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Let $\mathcal{A}$ be 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 $X$.

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I've been told that the answer is no in the more general case of just a monoidal category (without additive structure): mathoverflow.net/questions/34981/… I am very much interested in this question, however, and maybe just getting a non-commutative graded ring with nice properties could allow one to do non-commutative projective geometry in the Artin-Zhang sense. –  Chris Brav Dec 5 '10 at 12:20
    
What about graded modules over a graded-commutative ring, where the tensor structure switches signs? –  Tyler Lawson Dec 5 '10 at 18:10
    
Thanks for the comments. @Chris: There is no doubt that $\Gamma_*(\mathcal{L})$ is a (noncomm.) graded $\text{End}(\mathcal{O})$-algebra. I don't understand the answers in your question (perhaps my one is a duplicate?). So the claim is true if $\mathcal{A}$ is symmetric? How do we prove that? What are counterexamples in the general case? @Tyler: What do you mean by "switches signs"? –  Martin Brandenburg Dec 5 '10 at 21:16
    
@Martin: Let $R$ be a graded exterior algebra on x and y, with |x| = |y| = 1, and consider the category of graded left $R$-modules. The tensor structure is given by $M \otimes_R N$, where any left $R$-module $M$ inherits a right action by $m \cdot r = (-1)^{|m| |r|} rm$. (This is the sign switch I referred to.) If $\mathcal{L} = R[1]$ (a shifted copy of $R$), then the graded ring $\Gamma_*(\mathcal{L})$ is isomorphic as a graded ring to $R$ and is noncommutative. –  Tyler Lawson Dec 5 '10 at 21:33
    
(And this category is symmetric monoidal.) –  Tyler Lawson Dec 5 '10 at 21:35

1 Answer 1

The following arose originally as a comment above and is being moved to an answer (per suggestion).

Let $R = R^*$ be any graded ring which is graded-commutative in the sense of homological algebra, i.e. for homogeneous elements $x$ and $y$ we have $xy = (-1)^{|x| |y|} yx$. Consider the category of graded left $R$-modules. This has a tensor structure as follows. Any left $R$-module $M$ inherits a right action by $R$ via the formula $m\cdot r=(−1)^{∣m∣∣r∣}rm$. Using this, we can define a monoidal structure on left $R$-modules using the graded tensor product $M \otimes_R N$.

(Note that all of this really comes because graded abelian groups form a symmetric monoidal category under tensor product, using twist isomorphism $\tau(x \otimes y) = (-1)^{|x| |y|} y \otimes x$. In this category, $R$ is a commutative monoid object and the tensor is just defined by a standard coequalizer on modules.)

Now let $\mathcal{L} = R[1]$, by which I mean a shifted copy of $R$ so that the degree $n$ group $R[1]^n$ is $R^{n+1}$ (grading cohomologically in order to align with the delicate sensibilities of the ag.algebraic-geometry tag). As a left $R$-module, it is free on a generator $e$ with $|e|=-1$. Then this object is invertible, and tensor powers $\mathcal{L}^{\otimes n} = R[n]$ are free on generators $e^{n}$ for $n \in \mathbb{Z}$.

At this point, one should verify for themselves that the ring $\Gamma_*(\mathcal L)$ is isomorphic to $R$ as a graded ring. (Seriously, you should check this, especially if you usually take the attitude that "the signs just work themselves out". There may be a clever perspective that avoids sign issues, but the straightforward perspective is not so.)

However, if you choose not to verify this:

One then gets an identification $Hom_{gr. R-mod}(R, \mathcal{L}^{\otimes n}) \cong R[n]^{0} = R^n \cdot e_n$, and the multiplicative structure is given by $$r e^{|r|} \cdot s e^{|s|} = (-1)^{|r| |s|} (rs) e^{|rs|}.$$ In particular, this graded ring is noncommutative precisely when $R$ is noncommutative, (which is most of the time).

ADDED: You can verify that there is an isomporphism between $R$ and this ring, given by the formula: $$ \phi(r) = (-1)^{\binom{|r|}{2}} r \cdot e^r $$ There is no canonical sign switch if you use $\mathbb{Z}/2$-graded objects rather than $\mathbb{Z}$-graded objects (although mod-4 gradings are fine).

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