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



Symmetric monoidal structure on algebras
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 $\{A\_i(o\_j)\}\_{(i,j)\in I\times J}$.


