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Suppose $X$ is a variety defined over $\mathbb{Q}$, and $\omega$ is an algebraic form defined on $X$ which induces a nonzero element of the algebraic de Rham cohomology $\mathbb{H}^n(X)$. If $C$ is a nonzero element of $H_n(X(\mathbb{C}),\mathbb{Q})$, the following integration defines a period \begin{equation} P=\int_C \omega \end{equation} I do not understand Kontsevich-Zagier period conjecture very well, and does it imply that $P$ is non-zero?

If so, is there a proof to this "simple" (I am not sure whether it is simple) case? Any useful references?

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    $\begingroup$ I do not know the Kontsevich-Zagier period conjecture, but if you take $X$ to be a product $Y\times Z$, if you define $\omega$ to be the pullback of a class from $Y$, and if you define $C$ to be a cycle in a fiber of projection to $Y$, then $P$ equals $0$. $\endgroup$
    – Jason Starr
    Commented Mar 2, 2018 at 10:48
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    $\begingroup$ @JasonStarr Thank you. Do you know any examples in case $X$ is not the product of two varieties? $\endgroup$
    – Wenzhe
    Commented Mar 2, 2018 at 11:42
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    $\begingroup$ A non-product counterexample is $X=\mathbb{A}^1\backslash\{0,1\}$, $\omega=dz/z$, and $\gamma$ is a loop around $1$. $\endgroup$
    – Julian Rosen
    Commented Mar 2, 2018 at 14:24
  • $\begingroup$ There is no single Kontsevich-Zagier period conjecture, rather there are a collection of conjectures which can be divided into two categories $\endgroup$
    – shehryar sikander
    Commented Jun 27, 2019 at 20:40
  • $\begingroup$ 1) Given a number can one figure out if it is a period in the sense you outlined above. Which naturally occurring numbers are periods, for example special values of all motivic L-functions are conjectured to be periods(this is explicitly stated on page 21 of their paper and attributed to Beilinson-Deligne-Scholl) 2) What does the set of all periods look like (it is a subset of complex numbers and contains all algebraic numbers). Can one find all the relations amongst periods. $\endgroup$
    – shehryar sikander
    Commented Jun 27, 2019 at 20:59

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