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Jan 13, 2021 at 14:44 comment added amator2357 I see. Thank you, Jan.
Jan 13, 2021 at 12:33 comment added Jan Grabowski (2,n) Grassmannians only provide a model of a type A cluster algebra with a particular choice of coefficients. The coordinate ring of its big Schubert cell is another, with different coefficients and different relations. It's not clear to me that using models in this way will give an answer to the question as asked. Nevertheless, I think that this should be true (being finite type A is very strong) but I don't know of a reference or how to prove it directly.
Jan 13, 2021 at 9:55 comment added amator2357 Yes, when we don't invert the coefficients, this is true. I am still not sure if it holds if we do invert them (the coefficients).
Jan 12, 2021 at 23:52 comment added Sam Hopkins Since these are coordinate rings of Grassmannians of the form $\mathrm{Gr}(2,n)$, for which the defining equations are the 3-term Plucker relations, this should be true.
Jan 12, 2021 at 23:50 history asked amator2357 CC BY-SA 4.0