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Let $A$ be basic finite-dimensional $K$-algebra and $K$ algebraically closed.

Let $F$ be the free abelian group generated by representatives of the isomorphism classes of objects in $mod A$. We denote by $[X]$ such a representative. Let $F_0$ be the subgroup generated by $[X]-[Y]+[Z]$ for all exact sequences $$0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0 $$ in $mod A$. The Grothendieck group $K_0(A)$ is by definition the factor group $F/F_0$.

Let $P(1), \dotsc, P(n)$ be a complete set of representatives of the isomorphism classes of indecomposable projective $A$-modules. For an $A$-module $X$ the dimension vector is defined by $dim X = (dim_k Hom_A (P(i),X))$. The map $X \mapsto dim X$ induces an isomorphism of $K_0 (A)$ with $\mathbb{Z}^n$

A $A$-module $X$ will be called sincere if every simple $A$-module is a composition factor of $X$. So the $A$-module $X$ is called sincere, if $(dim X)_i \neq 0$ for all $i$.

I was studying triangulated equivalence between derived categories of the $mod A$ and $mod H$, $H$ be basic finite-dimensional hereditary $K$-algebra. There exists an isometry between $K_0 (A)$ and $K_0(H)$.

Let $F : D^b(H) \rightarrow D^b(A)$ the triangle-equivalence functor. Let $P_1, \dotsc ,P_n$ be a complete set of representatives of the isomorphism classes of indecomposable projective $H$-modules. Let $X_i = F(P_i)$, $1 \leq i \leq n$. We obtain a isometry $f : K_0(H) \rightarrow K_0(A)$ by defining for $$X = \sum_i \lambda_i \: dim P(i) \text{ the value for } fX = \sum_i \lambda_i \: dim X_i$$

I would like to the triangle-equivalence functor associated sincere in sincere. Let $Z$ a sincere $H$-module. Does $F(Z)$ is sincere?

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  • $\begingroup$ What do you mean by «if this definition of sincere module is invariant group basis»? The English does not add up in that phrase :-) (If you want, you can write the portuguese version and I or someone else can translate) $\endgroup$ – Mariano Suárez-Álvarez Nov 11 '15 at 22:27
  • $\begingroup$ I expect i made it clear now. Anything, I will write in Portuguese. :-) Thanks @MarianoSuárez-Alvarez $\endgroup$ – Vasco Nov 11 '15 at 23:12
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If $H$ is the path algebra of the quiver $\bullet\to\bullet$ over a field $k$, then there is a triangle-equivalence $F:D^b(H)\to D^b(H)$ with $F(k\to k)=(k\to0)$ and $F(0\to k)=(k\to k)$, so $F$ doesn't preserve sincerity.

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