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The question is why the statement in the title is true (is it?).

To elaborate, recall that Grassmannian cluster algebra, according to Scott`s paper Grassmannians and Cluster Algebras, is the cluster structure on $\mathbb{C}[\operatorname{Gr}(k,n)]$, corresponding to the following quiver, called $\Gamma_{k,n}$:

The quiver corresponding to a special Postnikov diagram, see Section 5

In case $k=2$ this is a chain of length $n-3$, so the corresponding cluster algebra is of type $\mathsf{A}_{n-3}$. It is also not hard to show that for $k=3$ and $n=6,7,8$ a suitable sequence of mutations transforms $\Gamma_{k,n}$ into a tree, which happens to be a Dynkin diagram of type $\mathsf{D}_4$, $\mathsf{E}_6$ or $\mathsf{E}_8$, respectively. In all other cases this cluster algebras is of infinite type.

Now I ran some computer experiments and it seems that in the infinite case the mutation class of $\Gamma_{k,n}$ contains no trees at all.

Q1: Does this experimental observation hold in all cases? If so, why?

Q2: Can one at least find some nice representative in each mutation class? Say, having the smallest number of (unoriented) cycles.

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    $\begingroup$ Pretty sure that Grassmannian cluster algebras, outside of the few finite type examples you mentioned, are not acyclic, which would imply what you're asking, but I can't find a reference for this statement right away. By the way, thanks to the theory of plabic graphs, the grid quiver you drew is usually considered a nice representative in its mutation class. $\endgroup$ Commented Jan 17, 2021 at 16:08
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    $\begingroup$ To support @Sam Hopkins' comment, very little is known about the clusters of the Grassmannian in infinite types, and even less for those clusters where the cluster variables are not just Plücker coordinates (but can be of higher degree). I'm struggling to think of a (known, proved) property of the Grassmannian CA that is inconsistent with being acyclic, but it's also possible I'm missing something obvious. And actually I prefer the clusters with triangles not just squares :-). This comment is too small to say all the things I am aware of for Gr(k,n) but we could discuss if you like? $\endgroup$ Commented Jan 17, 2021 at 18:17
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    $\begingroup$ We can answer for $Gr(4,8)$ and $Gr(3,9)$ since they are finite mutation type. Then maybe the representative you want is $E^{(1,1)}_7$ and $E^{(1,1)}_8$. I am not sure what to do for infinite mutation type... Adding to what @JanGrabowski said cluster algebras of Grassmannians are locally acyclic which implies they share many properties with acyclic cluster algebras. $\endgroup$ Commented Jan 17, 2021 at 18:34

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Regarding question Q2, one can go a little bit further and describe simple diagrams with few edges for some more cases.

Let us talk about $Gr(p, p+q)$, so that there is a symmetry between $p$ and $q$. Assume moreover that $p\geq 2$ and $q\geq 2$.

For $p=2$ and $2\leq q$, one can mutate to the tree $A_{q-1}$.

For $p=3$ and $2\leq q\leq 7$, one can mutate to $A_2, D_4$, $E_6$, $E_8$, $E^{(1,1)}_8$ and $K_{12}$.

for $p=4$ and $2\leq q\leq 5$, one can mutate to $A_3, E_6$, $E^{(1,1)}_7$ and $W_{12}$.

The symbols $E^{(1,1)}_8$ and $E^{(1,1)}_7$ stand for quivers related to elliptic singularities. These quivers are obtained from a tree by replacing the unique triple point by a double edge surrounded by three oriented triangles.

The symbols $K_{12}$ and $W_{12}$ stands for quivers related to two of the 14 unimodal singularities of Arnold. Unoriented graphs describing these singularities can be found in many references about singularity theory. For a picture, see page 3 of arXiv:0708.0210. The quivers are almost trees, and only have one double edge and three oriented triangles.

EDIT (2022-08)

On may expect a similar relationship between the case $(p,q)=(3,11)$ and the exceptional bimodal singularity $E_{20}$ defined by $x^2+y^3+z^{11}$, for which a simple diagram is known by works of Ebeling. In this case, a sequence of mutations is yet to be found.

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