Grothendieck groups of triangulated categories via cluster tilting subcategories (1812.08493v3)

Abstract: Let $k$ be a field and $\mathcal{C}$ a $k$-linear, Hom-finite triangulated category with split idempotents. In this paper, we show that under suitable circumstances, the Grothendieck group of $\mathcal{C}$, denoted $K_0(\mathcal{C})$, can be expressed as a quotient of the split Grothendieck group of a higher-cluster tilting subcategory of $\mathcal{C}$. Assume that $n\geq 2$ is an even integer, $\mathcal{C}$ is $n$-Calabi Yau and has an $n$-cluster tilting subcategory $\mathcal{T}$. Then, for every indecomposable $M$ in $\mathcal{T}$, there is an Auslander-Reiten $(n+2)$-angle in $\mathcal{T}$ of the form $M\rightarrow T_{n-1}\rightarrow\dots\rightarrow T_0\rightarrow M$ and \begin{align*} K_0(\mathcal{C})\cong K_0^{sp}(\mathcal{T})\big/\big \langle \sum_{i=0}^{n-1}(-1)^i[T_i]\mid M\in\mathcal{T} \text{ indecomposable } \big\rangle. \end{align*} Assume now that $d$ is a positive integer and $\mathcal{C}$ has a $d$-cluster tilting subcategory $\mathcal{S}$ closed under $d$-suspension. Then $\mathcal{S}$ is a so called $(d+2)$-angulated category whose Grothendieck group $K_0(\mathcal{S})$ can be defined as a certain quotient of $K_0^{sp}(\mathcal{S})$. We will show \begin{align*} K_0(\mathcal{C})\cong K_0(\mathcal{S}). \end{align*} Moreover, assume that $n=2d$, that all the above assumptions hold, and that $\mathcal{T}\subseteq \mathcal{S}$. Then our results can be combined to express $K_0(\mathcal{S})$ as a quotient of $K_0^{sp}(\mathcal{T})$.