CORRIGENDUM TO “CLUSTER CATEGORIES FROM GRASSMANNIANS AND ROOT COMBINATORICS”

Abstract In this note, we correct an oversight regarding the modules from Definition 4.2 and proof of Lemma 5.12 in Baur et al. (Nayoga Math. J., 2020, 240, 322–354). In particular, we give a correct construction of an indecomposable rank 
$2$
 module 
$\operatorname {\mathbb {L}}\nolimits (I,J)$
 , with the rank 1 layers I and J tightly 
$3$
 -interlacing, and we give a correct proof of Lemma 5.12.


§1. Indecomposable rank 2 modules with tightly 3-interlacing layers
In [1], we studied the category CM(B k,n ) of Cohen-Macaulay modules over the completion of an algebra B k,n , which is a quotient of the preprojective algebra of type A n−1 . The category CM(B k,n ) is important in a categorification of the cluster algebra structure on the homogeneous coordinate ring C[Gr(k, n)] of the Grassmannian variety of k -dimensional subspaces in C n (see [3]- [5]).
For the notation and background results used in this note, we refer the reader to [1 ,Sect. 1]. We thank Karin Erdmann and Alastair King for useful conversations about indecomposable modules.
In [1 ,Def. 4.2], we constructed a Cohen-Macaulay module of an arbitrary rank. In the case of rank 2, in [1 ,Lem. 5.12], we claimed that the constructed module is indecomposble. In fact, the rank 2 module from this lemma is not indecomposable. The aim of this note is to correct this mistake, that is, for given k -subsets I and J that are tightly 3-interlacing, to construct explicitly an indecomposable rank 2 Cohen-Macaulay module with filtration We show that this module is indecomposable by proving that its endomorphism ring does not have nontrivial idempotents.
Assume that we are in the case when I and J are tightly 3-interlacing k -subsets (|I \ J| = |J \ I| = 3 and noncommon elements of I and J interlace). Write I \ J as {i 1 , i 2 , i 3 } and The following construction covers all indecomposable rank 2 modules in case when the category CM(B k,n ) is tame and (k, n) = (3,9).
We want to define a rank 2 module L(I, J) in CM(B k,n ) in a similar way as rank 1 modules are defined in [5]. Let V i := C[|t|] ⊕ C[|t|], i = 1, . . . , n. The module L(I, J) has V i at each vertex 1, 2, . . . , n of Γ n , where Γ n is the quiver of the boundary algebra, that is, with vertices 1, 2, . . . , n on a cycle and arrows x i : i − 1 → i, y i : i → i − 1. Observe the following matrices: Note that these are all matrix factorisations of t 0 0 t : At the vertices of Γ n , L(I, J) has the spaces V 1 , . . . , V n . We define the maps x i , y i as follows: One easily checks that xy = yx and x k = y n−k at all vertices and that L(I, J) is free over the center of the boundary algebra. Hence, the following proposition holds.
∈ Hom(L(I, J), L(I, J)), then We check the relations which arise when we go from a peak of the rim of L(I, J) to a valley of the rim: If we consider matrices x i and y i as elements of the ring M 2 (C((t))), where all of them are units, then from x 2 ϕ 1 = ϕ 2 x 2 follows that Thus, t | b, so if we replace b by bt, this yields Similarly, from x 4 ϕ 3 = ϕ 4 x 4 , we have and from x 6 ϕ 5 = ϕ 6 x 6 , we have The statement about the divisibility follows since we have the two properties t | (d − c − a) and t | (d − a − 2c). Combined, they imply t | c and t | d − a as claimed.
In the general case, the proof is almost the same as in the case n = 6. The only thing left to note is that if i ∈ (I c ∩ J c ) ∪ (I ∩ J), then x i is a scalar matrix (either identity or t times identity), so the equality Proof. We first consider n = 6. In this case, we can assume I = {1, 3, 5} and J = {2, 4, 6}. Take ϕ = (ϕ i ) i ∈ End (L(I, J)) as in the previous proposition.
To show the indecomposability, we assume that ϕ is an idempotent endomorphism of L(I, J) and show that ϕ is trivial (the identity or the zero endomorphism).
Assume that ϕ 2 2 = ϕ 2 , that is The Assume first a = d. If b = 0, we get a = 1 2 , which contradicts to t | a − a 2 . Analogously for c = 0. Thus b = c = 0 and a = d = 0 or a = d = 1, the two trivial cases (note that if ϕ 2 is trivial, then x i ϕ i−1 = ϕ i x i yields ϕ 2 = ϕ i , for all i ).
So assume that a = d and d = 1 − a. Combining t | a(1 − a) with the fact that t divides a − d = 2a − 1 implies that t | 1, which is a contradiction.
For a general n, since ϕ i = ϕ i+1 for i + 1 ∈ (I c ∩ J c ) ∪ (I ∩ J), the proof follows as for n = 6.
The question of uniqueness of such a rank 2 indecomposable module is studied in [2]. For given tightly 3-interlacing I and J, there is a unique indecomposable rank 2 module with filtration L I | L J . This statement is clear in case when the category CM(B k,n ) is of finite representation type and in case when CM(B k,n ) is tame, with (k, n) ∈ { (3,9), (4, 8)}. Consequently, we have the following theorem.