# How to represent real algebraic numbers with period integrals

Background: A real period is defined to be the value of an integral of the form

$$\int_D R(x_1,\cdots,x_n)dV$$

where $R$ is a rational function with rational ceofficients, and $D\subseteq\Bbb R^n$ is a set of points satisfying a system of inequalities involving rational functions also with rational coefficients (combined via the Boolean operators and and or), and the integral converges absolutely. A period is then a complex number whose real and imaginary parts are real periods. The set of periods is called $\cal P$. In my other question here, I've asked why we can simplify this definition to only volume integrals over domains defined by polynomial inequalities, and why we can generalize the definition by replacing both instances of the phrase "rational coefficients" with "(real) algebraic coefficients."

So, question: how do we express an arbitrary (wlog real) algebraic number as a period integral?

I know using the argument principle from complex analysis, if $w$ is a root of $f(z)$ then

$$\frac{1}{2\pi i}\oint_{\partial D}z\frac{f'(z)}{f(z)}dz=w$$

where $D\subset\Bbb C$ is a disc of small enough radius around $w$ to not include any other zeros. (This is assuming that $w$ has multiplicity one, for instance take $f$ to be its minimal polynomial, and the boundary of the disc $\partial D$ is oriented clockwise.) But it's not obvious to me how to convert this into a real (multidimensional) integral. Perhaps I could invoke a continuum of different $D$ whose $\partial D$s form there own disk (like an onion), and then convert via polar coordinates? There's also the issue of $2\pi i$ to contend with as well. (Conjecturally, $\frac{1}{2\pi i}$ is not a period.)

(I can figure out why $\cal P$ is closed under addition: we can always write

$$\int_A R(\vec{x})dV+\int_B S(\vec{x})dV=\int_{\large A\sqcup(B+\vec{v})} \hskip -0.5in R(\vec{x})+S(\vec{x}-\vec{v})\,dV$$

for some rational-coordinate displacement vector $\vec{v}$ of sufficiently large magnitude, in which case $A\sqcup(B+\vec{v})$ can be described in the obvious way. Furthermore we can write

$$\int_A \hskip -0.05in R(x_1,\cdots,x_n)dV \hskip -0.05in \times \hskip -0.05in \int_BS(x_1,\cdots,x_k)dV \hskip -0.05in = \hskip -0.05in \int_{A\times B} \hskip -0.2in R(x_1,\cdots,x_n)S(x_{n+1},\cdots,x_{n+k})dV$$

describing $A\times B$ in the obvious way as well, so I know how $\cal P$ is closed under multiplication.)

Let $\alpha \in \bar{\mathbb{Q}}$ and $f(T)$ be its minimal polynomial, let $X = \mathbb{A}^1_\mathbb{Q}$ and $D = Spec( \mathbb{Q}[T]/ (T \cdot f(T)))$. Then, if we considering the singular 1-chain \begin{array}{rrl} \gamma: [0,1] & \rightarrow & X^{an} = \mathbb{C} \\ s & \mapsto & s \alpha \end{array} consisting on a path from 0 to $\alpha$. Since $D^{an}$ consists on the set of roots of $T \cdot f(T)$, both 0 and $\alpha$ belong to $D^{an}$, and this 1-chain $\gamma$ defines a class in $H_{1, sing}(X^{an}, D^{an}; \mathbb{Q})$. Taking the period pairing together with the 1-form $dT$ we get $$\langle [dT], [\gamma] \rangle = \int_\gamma dT = \int_0^\alpha dT = \alpha.$$