1
$\begingroup$

I am trying to show, or find a reference, for the following fact:

"Given O,P two infinity operads [in the sense of Lurie, HA, Definition 2.1.1.10], there always exist a tensor product".

In other words, this means the following. One can consider an infinity operad as a simplicial set over $K=N(Fin_*)$ with some marked edges, namely the inert maps. We call the ($\infty$) category of pairs $(X,\Gamma)$ where $X$ is a simplicial set and $\Gamma$ is a subset of edges that contains degenerate edges a marked simplicial set, denoted by $sSet^+_{/K}$.

Then, by the very definition, a map of infinity operads is a morphism as marked simplicial sets. Note that the product of two infinity operads, as marked simplicial set, is not necessarily an infinity operad. Indeed, one can show that there exist a model category on $sSet^+_{/K}$ such that infinity operads are exactly the fibrant objects, and every object is cofibrant.

Define bifunctors from $O,P$ to $Q$ (everything being an infinity operad) as $$Map_{sSet^+_{/K}}(O\times P, Q) $$ We say that $\alpha: O \times P \to Q$ exhibit Q as the tensor product of O,P if, for any other ifninity operad R, the composition map

$$ \alpha^*: Map_{sSet^+_{/K}}(Q,R) \to Map_{sSet^+_{/K}}(O \times P,R) $$ is an equivalence of sSets.

In literature is often said that it is enough to take a fibrant replacement of $O\times P$ in the category of marked sSets over K. I am trying to show, and here there are my attempts:

  1. Note that $\alpha^*$ is surjective in every degree. Indeed the diagram determined by $$O \times P \times (\Delta^n)^{\#} \to R$$ $$ O \times P \times (\Delta^n)^{\#} \to Q \times (\Delta^n)^{\#}$$ $$Q\times (\Delta^n)^{\#} \to 0,R \to 0$$

admits a lifting, because $R \to 0$ is a fibration and $O \times P \to Q$ is an acyclic cofibration. Recall that $(\Delta^n)^{\#}$ is the (left) canonical marked simplicial set associated to the n-cell, with just degenerate edges marked. This is the cosimplicial object we use for maps in degree n.

  1. For every marked S, the sequence

$$ S \times (\Delta^n)^{\#} \coprod S \times (\Delta^n)^{\#} \to S \times (\Delta^{n+1})^{\#} \to S \times (\Delta^n)^{\#} $$

exhibits $ S \times (\Delta^{n+1})^{\#}$ as a canonical cylinder object of $S \times (\Delta^n)^{\#}$ (canonical meaning that the last is a fibration). This is true because it is at the level of $\Delta^n$, and multiplying by $Q$ preserve at least cofibrations (~injectives) and equivalences.

  1. I would hope that, with some being fibrant argument, one can show that every left homotopy can be done via tha cylinder object above.

  2. As the map $O\times P \to Q$ is a weak equivalence, both are cofibrant, and $R$ is fibrant, we have that $\alpha^*$ is an equivalence between $Map(O \times P, R)_n/ homotopy$ and $Map(Q, R)_n /homotopy$.

Even if everything goes in the good direction, it seems that these facts point at $\alpha_*$ being a weak equivalence, and not an equivalence. In other words, it hints at the fact that it may be a simplicial model category (see https://ncatlab.org/nlab/show/simplicial+model+category, definition 2.1, point 3). But how the hell showing that fibers are contractible and not weakly contractible?

$\endgroup$
6
  • 2
    $\begingroup$ I am a bit confused by your definition of tensor product. How is $O\times P$ lying above $K$? Are you using the smash product map $K\times K\to K$? Otherwise I don't know how to make it work. Moreover, can you point out where in the literature is the statement you make claimed? I cannot find it in HA (the construction that Lurie gives for the tensor product is different). It's not clear at all to me why the fibrant replacement should be an $\infty$-operad... $\endgroup$ Commented Jun 14, 2019 at 12:51
  • $\begingroup$ Yes, I am using the smash product. For the definition and the claim, see HA, 2.1.4.2, 2.1.4.3 and 2.2.5.6. $\endgroup$ Commented Jun 14, 2019 at 13:14
  • 4
    $\begingroup$ I am sorry, but I think you have misread what Lurie is claiming. He's not taking the fibrant replacement in the marked model structure, but in the preoperadic model structure, where the statement is automatic since it is simplicial (see HA.2.1.4.6) $\endgroup$ Commented Jun 14, 2019 at 13:19
  • 4
    $\begingroup$ Also, sorry if I presume but from the last questions you asked you seem to be a bit lost in a sea of technicalities. This subject can be hard to learn! If you think it'd be helpful you are welcome to come to the homotopy theory chatroom for more informal conversations $\endgroup$ Commented Jun 14, 2019 at 13:36
  • 4
    $\begingroup$ No, the model structures (and so the weak equivalences and fibrant replacement) are very different, only the underlying categories are the same $\endgroup$ Commented Jun 14, 2019 at 14:12

0

You must log in to answer this question.