(A)$.

5. Supertrace on $W_{n}$

5.1. Two general facts ([S])

1. If there is no (super)trace on an associative (super)algebra $A$, then there is no (super)trace on any product $A\otimes B$ for any associative (super)algebra $B$ with unit: if $[a_{1},a_{2}]=a$, then $[a_{1}\otimes 1,a_{2}\otimes b]=a\otimes b$ for any $a_{1},a_{2}\in A$ and $b\in B$, i.e., if any element of $A$ can be represented as a linear combination of (super)commutators, then the same is true for any element of $A\otimes B$.

2. Let $\text{tr}_{i}$ be a (super)trace on the associative (super)algebra $A_{i}$ for $i=1,2$. Then,

[TABLE]

is a (super)trace on $A_{1}\otimes A_{2}$.

Lemma**.**

Consider $W_{n}$ as a superalgebra with $p(x_{i})=p(\partial_{x_{i}})=\bar{1}$.

Then, on $W_{n}$, there is an even trace.

Observe that if we consider $W_{n}$ as an algebra, not a superalgebra, no analog of Lemma 5.2.   Lemma takes place since the associative algebra $W_{n}$ is simple; on the other hand, the center (constants) is given by a non-trivial cocycle.

Proof.

(A. Lebedev) Actually, $W_{n}=\mathbb{K}1\oplus[W_{n},W_{n}]$, where $[W_{n},W_{n}]$ is the supercommutant. The proof below demonstrates existence of the trace, but its uniqueness (up to a non-zero factor) should be proved separately. For the proof of uniqueness, see [Va3, Mon].

Let $n=1$. Let $d:=\frac{d}{dx}$, let the weight wht of $x$ be equal to $1$, let $\text{wht}(d):=-1$. On $W_{n}$, define the following linear function $T$:

[TABLE]

Let us prove that $T$ is a supertrace on $W_{n}$, i.e.,

[TABLE]

Clearly, it suffices to prove this for the case where $P$ and $Q$ are monomials whose weights are opposite.

Case 1: $P=x^{n+1}d^{n}$ and $Q=d$. Then,

[TABLE]

This implies the answer for the case where $Q=d$ and any $P$ of weight $1$, because any such $P$ can be represented as a linear combination of operators of the form $x^{n+1}d^{n}$.

Case 2: $\text{wht}(P)=-1$ and $Q=x$. Then,

[TABLE]

since $P(1)=0$ because $\text{wht}(P)<0$.

This implies the general case where $P$ and $Q$ are monomials of opposite weights, because we can transplant $x$ and $d$, one by one, from the end of $PQ$ to the beginning until we get $QP$, and each transplantation changes the sign by the opposite; by $(-1)^{deg(Q)}=(-1)^{p(P)p(Q)}$ altogether.

Since $T(1)=\frac{1}{2}$, it follows that 1 can not be represented as a linear combination of supercommutators.

For $n>1$, recall that $W_{n}\simeq W_{n-1}\otimes W_{1}$ and apply general fact 2), see Subsection 5.1. ∎

Comment: how to guess the form of $T$. Let $P$ be a differential operator of weight  0. In the basis $1$, $x$, $x^{2}$, …, consider $P$ as a linear operator on the space of polynomials and consider the matrix of $P$. Naively, ignoring possible divergence, the supertrace of this matrix is equal to

[TABLE]

Since $\text{wht}(P)=0$, then $Px^{n}$ is equal to the above-mentioned coefficient of $x^{n}$. In other words, the coefficient is equal to the value of $Px^{n}$ at $x=1$. Hence, the supertrace is equal to

[TABLE]

Acknowledgements

We are thankful to A. Lebedev for help. DL was supported by the grant AD 065 NYUAD.