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Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. The Auslander-Reiten quiver $\Gamma_A$ of $A$ is a means of presenting the category of finitely generated right $A$-modules. The Auslander-Reiten quiver is a locally finite quiver whose vertices are indecomposable modules (up to isomorphism) and whose arrows are irreducible morphisms between indecomposable modules. It is given the structure of a translation quiver via the Auslander-Reiten translate $\tau$.

Two distinct vertices/modules of $\Gamma_A$ are said to belong to the same component if there exists a finite path/composition of irreducible morphisms between them. If $A$ is representation-finite (i.e. the category of finitely generated right $A$-modules contains finitely many non-isomorphic indecomposable objects), then $\Gamma_A$ consists of finitely many components, which each have finitely many vertices. However if $A$ is representation-infinite, then necessarily $\Gamma_A$ is infinite.

It is easy to construct an example of a representation-infinite algebra $A$ for which there exists a finite component of $\Gamma_A$. (Edit: the example I provide in this question is false, as indicated by the answer given by Jeremy Rickard.) For example, consider the algebra $KQ/I$, where $KQ$ is the path algebra of the quiver $$ Q\colon \; 1 \begin{smallmatrix} \alpha \\ \rightarrow \\ \rightarrow \\ \beta \end{smallmatrix} 2 \begin{smallmatrix} \gamma \\ \rightarrow \\ \color{white} \gamma \end{smallmatrix} 3 \begin{smallmatrix} \delta \\ \rightarrow \\ \color{white} \delta \end{smallmatrix} 4$$ and $I$ is the ideal generated by the set $\{\beta\gamma, \gamma\delta\}$. This is representation-infinite because it contains the Kronecker quiver as a subquiver. But the relation $\gamma\delta$ 'cuts off' the right-hand side of the quiver, which means we have a finite Auslander-Reiten component $$\begin{matrix} S(4) & & & & S(3) \\ & \searrow & & \nearrow & \\ & & P(3)=I(4) & & \end{matrix}$$ (which is essentially $\mathrm{mod}\;K\mathbb{A}_2$) in $\Gamma_A$, where $S(v)$, $P(v)$ and $I(v)$ are the simple, indecomposable projective, and indecomposable injective modules corresponding to the vertex $v$ in $Q$ respectively. (One can also use the fact that this is a string algebra to easily compute $\Gamma_A$.)

My question is this: When is it the case that for a representation-infinite algebra $A$, every component of $\Gamma_A$ contains infinitely many vertices? Is this for example true if the algebra is self-injective? (I imagine the proof of the latter question is straightforward if true, but please point me to a reference if one exists.)

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If $A$ is connected and has infinite representation type, then every component of its Auslander-Reiten quiver is infinite. See, for example, Theorem 5.4 in Assem, Simson and Skowronski’s Elements of the Representation Theory of Associative Algebras, Volume 1.

The example you give is not a complete component. It continues (infinitely) to the right, starting with an irreducible map from $S(3)$ to $P(2)$.

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  • $\begingroup$ Ah yes, there is of course a continuation to the right. I think this confirms that I shouldn't be doing maths at 1am! I shall leave the example in my question, but add an edit indicating that the example is false for future searchers. $\endgroup$ – Iteraf Jun 18 '18 at 9:54
  • $\begingroup$ @Iteraf If you shouldn’t be doing maths at 1am, then I definitely shouldn’t be answering MathOverflow questions at 5am! $\endgroup$ – Jeremy Rickard Jun 18 '18 at 12:18

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