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Let $\mathcal{T}$ be a $\Bbbk$-linear triangulated category which is Hom-finite and Krull-Schmidt, with translation functor $\Sigma$ satisfying $\Sigma^2 = \text{id}$. Suppose that $\mathcal{T}$ has finitely many indecomposables $$ U_1^{(i)}, U_2^{(i)}, \dots , U_{n_i}^{(i)}$$ for $i = 1, \dots , m$, where $m,n_1, \dots n_m \in \mathbb{N}$. Write $\mathcal{F}_i = \{ U_1^{(i)}, \dots , U_{n_i}^{(i)} \}$. Suppose further that $$ \text{Hom}_{\mathcal{T}} ( U_k^{(i)} , U_\ell^{(j)}) = 0$$ whenever $i \neq j$, while $$ \text{Hom}_{\mathcal{T}} ( U_k^{(i)} , U_\ell^{(i)}) \neq 0$$ for all $i,k, \ell$.

Does it follow that restricting $\Sigma$ to the elements of $\mathcal{F}_i$ induces a permutation of $\mathcal{F}_i$? That is, for each $i,k$, does there exist $\ell$ with $\Sigma U_k^{(i)} = U_\ell^{(i)}$?

It's easy to see that $\Sigma$ induces a permutation of the $U_k^{(i)}$ and that, for each $i$, $\Sigma \mathcal{F}_i \subseteq \mathcal{F}_j$ for some $j$, with equality only if $\mathcal{F}_i$ and $\mathcal{F}_j$ have the same size. Therefore, if the $\mathcal{F}_i$ have sizes which are pairwise distinct then the result holds.

However, I run into trouble when $|\mathcal{F}_i| = |\mathcal{F}_j|$ for some $i \neq j$. I can't even do the case when there are four indecomposables which split into two families $\mathcal{F}_1$ and $\mathcal{F}_2$. My current approach has been to consider long exact sequences arising from applying certain $\text{Hom}$ functors to triangles, but this has only got me so far.

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There's a trivial counterexample with two indecomposables $U$ and $V$, where $U$ and $V$ have endomorphism ring $\mathbb{k}$, $\operatorname{Hom}(U,V)=0$, $\operatorname{Hom}(V,U)=0$, $\Sigma U=V$, $\Sigma V=U$, and the only distinguished triangles are finite direct sums of $$U\stackrel{\text{id}}{\to}U\to 0\to \Sigma U=V$$ and rotated versions.

But this is essentially the only counterexample. Suppose $\mathcal{F}_i$ is a family of indecomposables as in the question, with $\Sigma\mathcal{F}_i\subseteq\mathcal{F}_j\neq\mathcal{F}_i$, let $\alpha:X\to Y$ be a nonzero map between two objects of $\mathcal{F}_i$ and complete to a triangle $$X\stackrel{\alpha}{\to}Y\stackrel{\beta}{\to}Z\stackrel{\gamma}{\to}\Sigma X.$$ Then the restriction of $\gamma$ to every indecomposable summand of $Z$ must be non-zero, or else the triangle has a nontrivial direct sum decomposition, contradicting the indecomposability of $X$ and $Y$ and the fact that $\alpha\neq0$. So every indecomposable summand of $Z$ is isomorphic to an object of $\mathcal{F}_j$, since it has a non-zero map to $\Sigma X$, and therefore $\beta=0$. But then $\alpha$ is a split epimorphism, and since $X$ is indecomposable, $\alpha$ must be an isomorphism.

Therefore $\mathcal{F}_i$ has only one object $X$, up to isomorphism, with $\operatorname{End}(X)$ a division ring.

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  • $\begingroup$ Presumably in the first triangle one can replace $\text{id}$ by any element of $\Bbbk$. In the trivial counterexample, can you not also have triangles of the form $U \oplus U \xrightarrow{\sim} U \oplus U \to 0 \to V \oplus V$, where the map from $U \oplus U$ to itself is not necessarily the identity? $\endgroup$
    – user324795
    Commented Apr 29, 2016 at 13:36
  • $\begingroup$ @user324795 Yes, but replacing $\text{id}$ by another nonzero element of $\mathbb{k}$ gives an isomorphic triangle, and whatever isomorphism $U\oplus U\to U\oplus U$ you use will give a triangle isomorphic to the direct sum of two copies of the triangle in my answer. $\endgroup$ Commented Apr 30, 2016 at 8:57
  • $\begingroup$ Ah, of course. I must be missing something obvious, but I don't quite understand the sentence "Then the restriction of $\gamma$...". Why does the triangle decompose? $\endgroup$
    – user324795
    Commented May 2, 2016 at 9:09
  • $\begingroup$ @user324795 Because the map $\gamma$ decomposes. If $Z=Z'\oplus Z''$, where the restriction of $\gamma$ to $Z''$ is zero, then the triangle would be isomorphic to a direct sum of two triangles $X\to Y'\to Z'\to\Sigma X$ and $0\to Y''\to Z''\to0$. $\endgroup$ Commented May 2, 2016 at 9:21
  • $\begingroup$ I'm still being slow; why does $Y$ also decompose in this situation? $\endgroup$
    – user324795
    Commented May 4, 2016 at 10:33

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