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I am getting myself acquainted to superalgebras. One often comes across odd polynomial rings of the form

$$k\langle x_i \rangle_{i\in I} / (x_ix_j -(-1)^{|x_i||x_j|} x_jx_i)$$

for some index set $I$, where the relation divided out sometimes is required for all choices of $i, j$ (which implies $x_i^2=0$ if $x_i$ is odd), and sometimes this is only required for $i\neq j$ (for instance in the definition 2.1.1 of the odd NilHecke Algebra in https://arxiv.org/abs/1111.1320).

Question: How do I know when $x_i^2$ should be assumed to vanish?

Edit: Maybe I should clarify why I care. For perfectly supersymmetric algebras, I have nice identities how to drag tensor factors and morphisms past elements. It seems that this gets highly problematic if some elements are not assumed to anticommute with themselves. So why should one consider this?

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Why should one consider both cases? Just as Lie algebras are "built out of copies of $\mathfrak{sl}(2)$", (basic classical) Lie superalgebras are essentially built out of copies of three elementary algebras: $\mathfrak{sl}(2)$, $\mathfrak{sl}(1|1)$, and $\mathfrak{osp}(1|2)$. (This essentially corresponds to the three flavors of roots: even, odd isotropic, and odd anisotropic.) If we look at the positive parts of the enveloping algebras of direct sums of these pieces over some index set $I$, we have

$$U^+(\mathfrak{sl}(2)^{\oplus I})\cong k[e_i\ |\ i\in I]\qquad U^+(\mathfrak{sl}(1|1)^{\oplus I})\cong k[e_i\ |\ i\in I]/(e_ie_j=-e_je_i)\qquad U^+(\mathfrak{osp}(1|2)^{\oplus I})\cong k[e_i\ |\ i\in I]/(e_ie_j=-e_je_i\text{ for }i\neq j)$$

where $k$ is the base field, and the $e_i$ in the skew-commuting versions are odd.

In other words, all three flavors (non-super; super with all odd generators anticommuting; and super with all distinct odd generators anticommuting) arise naturally from the elementary Lie superalgebras, so one is led to consider all three possibilities.

Note, however, that odd elements which do not anticommute with themselves should still anticommute when living in different tensor factors: that is, if $e$ is odd, then $$(1\otimes e)(e\otimes 1)=-(e\otimes 1)(1\otimes e)=-e\otimes e$$ regardless of whether $e^2$ is zero or not. One way to see that this ought to be the case is that we would still like e.g. $U^+(\mathfrak{osp}(1|2)^{\oplus I})\cong U^+(\mathfrak{osp}(1|2))^{\otimes I}$ to be true.

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  • $\begingroup$ Does your last statement hold true if, let aside Lie superalgebras for a moment, $e$ is an element of the algebra the tensor product is taken over? Certainly not, or am I mistaken? $\endgroup$
    – Bubaya
    Commented Jun 21, 2017 at 17:05
  • $\begingroup$ I am not sure what you mean: in the statement that $(1\otimes e)(e\otimes 1)=-e\otimes e$, $e$ is an odd element of some unital superalgebra $A$ (say, $A=k[e]$, with $\mathbb Z/2\mathbb Z$ grading induced by degree), and $1\otimes e,e\otimes 1\in A\otimes A$. $\endgroup$
    – Sean Clark
    Commented Jun 21, 2017 at 17:33
  • $\begingroup$ The point being: given superalgebras $A,B$, the algebra $A\otimes B$ is given a multiplication via the maps $$a\otimes b\otimes a'\otimes b'\mapsto (-1)^{p(a')p(b)} a\otimes a'\otimes b\otimes b'\mapsto aa'\otimes bb'$$ where the first map is the braiding on the middle factors, and the second map is the tensor product of multiplication maps. $\endgroup$
    – Sean Clark
    Commented Jun 21, 2017 at 17:37
  • $\begingroup$ What I mean is: Let $B\supset A$ be a superalgebra over a superalgebra $A$. I can consider $B$ as an $A$-$A$-superbimodule. Let $e\in A$. Consider the tensor product $B\otimes_A B$. Then $$(1\otimes e)(e\otimes 1)(1\otimes_A 1) = (e\otimes 1)(1\otimes_A e) = (e\otimes 1)(e\otimes_A 1) = (e^2\otimes 1)$$ where the last factor (with $\otimes_A$) means an element of a tensor product of bimodules. Hence in this case, the first two factors (elements of $A\otimes A$) should not anticommute, should they? How to make this rigorous? $\endgroup$
    – Bubaya
    Commented Jun 22, 2017 at 10:36
  • $\begingroup$ You seem to have switched the order of multiplication in your equation. In any case, that just shows that $(a\otimes a')(b\otimes b')=ab\otimes b'a'$ is not a well-defined $A\otimes A$ module structure on $B\otimes_A B$. Again, this makes sense from the point of view of the braiding: $a'$ has to braid past $b,b'$ so the correct action should carry a sign of $(-1)^{p(b)p(a')+p(b')p(a')}$. This can be done naturally if e.g. we define the right module structure in our bimodules to be $v\cdot a=(-1)^{p(v)p(a)}va$ $\endgroup$
    – Sean Clark
    Commented Jun 22, 2017 at 13:56

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