show/hide this revision's text 3 added 63 characters in body; deleted 56 characters in body

Let $\mathcal{A}$ be a an abelian tensor category with unit $\mathcal{O}$. An object $\mathcal{L}$ is called invertible or a line bundle if there is some $\mathcal{L}^{-1}$ such that $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{L}^{-1} \otimes \mathcal{L} \cong \mathcal{O}$. Equivalently, $\mathcal{L} \otimes -$ is an equivalence of categories. Now define a graded ring $\Gamma_*(\mathcal{L})$ as follows:

As an abelian group, take the direct sum of the $\text{Hom}(\mathcal{O},\mathcal{L}^{\otimes n})$, where $n \geq 0$. The product of homogenuous elements $s : \mathcal{O} \to \mathcal{L}^{\otimes n}, t : \mathcal{O} \to \mathcal{L}^{\otimes m}$ is defined by $s \otimes t$, where we identify $\mathcal{O} \otimes \mathcal{O} \cong \mathcal{O}$ and $\mathcal{L}^{\otimes n} \otimes \mathcal{L}^{\otimes m} \cong \mathcal{L}^{\otimes (n+m)}$.

Question Is $\Gamma_*(\mathcal{L})$ commutative? Note that this is known in degree $0$ since $\text{End}(\mathcal{O})$ is commutative, even if we do not assume that $\mathcal{A}$ is symmetric. Actually it's not hard to see that $\text{End}(\mathcal{O})$ is central in $\Gamma_*(\mathcal{L})$. Remark that all this is well-known in the case of $\mathcal{A} = \text{Qcoh}(X)$ for a scheme $X$.
show/hide this revision's text 2 added 121 characters in body; added 78 characters in body

Let $\mathcal{A}$ be a tensor category with unit $\mathcal{O}$. An object $\mathcal{L}$ is called invertible or a line bundle if there is some $\mathcal{L}^{-1}$ such that $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{L}^{-1} \otimes \mathcal{L} \cong \mathcal{O}$. Equivalently, $\mathcal{L} \otimes -$ is an equivalence of categories. Now define a graded ring $\Gamma_*(\mathcal{L})$ as follows:

As an abelian group, take the direct sum of the $Hom(\mathcal{O},\mathcal{L}^{\otimes \text{Hom}(\mathcal{O},\mathcal{L}^{\otimes n})$, where $n \geq 0$. The product of homogenuous elements $s : \mathcal{O} \to \mathcal{L}^{\otimes n}, t : \mathcal{O} \to \mathcal{L}^{\otimes m}$ is defined by $s \otimes t$, where we identify $\mathcal{O} \otimes \mathcal{O} \cong \mathcal{O}$ and $\mathcal{L}^{\otimes n} \otimes \mathcal{L}^{\otimes m} \cong \mathcal{L}^{\otimes (n+m)}$.

Question Is $\Gamma_*(\mathcal{L})$ commutative? Note that this is known in degree $0$ since $End(\mathcal{O})$ \text{End}(\mathcal{O})$ is commutative, even if we do not assume that $\mathcal{A}$ is symmetric. Also it Actually it's not hard to see that $\text{End}(\mathcal{O})$ is central in $\Gamma_*(\mathcal{L})$. Remark that all this is well-known in the case of $\mathcal{A} = \text{Qcoh}(X)$ for a scheme $X$.
show/hide this revision's text 1

graded ring associated to a line bundle in a tensor category

Let $\mathcal{A}$ be a tensor category with unit $\mathcal{O}$. An object $\mathcal{L}$ is called invertible or a line bundle if there is some $\mathcal{L}^{-1}$ such that $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{L}^{-1} \otimes \mathcal{L} \cong \mathcal{O}$. Now define a graded ring $\Gamma_*(\mathcal{L})$ as follows:

As an abelian group, take the direct sum of the $Hom(\mathcal{O},\mathcal{L}^{\otimes n})$, where $n \geq 0$. The product of homogenuous elements $s : \mathcal{O} \to \mathcal{L}^{\otimes n}, t : \mathcal{O} \to \mathcal{L}^{\otimes m}$ is defined by $s \otimes t$, where we identify $\mathcal{O} \otimes \mathcal{O} \cong \mathcal{O}$ and $\mathcal{L}^{\otimes n} \otimes \mathcal{L}^{\otimes m} \cong \mathcal{L}^{\otimes (n+m)}$.

Question Is $\Gamma_*(\mathcal{L})$ commutative? Note that this is known in degree $0$ since $End(\mathcal{O})$ is commutative, even if we do not assume that $\mathcal{A}$ is symmetric. Also it is well-known in the case of $\mathcal{A} = \text{Qcoh}(X)$ for a scheme $X$.