Let $\Sigma=(X,ex,B)$ be a seed, $\mathcal{A}(\Sigma)$ a corresponding geometric cluster algebra and $\mathcal{X}_{\Sigma}$ the set of all cluster variables of $\mathcal{A}(\Sigma)$. We call a sequence $m=(x_1,\dots,x_k)$, $k \geq 1$, consisting of distinct exchangeable variables from $ex$ a simple sequence and let $\mu_m:=\mu_{x_k} \circ \cdots \circ \mu_{x_1}$. Is the equality \begin{equation} \mathcal{X}_{\Sigma}=\{y \in \mathcal{X}_{\Sigma}|\mu_m(x)=y \;\text{for some simple sequence}\; m \;\text{and some} \; x \in ex\} \end{equation} true in general? The $\supseteq$ part is trivial, but the other way around confuses me. So for instance, for a type $A$ cluster algebra this should be true, but I am not sure about other types. Any help in a form of a hint/explanation/reference/counter example would be much appreciated.
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$\begingroup$ Are you asking if we get every cluster variable by mutating at most once at each vertex? That doesn't seem possibly true: there are usually infinitely many cluster variables. Maybe I'm confused about what you are asking though... $\endgroup$– Sam HopkinsCommented Aug 16, 2021 at 21:10
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$\begingroup$ Hi @SamHopkins, thanks for the comment. I am asking if we can get every cluster variable by mutating at variables from the initial seed only. So say if $ex=\{x_1,\dots,x_k\}$ then, for instance, we can do $\mu_{x_k}\circ \cdots \circ \mu_{x_1}$ or just $\mu_{x_1} \circ \mu_{x_2}$, etc . Yeah, that's too much to ask, as you pointed out. Maybe it is true for finite type guys. $\endgroup$– amator2357Commented Aug 16, 2021 at 21:22
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1$\begingroup$ As I said before, there are clearly only finitely many sequences of the kind you have in mind, which means for infinite type cluster algebras there's no chance of reaching every cluster variable. But even for finite type (like Type A which you mentioned), I don't think you'll be able to get to every cluster variable in that few mutations. Indeed the diameter of the associahedron apparently grows like $2n$ (see arxiv.org/abs/1207.6296), which is more than the number of flips you're allowing. $\endgroup$– Sam HopkinsCommented Aug 17, 2021 at 4:00
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1$\begingroup$ Thanks for explaining @amator2357; because you'd not named the variables in your fixed ex, I misunderstood. I agree with @SamHopkins: this will almost never happen. In particular if you can find a cluster with no variables from ex such that no mutations of any of these are in ex. If you're working on surfaces, this might not be too hard, since you just have to construct a triangulation that looks "transverse" to your initial one (i.e. has its edges as different as possible). For general CAs, I'd conjecture your equality only holds for a small number of cases in very small rank. $\endgroup$– Jan GrabowskiCommented Aug 17, 2021 at 6:58
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1$\begingroup$ Thank you both for your comments, you have been very helpful. $\endgroup$– amator2357Commented Aug 17, 2021 at 8:20
1 Answer
Not really an answer, but a comment about the related question:
Can one reach every cluster in that way ?
It is rather easy to check (using sagemath or Keller's applet) that for the cyclic quiver in type $D_n$, not every cluster can be reached in that way. For example, only 13 among the 14 clusters in type $D_3$ are reached.
A conjecture for the number of reached clusters for this initial quiver in type $D_n$ can be found in this OEIS sequence.
EDIT:
Even in type $A_4$, not all clusters are reached if the initial cluster is not the equioriented Dynkin quiver.
Here is a small sage program that generates the set of reached clusters:
def recursive_comb(carquois):
"""
EXAMPLES::
sage: Q = ClusterQuiver(DiGraph({0:[1]}))
sage: len(list(recursive_comb(Q)))
5
"""
n = carquois.n()
seed = ClusterSeed(carquois)
initial_cluster = seed.cluster()
def voisins(s):
c = s.cluster()
for i in range(n):
if c[i] in initial_cluster:
yield s.mutate(i, inplace=False)
return RecursivelyEnumeratedSet([seed], voisins, structure='graded')
EDIT:
Using the same program, one can see that some cluster variables are missing already for some choices of initial quiver in type $D_4$:
sage: Q1 = ClusterQuiver(DiGraph({'b':['c'],'c':['d','e']}))
sage: Q2 = ClusterQuiver(DiGraph({'c':['b','d','e']}))
sage: len(set(v for c in recursive_comb(Q1) for v in c.cluster()))
16
sage: len(set(v for c in recursive_comb(Q2) for v in c.cluster()))
15
EDIT:
Even in type $A$, some clusters can be out of reach:
sage: Q1 = ClusterQuiver(DiGraph({'b':['c','a'],'c':['d']}))
sage: len(list(recursive_comb(Q1)))
37
instead of 42, the full number of clusters.
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$\begingroup$ Very interesting, thank you for sharing! I'd imagine that you could reach every cluster in that way when working with type $A_n$ guys. Do you know if it's true? $\endgroup$ Commented Dec 29, 2021 at 23:10
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$\begingroup$ Great to know this, thank you for that insight! $\endgroup$ Commented Dec 31, 2021 at 9:57