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Sep 14, 2023 at 0:01 vote accept Mare
Sep 12, 2023 at 19:18 history became hot network question
Sep 12, 2023 at 17:37 comment added David E Speyer Slide 19 at pi.math.cornell.edu/~bazse/minusculemultiples_slides.pdf states that $\mathbb{C}H(w^p)$ is $P_p$ when $w^p$ is minuscule. The heaps corresponding to (co)-minuscule weights have been extensively studied, there are analogues of partitions, SYT, promotion, etc for them. See Figures 3-7 in arxiv.org/abs/math/0608276 for pictures of these heaps in types ADE.
Sep 12, 2023 at 17:22 comment added David E Speyer Yes, in type $A_n$ I know what the indecomposable projectives look like, and they are these $\mathbb{C} H(w^i)$ modules. The heap poset in this case is the product of an $i-1$ chain and an $n-i$ chain, so the distributive lattice in question is the lattice of partitions contained in an $i \times (n+1-i)$ box.
Sep 12, 2023 at 17:10 comment added Mare @DavidESpeyer In type $A_n$ for small $n$, the computer gives me a bijection on the points of this lattice for a given indecomposable projective $P$. I wonder what this bijection is (it is not rowmotion on the points of the lattice as the computer bijection has fixed points).
Sep 12, 2023 at 17:07 comment added Mare @DavidESpeyer thanks, I think you are right that it might not be finite in Dynkin type $D_n$ but it should be finite for type $A_n$. I edited my question.
Sep 12, 2023 at 17:07 history edited Mare CC BY-SA 4.0
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Sep 12, 2023 at 16:38 comment added David E Speyer So I suspect the answer to (1) is "yes, the submodule lattice of $P_i$ is the lattice of order ideals in the heap $H(w^i)$, in the case that $w^i$ is minuscule, and it isn't distributive when $w^i$ is not minuscule". But, if you are sure that this lattice is always distributive, then I messed up in my computation above.
Sep 12, 2023 at 16:37 comment added David E Speyer See Stembridge 2001 doi.org/10.1006/jabr.2000.8488 for the notions of "minuscule" and "heap", see Section 3 of arxiv.org/abs/2202.02490 for a recipe associating a preprojective module $\mathbb{C}H(w)$ to any minuscule element $w$ and see Prop 3.13 for the statement that the submodule lattice of $\mathbb{C}H(w)$ is isomorphic to the lattice of order ideals in $H(w)$. I haven't yet found a reference that says that $\mathbb{C} H(w^i) \cong P_i$, but I think I can show this.
Sep 12, 2023 at 16:34 comment added David E Speyer Here is what I suspect is going on. Let $W$ be the Weyl group, let $W_i$ be the parabolic subgroup generated by $s_j$ for $j \neq i$. Let $w_0$ and $(w_0)_i$ be the maximal elements of $W$ and $W_i$ and put $w^i = w_0 (w_0)_i$. I suspect that $P_i$ has distributive submodule lattice iff $w^i$ is minuscule. In this sense, I suspect that the submodule lattice of $P_i$ is isomorphic to the lattice of order ideals in the heap of $w^i$. (continued)
Sep 12, 2023 at 16:24 comment added David E Speyer Wait, are you sure that the submodule lattice of any indecomposable projective is distributive? Look at $D_4$, with the central vertex numbered as $0$. I get that the indecomposable projective generated at the central vertex has a composition series with subquotients $(S_0, S_1, S_2, S_3, S_0, S_0, S_1, S_2, S_3, S_0)$. In particular, it has $S_0^2$ as a subquotient.
Sep 12, 2023 at 14:22 answer added David E Speyer timeline score: 10
Sep 12, 2023 at 11:53 comment added Dave Benson Just a comment: as long as you're working over an infinite field, the submodule lattice of a module over a finite dimensional algebra is finite if and only if it is distributive, and this happens if and only if there is no subquotient isomorphic to a direct sum of two isomorphic simple modules.
Sep 12, 2023 at 11:16 history asked Mare CC BY-SA 4.0