3
$\begingroup$

I stuck at a relatively simple thing of formalization in infinity setting. I use here the formalism of quasi categories, i.e. simplicial sets with inner horn fillings.

Suppose $O^{\otimes}$ is an infinity operad and $C^{\otimes}$ a symmetric monoidal infinity category. Then I read that $Alg_O(C^{\otimes})$ can be endowed with a symmetric monoidal structure by "pointwise tensor product".

If everything was strict, it would be clear that one should put $(F \otimes G)(C) = F(C) \otimes G(C)$. With higher homotopies, I don't know how to move.

The first attempt I did is to look for a functor $Alg_O(C^{\otimes}) \to N(Fin_*)$ that encodes the pointwise thing, hoping that somehow the underlying category would be $Fun(O, C)$ (functors between the underlying categories). But if we think about the projection to $\to N(Fin_*)$ as the "degree" or the function "how many objects are there", then an algebra $O^{\otimes} \to C^{\otimes}$ does not have a precise degree: it simply preserves the degree of the involved objects.

The second attempt I made is to construct $Alg_O(C^{\otimes})^{\otimes}$ such that the underlying category is $Alg_O(C^{\otimes})$. In case $Alg_O(C^{\otimes})$ was strict, i know that the $\otimes$ construction is something like the set of tuples $(\langle n \rangle, x_1, \ldots, x_n)$ with morphisms given by $$ Hom ( (\langle n \rangle, x_1, \ldots, x_n), (\langle m \rangle, y_1, \ldots, y_m) ) = \coprod_{\alpha: \langle n \rangle \to \langle m \rangle } \prod_{i=1}^m Hom( \otimes_{\alpha(j)=i} x_j, y_i) $$

Also, in case $Alg_O(C^{\otimes})$ was a simplicial category, we could promote this definition to a simplicial categorical one by substituting hom-sets with map-ssets.

Nevertheless, in case the higher information is described as a simplicial set (as it is in my case), I don't know how to put it into the picture. Any hints? Thanks!

Andrea

$\endgroup$

1 Answer 1

7
$\begingroup$

This is worked out in Higher Algebra, example 3.2.4.4.

Concretely, $\mathrm{Alg}_{\mathcal{O}}(\mathcal{C})^\otimes$ is defined as follows: it is the simplicial set over $\mathrm{Fin}_\ast$ such that for any simplicial set $K\to \mathrm{Fin}_\ast$, the set of maps $K\to \mathrm{Alg}_{\mathcal{O}}(\mathcal{C})^\otimes$ over $\mathrm{Fin}_\ast$ is equivalent to the set of diagrams $$ \require{AMScd} \begin{CD} K\times\mathcal{O}^\otimes @>>> \mathcal{C}^\otimes\\ @VVV @VVV \\ \mathrm{Fin}_\ast\times\mathrm{Fin}_\ast @>{\wedge}>> \mathrm{Fin}_\ast \end{CD} $$ such that for every $k\in K$ restriction of the top arrow to $\{k\}\times \mathcal{O}^\otimes$ sends inert morphisms to inert morphisms. Here the bottom arrow is the smash product of pointed sets, sending $(I_+,J_+)$ to $(I\times J)_+$.

If you consider the fiber over $1_+$, you see that it is exactly $\mathrm{Alg}_\mathcal{O}(\mathcal{C})$, moreover for every $o\in\mathcal{O}$ there is a canonical symmetric monoidal functor $\mathrm{Alg}_\mathcal{O}(\mathcal{C})^\otimes\to \mathcal{C}^\otimes$, given by taking the fiber over $o$ of the previous diagram, thus showing that this deserves the name "pointwise tensor product".

This might seem an abstract way of defining it, and it is, but if you reflect a bit on what it is doing you'll see that it makes sense: the smash product of pointed sets is precisely encoding the multiplication you want on the algebra $A\otimes B$. For example, for an associative algebra, the product on $A\otimes B$ is given by $$(A\otimes B)\otimes (A\otimes B)\cong (A\otimes A)\otimes (B\otimes B)\to A\otimes B$$ as desired.

Another way of thinking about this is that $\mathrm{Alg}_\mathcal{O}(\mathcal{C})^\otimes$ is the universal object $\mathcal{A}^\otimes$ together with an "evaluation map" $\mathcal{A}^\otimes\times\mathcal{O}^\otimes\to C^\otimes$ sending $(\{A_i\}_{i\in I},\{o_j\}_{j\in J})$ to $\{A_i(o_j)\}_{(i,j)\in I\times J}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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