# Derived category and L-function

For abelian varieties over $$\mathbb{Q}$$ $$\mathscr{A}$$ and $$\mathscr{A}'$$, if derived categories $$D(\mathscr{A})$$ and $$D(\mathscr{A}')$$ are equivalent then L-functions are same $$L(s,\mathscr{A})=L(s,\mathscr{A}')$$.

Is this true for any smooth projective varieties? In other words, for smooth projective varieties $$X$$ and $$Y$$, $$D(X)\simeq D(Y)$$ then $$L(s,H^i(X))=L(s,H^i(Y))$$ ?

• It would be nice if you explain what are $L$-functions of abelian categories and what are $L$-functions of cohomology groups. Oct 9, 2020 at 15:54
• Thank you Sasha. Definition is here mathoverflow.net/questions/144285/…, and for an Abelian variety $\mathscr{A}$, we set $L(s,\mathscr{A}):= L(s,H^1(\mathscr{A}))$. Oct 9, 2020 at 16:03
• Thanks for the clarification. In fact, I am sure this is not known. One of the reasons is the following. I guess (correct me if I am wrong) one can extract the Betti number $b_i(X)$ from the $L$-function $L(s,H^i(X))$ (as its value at some point, or as the value of its derivative at some point), but it is not known (although conjectured) that $b_i(X)$ is a derived invariant. Oct 9, 2020 at 17:52
• Is it possible for the infinity version of the derived category? If you can compute the Hochschild homology of the category over Q with the S1 action it may be possible to calculate algebraic de rham cohomology from it? Oct 10, 2020 at 7:11
• @AndyJiang: I have no idea about that. Do you have some reference? Oct 10, 2020 at 8:27