The homological quadratic form of a biextension algebra

Abstract

Let A be a finite dimensional algebra over an algebraically closed field k. It is conjectured that A has to be of tame representation type provided A is strongly simply connected and its Tits quadratic form is weakly non-negative.¶In the paper a partial result in this direction is proved. Instead of the Tits form the Euler form $\chi _A$ is considered. Let $R_1,\ldots ,R_t$ and $R^\prime _1,\ldots ,R^\prime _s$ be two sequences of modules over an algebra B. We consider $R=\mathop\oplus\limits ^t_{i=1}R_i$ as a $B-k^t$ -bimodule and $R^\prime =\mathop\oplus\limits ^s_{j=1}R^\prime _j$ as a $B-k^s$ -bimodule. The biextension $[R^\prime ]B[R]$ of B by the two sequences is by definition the matrix algebra¶¶ $\left (\matrix {k^t&0&0\cr R&B&0\cr DR^\prime \mathop\otimes _BR&DR^\prime&k^s}\right)$ ¶¶equipped with the obvious addition and multiplication, where we denote by $D=\hbox {Hom}_k(-,k)$ the usual duality. For any set of pairs of indices $L\subset \{1,\ldots ,t\}\times \{1,\ldots ,s\}$ , consider the subspace $V=\mathop\oplus\limits_{(i,j)\in L} DR_{j} \otimes_{B} R_j$ of the space $DR^\prime \otimes _B R$ and the associated ideal J(V) in $[R^\prime ]B[R]$ . The algebra $A=[R^\prime ]B[R]/J(V)$ is called a truncated biextension of B.¶The main result of the paper says: If B is a strongly simply connected algebra with at least 6 vertices and $R_1,\ldots ,R_t;R^\prime _1,\ldots ,R^\prime _s$ are two sequences of indecomposable B-modules such that $\chi_A$ is non-negative with $\hbox {corank}\,\chi _{_A}=1+s+t$ and $\hbox {corank}\, \chi _{_B}=1$ , then $A=[R^\prime ]B[R]$ is of tame representation type.