# Polynomial identities of supercommutative-gradable algebras

All algebras below are associative, and not assumed unital, and, to fix ideas, over the complex numbers.

An algebra $$A$$ is supercommutative-gradable if it admits a grading $$A=A_0\oplus A_1$$ in $$\mathbf{Z}/2\mathbf{Z}$$ ($$A_iA_j\subset A_{i+j}$$ for $$i,j\in\mathbf{Z}/2\mathbf{Z}$$) that makes it supercommutative: for $$a,b$$ homogeneous $$ab=ba$$ if either $$a,b$$ has even degree, and $$ab=-ba$$ for $$a,b$$ of odd degree.

I insist that by supercommutative-gradable, I assume that such a grading exists, but do not endow $$A$$ with it: I still view $$A$$ as a bare algebra, with no fixed grading.

What are the polynomial identities satisfied by supercommutative-gradable algebras? More precisely, in universal algebra terms: what is the variety generated by supercommutative-gradable algebras? [In particular, is it finitely generated? (Edit: Yes!)]

(For readers not familiar with universal algebra or polynomial identities, see addendum below to make the question precise.)

For instance, the class of supercommutative-gradable algebras satisfies the identities $$(xy-yx)z-z(xy-yx)$$ and $$x^2y^2-2xyxy+2yxyx-y^2x^2$$, and none of these two follows from the other one. (The identity $$(xy-yx)z-z(xy-yx)$$ holds because $$xy-yx$$ always has even degree, hence is central.)

Note: (about the above convention above for the meaning of supercommutative gradable: $$\mathbf{Z}$$-gradings vs $$\mathbf{Z}/2\mathbf{Z}$$-gradings)

Let $$\mathcal{A}$$ be the class of supercommutative-gradable algebras. Some subclasses of $$\mathcal{A}$$ could compete for being called "supercommutative-gradable algebras", namely the class $$\mathcal{A}_{\mathbf{Z}}$$ (resp. $$\mathcal{A}_{\mathbf{N}}$$, resp. $$\mathcal{A}_{\mathbf{N}_{>0}}$$), those algebras admitting an algebra grading in $$\mathbf{Z}$$ (resp...) satisfying the supercommutativity rule. Also we have smaller classes $$\mathcal{A}^1_{\mathbf{Z}}$$, $$\mathcal{A}^1_{\mathbf{N}}$$ in which we assume the algebra unital with unit of degree $$0$$. All the obvious inclusions between these classes are strict. However, the question is not sensitive to the choice of class: indeed, if $$A\in\mathcal{A}$$, then it is quotient of an algebra in $$\mathcal{A}_{\mathbf{N}_{>0}}$$, which itself (adding a unit) is subalgebra of an algebra in $$\mathcal{A}^1_{\mathbf{N}}$$. For the former quotient assertion: write $$A=A_1\oplus A_2$$ (writing $$A_2$$ rather than $$A_0$$) and consider the free $$\mathbf{Z}$$-graded supercommutative algebra $$\tilde{A}$$ over the vector space $$A_1\oplus A_2$$ with $$A_1,A_2$$ of degree $$1,2$$: then $$A$$ is canonically quotient of $$\tilde{A}$$.

Addendum (basic definitions of identities in algebras, varieties)

Fix the associative (non-unital) free $$\mathbf{C}$$-algebra $$\mathbb{F}=\mathbf{C}\langle X_n:n\in\mathbf{N}\rangle$$. An element $$P\in \mathbb{F}$$ is a polynomial identity of a class $$\mathcal{C}$$ of algebras if $$P$$ vanishes in every $$A\in\mathcal{C}$$, that is, if $$P$$ belongs to the kernel of every homomorphism $$\mathbb{F}\to A$$ for every $$A\in\mathcal{C}$$.

The set of polynomial identities of $$\mathcal{C}$$ forms a 2-sided ideal $$I_\mathcal{C}$$ of $$F$$ satisfying strong conditions: it is fully invariant (=stable under all endomorphisms); it is strongly graded, in the sense that it is a graded ideal for the unique algebra grading of $$\mathbb{F}$$ in the free abelian group $$\mathbf{Z}^{(\mathbf{N})}$$ (with basis $$(e_n)$$) for which $$X_n$$ has degree $$e_n$$ for every $$n$$ (for instance $$x_1x_2x_1^4x_2-x_2^2x_1^5$$ has degree $$5e_1+2e_2$$, while $$x_1^2+x_2^2$$ is not strongly homogeneous). Describing polynomial identities of $$\mathcal{C}$$, in practice, means exhibiting generators of $$I_\mathcal{C}$$ as a fully invariant 2-sided ideal.

For instance, for $$\mathcal{C}$$ the class of commutative algebras: the polynomial identities of $$\mathcal{C}$$ are generated by $$X_0X_1-X_1X_0$$.

The variety generated by $$\mathcal{C}$$ is the class of all algebras in which all $$P\in I_{\mathcal{C}}$$ are polynomial identities. It is also the smallest class of algebras containing $$\mathcal{C}\cup\{\{0\}\}$$ and stable under taking quotients, subalgebras, and arbitrary (unrestricted) direct products. The mapping $$\mathcal{V}\mapsto I_\mathcal{V}$$ is a canonical bijection between the "set" of varieties (of associative algebras) and fully invariant 2-sided ideals of $$\mathbb{F}$$. [It's not properly a set: to make it a set, cheat by fixing a set $$X$$ of cardinal $$2^{\aleph_0}$$ and consider $$\mathbf{C}$$-algebra structures with underlying set $$X$$.]

A variety of associative algebras $$\mathcal{V}$$ is finitely based if the ideal $$I_\mathcal{V}$$ is finitely generated as fully invariant ideal (it's not always the case). To my surprise it's always the case (I expected the contrary, by analogy with groups or Lie algebras in finite characteristic).

I believe that the identity $$(xy-yx)z-z(xy-yz)$$ generates everything (in characteristic 0, at least). To show that no further identities are needed, it is enough to exhibit one algebra that has no further identities. It follows from an old theorem of Krakowski and Regev that the Grassmann algebra of a countably dimensional vector space works for that purpose.
• Actually I was performing computations leading to the opposite conclusion, and precisely that the identity $wxyz+wyxz+zyxw-xwyz-ywxz-zwyz$ does not follow from the identity $[[x,y],z]$. Indeed (with the help of Sagemath) I computed that in the consequences of $[[x,y],z]$ in degree $(1,1,1,1)$ (i.e., in a space of dimension 24) has dimension 14, while the actual set of relations in this degree has dimension 16 (and the above relation is an explicit element testifying this). I'll double check. – YCor May 31 '20 at 12:53
• I didn't take into account the new relations such as $[[wx,y],z]$ (only those such as $[[x,y],z]w$ or $w[[x,y],z]$). Taking all of them into account, I get all 16 dimensions, and in particular the above relation follows from $[[a,b],c]$. The paper you link seems to yield the full conclusion. Thanks a lot! – YCor May 31 '20 at 14:16
• I see, thanks! By the way, there is a fun geometric interpretation of free algebras in this variety: the free algebra generated by V is isomorphic to the space of even polynomial differential forms on $V^*$ with the "quantised product" $\alpha\cdot \beta=\alpha\wedge \beta+ d\alpha \wedge d\beta$ (see Proposition on p. 782 in the paper intlpress.com/site/pub/pages/journals/items/mrl/content/vols/…). – Vladimir Dotsenko Jun 2 '20 at 11:22