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In the following link$^{[1]}$, page $2$, we find the following question :

Let $X$ be a smooth $ \mathbb{Q} $ - variety and let $\mathcal{P} \mathrm{er} (X)$ be the subfield of $\mathbb{C}$ generated by the image of the pairing $(3)$. What is the transcendance degree of the finitely generated extension $ \mathcal{P} \mathrm{er} (X) / \mathbb{Q} $ ?

My questions are :

What is explitcitly, the definition of the set : $ \mathcal{P} \mathrm{er} (X) / \mathbb{Q} $ ? Is it defined by : $$ \mathcal{P} \mathrm{er} (X) / \mathbb{Q} = \left\{ \ P \left( \int_{ \gamma_{1} } \omega_1 , \dots , \int_{ \gamma_{m} } \omega_m \right) \;\middle|\; P \in \mathbb{Q} [t_1 , \dots , t_m ] \ \right\} $$ and if it's right, what is $m$ ? why ?

Why is $ \mathcal{P} \mathrm{er} (X) / \mathbb{Q} $ a finitely generated extension of $\mathbb{Q}$ ?

Thanks in advance for your help.


$^{[1]}$ Periods and the conjectures of Grothendieck and Kontsevich–Zagier by Joseph Ayoub (Universität Zürich, Switzerland) 2014

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    $\begingroup$ Strange questions. The left-hand side of the pairing (3) is a finite-dimensional vector space over $\mathbb{Q}$, so its image is finite-dimensional, thus generates a finitely generated extension of $\mathbb{Q}$. You can take for $m$ the dimension of this image. $\endgroup$ – abx Aug 18 '16 at 20:18
  • $\begingroup$ Same question at math.stackexchange.com/questions/1896179/… $\endgroup$ – Watson Dec 3 '16 at 22:12
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In your definition of $\mathcal P er(X)/\mathbb Q$, replace $\mathbb Q[t_1,\dots,t_m]$ with $\mathbb Q(t_1,\dots,t_m)$. Since periods are often transcendental, you need denominators to make it a field. It is a finitely generated extension of $\mathbb Q$ because $m$ is finite. This brings us to your second question.

A period is the integral of an algebraic de Rham class on $X$ against a singular homology class in $H_k(X(\mathbb C),\mathbb Z)$. There are $b_k(X)$ independent homology classes and $b_k(X)$ independent algebraic de Rham classes, so you get $b_k(X)^2$ period integrals from each cohomological degree.

The first case to consider is when $X/\mathbb Q$ is a smooth genus $g$ curve. Here, $H^0(\Omega^1_X)$ is a $g$-dimensional $\mathbb Q$-vector space. Every holomorphic 1-form is automatically closed, so this gives a subspace of the de Rham cohomology $H^1_{dR}(X(\mathbb C),\mathbb C)$, defined over $\mathbb Q$. But $b_1(X)=2g$, so where are the remaining algebraic de Rham classes? Recall that $$H^1_{AdR}(X) = \mathbb H^1_{Zar}(X,\Omega_X^\bullet)$$ and there is a spectral sequence $E_1^{p,q} = H^p(X,\Omega^q_X)\Rightarrow \mathbb H^{p+q}_{Zar}(X,\Omega_X^\bullet)$, which degenerates at $E_1$. In other words, $H^1_{AdR}(X)\simeq H^0(\Omega^1_X)\oplus H^1(\mathcal O_X)$ as $\mathbb Q$-vector spaces. Under the isomorphism $$H^1_{AdR}(X)\otimes_{\mathbb Q}\mathbb C \simeq H^1_{dR}(X(\mathbb C),\mathbb C),$$ we have a full dimensional rational subspace of algebraic de Rham classes.

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    $\begingroup$ A small complement: in the curve case, the "remaining algebraic de Rham classes" can be represented by meromorphic differentials forms "of the second kind", that is, with no residues. Thus all the periods are of the form $\int_{\gamma }\omega $ for some meromorphic $\omega $. In the case $g=1$ this gives the famous periods $\omega _1,\omega _2,\eta _1,\eta _2$ with the Legendre relation $\omega _1\eta _2-\omega _2\eta _1=2\pi i$. $\endgroup$ – abx Aug 19 '16 at 7:31

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