how do we prove that a sum of two periods is still a period?

Question

Kontsevich and Zagier define periods as the values of absolutely convergent integrals $\int_\sigma f$ where $f$ is a rational function with rational coefficients and $\sigma$ is a semi-algebraic subset of $\mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...

Answer 1

Let $\alpha$ and $\beta$ be two periods corresponding respectively to two absolutely convergent integrals $\int_\sigma f(x)dx$ and $\int_\tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $\Bbb Q$ with $r$ (resp. $s$) variables and $\sigma$ (resp. $\tau$) is a semi-algebraic subset of $\Bbb R^r$ (resp. $\Bbb R^s$).

Setting $\omega:=\sigma\times(0,1)\times(0,1)^s\coprod(0,1)^r\times(1,2)\times\tau$, one immediately gets that $$\alpha+\beta=\int_\omega \left[f(x)+g(y)\right]dxdydt$$which is again an absolutely convergent integral, so that $\alpha+\beta$ is a period.

Answer 2

For a given $n \in \mathbb{N}$, consider the algebraic function:

$$ f_n : \mathbb{R}^{n+2} \rightarrow \mathbb{C} $$

$$ f_n(x,y,z_1,\dots,z_n) = x + iy $$

We're going to define a 'simple period' to be anything that can be written as an integral of some $f_n$ over a semialgebraic set.

Lemma 1: Simple periods are closed under addition.

Proof: If we have $\alpha = \int_{\sigma} f_n$ and $\beta = \int_{\tau} f_m$, then we have:

$$ \alpha + \beta = \int_{\sigma} f_n + \int_{\tau} f_m $$

$$ = \int_{\sigma \times [0,1]^{k - n}} f_k + \int_{\tau \times [2,3]^{k - m}} f_k $$

$$ = \int_{\sigma \times [0,1]^{k - n} \cup \tau \times [2,3]^{k - m}} f_k $$

where $k = \max(m, n) + 1$.

Lemma 2: If $p,q$ are rationals and $\omega \subseteq \mathbb{R}^n$ is any semialgebraic set, then $(p + iq) \textrm{ Vol}(\omega)$ is a simple period.

Proof: Take the domain of integration to be $\varpi = [p-\frac{1}{2},p+\frac{1}{2}] \times [q-\frac{1}{2},q+\frac{1}{2}] \times \omega$. Then we have:

$$ \int_{\varpi} f_n = \textrm{ Vol}(\omega) \int_{x = p-\frac{1}{2}}^{p+\frac{1}{2}} \int_{y = q-\frac{1}{2}}^{q+\frac{1}{2}} (x + iy) $$

$$ = (p + iq) \textrm{ Vol}(\omega) $$

Proposition: Every period can be written as a sum of expressions of the form $(p + iq) \textrm{ Vol}(\omega)$, so every period is a sum of simple periods, and therefore is itself a simple period.

Proof: The real and imaginary parts of an algebraic function are themselves algebraic, so given any algebraic function $g$ defined on a semialgebraic set $\omega$ we can write:

$$ \int_{\omega} g = \int_{\omega} \Re(g) + i \int_{\omega} \Im(g) $$

and consider the two terms separately. Consequently, it suffices to prove this statement for real algebraic functions $g$. Given such a function, we can define the semialgebraic sets:

$$ \omega_{+} = \{ x \in \omega : g(x) \geq 0 \} $$

$$ \omega_{-} = \{ x \in \omega : g(x) \leq 0 \} $$

Then we have:

$$ \int_{\omega} g = \int_{\omega_{+}} g - \int_{\omega_{-}} -g $$

By again considering these two terms separately, we can further restrict our attention to integrals of nonnegative real functions $g$ over a semialgebraic set $\omega$.

But then we can express the integral of $g$ as the 'volume under the graph of the function':

$$ \int_{\omega} g = \textrm{Vol}(\{ (x_1, \dots, x_n, y) : (x_1, \dots, x_n) \in \omega \textrm{ and } 0 \leq y \leq g(x_1, \dots, x_n) \}) $$

This completes the proof of the proposition. As a corollary of this and Lemma 1, it follows that periods are closed under addition.