9
$\begingroup$

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 )?

$\endgroup$
1
  • 1
    $\begingroup$ 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

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.