In the [following link](http://user.math.uzh.ch/ayoub/PDF-Files/periods-GKZ.pdf)$^{[1]}$, 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.




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



What is explicitly, $ \mathcal{P} \mathrm{er} ( X ) / \mathbb{Q} $?
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.


