# M(5,1,5)

M(5,1,5) - $B_0(kA_7)$
Representative: $B_0(kA_7)$ $C_5$ $C_4$ 5 4 1 $\left( \begin{array}{cccc} 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2 \\ \end{array} \right)$ Yes Yes Yes $B_0(\mathcal{O}A_7)$ $\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$ 1 {{{PIgroup}}} Yes Yes M(5,1,4), M(5,1,6) Yes {{{coveringblocks}}} {{{coveredblocks}}} {{{pcoveringblocks}}}

## Basic algebra

Quiver: a:<1,2>, b:<2,3>, c:<3,4>, d:<4,3>, e:<3,2>, f:<2,1>

Relations w.r.t. $k$: ab=bc=de=ef=0, fa=be, eb=cd

## Projective indecomposable modules

Labelling the simple $B$-modules by $S_1, S_2, S_3, S_4$, the projective indecomposable modules have Loewy structure as follows:

$\begin{array}{cccc} \begin{array}{c} S_1 \\ S_2 \\ S_1 \\ \end{array}, & \begin{array}{ccc} & S_2 & \\ S_1 & & S_3 \\ & S_2 & \\ \end{array}, & \begin{array}{ccc} & S_3 & \\ S_2 & & S_4 \\ & S_3 & \\ \end{array}, & \begin{array}{c} S_4 \\ S_3 \\ S_4 \\ \end{array} \end{array}$

## Irreducible characters

All irreducible characters have height zero.