What is the categorical operad whose algebras are the (based) symmetric monoidal categories?

Note that the definition of symmetric monoidal category can be found here. https://ncatlab.org/nlab/show/symmetric+monoidal+category#:~:text=A%20symmetric%20monoidal%20category%20is,categorical%20products%20may%20be%20commutative.

A symmetric monoidal category is based if the unit is strict.

I tried defining it myself but have had no luck. Can someone please tell me or point me in the right direction?

  • $\begingroup$ Sorry what do you mean by based symmetric monoidal categories? $\endgroup$ – trujello Jun 30 at 1:30
  • $\begingroup$ See this, ncatlab.org/nlab/show/…. $\endgroup$ – Johnathon Taylor Jun 30 at 3:04
  • $\begingroup$ Also, based means the unit is strict. $\endgroup$ – Johnathon Taylor Jun 30 at 3:04
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    $\begingroup$ I'm not quite sure what you mean by the unit being strict (i.e. perhaps $I \otimes A = A$) but one reference you might consult is Infinity Operads and Monoidal Categories with Group Equivariance. In that text Donald Yau outlines a general procedure for constructing all kinds of categorical operads that are in a coherent sense acted upon by a "group $G$ action operad". Then the algebras over those categorical operads "acted on" by a $G$-action operad give you braided, symmetric, ribbon monoidal categories by varying $G = B_n$ (braids), $G = S_n$, etc (I'm paraphrasing a lot here) $\endgroup$ – trujello Jun 30 at 3:32
  • $\begingroup$ That is exactly what I mean by strict. $\endgroup$ – Johnathon Taylor Jun 30 at 3:54

I was looking and found the answer. The answer I found comes from the link https://ncatlab.org/nlab/show/symmetric+monoidal+category#:%7E:text=A%20symmetric%20monoidal%20category%20is,categorical%20products%20may%20be%20commutative . Apparently, a symmetric monoidal category is equivalently a a category that is equipped with the structure of an algebra over the little k-cubes operad for k≥3.

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  • $\begingroup$ The algebras are actually symmetric monoidal $\infty$-categories. $\endgroup$ – Johnathon Taylor Jul 2 at 20:48

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