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 )?

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

1 Answer 1


Assume that every dg-category is over a field $k$. My guess is that there is a natural (I believe fully faithful) dg-functor \begin{equation} \Phi \colon \mathrm{Perf}(A) \otimes \mathrm{Perf}(B) \to \mathrm{Perf}(A \otimes B) \end{equation} 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 $F(i,j) \to F(i',j')$.

Disclaimer: I have not checked every claim thoroughly, so this is more a guess of mine than a precise answer, but hopefully the ideas will work.


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