3
$\begingroup$

Say you have a quadratic algebra $A$ , that is, an algebra defined by a finite list of generators over a ground ring $k$ (either a field or a direct product of fields, which will be assumed $\mathbb{Z}/2$ for simplicity in this question) such that the corresponding ideal of relations $R$ is generated by its quadratic elements.

If $V$ is the $k$-module formally spanned by the chosen generators, then $A$ is $T(V)/R$, where $T(V)$ is the tensor algebra. The quadratic dual of $A$, denoted $A^!$, is then defined to be the algebra $T( V^* ) / R^{\perp}$, where $R^{\perp}$ is the annihilator in $V^* \otimes V^*$ of the original relation ideal $R \subset V \otimes V$.

My question is: is there an alternative way to think about the quadratic dual, along the following graphical lines?

For each quadratic monomial in $A$, draw a vertex of a simplicial complex $C$. For a relation in $R$ that sets two monomials equal, connect the corresponding vertices with an edge. In general, for a relation that sets the sum of some monomials equal to zero, draw a simplex spanned by the corresponding vertices.

Simple topological features of $C$ correspond to different types of relations in the quadratic dual $A^!$. For example, an isolated vertex is a quadratic monomial in $A$ involved in no relations. The corresponding dual monomial in $A^!$ therefore annihilates $R$ and so is contained in the ideal $R^{\perp}$. Since the generator is in the relation ideal, it's zero in $A^!$ so we may omit it from the complex $C^!$ of $A^!$.

An isolated edge in $C$ means two monomials of $A$ are set equal and take part in no other relations. The corresponding difference of dual monomials in $V^*$ is thus contained in $R^{\perp}$, so $C^!$ gets an edge between the two corresponding vertices.

More interestingly, an isolated complete graph $K_n$ in $C$ dualizes to an $n-1$-simplex in $C^!$, and vice-versa. So I'm tempted to think this is part of a duality on complexes, or in other words that you can accurately define $C^!$ topologically from $C$, without mentioning the quadratic algebras $A$ and $A^!$.

My question, more precisely: does this construction exist somewhere? I'm not quite sure how you deal with more complicated topological features of $C$ than the ones I just mentioned. A reference would be great, if it exists. Or maybe the duality is some natural topological thing I've been missing? At any rate, I'm interested to hear what people have to say.

$\endgroup$
6
$\begingroup$

It seems to me like the concept you're searching for is duality of matroids. Your simplices exactly correspond to the dependent sets of the matroid given by your monomials, assuming I understand your definition correctly. Incidentally, it's topologically better to look at the sets of variables that don't fit into a relation together (the independent sets), since these form an abstract simplicial complex. Knowing just these sets, you can figure out which sets are not in relations for $R^\perp$ very easily: those whose complements contain an independent set of maximal size for $R$. Anyways, lots is known about duality for matroids, though I'm not sure what it's likely to tell you about quadratic duality.

$\endgroup$
  • $\begingroup$ That's a really interesting answer, thanks! $\endgroup$ – Andy Manion Jun 16 '13 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.