# Certain exact functor on the Grothendieck group of a module category

Let $$A$$ be a be a finite dimensional, associative, and unital $$\mathbb{C}$$-algebra. Let $$\mathcal{A}$$ be the category of finitely generated $$A$$-modules. Since $$A$$ is an Artinian ring, there are only finitely many non-isomorphic simple $$A$$-modules in $$\mathcal{A}$$. For our purposes assume that there are only two non-isomorphic simple $$A$$-modules, $$\left\{S_{1}, S_{2}\right\}$$.

Recall that the Grothendieck group $$K_{0}(\mathcal{A})$$ of $$\mathcal{A}$$ is the free abelian group generated by isomorphism classes $$[M]$$ of $$A$$-modules satisfying the relation that $$[M_{2}] = [M_{1}] + [M_{3}]$$ whenever there exists a short exact sequence $$\begin{equation*} 0 \rightarrow M_{1} \rightarrow M_{2} \rightarrow M_{3} \rightarrow 0 \end{equation*}$$ We also have that $$\left\{[S_{1}], [S_{2}]\right\}$$ is a basis of $$K_{0}(\mathcal{A})$$. Also recall that exact functors induce group homomorphisms on Grothendieck groups by $$\begin{equation*} [F] : K_{0}(\mathcal{C}) \rightarrow K_{0}(\mathcal{D})\\ [F] : [X] \mapsto [FX] \end{equation*}$$ where $$F$$ is an exact functor between essentially small abelian categories $$\mathcal{C}$$ and $$\mathcal{D}$$.

Given this setup, does there exist an exact endofunctor $$\begin{equation*} F: \mathcal{A} \rightarrow \mathcal{A} \end{equation*}$$ such that \begin{align*} [F]\left([S_{1}]\right) &= [FS_{1}] = [S_{1}]\\ [F]\left([S_{2}]\right) &= [FS_{2}] = [S_{1}] - [S_{2}] \end{align*}

Is there an endofunctor such that $$[F]([S_1]) = [FS_1] = -[S_1]$$

• No. Look at the composition series of $F(S_2)$. Apr 6, 2019 at 22:19
• @QiaochuYuan Is there an exact endofunctor such that $[F]([S_1]) = [FS_1] = -[S_1]$? Apr 7, 2019 at 14:38