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 $5\rightarrow 2$?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.