Cluster algebras from quivers vs. cluster algebras from skew-symmetrizable matrices

Question

In a course on cluster algebras we first defined the cluster algebra arising from a quiver. Next we saw that each quiver gives rise to an exchange matrix that is skew-symmetric. So we generalized to cluster algebras arising from skew-symmetrizable matrices. However, does this generalization actually give rise to more/new cluster algebras?

If this is the case, is there an easy example of a cluster algebra that arises from a skew-symmetrizable matrix that does not arise from a quiver?

If this is not the case, why are we interested in cluster algebras arising from skew-symmetrizable matrices?

Answer 1

First note that, for most of cluster algebras, it is not the actual resulting cluster algebra (as a space) that we care about, but instead the cool elements and relations between them that the combinatorics gives us (these are seeds and the exchange relations). You can give any reasonably big enough space a cluster algebra structure and there are usually a lot of choices. (I'm aware of examples where the cluster algebra has both wild and finite type structures simultaneously.)

With that out of the way, we can pick an example from Andrew Hubery's comment. The smallest example is to compare $B_2$ with $A_2$ (the only finite type, skew-symmetric, rank 2 cluster algebra. Hopefully it's clear that we don't need to consider infinite type cluster algebras). $A_2$ has just $5$ seeds, $B_2$ has $6.$ I list them using the sage shell.

Code to generate:

A = ClusterAlgebra(['A', 2])
print("seeds for type A_2")
for s in A.seeds():
    print(s.cluster_variables())

Which gives:

seeds for type A_2
[x0, x1]
[x0, (x0 + 1)/x1]
[(x1 + 1)/x0, x1]
[(x0 + x1 + 1)/(x0*x1), (x0 + 1)/x1]
[(x1 + 1)/x0, (x0 + x1 + 1)/(x0*x1)]

Note there are 5 non-isomorphic seeds.

For $B_2,$ on the other hand:

B = ClusterAlgebra(['B', 2])
print("seeds for type B_2")
for s in B.seeds():
    print(s.cluster_variables())

gives:

seeds for type B_2
[x0, x1]
[x0, (x0 + 1)/x1]
[(x1^2 + 1)/x0, x1]
[(x0^2 + x1^2 + 2*x0 + 1)/(x0*x1^2), (x0 + 1)/x1]
[(x1^2 + 1)/x0, (x1^2 + x0 + 1)/(x0*x1)]
[(x0^2 + x1^2 + 2*x0 + 1)/(x0*x1^2), (x1^2 + x0 + 1)/(x0*x1)]

Now there are 6 seeds.