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

Question

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.

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

Answer 1

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.