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Consider two quadratic algebras based on the vector space $\mathbb{R}^3$ with basis $x,y,z$. The antisymmetric tensor algebra $\Lambda \mathbb{R}^3$ obviously has zero divisors, e.g. $(x)(xy)=0$, but the symmetric tensor algebra $S \mathbb{R}^3$ has not, as it is isomorphic to the polynomial functions on $x,y,z$. For $\Lambda \mathbb{R}^3$ the existence of zero divisors is obvious at degree 1, as $x^2=0$.

We could ask this more generally for a possibly noncommutative quadratic algebra with relations $R\subset V\otimes V$ - when is the quadratic algebra $Q(V,R)$ an integral domain? To give a more specific question, if $Q(V,R)$ an not integral domain, then are there always examples of zero divisors in degree 1, i.e. $x,y\in V$ so that $xy\in R$?

Apologies if this is well known - could you give me a reference for a proof or counterexample if it is! The question arose from trying to generalise the behaviour of the symbols of differential operators, where classically $SV$ is an integral domain, to operators with other relations.

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Partial answer: when $V$ is 4-dimensional there is a commutative counter-example. In other words, there is a subset of relations $R \subseteq V \otimes V$ containing the commutators and such that the associated quadratic algebra has zero divisors but not in degree 1.

Let $V$ be spanned by $W,X,Y,Z$, and let $R \subseteq V\otimes V$ be the subspace spanned by the commutators and by the two polynomials $Y^2-XZ$ and $XW-YZ$. The ideal generated by $R$ corresponds geometrically to the intersection of the two quadrics, in this case the union of a line and a twisted cubic. So the quadratic algebra has zero divisors, namely $$X(YW-Z^2) = (Y^2 - XZ)Z + (XW-YZ)Y = 0.$$

Now suppose that $UV = 0$ for $U, V$ both of degree 1. This would imply that there is a hyperplane containing the twisted cubic, which is not the case.

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  • $\begingroup$ Looks good, though your last comment takes a bit of thought. Thanks! $\endgroup$ – Edwin Beggs Jun 3 '16 at 13:20

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