I wanted to know if it is possible to construct an indecomposable self-injective finite-dimensional algebra $\Lambda$ whose Auslander-Reiten quiver $\Gamma_\Lambda$ is not connected. I'd love to see examples as well as construction methods, if they exist, since I may later want to obtain examples satisfying further properties.
1 Answer
Probably any example you try (that is not of finite representation type) will work. It is conjectured that the Auslander-Reiten quiver of a finite dimensional algebra $A$ is never connected, and in fact has infinitely many connected components, unless $A$ has finite representation type. See (2) and (3) in the list of conjectures at the end of
Auslander, Maurice; Reiten, Idun; Smalø, Sverre O., Representation theory of Artin algebras, Cambridge Studies in Advanced Mathematics. 36. Cambridge: Cambridge University Press. xiv, 423 p. (1995). ZBL0834.16001.
For an algebra over an uncountable field with at least a one-parameter family of indecomposables, this is obviously true for unenlightening reasons: there are uncountably many indecomposable modules, but each component of the Auslander-Reiten quiver is countable.
If you want explicit examples, Benson gives details for some group algebras ($C_2\times C_2$, $A_4$ and $D_{2^n}$ in characteristic two) in Section 4.17 of
Benson, D.J., Representations and cohomology. I: Basic representation theory of finite groups and associative algebras, Cambridge Studies in Advanced Mathematics, 30. Cambridge etc.: Cambridge University Press. xi, 224 p. (1991). ZBL0718.20001.
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$\begingroup$ Great answer! BTW, I guess the answer is the same if we replace the AR-quiver with the stable AR-quiver, right? $\endgroup$ Commented May 26, 2018 at 12:52
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2$\begingroup$ @FernandoMuro Yes, that’s right. $\endgroup$ Commented May 26, 2018 at 12:54