I have a question:
Suppose we have the quiver \begin{matrix} & 1 & \overset{}{\longrightarrow} && 2 \\ && \searrow_{} & & \downarrow_{} \\ & & && 3 \end{matrix} and let us to associate it the following representation $M$: \begin{matrix} & K & \overset{\tiny{\left[\begin{array}{c} \hspace{-0.15cm}1\\0\end{array}\right]}}{\longrightarrow} && K^2 \\ && \searrow_{1} & & \downarrow_{[0 ~ ~ 1]} \\ & & && K \end{matrix} Now the question is that " Does this representation indecomposable?"
What can I say is no. Because it can be decompose as follows
Let us call the following representation $M_1$: \begin{matrix} & K & \overset{1}{\longrightarrow} && K \\ && \searrow_{1} & & \downarrow_{1} \\ & & && K \end{matrix} And let us call the following representation $M_2$: \begin{matrix} & 0 & \overset{0}{\longrightarrow} && K \\ && \searrow_{0} & & \downarrow_{0} \\ & & && 0 \end{matrix} And then we have $M=M_1\oplus M_2$.
Can someone let me know if I am wrong? Why? And also can someone help me in drawing those diagrams with their vertices shown by a small circle as it is usual in quivers?
Many thanks!
You can show that this representation V is indecomposable by computing the idempotents in the endomorphism ring End(V) since there is a theorem that says that a representation V is indecomposable if and only if the only idempotents in End(V) are 0 and 1.
Sketch of proof for the fact that 0 and 1 are the only idempotents in End(V):
1) You take an arbitrary endomorphism f of this representation and compute the relations between its three components (use the definition of a morphism of quiver representations) and see that you do not have much choice.
2) Then you assume that this endomorphism f is an idempotent (f^2 = f). Then its components must be idempotents. If you then think about the possible idempotent linear maps K --> K (which you will need in two of the components), step 1 should do the trick.