Which cluster algebras are coordinate rings of double Bruhat cells?

Question

Background

A uselessly vague paragraph follows. A cluster algebra is a commutative algebra $A$ with a distinguished set of generators called cluster variables. These cluster variables are grouped into clusters: subsets of cardinality $n$ (for some $n$). Each cluster is equipped with some combinatorial data which allows the complete set of cluster variables (and hence $A$) to be recovered from any given cluster. There is a related extension algebra $U$, called the upper cluster algebra, which coincides with the cluster algebra $A$ in small examples.

A fundamental example of this structure is a double Bruhat cell. Given a complex, reductive group $G$ with Borel $B$ and Weyl group $W$, there is a Bruhat decomposition $$ G= \coprod_{w\in W} BwB$$ Choosing the opposite Borel $B^-$ instead gives the opposite Bruhat decomposition $$ G=\coprod_{v\in W} B^-vB^-$$ and so $$G = \coprod_{(w,v)\in W^2} BwB\cap B^-vB^-$$ The pieces $G_{w,v}$ of this disjoint union are called double Bruhat cells. Then Berenstein, Fomin and Zelevinsky have shown the following.

**Theorem ([BFZ05][1])** The coordinate ring of a double Bruhat cell is naturally an upper cluster algebra.

The Question

Roughly, my question is this. If I have an upper cluster algebra, how can I tell if it is the coordinate ring of a double Bruhat cell?

Note that there may be many different choices of $G,v,w$ which give isomorphic double Bruhat cells $G_{w,v}$, so it seems unlikely there is a constructive way of producing an example.

A more technical aspect of the question is the presence of frozen variables. These are cluster variables which cannot be mutated, and so they are in every cluster. Frozen variables may always be eliminated by setting them equal to 1, without affecting the combinatorics of the clusters.

The coordinate rings of double Bruhat cells will have many frozen variables, and I want to be able to eliminate or add frozen variables without changing the 'class' of cluster algebra. Though this is not a standard terminology, let us call two cluster algebras equivalent if they are related by the equivalence relation generated by 'setting some frozen variables' equal to 1.

Then I ask, If I have an upper cluster algebra, how can I tell if it is equivalent to the coordinate ring of a double Bruhat cell?

This second question is harder, but it does a better job of getting to the combinatorial type of the cluster algebra.

Answer 1

I think this is a hard problem. As evidence, it came up briefly in a conversation I was having with Lauren Williams and Sergey Fomin the other day, and neither of them knew a quick answer. I'll repeat here the one observation which I made at the time:

The cluster algebra of any double Bruhat cell of type $A$ which looks like $G^{e,w}$ is rigid, in the sense of Quivers with Potential I.

I suspect that it might be true that all double Bruhat cells are rigid, but I am too lazy to check.

Here is how to read this fact off from the Quivers with Potential paper. First, look at Section 6, where they define rigidity and prove that it is a mutation invariant. Next, look at Example 8.7. The authors check that the quiver in that example is acyclic -- you can check that this is the quiver corresponding to the reduced word $(s_1 s_2 \ldots s_{n-1})(s_1 s_2 \ldots s_{n-2}) \cdots (s_1 s_2 ) s_1$ for the longest element in $S_n$. So this shows that $G^{e, w_0}$ is rigid. Prop. 8.9 shows that, if a cluster algebra is rigid, then its restriction to an induced subquiver is rigid. I think (check this!) that, for any $w \in S_n$, if you choose a reduced word for $w_0$ which has $w$ as an initial segment, then you can get the quiver for $G^{e,w}$ by taking the quiver for $G^{e, w_0}$ corresponding to that reduced word and restricting to the subquiver coming from the word for $w_0$.

I'm looking forward to seeing other answers to this. Greg, I hope you'll report back on anything you work out as well!