In the sec 1.11. "Delignes' tensor product of locally finite abelian categories" of the book "Tensor Categories" of EGNO, the deligne's tensor product $C \boxtimes D$ of two k-linear locally finite abelian categories $C$ ad $D$ is defined by the universal property: $$ \matrix{ C \times D & \xrightarrow{\boxtimes} & C \boxtimes D\\ & F \searrow & \downarrow{ \exists ! G} \\ & & A } $$ i.e., $\boxtimes: C \times D \rightarrow C \boxtimes D$ is a bifuntor which is right exact in both variables and is such that for any right exact bifunctor $F: C \times D \rightarrow A$, where $A$ is k-linear locally finite abelian category, there exists a unique right exact functor $G: C \boxtimes D \rightarrow A$ such that $$ G \circ \boxtimes = F. $$

Let $Vec$ be the category of finite dimensional $k$-vector spaces. By the proof provided in the book, $Vec \boxtimes Vec \cong Vec$ and $\boxtimes$ is give by the tensor product of vector spaces. To my understanding, if $B$ is a category which is equivalent to $C \boxtimes D$, then $B$ can also be viewed as the Delignes tensor product of $C$ and $D$.

Let $sVec$ be the skeleton of $Vec$. Then there should exists a functor $G: sVec \rightarrow Vec$ such that the following diagram commutes: $$ \matrix{ Vec \times Vec & \xrightarrow{\otimes} & sVec \\ & \otimes \searrow & \downarrow{ G} \\ & & Vec }. $$ However, this is clearly impossible. So my question is if we can view $sVec$ as $Vec \boxtimes Vec$?

  • $\begingroup$ The universal property, that $G \circ \boxtimes = F$, may be too strong. From a 2-categorical point of view you may want there to be a specified isomorphism, not an equality. $\endgroup$ – David Roberts Nov 3 '17 at 6:54

The universal property isn't a characterization of $C \boxtimes D$, per se: the universal property is a property of the pair $(C \boxtimes D, \boxtimes)$ as an object of the coslice 2-category whose objects are pairs consisting of a category $B$ and a functor $C \times D \to B$. Now, $\mathsf{Vec}$ and $\mathsf{sVec}$ are equivalent just as categories, but there's no way to make them equivalent in this coslice 2-category (being equivalent here is a stronger condition). Does this help?

| cite | improve this answer | |
  • $\begingroup$ Yes, thank you for your remark. Is that the case that even in the coslice 2-category, as mentions by David above, we can only require that $G \circ \boxtimes \cong F$ (not equals)? $\endgroup$ – heller Nov 3 '17 at 7:26
  • $\begingroup$ @heller Yes: the right level of uniqueness should be that $C \boxtimes D$ is defined up to a unique equivalence in this coslice category, and morphisms in the coslice 2-category are pairs (functor,natural isomorphism). So there should be a unique $G$ with a unique natural equivalence $G \circ \boxtimes \cong F$. I would expect Deligne to be very careful with this sort of details, so it could be useful to read Deligne rather than the EGNO book. $\endgroup$ – Dan Petersen Nov 3 '17 at 8:05
  • $\begingroup$ I actually checked the sec 5 "Produit tensoriel de categories abeliennes" of Deligne. He did not mention the universal property in the definition of the "Deligne" tensor product. Instead he require that $F \in Fun(C \boxtimes D, A) \rightarrow F \circ \boxtimes \in Fun(C \times D, A)$ induces an equivalence of the two categories. That is why I think that $sVec$ can also be viewed as $Vec \boxtimes Vec$. $\endgroup$ – heller Nov 3 '17 at 8:21
  • $\begingroup$ @heller But that's equivalent to the universal property, suitably weakened. Essential surjectivity means any functor on the RHS factors through $\boxtimes$ up to a natural equivalence, etc. $\endgroup$ – Dan Petersen Nov 3 '17 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.