Throughout, we assume our algebras are basic. For a representation-finite (RF) selfinjective algebra $A$ over algebraically closed field $K$, we say $A$ is standard if $K(\Gamma\_A) \simeq \mathrm{ind}-A$. Here $\Gamma\_A$ is the Auslander-Retien (AR) quiver of $A$, $K(\Gamma\_A)$ is the mesh category, and $\mathrm{ind}-A$ is the full subcategory of $\mathrm{mod}-A$ with objects being indecomposable (f.d.) $A$-modules. This definition is used throughout the classification of RF selfinjective algebras and related works by Riedtmann, Bongartz-Gabriel and Bretscher-Läser-Riedtmann.


However, it seems to me a different definition is used for representation-infinite (RI) selfinjective algebras. This definition, I believe, is originated from Skowronski in [Selfinjective algebras of polynomial growth.
Math. Ann. 285 (1989), no. 2, 177–199]: we say a selfinjective algebra $A$ is standard, if there exists a Galois covering $R\to R/G=A$ such that (1) $R$ is simply connected locally bounded category and $G$ is an admisible torsion-free group of $K$-linear automorphisms of $R$.


A detailed description for the terminologies in this definition can be found in the same paper or even better in the survey by Skowronski in [Selfinjective algebras: finite and tame type. Trends in representation theory of algebras and related topics, 169–238, Contemp. Math., 406, Amer. Math. Soc., Providence, RI, 2006] It is remarked in Skowronski's paper that such notion of standard is the same as the one described in the first paragraph, when $A$ is representation-finite. My question is:


When $A$ is RI selfinjective and standard, do we have $K(\Gamma\_A)\simeq \mathrm{ind}-A$?
In fact, it seems to me the new definition is implicitly saying that "$K(\Gamma\_A)\simeq \mathrm{ind}-A$" does not make sense, am I correct or wrong? If so, what's wrong with such statement in RI selfinjective algebras?



Standard selfinjective algebras and Auslander-Reiten quiver
If $A$ is representation-infinite, then $\Gamma\_A$ has more than one component, and so if its mesh category were equivalent to the category of indecomposable modules, the indecomposable modules could be partitioned into two sets with no non-zero maps between them, which is not the case if $A$ is connected.


