# When should I expect a quiver with potential to be rigid?

This question is pretty technical, but there are some very smart people here.

Fix a quiver Q, WITH oriented cycles. Let k[[Q]] be the completed path algebra. (Like the path algebra, but we allow formal infinite sums.) My question is about what behavior I should expect when I equip Q with a generic choice of potential.

A potential, denoted S, is a formal sum of (nontrivial) closed cycles in k[[Q]]. I'm interested in the case where the coefficients of that sum are chosen generically. Let J(S) denote the Jacobian ideal of S. This is a two sided ideal which is, roughly speaking, generated by the derivatives of S. (See Derksen-Weyman-Zelevinsky for the precise definitions.) [DWZ] show that the space of deformations of (Q,S) is k[[Q]]/(J(S) + [,]) where [,] is the vector space (NOT usually an ideal) spanned by commutators. They define (Q,S) to be rigid if this vector space is spanned by the empty cycle.

Heuristic arguments seem to suggest that a generic potential is always rigid. But example 8.6 in [DWZ] shows that this is false. Can someone give me a better heuristic or, better yet, a theorem?

EDITED to fix errors involving cycles of length 0.

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Shoot. My answer was totally going to be "when the potential is generic." –  Ben Webster Oct 21 '09 at 17:27
If any potential is rigid, a generic potential is rigid. So a reasonable reformulation of the question is "which quivers have rigid potentials?" –  Hugh Thomas Nov 4 '09 at 17:50

As I understand it, the quotient C(Q) = k[[Q]]/[,] is the (completed) vector space of formal linear combinations of oriented circuits in Q. Define a detour in Q to be an edge e and an oriented path p with the same source and sink as e. For every detour (e,p), there is a detour operator D(e,p). Given a circuit s, D(e,p)(s) is a sum of terms for each occurrence of e in s; each term is obtained by replacing that occurrence of e by p. Then the image I(S) of J(S) appears to be the span of all D(e,p)(S).

If I have all of that straight, then I can't think of a significant condition to guarantee that there is a rigid potential S. However, I can think of a significant condition to guarantee that there isn't one. Suppose for simplicity that the quiver Q has no parallel edges. Consider further just the shortest non-trivial circuits in C(Q); they span C(Q)g, where g is the oriented girth of Q. For these circuits, the only detour operators D(e,p) that matter are the trivial ones with p = e. In other words, you have no choice but to replace an edge e with itself. But often C(Q)g is bigger than the space of edges, i.e., Q could have more shortest cycles than it has edges. In this case, just for dimension reasons, I(S) can't contain C(Q)g.

What bothers me is that if this is correct, then Derksen, Weyman, and Zelevinsky worked a little harder than necessary to make their Example 8.6.

Now that I read on a bit more of the paper, the authors give some constructions of quivers with rigid potentials. For instance, they establish that the existence of a rigid potential is a mutation-invariant property of quivers. They also provide an example of a rigid quiver which does not mutate to an acyclic quiver. You can go a little further: In studying the cyclic part of the path algebra of a quiver, you can restrict attention to its strongly connected components. So you can start with any collection of strongly connected, rigid quivers, then connect them any way you like acyclically, and then mutate that.

It sounds like the status quo of the question is that there are several constructions of quivers that do not have rigid potentials, and several constructions of quivers that have rigid potentials, but that finding a good characterization of the dichotomy is an open problem.

Also, clearly the quivers that do not have rigid potentials are recursively enumerable. (I.e., there is an algorithm to confirm that a quiver does not have a rigid potential, but it might not terminate if it does have one.) Maybe a good remaining question is whether quivers with rigid potentials are also recursively enumerable, i.e., that the dichotomy is recursive.

One way to confirm that a quiver has a rigid potential is to find a J(S) that contains all paths of some length. I(S) then automatically contains all loops of that length or greater. Moreover, it does not matter whether S includes any terms beyond the cutoff length. If a rigid potential always has this property, then determining whether there is one is recursive.

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