# Is the tensor product of pretriangulated dg-categories a pretriangulated dg-category?

In "Grothendieck ring of pretriangulated categories", Bondal, Larsen and Lunts define a product of perfect (pretriangulated with Karoubian homotopy category) dg-categories as $$A\bullet B:=Perf(A\otimes B)$$ where $$\otimes$$ is the usual (non derived) tensor product of dg-categories and Perf(A) is the full subcategory of semifree A-modules homotopy equivalent to a direct summand of a $$A^{pre-tr}$$-module coming from A (Definition 3.13 in the paper). Alternatively in remark 4.9 they mention that taking $$(A\otimes B)^{pre-tr}$$ would define a product for their Grothendieck ring.

My question is why is taking Perf again necessary? Are there examples of pairs of perfect ( or pretriangulated ) dg-categories whose tensor product is not perfect ( or pretriangulated )?

• Almost any two given categories you take will fail to have a tensor product with the desired property. Feb 26, 2019 at 14:59

Assume that every dg-category is over a field $$k$$. My guess is that there is a natural (I believe fully faithful) dg-functor $$$$\Phi \colon \mathrm{Perf}(A) \otimes \mathrm{Perf}(B) \to \mathrm{Perf}(A \otimes B)$$$$ which induces the equivalence $$\mathrm{Perf}(\mathrm{Perf}(A) \otimes \mathrm{Perf}(B)) \xrightarrow{\sim} \mathrm{Perf}(A \otimes B)$$, and $$\mathrm{Perf}(A) \otimes \mathrm{Perf}(B)$$ is perfect if and only if $$\Phi$$ is a quasi-equivalence.
I think taking $$A=B=\Delta^1$$ (the dg-category freely generated on $$0 \to 1$$) already gives an example where $$\Phi$$ is not essentially surjective in $$H^0$$. In fact, $$\Phi$$ should map $$(X,Y)$$ to the dg-module $$(a,b) \mapsto X(a) \otimes Y(b)$$. In our case, take $$F \in \mathrm{Perf}(A \otimes B)$$ such that \begin{align} F(0,0) &= 0, \\ F(0,1) &= k, \\ F(1,0) &= k, \\ F(1,1) &=k, \end{align} with the obvious maps $$F(i,j) \to F(i',j')$$.