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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 $L(s,H^i(X))=L(s,H^i(Y))$ ?

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    $\begingroup$ It would be nice if you explain what are $L$-functions of abelian categories and what are $L$-functions of cohomology groups. $\endgroup$
    – Sasha
    Commented Oct 9, 2020 at 15:54
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    $\begingroup$ Thank you Sasha. Definition is here mathoverflow.net/questions/144285/…, and for an Abelian variety $\mathscr{A}$, we set $L(s,\mathscr{A}):= L(s,H^1(\mathscr{A}))$. $\endgroup$
    – user145752
    Commented Oct 9, 2020 at 16:03
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    $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented Oct 9, 2020 at 17:52
  • $\begingroup$ 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? $\endgroup$
    – Andy Jiang
    Commented Oct 10, 2020 at 7:11
  • $\begingroup$ @AndyJiang: I have no idea about that. Do you have some reference? $\endgroup$
    – Sasha
    Commented Oct 10, 2020 at 8:27

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