Line bundles in abelian $\otimes$-categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:31:24Z http://mathoverflow.net/feeds/question/47079 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47079/line-bundles-in-abelian-otimes-categories Line bundles in abelian $\otimes$-categories Martin Brandenburg 2010-11-23T10:35:10Z 2010-12-05T09:05:16Z <p>By an abelian $\otimes$-category I mean a symmetric monoidal category $(\mathcal{A},\otimes,\mathcal{O})$, such that $\mathcal{A}$ also is an abelian category and for every $M \in \mathcal{A}$ the functor $M \otimes -$ is cocontinuous (i.e. right exact and preserves coproducts; in particular additive). A line bundle is defined as an object $\mathcal{L}$ of $\mathcal{A}$ such that there is some object $\mathcal{L}'$ such that $\mathcal{L} \otimes \mathcal{L}' \cong \mathcal{O}$. An example is the category of (quasi-coherent) modules on a locally ringed space. The line bundles then coincide with the modules which are locally free of rank $1$ (see <a href="http://mathoverflow.net/questions/33489/justification-of-the-term-invertible-sheaf" rel="nofollow">here</a>).</p> <p>Now I want to show that in general these line bundles have similar properties as in the case of the module category. For example it's not hard to show that if $\mathcal{L}$ is a line bundle, then it is flat in the sense that $\mathcal{L} \otimes -$ is exact (it is even an automorphism of $\mathcal{A}$ with inverse $\mathcal{L}^{-1} \otimes -$). The isomorphism classes of line bundles yield a group, which may be denoted as $\text{Pic}(\mathcal{A})$.</p> <p><strong>Question 1.</strong> Is there any literature about these abelian $\otimes$-categories which treats them systematically? Perhaps the "usual" definition differs a little from mine, this does not matter.</p> <p><strong>Question 2.</strong> Let $\mathcal{L}$ be a line bundle and $\phi : \mathcal{L} \to \mathcal{L}$ an epimorphism. Does it follow that $\phi$ is an isomorphism?</p> <p><strong>Question 3.</strong> Let $\mathcal{L}$ be a line bundle and assume $\phi : \mathcal{L} \to \mathcal{L}$ is an epimorphism. Does it follow that there is an epimorphism $\psi : \mathcal{L}^{-1} \to \mathcal{L}^{-1}$ such that $\phi \otimes \psi$ corresponds to the identity of $\mathcal{O}$ under the isomorphism $\mathcal{L} \otimes \mathcal{L}^{-1} \cong \mathcal{O}$?</p> <p><strong>Question 4.</strong> Assume we also have a $\lambda$-structure on $\mathcal{A}$ which is compatible with the given data. Is it possible to give a reasonable definition of a locally free object of rank $n$? See also <a href="http://math.stackexchange.com/questions/11012/characterizing-free-modules-by-exterior-power" rel="nofollow">this</a> question.</p> http://mathoverflow.net/questions/47079/line-bundles-in-abelian-otimes-categories/47160#47160 Answer by Steve Lack for Line bundles in abelian $\otimes$-categories Steve Lack 2010-11-23T23:14:56Z 2010-11-23T23:14:56Z <p>Question 2: this does not really depend on line bundles. If $\phi:\mathcal {L\to L}$ is a non-invertible epimorphism, then $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is epi, since $-\otimes\mathcal L'$ is exact, and non-invertible, since otherwise $\phi\otimes 1\otimes 1:\mathcal{L\otimes L'\otimes L\to L\otimes L'\otimes L}$ would be invertible. Thus there is a non-invertible epimorphism $\mathcal {O\to O}$.</p> <p>Question 3: this has the same answer as Question 2. If the answer to Q2 is yes, then we can take $\psi$ to be the identity and get a positive answer to Q3. If the answer to Q2 is no, then (as above) if $\phi:\mathcal{ L\to L}$ is a non-invertible epimorphism, also $\phi\otimes 1:\mathcal{ L\otimes L'\to L\otimes L'}$ is a non-invertible epimorphism. But now if $\psi:\mathcal {L'\to L'}$ is any map, then $\phi\otimes\psi=(1\otimes\psi)\circ(\phi\otimes 1)$ and if this is invertible then $\phi\otimes 1$ is split monic and so invertible (since it is already known to be epi).</p>