Consider Weyl algebra, i.e. the algebra of $x^i$ and $p_i=\frac{\partial}{\partial x^i}$, its elements are differential operators $F(x,p)$. Weyl algebra is $\mathbb{Z}_2$ graded, hence one ask if there exists a supertrace. It turns out that there is one $$str\, F(x,p)=F(0,0)$$ I am looking for original references where this fact was established, some general comments are also welcome. Say, Pinczon et al in a 2005 paper "Supertrace and superquadratic Lie structure on the Weyl algebra, with applications to formal inverse Weyl transform" mention this fact without any refs.
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$\begingroup$ I found the linked paper very confusing. At some point they say: "The Weyl algebra W is a Lie algebra with bracket denoted by [.,.]L , and since it is naturally Z2-graded, W is a Lie superalgebra with bracket denoted by [.,.]." That may work for the Weyl algebra because [W,W] is central, but is this true in general? The Jacobi identity that has to be satisfied is different in both cases. Writing $\mathfrak{g} = \mathfrak{g}_0 \oplus \mathfrak{g}_1$ for the 2-graded Lie algebra, the Jacobi identity has 4 graded components: $000$, $001$, $011$ and $111$. (continued below) $\endgroup$– José Figueroa-O'FarrillCommented Aug 3, 2014 at 17:30
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$\begingroup$ The first three are the same if we think of $\mathfrak{g}$ as a Lie superalgebra, but the last component is different. In the Lie algebra case it defines an element in $\textrm{Hom}_{\mathfrak{g}_0}(\Lambda^3\mathfrak{g}_1,\mathfrak{g}_1)$ whereas in the Lie superalgebra case it defines an element in $\textrm{Hom}_{\mathfrak{g}_0}(\textrm{Sym}^3\mathfrak{g}_1,\mathfrak{g}_1)$ $\endgroup$– José Figueroa-O'FarrillCommented Aug 3, 2014 at 17:32
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$\begingroup$ I think they had the following simple fact in mind. If a given associative algebra is $\mathbb{Z}_2$-graded then one can construct a Lie algebra and a Lie superalgebra, where the bracket is (graded) commutator. For example, one can ask about traces and supertraces at the same time $\endgroup$– JohnCommented Aug 3, 2014 at 19:05
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$\begingroup$ Thanks. I see. That's indeed true, just not exactly what is stated in the paper. $\endgroup$– José Figueroa-O'FarrillCommented Aug 3, 2014 at 20:29
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