I'm interested in the Grothedieck group of the triangulated category $K^b(\text{proj}A)$ when $A$ is a finite dimensional algebra over a field of infinite global dimension.
For this purpose, It would be helpful to know if $K^b(\text{proj}A)$ is Krull-Schmidt. I couldn't find any proof myself nor in the literature.


It could be helpful to know if $D^b(A)$ is Krull-Schmidt as $K^b(\text{proj}A)$ is a subcategory closed under direct sums.


I would also appreciate any literature recommendations about the Grothendieck group of a arbitrary triangulated category or $K^b(\text{proj}A)$ when $A$ is of infinite global dimension.


Thank you:)



Literature request: $K^b(\text{proj} A)$ Krull-Schmidt for $\text{gl dim}A = \infty$ and general results about its Grothendieck group
I'm assuming your notation denotes the homotopy category of perfect complexes over $A$. In order for this to be Krull-Schmidt, at least the category of finitely generated projective modules has to be Krull-Schmidt, which is often not the case. The bounded derived category is also usually not Krull-Schmidt.


On the other hand, if $A$ is a finite dimensional algebra over a field, or more generally an Artin algebra, then Krull-Schmidt holds for finitely generated modules and for bounded complexes of them.
In this case, the Grothendieck group of the perfect complexes is free abelian of rank equal to the number of isomorphism classes of projective modules. The Grothendieck group of the bounded derived category is also a free abelian group of rank equal to the number of isomorphism classes of simple modules. These have the same rank, but the natural map from the former into the latter is represented by the Cartan matrix. If $A$ has finite global dimension this is an isomorphism. If $A$ is the group algebra of a finite group it is injective but not surjective. In general for finite dimensional algebras it isn't necessarily injective, even if the algebra is symmetric.


