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As already brought up in this question, I have some difficulties understanding the the modules in the parabolic category $\mathcal O_0^\mathfrak p$. Although I got a lot of comments that helped me understand the composition series of the modules in question, I still wonder why my original approach failed.

Assuming we work in $\mathfrak{sl}_3$, denote the simple reflections that generate the Weyl group by $s, t$ and let $W_\mathfrak p=\{e, t\}$ such that $W^\mathfrak p=\{e, s, st\}$. What is the projective cover $P^\mathfrak p(s)$ of highest weight $s$? By all the theory mentioned in the previous question, we know it has composition factors $L(s), L(st), L(e)$ and $L(s)$ in this order.

What I originally tried to do: We know that $\mathcal O_0$ is equivalent to $\operatorname{Mod}_{\operatorname{End}(P)}$ where $P=\bigoplus_{w\in W}P(w)$ is a projective generator of $\mathcal O_0$. Now $\operatorname{End}(P)$ is isomorphic to the path algebra $A$ of the quiver given in this paper, section 5.1.2. Then $P(s)=e_2A$ is the quiver representation $$ \begin{array}{c} & \boxed{1\rightarrow 2}\\ \boxed{\begin{matrix}e_2 \\ 2\rightarrow1\rightarrow2\end{matrix}} && \boxed{3\rightarrow 1\rightarrow 2}\\ \boxed{\begin{matrix}4\rightarrow 2\\4\rightarrow 2 \rightarrow 1 \rightarrow 2\end{matrix}} && \boxed{\begin{matrix}5\rightarrow 2\\ 5\rightarrow 2\rightarrow 1 \rightarrow 2\end{matrix}}\\ & \boxed{\begin{matrix}6\rightarrow 4 \rightarrow 2\\6\rightarrow 4\rightarrow 2\rightarrow 1\rightarrow 2\end{matrix}} \end{array} $$

What is the largest quotient of this containing only the simples to $e, s, st$, i. e. not containing the simple quiver representations $L(3), L(5), L(6)$ in its composition series? Well, let's see what is generated by the respective paths:

Under the relations mentioned in the article, $3\rightarrow 1\rightarrow 2$ generates (let's omit the arrows):

  • $5312 \propto 5212$,
  • $65312 \propto 64212$
  • $4312 \propto 4212$

and $52$ generates:

  • $252 \propto 212$ (*)
  • $652 \propto 642$

Hence the only paths in the representation that survive are $e_2, 42$ and $12$, corresponding to the simples $L(e), L(st)$ and $L(s)$. In particular, $212$ is lost due to (*), i. e. the socle $L(s)$ of the quotient is lost. This should not happen.

Question: Where in my computation of the submodule did I make a mistake? Why is the path $2\rightarrow 1\rightarrow 2$ not contained in the submodule generated by $5\rightarrow 2$?

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  • $\begingroup$ I will think about this approach to the problem next time I am on a computer. Maybe someone else who is more familiar with it can answer in the mean time. $\endgroup$ – Tobias Kildetoft Apr 30 '17 at 13:57
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The arrangement of the weights into that hexagon shape in the cited paper "quivers and endomorphism rings" simple is other than expected: I assumed the corners to stand for

$$\begin{array}{c} e\\ s\qquad t\\ st\qquad ts\\ w_0 \end{array}$$

However, the actual arrangement employed in that paper is $$\begin{array}{c} e\\ s\qquad t\\ ts\qquad st\\ w_0 \end{array}$$

With the same numbering, the "forbidden" weights $t, ts, w_0$ correspond to corners $3,4,6$. (*) Needs to be replaced by $424=0$, so this composition factor does not die.

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