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 ${\textrm {CM}}(B_{k,n})$ 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 ${\textrm {CM}}(B_{k,n})$ is important in a categorification of the cluster algebra structure on the homogeneous coordinate ring $\mathbb C[Gr(k, n)]$ of the Grassmannian variety of k-dimensional subspaces in $\mathbb C^n$ (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 $L_I\mid L_J$ .

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 ( $|I\setminus J|=|J\setminus I|=3$ and noncommon elements of I and J interlace). Write $I\setminus J$ as $\{i_1,i_2,i_3\}$ and $J\setminus I=\{j_1,j_2,j_3\}$ so that $1\le i_1<j_1<i_2<j_2<i_3<j_3\le n$ .

The following construction covers all indecomposable rank 2 modules in case when the category ${\textrm {CM}}(B_{k,n})$ is tame and $(k,n)=(3,9)$ .

We want to define a rank 2 module $\operatorname {\mathbb {L}}\nolimits (I,J)$ in ${\textrm {CM}}(B_{k,n})$ in a similar way as rank 1 modules are defined in [5]. Let $V_i:=\operatorname {\mathbb {C}}\nolimits [|t|]\oplus \operatorname {\mathbb {C}}\nolimits [|t|]$ , $i=1,\dots , n$ . The module $\operatorname {\mathbb {L}}\nolimits (I,J)$ has $V_i$ at each vertex $1,2,\dots , n$ of $\Gamma _n$ , where $\Gamma _n$ is the quiver of the boundary algebra, that is, with vertices $1,2,\dots , n$ on a cycle and arrows $x_i: i-1\to i$ , $y_i:i\to i-1$ . Observe the following matrices:

$$ \begin{align*} \begin{array}{llll} A_1:=\begin{pmatrix} t & -2 \\ 0 & 1 \end{pmatrix}, & B_1:=\begin{pmatrix} t & 0 \\ 0 & 1 \end{pmatrix}, & C_1:= \begin{pmatrix} t & -1 \\ 0 & 1 \end{pmatrix}, & D_1:= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \\ A_2:=\begin{pmatrix} 1 & 2 \\ 0 & t \end{pmatrix}, & B_2:= \begin{pmatrix} 1 & 0 \\ 0 & t \end{pmatrix}, & C_2:= \begin{pmatrix} 1 & 1 \\ 0 & t \end{pmatrix}, & D_2:= \begin{pmatrix} t & 0 \\ 0 & t \end{pmatrix}. \end{array} \end{align*} $$

Note that these are all matrix factorisations of $ \begin {pmatrix} t & 0 \\ 0 & t \end {pmatrix}$ : $A_1A_2=B_1B_2=C_1C_2=D_1D_2=\begin {pmatrix} t & 0 \\ 0 & t \end {pmatrix}$ .

Definition 1.1. Let $I, J$ be tightly 3-interlacing k-subsets of $\{1,2,\dots ,n\}$ . At the vertices of $\Gamma _n$ , $\operatorname {\mathbb {L}}\nolimits (I,J)$ has the spaces $V_1,\dots , V_n$ . We define the maps $x_i,y_i$ as follows:

$$ \begin{align*} x_i:V_{i-1}\to V_i \mbox{ acts as } \left\{ \begin{array}{rl} A_1, & \mbox{if}\ i=i_1, \\ B_2, & \mbox{if}\ i=j_1, \\ B_1, & \mbox{if}\ i=i_2, \\ C_2, & \mbox{if}\ i=j_2, \\ C_1, & \mbox{if}\ i=i_3, \\ A_2, & \mbox{if}\ i=j_3, \\ D_1, & \mbox{if}\ i\in I\cap J,\\ D_2, & \mbox{if}\ i\in I^c\cap J^c. \end{array}\right. \hskip.5cm y_i:V_{i}\to V_{i-1} \mbox{ acts as } \left\{ \begin{array}{rl} A_2, & \mbox{if}\ i=i_1, \\ B_1, & \mbox{if}\ i=j_1, \\ B_2, & \mbox{if}\ i=i_2, \\ C_1, & \mbox{if}\ i=j_2, \\ C_2, & \mbox{if}\ i=i_3, \\ A_1, & \mbox{if}\ i=j_3, \\ D_2, & \mbox{if}\ i\in I\cap J,\\ D_1, & \mbox{if}\ i\in I^c\cap J^c. \end{array}\right. \end{align*} $$

One easily checks that $xy=yx$ and $x^k=y^{n-k}$ at all vertices and that $\operatorname {\mathbb {L}}\nolimits (I,J)$ is free over the center of the boundary algebra. Hence, the following proposition holds.

Proposition 1.2. The module $\operatorname {\mathbb {L}}\nolimits (I,J)$ as constructed in Definition 1.1 is in ${\textrm {CM}}(B_{k,n})$ .

For the remainder of the paper, if $w=tv$ , then $t^{-1}w$ stands for v.

Proposition 1.3. Let I and J be tightly 3-interlacing, $n\ge 6$ arbitrary, $I\setminus J=\{i_1,i_2,i_3\}$ and $J\setminus I=\{j_1,j_2,j_3\}$ where $1\le i_1<j_1<i_2<j_2<i_3<j_3\le n$ . If $\varphi =( \varphi _i)_{i=1}^n\in $ Hom $(\operatorname {\mathbb {L}}\nolimits (I,J),\operatorname {\mathbb {L}}\nolimits (I,J))$ , then

$$ \begin{align*} \varphi_{i_1}=\varphi_{i_2}=\varphi_{i_3}&=\begin{pmatrix}a & bt \\ c & d \end{pmatrix},\\ \varphi_{j_1}&=\begin{pmatrix} a & b \\ ct & d \end{pmatrix},\\ \varphi_{j_2}&=\begin{pmatrix} a+c & b+t^{-1}(d-a-c) \\ ct & d-c \end{pmatrix},\\ \varphi_{j_3}&=\begin{pmatrix} a+2c & b+2t^{-1}(d-a-2c) \\ ct & d-2c \end{pmatrix},\\ \varphi_{i}&=\varphi_{i-1}, \,\, \text{for } i\in (I^c \cap J^c)\cup (I\cap J), \end{align*} $$

with $a,b,c,d\in \operatorname {\mathbb {C}}\nolimits [|t|]$ . Furthermore, $t\mid c$ and $t\mid (a-d)$ .

Proof. First we prove the statement for $n=6$ , $I=\{1,3,5\}$ , $J=\{2,4,6\}$ , and $\varphi \in $ End $(\operatorname {\mathbb {L}}\nolimits (I,J))$ . Then $\varphi =\left ( \varphi _1, \varphi _2, \ldots , \varphi _6 \right ),$ where each $\varphi _i$ is an element of $M_2(\operatorname {\mathbb {C}}\nolimits [[t]])$ (matrices over the center).

We check the relations which arise when we go from a peak of the rim of $\operatorname {\mathbb {L}}\nolimits (I,J)$ to a valley of the rim:

$$ \begin{align*} \begin{array}{clccl} (i) & x_2\varphi_1 = \varphi_2x_2, & \quad & (ii) & x_3\varphi_2 = \varphi_3x_3, \\ (iii) & x_4\varphi_3 = \varphi_4x_4, & & (iv) & x_5\varphi_4 = \varphi_5x_5, \\ (v) & x_6\varphi_5 = \varphi_6x_6, & & (vi) & x_1\varphi_6 = \varphi_1 x_1. \end{array} \end{align*} $$

Let $\varphi _1=\begin {pmatrix} a & b \\ c & d \end {pmatrix}$ . The equalities $\varphi _1=\varphi _3=\varphi _5$ follow immediately from $t\varphi _3=\varphi _3B_1B_2=B_1B_2\varphi _1=t\varphi _1$ and $t\varphi _5=\varphi _5C_1C_2=C_1C_2\varphi _3=t\varphi _3$ .

If we consider matrices $x_i$ and $y_i$ as elements of the ring $M_2(\operatorname {\mathbb {C}}\nolimits ((t)))$ , where all of them are units, then from $ x_2\varphi _1 = \varphi _2x_2$ follows that

$$ \begin{align*}\varphi_2=x_2\varphi_1x_2^{-1}=\begin{pmatrix} 1 & 0 \\ 0 & t \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & t^{-1} \end{pmatrix} = \begin{pmatrix} a & t^{-1}b \\ ct & d \end{pmatrix}.\end{align*} $$

Thus, $t\mid b$ , so if we replace b by $bt$ , this yields

$$ \begin{align*}\varphi_1=\begin{pmatrix} a & bt \\ c & d\end{pmatrix}, \quad \varphi_2=\begin{pmatrix} a & b \\ ct & d\end{pmatrix}.\end{align*} $$

Similarly, from $x_4\varphi _3 = \varphi _4x_4$ , we have

$$ \begin{align*}\varphi_4=x_4\varphi_3x_4^{-1}= \begin{pmatrix} 1 & 1 \\ 0 & t \end{pmatrix} \begin{pmatrix} a & bt \\ c & d\end{pmatrix} \begin{pmatrix} 1 & -t^{-1} \\ 0 & t^{-1} \end{pmatrix} = \begin{pmatrix} a+c & b+t^{-1}(d-a-c) \\ ct & d-c \end{pmatrix} ,\end{align*} $$

and from $x_6\varphi _5 = \varphi _6x_6$ , we have

$$ \begin{align*}\varphi_6=x_6\varphi_5x_6^{-1}=\begin{pmatrix} 1 & 2 \\ 0 & t \end{pmatrix} \begin{pmatrix} a & bt \\ c & d\end{pmatrix} \begin{pmatrix} 1 & -2t^{-1} \\ 0 & t^{-1} \end{pmatrix} =\begin{pmatrix} a+2c & b+2t^{-1}(d-a-2c) \\ ct & d-2c \end{pmatrix}.\end{align*} $$

The statement about the divisibility follows since we have the two properties $t\mid (d-c-a)$ and $t\mid (d-a-2c)$ . Combined, they imply $t\mid c$ and $t\mid d-a$ 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 $i\in (I^c \cap J^c)\cup (I\cap J)$ , then $x_i$ is a scalar matrix (either identity or t times identity), so the equality $x_{i}\varphi _{i-1}=\varphi _{i}x_{i}$ yields $\varphi _{i-1}=\varphi _{i}$ .

Proposition 1.4. Let $I,J$ be tightly 3-interlacing, $n\ge 6$ arbitrary. Then the module $\operatorname {\mathbb {L}}\nolimits (I,J)$ is indecomposable.

Proof. We first consider $n=6$ . In this case, we can assume $I=\{1,3,5\}$ and $J=\{2,4,6\}$ . Take $\varphi =(\varphi _i)_i\in $ End $(\operatorname {\mathbb {L}}\nolimits (I,J))$ as in the previous proposition.

To show the indecomposability, we assume that $\varphi $ is an idempotent endomorphism of $\operatorname {\mathbb {L}}\nolimits (I,J)$ and show that $\varphi $ is trivial (the identity or the zero endomorphism).

Assume that $\varphi _2^2=\varphi _2$ , that is

$$ \begin{align*} \varphi_2^2=\begin{pmatrix} a^2+bct & (a+d)b \\ (a+d)ct & d^2+bct\end{pmatrix}= \begin{pmatrix}a & b \\ ct & d \end{pmatrix}. \end{align*} $$

The equations $a^2+bct=a \quad \mbox {and} \quad d^2+bct=d$ on the diagonal entries give $a-a^2=d-d^2$ , that is, $a-d=a^2-d^2=(a-d)(a+d)$ and hence $a=d$ or $a+d=1$ . The equations also show that $t\mid a(1-a)$ and that $t\mid d(1-d)$ .

Assume first $a=d$ . If $b\ne 0$ , we get $a=\frac {1}{2}$ , which contradicts to $t\mid a-a^2$ . Analogously for $c\ne 0$ . Thus $b=c=0$ and $a=d=0$ or $a=d=1$ , the two trivial cases (note that if $\varphi _2$ is trivial, then $x_{i}\varphi _{i-1}=\varphi _{i}x_{i}$ yields $\varphi _2=\varphi _i$ , for all i).

So assume that $a\ne d$ and $d=1-a$ . Combining $t\mid a(1-a)$ with the fact that t divides $a-d=2a-1$ implies that $t\mid 1$ , which is a contradiction.

For a general n, since $\varphi _i=\varphi _{i+1}$ for $i+1\in (I^c\cap J^c)\cup (I\cap J)$ , 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 $L_I\mid L_J$ . This statement is clear in case when the category ${\textrm {CM}}(B_{k,n})$ is of finite representation type and in case when ${\textrm {CM}}(B_{k,n})$ is tame, with $(k,n)\in \{(3,9),\, (4,8)\}$ . Consequently, we have the following theorem.

Theorem 1.5 [2, Th. 1.2].

Let $M\in {\textrm {CM}}(B_{k,n})$ be an indecomposable module with profile $I\mid J$ . Then, up to isomorphism, M is the unique indecomposable rank $2$ module with filtration $I\mid J$ if and only if its poset is $1^3\mid 2$ and I and J are almost tightly $3$ -interlacing.