On wildness of idempotent generated algebras associated with extended Dynkin diagrams

Vitalij M. Bondarenko

Algebra and discrete mathematics, no. 3 (2004), pp. 1-11.

Résumé

Let $\mathrm{\Lambda }$ denote an extended Dynkin diagram with vertex set ${\mathrm{\Lambda }}_0=\{0,1,\dots ,n\}$. For a vertex $i$, denote by $S(i)$ the set of vertices $j$ such that there is an edge joining $i$ and $j$; one assumes the diagram has a unique vertex $p$, say $p=0$, with $|S(p)|=3$. Further, denote by $\mathrm{\Lambda }\setminus 0$ the full subgraph of $\mathrm{\Lambda }$ with vertex set ${\mathrm{\Lambda }}_0\setminus \{0\}$. Let $\mathrm{\Delta }=({\delta }_i|i\in {\mathrm{\Lambda }}_0)\in {\mathbb{Z}}^{|{\mathrm{\Lambda }}_0|}$ be an imaginary root of $\mathrm{\Lambda }$, and let $k$ be a field of arbitrary characteristic (with unit element 1). We prove that if $\mathrm{\Lambda }$ is an extended Dynkin diagram of type ${\tilde{D}}_4$, ${\tilde{E}}_6$ or ${\tilde{E}}_7$, then the $k$-algebra ${\mathcal{Q}}_k(\mathrm{\Lambda },\mathrm{\Delta })$ with generators $e_i$, $i\in {\mathrm{\Lambda }}_0\setminus \{0\}$, and relations $e_i^2=e_i$, $e_i{e}_j=0$ if $i$ and $j\ne i$ belong to the same connected component of $\mathrm{\Lambda }\setminus 0$, and $\sum _{i=1}^n{\delta }_i{e}_i={\delta }_{0}1$ has wild representation type.

Détail

Mots-clés : idempotent, extended Dynkin diagram, representation, wild typ.