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Let $Q$ be a wild quiver without oriented cycles and let $V$ be an indecomposable representation of $Q$. Assume that $V_i\neq 0$ for each vertex $i$ of $Q$. The base field $k$ is algebraically closed. If $V$ is not a Schur representation, $\operatorname{End}V$ is a local $k$-algebra different from $k$, so there is a non-trivial nilpotent endomorphism. What I would like to know is the following: does there exist a nilpotent endomorphism $\phi$ such that for each vertex $i$ $\phi$ is not zero at $V_i$?

I have a large body of examples in which this question has positive answer, however I still can't prove it in full generality... am I missing some key example?

This question arised while solving problems for a homework sheet for a course on quiver representations, however it does not help me in any way to solve that problem, which was simply to prove that $V$ is indecomposable iff $End V$ is local.

Edit: forgot the key word, I want $\phi$ to be nilpotent.

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Take the quiver $1 \rightarrow 2 \rightrightarrows 3$.

Let $V$ be the representation with $V_1=k$, $V_2=k^2$, $V_3=k^2$, with arrows acting by $\pmatrix{0&1}$, $\pmatrix{0&1\\0&0}$ and $\pmatrix{1&0\\0&1}$.

Then, up to scalar multiplication, the only nilpotent endomorphism is zero at vertex $1$ and $\pmatrix{0&1\\0&0}$ at vertices $2$ and $3$.

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  • $\begingroup$ I guess you can also find them where the dimension of the vector spaces are all bigger than $1$ right? $\endgroup$
    – Shana
    Commented Dec 4, 2017 at 0:06

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