I am reading Persistence Theory: From Quiver Representations to Data Analysis by Steve Y. Oudot and have the following question on Gabriel's theorem. Consider the quiver
$$\bullet \longrightarrow \bullet$$
and its representation over $\mathbb R$
$$ \mathbb R^2 \xrightarrow{\left(\begin{smallmatrix}1 & 1 \\ 0 & 1\end{smallmatrix}\right)} \mathbb R^2.$$
According to Krull, Remark, Schmidt theorem (p. 16), this representation decomposes as a direct sum of indecomposable representations. According to Gabriel's theorem (p. 17), every indecomposable finite-dimensional representation for such a quiver is isomorphic to some interval representation $\mathbb I[b,d]$. In our case, we have only two vertices of the quiver, so the only nontrivial interval representation is
$$\mathbb R \xrightarrow{\;\;\mathbb{1}\;\;} \mathbb R.$$
However, the mapping in our representation is Jordan cell, it's not diagonalizable. Therefore, it cannot be decomposed as a direct sum of one-dimensional mappings. It seem to contradict the aforementioned theorems.
What is my mistake?
Best Answer
I don't know what the notation $\Bbb I[a,b]$ means, but every representation of $\bullet \longrightarrow \bullet$ is a direct sum of copies of $\Bbb R\xrightarrow{1}\Bbb R$, $\Bbb R\rightarrow0$ and $0\rightarrow\Bbb R$. That's because we can change bases independently in the two vector spaces so that the matrix for the arrow becomes of the shape $$\pmatrix{I&O\\O&O}.$$