6
$\begingroup$

Consider a compact closed category, i.e., a symmetric monoidal category with a unit $\eta$ and co-unit $\epsilon$. It seems natural to demand that the tensor product of two units (for different objects) is a unit again (for the tensor product of the objects). That is, we have an equality between the morphisms $$ 1\xrightarrow{\simeq} 1\otimes 1\xrightarrow{\eta_A\otimes \eta_B} (A^*\otimes A)\otimes (B^*\otimes B) \xrightarrow{\simeq} (A^*\otimes B^*)\otimes (A\otimes B) $$ and $$ 1\xrightarrow{\eta_{A\otimes B}} (A^*\otimes B^*)\otimes (A\otimes B) $$ $\simeq$ stands for some combination of unitors, associators and braidings, I can spell it out if you insist. The analogous equation should hold for the co-unit.

Can this equation be derived from the axioms for compact closed categories? Or is it somehow trivially implied from the way closed compact categories are defined? If not, is there a name for this property, or is it equivalent to some other known property?

Are there closed compact categories for which this equation doesn't hold? I know that it holds for the compact closed categories of finite vectorspaces, and sets and relations. If I'm trying to define a unit and co-unit for the symmetric monoidal category of super vector spaces, imposing the above equation seems to prevent me from doing that (however, I'm also not sure if super vector spaces can be extended to a compact closed category).

Improve this question
$\endgroup$
1
  • $\begingroup$ I just realize that in the second morphism it should be $(A\otimes B)^*$ instead of $A^*\otimes B^*$ (in the examples I had in mind, $A^*$ was the same as $A$ for any $A$). So in general it makes only sense to demand that the two morphisms are equivalent up to composing with an isomorphism from $(A\otimes B)^*$ and $A^*\otimes B^*$. For all choices of $*$, units and co-units, we can find such isomorphisms. $\endgroup$
    – Andi Bauer
    Commented May 20, 2021 at 11:38

1 Answer 1

Reset to default
5
$\begingroup$

The definition of compact closed category merely says that for every object $A$ there exists an object $A^\star$ and a unit $\eta_A$ and counit $\varepsilon_A$ satisfying the snake equations. That is, it is a property, not structure. Since this data is unique up to isomorphism, you can choose different units and counits for different objects as you please, as long as they satisfy the snake equation. If you choose $\eta_A$ for $A$ and $\eta_B$ for $B$, then their combination $1 \to (A^* \otimes B^*) \otimes (A \otimes B)$ will satisfy the snake equation for $A \otimes B$, and hence you could use it for $\eta_{A \otimes B}$. See chapter 3 of Heunen&Vicary.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, that makes sense! Does that mean that it's always possible to consistently choose an $\eta$ for every object, such that all the equations in my question hold? $\endgroup$
    – Andi Bauer
    Commented Jul 23, 2020 at 22:33
  • $\begingroup$ Also, there was this hidden subquestion about whether super vectorspaces are compact closed. If you know the answer, it would be helpful if you could comment on that. $\endgroup$
    – Andi Bauer
    Commented Jul 23, 2020 at 22:37
  • $\begingroup$ I don't see anything going wrong with the snake equations (for super vector spaces with cap being the usual trace), but it doesn't seem possible to consistently choose the $\eta$s as mentioned above. $\endgroup$
    – Andi Bauer
    Commented Jul 23, 2020 at 22:38
  • 1
    $\begingroup$ Yes, the category of super vector spaces is compact, see also that chapter. I don't know if in general you can fix a consistent choice of eta. I suspect not, it seems like a choice principle. $\endgroup$
    – Chris Heunen
    Commented Jul 25, 2020 at 9:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .