Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: 'center' of monoidal categories

Matteo Capucci (he/him) (Mar 09 2022 at 17:41):

I'm aware that the correct definition of 'center' for monoidal categories is [[Drinfeld center]], where each object comes equipped with its own way to 'braid' with every other object.
My question is, does this coincide with the pseudocoequalizer image.png

Matteo Capucci (he/him) (Mar 09 2022 at 17:41):

where $\otimes$ is $M$'s product and $\otimes'$ is the same product but with swapped arguments?

Matteo Capucci (he/him) (Mar 09 2022 at 17:42):

AFAIU, $M'$ would be exactly like $M$ but with a bunch of formal isomorphisms $m \otimes m' \cong m' \otimes m$ thrown in, satisfying naturality-like coherence. Is this correct?

Matteo Capucci (he/him) (Mar 09 2022 at 17:43):

In that case, it doesn't feel it coincides with $Z(M)$ because the latter has 'all possible braidings', whereas this category has only a formal one (and I'm not even sure it actually is a braiding)

Mike Shulman (Mar 09 2022 at 17:49):

Yes: the center should be a limit construction, not a colimit.

Matteo Capucci (he/him) (Mar 10 2022 at 09:27):

I see! Is there a name for the above coequalizer? Like the 'braidization'?

Matteo Capucci (he/him) (Mar 10 2022 at 09:35):

Also, do you have any intuition in why this is the universal property of $Z$? It seems very arbtirary to me: image.png

Joe Moeller (Mar 10 2022 at 14:07):

Looks like Z is right adjoint to including braided into planar.

Martti Karvonen (Mar 10 2022 at 16:10):

How does the "bijective on objects and surjective on morphisms" part relate to that?

Joe Moeller (Mar 10 2022 at 16:12):

Oh, I just didn't read that :sweat_smile:

Mike Shulman (Mar 10 2022 at 16:59):

I doubt that that coequalizer is actually braided, since among other things you didn't impose the braid relations.

Mike Shulman (Mar 10 2022 at 17:00):

I also wouldn't expect any kind of center to have a nice universal property. The center of an ordinary group isn't even a functor! Perhaps the surjectivity requirement is to get around that?

Matteo Capucci (he/him) (Mar 10 2022 at 18:32):

Mike Shulman said:

I doubt that that coequalizer is actually braided, since among other things you didn't impose the braid relations.

uh, indeed!

Matteo Capucci (he/him) (Mar 10 2022 at 18:36):

Mike Shulman said:

I also wouldn't expect any kind of center to have a nice universal property. The center of an ordinary group isn't even a functor! Perhaps the surjectivity requirement is to get around that?

Maybe? I do have a nice universal property for it (first suggested to me by @Eigil Rischel) which is to classify actions of braided monoidal categories.
That is, if $\mathcal C$ is monoidal, $Z(\mathcal C)$ acts on it and it is terminal among actions of braided monoidal categories on $\mathcal C$.

It's the nicest UP for the center I've found so far but it is quite far from the original context so it's a bit mysterious.

Mike Shulman (Mar 10 2022 at 18:40):

I would suggest first looking for a universal property of the center of a group or a monoid.

Mike Shulman (Mar 10 2022 at 18:40):

Walk before we try to run...

Matteo Capucci (he/him) (Mar 10 2022 at 19:20):

Mike Shulman said:

I would suggest first looking for a universal property of the center of a group or a monoid.

For monoids it works like this: every action of a commutative monoid $M$ on a monoid $C$ (in the category of monoids) correspond to a monoid homomorphism $M \to Z(C)$, and viceversa.
If $\ast : M \times C \to C$ is such an action, $- \ast 1 : M \to C$ is a morphism of monoids because $(mn) \ast 1 = (mn) \ast 11 = (m \ast 1)(n \ast 1)$. Moreover it factors through the center because $M$ is commutative, hence $(m \ast 1)(n \ast 1) = (mn) \ast 1 = (nm) \ast 1 = (n \ast 1)(m \ast 1)$.
Viceversa, given a morphism $\phi : M \to Z(C)$, one can define the action $m \ast x = \phi(m)x$. It's easy to see it's a well-defined action: $m \ast (n \ast x) = \phi(m)\phi(n)x = \phi(mn)x = mn \ast x$, $1 \ast x =\phi(1)x = 1x = x$, and that $\ast$ is an homomorphism of monoids (here we use the fact it lands in $Z(C)$): $(mn)\ast(xy) = \phi(mn)xy=\phi(m)\phi(n)xy = \phi(m)x\phi(n)y = (m \ast x)(n \ast y)$.

Matteo Capucci (he/him) (Mar 10 2022 at 19:24):

Mike Shulman said:

Walk before we try to run...

That's good advice, this time I ran before walking though :sweat_smile: hopefully I didn't break a leg

Matteo Capucci (he/him) (Mar 10 2022 at 19:26):

Actually, we're coming out with a preprint on actegories in the coming days and it'd be precious to have some people have a look at it before we actually submit to a journal. I will post a plead when the time comes.