Timeline for Grassmannian cluster algebra of infinite type has no trees in its mutation class
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2021 at 10:44 | history | edited | F. C. | CC BY-SA 4.0 |
fix a typo and add one tag
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| Mar 22, 2021 at 8:10 | answer | added | F. C. | timeline score: 3 | |
| Jan 17, 2021 at 18:34 | comment | added | John Machacek | 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. | |
| Jan 17, 2021 at 18:17 | comment | added | Jan Grabowski | 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? | |
| Jan 17, 2021 at 16:08 | comment | added | Sam Hopkins | 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. | |
| Jan 17, 2021 at 14:47 | history | asked | Andrei Smolensky | CC BY-SA 4.0 |