Many triangulated categories which show up in mathematics, such as derived categories of various sorts, arise as the homotopy category of a stable $\infty$-category.

Stable $\infty$-categories give an important example of a symmetric monoidal $(\infty,2)$-category where the tensor product of stable $\infty$-categories $C$ and $D$ is defined to be the stable infinity category $C \otimes D$ which is universal for functors out of $C \times D$ which are exact in each variable separately.

I gather that taking the homotopy category gives us a functor to the (ordinary, i.e. "(2,2)-category") of triangulated categories, triangulated functors, and transformations.

I am wondering if there is some symmetric monoidal structure on the category of triangulated categories so that "taking the homotopy" category becomes a (strong) symmetric monoidal functor? Can you recover the triangulated category $ho(C \otimes D)$ from the triangulated categories $Ho(C)$ and $Ho(D)$?

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    $\begingroup$ I think it's unlikely. Wouldn't this mean that you can recover the derived category of $H\mathbb F_p\wedge H\mathbb F_p$ out of the category of $\mathbb F_p$-vector spaces? A possible way towards a counterexample would be taking a triangulated category with two models $C$ and $D$ and trying to check that the homotopy categories of the tensor squares $C\otimes C$ and $D\otimes D$ are different, but I don't have any simple example in mind to carry out this. $\endgroup$ – Fernando Muro Jul 29 '15 at 14:01
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    $\begingroup$ I agree that it is unlikely, but to me not obviously so, hence the question. Maybe a better question is what is the minimal additional structure that I need in order to recover the homotopy category of the tensor products from the two homotopy categories? For example probably having the structure of a stable derivator is sufficient, but that is a lot of data. Can we get away with less? For example what if I remember the action by the homotopy category of spectra $ho(Sp)$? $\endgroup$ – Chris Schommer-Pries Jul 29 '15 at 14:40
  • $\begingroup$ Sure, it's a very interesting question. One can learn a lot even from a counterexample. It might be easier to show that the tensor product of triangulated categories does not carry a triangulation, but the non-existence of a canonical triangulated envelop could be much more difficult. $\endgroup$ – Fernando Muro Jul 29 '15 at 19:15

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