M(16,10,2)

M(16,10,2) - $B_0(k(C_4 \times A_5))$
Representative: $B_0(k(C_4 \times A_5))$ $C_4 \times C_2 \times C_2$ $C_3$ 16 3 1 $\left( \begin{array}{ccc} 8 & 8 & 4 \\ 8 & 16 & 8 \\ 4 & 8 & 8 \\ \end{array} \right)$ Yes Yes Yes $B_0(\mathcal{O}(C_4\times A_5))$ $\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array}\right)$ 1 $D_8 \times C_2$[1] $D_8 \times S_4 \times C_2$[2] No Yes M(16,10,3) Yes

Basic algebra

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

Relations w.r.t. $k$: $ad=cb=0$, $bcda=dabc$, $e^4=f^4=g^4=0$, $af=ea$, $bg=fb$, $cf=gc$, $de=fd$

Irreducible characters

All irreducible characters have height zero.

Notes

1. ${\rm Pic}_{\mathcal{O}}(B_0(\mathcal{O}(C_4 \times A_5)))=\mathcal{L}(B_0(\mathcal{O}(C_4 \times A_5)))=\mathcal{L}(\mathcal{O}C_4) \times \mathcal{T}(B_0(\mathcal{O}A_5))$ by [EL18c], giving the isomorphism type of ${\rm Pic}_\mathcal{O}(B)$ in general.
2. By Theorem 3.7 of [EL18c].