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?

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    $\begingroup$ The ends of the arrow are different vertices. If the quiver was just one vertex with a loop then diagonalisability might be relevant. $\endgroup$ – Angina Seng Oct 13 '19 at 15:20
  • $\begingroup$ @LordSharktheUnknown Ah, indeed. We can change the basis in the image and the preimage independently and just make $(1, 0)$, $(1, 1)$ to be a new basis in the image. Then our mapping is just identity and everything works. Thanks, it seem to be clear now. Would you mind making an answer from your comment so I'll be able to accept it? $\endgroup$ – Ilya V. Schurov Oct 13 '19 at 15:23

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 $$\pmatrix{I&O\\O&O}.$$


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