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Let $P^{\bullet} = (P^i, d^i)_{i\leq 0}$ be an indecomposable object in category of complexes bounded above, where each $P_i$ is a finitely generated projective module over an finite dimensional algebra. Denote by $\tau^{\geq -n} (P^{\bullet})$ the brutal truncation of $P^{\bullet}$, $n \geq 1$, i.e., $$ \tau^{\geq -n} (P^{\bullet}) = \cdots \to 0 \to P^{-n} \to \cdots \to P^{-1} \to P^{0} \to 0 \to \cdots $$

Is it indecomposable $\tau^{\geq -n} (P^{\bullet})$ for all $n \geq 1$?

Thanks!

Edit:

– Suppose $P^{\bullet}$ is unbounded below, i.e., $P^{\bullet} \in \mathcal{C}^{-}(\text{proj}-A) \setminus \mathcal{C}^{b}(\text{proj}-A)$.

– Suppose $P^0 \neq 0$, in order to fix notation.

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No. Let $k$ be a field, and let $A = k[x,y]/(x^2, y^2, xy, yx)$. Then the morphism $A \to A\oplus A$ sending $1$ to $(x,y)$ gives an indecomposable complex $$P^\bullet: \ldots \to 0 \to A\to A\oplus A\to 0 \to \ldots $$ If you put $A\oplus A$ in degree $-1$, then $\tau^{\geq -1}(P^\bullet)$ is not indecomposable. Notice that the image of the map described above lies in the radical of $A\oplus A$.


Edit: An example which is unbounded below can be obtained using the same algebra $A$. Take the following complex: $$ Q^\bullet : \ldots \longrightarrow A\oplus A \stackrel{\begin{pmatrix}y & 0 \\ x & 0\end{pmatrix}}{\longrightarrow} A\oplus A \stackrel{\begin{pmatrix}y & 0 \\ x & 0\end{pmatrix}}{\longrightarrow} A\oplus A \stackrel{\begin{pmatrix}x & 0 \end{pmatrix}}{\longrightarrow} A \longrightarrow 0 \longrightarrow \ldots,$$ where $A$ is in degree $0$. Then $Q^\bullet$ is indecomposable, but $\tau^{\geq -n}(Q^\bullet)$ has a direct summand given by the complex with $A$ concentrated in degree $-n$, so it is not indecomposable.

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  • $\begingroup$ Thank you, @Pierre-Guy. I understand your example, but that was not what I thought when I wrote my question. For $\tau^{\geq -n} P^{\bullet}$ with $n\geq 1$ wanted to say: $$ \tau^{\geq -1} P^{\bullet} = \cdots \to 0 \to P^{-1} \to P^0 \to 0 \to \cdots $$ $$ \tau^{\geq -2} P^{\bullet} = \cdots \to 0 \to P^{-2} \to P^{-1} \to P^0 \to 0 \to \cdots $$ $$ (\cdots) $$ with $P^0 \neq 0 \neq P^{-1}$. Do you understand me? $\endgroup$ – Ogniov Oct 10 '13 at 0:05
  • $\begingroup$ @Chico I'm not sure I'm following you. Is the fact that $P_0 = 0$ the only problem with my example, or am I missing more conditions? $\endgroup$ – Pierre-Guy Plamondon Oct 10 '13 at 4:39
  • $\begingroup$ thank you for the attention and sorry the poor english. See I edited my question. $\endgroup$ – Ogniov Oct 10 '13 at 9:43
  • $\begingroup$ using your example: Let us denote by x and y the endomorphisms of $A$ given by the multiplications of $x$ and $y$ respectively. Let $$ P^{\bullet}: \cdots \to A \to A \to A \to A \to 0 \to \cdots $$ where $d^i =$ x when $i$ is odd integer and $d =$ y when $i$ is even integer. So $P^{\bullet}$ is indecomposable object in $\mathcal{C}^{-}$ and for each $n\geq 1$, $\tau^{\geq -n} P^{\bullet}$ is indecomposable object in $\mathcal{C}^{b}$. Notice that the image of the maps x and y lies in the radical of $A$. $\endgroup$ – Ogniov Oct 10 '13 at 10:13
  • $\begingroup$ @Chico I've edited my answer to include an example which is not bounded below. $\endgroup$ – Pierre-Guy Plamondon Oct 10 '13 at 15:30

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