By Sjoerd Beentjes

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By restricting multiplication in the path algebra, kQ1 obtains a kQ0 -bimodule structure. It is of finite length since it is finite dimesional, with basis the finite set Q1 of arrows in Q. So we may put Λ1 = kQ1 . The multiplication in the path algebra k Q corresponds to concatenation ∼ kQ1 ⊗kQ . . ⊗kQ kQ1 . Thus, k Q is a tensor ring. of tensors under the identification kQr = 0 0 The above projective resolution translates into the following somewhat more explicit statement, originally due to Ringel.

4, and consider its extended Hall algebra HAe = HA ⊗ C[K (A)]. For any object M of A, and for any positve integer r, denote by SM,r := {(L1 , . . , Lr ) | 0 = Lr Lr−1 ... 17) the set of strict r-step filtrations of M . Xiao’s antipode is the defined as follows. 29. The antipode of [M ]kα ∈ HAe is defined as S([M ]kα ) = kα−1 S([M ]). On basis objects, the antipode is given by −1 S( M ) := kM r (−1)r r∈N Li /Li+1 , Li+1 m M/L2 L2 /L3 . . Lr−1 /Lr Lr , L• ∈SM,r i=1 where we have used the alternative basis objects N = aN [N ].

9 of [29, p. 14], every left Λ-module M over a tensor ring Λ has both a projective and injective resolution of length one; in particular, tensor rings are hereditary. 3) where M (λ ⊗ m) = λ · m, and δM ((λ ⊗ µ) ⊗ m) = (λ ⊗ µ) ⊗ m − λ ⊗ (µ · m). Note that we interpret Λ+ as Λ ⊗Λ0 Λ1 here. 1 Generalities on quivers How does this apply to path algebras? Given a quiver Q, we find by a straightforward argument that the vertices ei form a complete set of primitive3 orthogonal idempotents, which implies that kQ0 ∼ = i kei as k-algebras.