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David E Speyer
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Now, let $\omega'_i$ be the unique fundamental rootweight so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.

Now, let $\omega'_i$ be the unique fundamental root so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.

Now, let $\omega'_i$ be the unique fundamental weight so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.

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David E Speyer
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IHere is the answer to question (1). The first thing to know is that the premise of the question is wrong: Indecomposable projectives do not always have some thoughts abouta distributive submodule lattice. Rather, $P_i$ has a distributive submodule lattice if and only if the other partsweight $\omega_i$ is minuscule. (Because we are in simply-laced type, but"minuscule" and "co-minscule" are synonyms.) In this case, the submodule lattice of $P_i$ is isomorphic to the lattice of order ideals in the heap $H(w^i)$.

The minuscule weights have been classified: See Table 1 in Thomas-Yong and ignore the rows for types $B$ and $C$. The corresponding heaps are also well known; see the Figures in Section 2.2 of the Thomas-Yong paper.


I needwill now explain how to work on other thingspiece these results together, using facts from Baumann and Kamnitzer and Dranowski, Elek, Kamnitzer and Morton-Ferguson.

We use the usual Coxeter notations: $\alpha_i$ are the simple roots, $s_i$ are the simple reflections and $\omega_i$ are the fundamental weights.

To every weight $\gamma$, Baumann and Kamnitzer define a preprojective module $N(\gamma)$, such that $N(\omega_i) = P_i$ (see Prop 10). Here are the key facts about $N(\gamma)$:

  • If $\gamma$ is antidominant, then $N(\gamma) = 0$.
  • If $s_j \gamma = \gamma + c \alpha_j$ for some $c \geq 0$, then we have a short exact sequence $0 \to N(\gamma) \to N(s_j \gamma) \to S_j^c \to 0$.
  • As a corollary of the proceeding two statements, let $\gamma'$ be the unique antidominant weight in the same orbit as $\gamma$ and let $\gamma - \gamma' = \sum d_i \alpha_i$. Then $N(\gamma)$ has dimension $(d_1, d_2, \ldots, d_n)$.
  • Let $\gamma$ and $\gamma'$ be as in the previous bullet point, and let $\gamma' = - \sum c_i \omega_i$. Then $N(\gamma)$ is uniquely characterized as the only preprojective module with dimension $(d_1, d_2, \ldots, d_n)$ and socle of dimension $(c_1, c_2, \ldots, c_n)$.

Now, let $\omega'_i$ be the unique fundamental root so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.

In particular, if any $c_k$ is $\geq 2$, then the submodule lattice of $P_i$ is not distributive.

If all the $c_k$ are $1$, then $w^i$ is what is called $\omega_i$-minuscule (see Definition 2.4 in Dranowski, Elek, Kamnitzer and Morton-Ferguson), and is therefore minuscule, and the results of that paper apply. I'll summarize those results now.

We define a poset $H(w^i)$, called the heap of $w^i$, as follows: The ground set of $H(w^i)$ is $[L]$. The order relation is the transitive closure of the following: $a \prec b$ if $a<b$ and $s_{j_a} s_{j_b} \neq s_{j_b} s_{j_a}$. We have the following:

  • $H(w^i)$ indexes a basis for $\mathbb{C} H(w^i)$. To understand the maps in $\mathbb{C} H(w^i)$, take the Hasse diagram of $H(w^i)$ (see the Thomas-Yong paper) and place $1$'s and $-1$'s on the edges in order to make the preprojective relations hold.

  • Order ideals of $H(w^i)$ correspond to submodules of $\mathbb{C} H(w^i)$. These also correspond to the weak order interval $[e, w^i]$, and to roots in the $W$ orbit of $\omega_i$.

  • The submodule lattice of $\mathbb{C}H(w^i)$ is the lattice of order ideals in $H(w^i)$.

  • Total orders of $H(w^i)$ correspond to reduced words for $w^i$, which correspond to composition series of $P_i$.

In type $A_n$, $H(w^i)$ is the product of an $i$-vertex chain and an $(n+1-i)$ vertex chain. Order ideals correspond to partitions fitting in an $i \times (n+1-i)$ box, and the submodule lattice is this part of Young's lattice. Total orders on $H(w^i)$ are standard Young tableaux of shape $i \times (n+1-i)$.

I have some thoughts about the other parts, but I need to work on other things for now.

 

Here is the answer to question (1). The first thing to know is that the premise of the question is wrong: Indecomposable projectives do not always have a distributive submodule lattice. Rather, $P_i$ has a distributive submodule lattice if and only if the weight $\omega_i$ is minuscule. (Because we are in simply-laced type, "minuscule" and "co-minscule" are synonyms.) In this case, the submodule lattice of $P_i$ is isomorphic to the lattice of order ideals in the heap $H(w^i)$.

The minuscule weights have been classified: See Table 1 in Thomas-Yong and ignore the rows for types $B$ and $C$. The corresponding heaps are also well known; see the Figures in Section 2.2 of the Thomas-Yong paper.


I will now explain how to piece these results together, using facts from Baumann and Kamnitzer and Dranowski, Elek, Kamnitzer and Morton-Ferguson.

We use the usual Coxeter notations: $\alpha_i$ are the simple roots, $s_i$ are the simple reflections and $\omega_i$ are the fundamental weights.

To every weight $\gamma$, Baumann and Kamnitzer define a preprojective module $N(\gamma)$, such that $N(\omega_i) = P_i$ (see Prop 10). Here are the key facts about $N(\gamma)$:

  • If $\gamma$ is antidominant, then $N(\gamma) = 0$.
  • If $s_j \gamma = \gamma + c \alpha_j$ for some $c \geq 0$, then we have a short exact sequence $0 \to N(\gamma) \to N(s_j \gamma) \to S_j^c \to 0$.
  • As a corollary of the proceeding two statements, let $\gamma'$ be the unique antidominant weight in the same orbit as $\gamma$ and let $\gamma - \gamma' = \sum d_i \alpha_i$. Then $N(\gamma)$ has dimension $(d_1, d_2, \ldots, d_n)$.
  • Let $\gamma$ and $\gamma'$ be as in the previous bullet point, and let $\gamma' = - \sum c_i \omega_i$. Then $N(\gamma)$ is uniquely characterized as the only preprojective module with dimension $(d_1, d_2, \ldots, d_n)$ and socle of dimension $(c_1, c_2, \ldots, c_n)$.

Now, let $\omega'_i$ be the unique fundamental root so that $-\omega'_i$ and $\omega_i$ are in the same orbit. Let $w^i$ be the Coxeter group element of minimal length such that $w^i \omega_i = - \omega'_i$ and let $s_{j_1} s_{j_2} \cdots s_{j_L}$ be a reduced word for $w^i$. Put $\gamma(k) = s_{j_k} \cdots s_{j_2} s_{j_1} (- \omega'_i)$, so $\gamma(0) = - \omega'_i$, $\gamma(L) = \omega_i$ and $\gamma(k) = \gamma(k-1) + c_k \alpha_{i_k}$ for some integer $c_k$; it turns out (proof omitted) that $c_k>0$. So the modules $N(\gamma(k))$ form a filtration of $N(\omega_i) = P_i$, with subquotients $S_{j_k}^{c_k}$.

In particular, if any $c_k$ is $\geq 2$, then the submodule lattice of $P_i$ is not distributive.

If all the $c_k$ are $1$, then $w^i$ is what is called $\omega_i$-minuscule (see Definition 2.4 in Dranowski, Elek, Kamnitzer and Morton-Ferguson), and is therefore minuscule, and the results of that paper apply. I'll summarize those results now.

We define a poset $H(w^i)$, called the heap of $w^i$, as follows: The ground set of $H(w^i)$ is $[L]$. The order relation is the transitive closure of the following: $a \prec b$ if $a<b$ and $s_{j_a} s_{j_b} \neq s_{j_b} s_{j_a}$. We have the following:

  • $H(w^i)$ indexes a basis for $\mathbb{C} H(w^i)$. To understand the maps in $\mathbb{C} H(w^i)$, take the Hasse diagram of $H(w^i)$ (see the Thomas-Yong paper) and place $1$'s and $-1$'s on the edges in order to make the preprojective relations hold.

  • Order ideals of $H(w^i)$ correspond to submodules of $\mathbb{C} H(w^i)$. These also correspond to the weak order interval $[e, w^i]$, and to roots in the $W$ orbit of $\omega_i$.

  • The submodule lattice of $\mathbb{C}H(w^i)$ is the lattice of order ideals in $H(w^i)$.

  • Total orders of $H(w^i)$ correspond to reduced words for $w^i$, which correspond to composition series of $P_i$.

In type $A_n$, $H(w^i)$ is the product of an $i$-vertex chain and an $(n+1-i)$ vertex chain. Order ideals correspond to partitions fitting in an $i \times (n+1-i)$ box, and the submodule lattice is this part of Young's lattice. Total orders on $H(w^i)$ are standard Young tableaux of shape $i \times (n+1-i)$.

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David E Speyer
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Here is an answer to question (2), strongly inspired by Dave Benson's comment:

Theorem Let $A$ be any ring and let $M$ be a finite length $A$-module. Then the lattice of $A$-submodules is distributive iff $M$ does not have a subquotient of the form $S^2$, for $S$ simple.

Proof First, suppose that $M$ has an $S^2$ subquotient, say $X \subset Y \subseteq M$ with $Y/X \cong S^2$. Then the interval $[X,Y]$ in the submodule lattice of $M$ is isomorphic to the submodule lattice of $S^2$. In particular, the submodule lattice of $S^2$ contains $S \oplus \boldsymbol{0}$, $\boldsymbol{0} \oplus S$ and $\Delta:= \{ (s,s) : s \in S \}$. Then $\{ \boldsymbol{0}, S \oplus \boldsymbol{0}, \boldsymbol{0} \oplus S, \Delta, S^2 \}$ form a copy of the $M_3$ diamond lattice, which is not distributive.

Conversely, suppose that the submodule lattice of $M$ is not distributive. Then it must contain either a copy of $M_3$ or $N_5$. But the submodule lattice is modular, so it cannot contain $N_5$, so it must contain $M_3$. In other words, there are submodules $X \subset W_1, W_2, W_3 \subset Y$ with $W_i \cap W_j = X$ and $W_i + W_j = Y$ for all $i \neq j$. Put $Z = Y/X$ and $V_i = W_i/X$. So $Z$ has submodules $V_i$ such that $W_i \cap W_j = \{ 0 \}$ and $W_i + W_j = Y/X$ for $i \neq j$. Then we have $$V_2 \cong Z/V_1 \cong V_3 \cong Z/V_2 \cong V_1 \cong Z/V_3$$ so $V_1 \cong V_2 \cong V_3$ and $Z \cong V_1^2$. Let $S$ be a simple quotient of $V$, then $S^2$ is a quotient of $Z$ and is a subquotient of $Y/X$. $\square$

Corollary: The submodule lattice of $M_1 \oplus M_2 \oplus \cdots \oplus M_r$ is distributive iff (1) the submodule lattice of each $M_i$ is individually distributive and (2) for $i \neq j$, the modules $M_i$ and $M_j$ have no common simple subquotient.

Now, in the preprojective Dynkin case, every simple module is a subquotient of every indecomposable projective module. So this simplifies to

Corollary: In the original Dynkin case that the OP asked about, if a projective module has distributive submodule lattice iffthen it is indecomposable.

I have some thoughts about the other parts, but I need to work on other things for now.

Here is an answer to question (2), strongly inspired by Dave Benson's comment:

Theorem Let $A$ be any ring and let $M$ be a finite length $A$-module. Then the lattice of $A$-submodules is distributive iff $M$ does not have a subquotient of the form $S^2$, for $S$ simple.

Proof First, suppose that $M$ has an $S^2$ subquotient, say $X \subset Y \subseteq M$ with $Y/X \cong S^2$. Then the interval $[X,Y]$ in the submodule lattice of $M$ is isomorphic to the submodule lattice of $S^2$. In particular, the submodule lattice of $S^2$ contains $S \oplus \boldsymbol{0}$, $\boldsymbol{0} \oplus S$ and $\Delta:= \{ (s,s) : s \in S \}$. Then $\{ \boldsymbol{0}, S \oplus \boldsymbol{0}, \boldsymbol{0} \oplus S, \Delta, S^2 \}$ form a copy of the $M_3$ diamond lattice, which is not distributive.

Conversely, suppose that the submodule lattice of $M$ is not distributive. Then it must contain either a copy of $M_3$ or $N_5$. But the submodule lattice is modular, so it cannot contain $N_5$, so it must contain $M_3$. In other words, there are submodules $X \subset W_1, W_2, W_3 \subset Y$ with $W_i \cap W_j = X$ and $W_i + W_j = Y$ for all $i \neq j$. Put $Z = Y/X$ and $V_i = W_i/X$. So $Z$ has submodules $V_i$ such that $W_i \cap W_j = \{ 0 \}$ and $W_i + W_j = Y/X$ for $i \neq j$. Then we have $$V_2 \cong Z/V_1 \cong V_3 \cong Z/V_2 \cong V_1 \cong Z/V_3$$ so $V_1 \cong V_2 \cong V_3$ and $Z \cong V_1^2$. Let $S$ be a simple quotient of $V$, then $S^2$ is a quotient of $Z$ and is a subquotient of $Y/X$. $\square$

Corollary: The submodule lattice of $M_1 \oplus M_2 \oplus \cdots \oplus M_r$ is distributive iff (1) the submodule lattice of each $M_i$ is individually distributive and (2) for $i \neq j$, the modules $M_i$ and $M_j$ have no common simple subquotient.

Now, in the preprojective Dynkin case, every simple module is a subquotient of every indecomposable projective module. So this simplifies to

Corollary: In the original Dynkin case that the OP asked about, a projective module has distributive submodule lattice iff it is indecomposable.

I have some thoughts about the other parts, but I need to work on other things for now.

Here is an answer to question (2), strongly inspired by Dave Benson's comment:

Theorem Let $A$ be any ring and let $M$ be a finite length $A$-module. Then the lattice of $A$-submodules is distributive iff $M$ does not have a subquotient of the form $S^2$, for $S$ simple.

Proof First, suppose that $M$ has an $S^2$ subquotient, say $X \subset Y \subseteq M$ with $Y/X \cong S^2$. Then the interval $[X,Y]$ in the submodule lattice of $M$ is isomorphic to the submodule lattice of $S^2$. In particular, the submodule lattice of $S^2$ contains $S \oplus \boldsymbol{0}$, $\boldsymbol{0} \oplus S$ and $\Delta:= \{ (s,s) : s \in S \}$. Then $\{ \boldsymbol{0}, S \oplus \boldsymbol{0}, \boldsymbol{0} \oplus S, \Delta, S^2 \}$ form a copy of the $M_3$ diamond lattice, which is not distributive.

Conversely, suppose that the submodule lattice of $M$ is not distributive. Then it must contain either a copy of $M_3$ or $N_5$. But the submodule lattice is modular, so it cannot contain $N_5$, so it must contain $M_3$. In other words, there are submodules $X \subset W_1, W_2, W_3 \subset Y$ with $W_i \cap W_j = X$ and $W_i + W_j = Y$ for all $i \neq j$. Put $Z = Y/X$ and $V_i = W_i/X$. So $Z$ has submodules $V_i$ such that $W_i \cap W_j = \{ 0 \}$ and $W_i + W_j = Y/X$ for $i \neq j$. Then we have $$V_2 \cong Z/V_1 \cong V_3 \cong Z/V_2 \cong V_1 \cong Z/V_3$$ so $V_1 \cong V_2 \cong V_3$ and $Z \cong V_1^2$. Let $S$ be a simple quotient of $V$, then $S^2$ is a quotient of $Z$ and is a subquotient of $Y/X$. $\square$

Corollary: The submodule lattice of $M_1 \oplus M_2 \oplus \cdots \oplus M_r$ is distributive iff (1) the submodule lattice of each $M_i$ is individually distributive and (2) for $i \neq j$, the modules $M_i$ and $M_j$ have no common simple subquotient.

Now, in the preprojective Dynkin case, every simple module is a subquotient of every indecomposable projective module. So this simplifies to

Corollary: In the original Dynkin case that the OP asked about, if a projective module has distributive submodule lattice then it is indecomposable.

I have some thoughts about the other parts, but I need to work on other things for now.

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David E Speyer
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David E Speyer
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