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...


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 $\omega:=\sigma\times\left\lbrace0\right\rbrace\times(0,1)^s\coprod(0,1)^r\times\left\lbrace1\right\rbrace\times\tau$, one immediately gets that $$\alpha+\beta=\int_\omega \left[(1-t)f(x)+tg(y)\right]dxdydt$$which is again an absolutely convergent integral, so that $\alpha+\beta$ is a period.

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    $\begingroup$ @periods: if you're not satisfied by the answer, please tell me how to improve it. $\endgroup$ – Gaussian Apr 2 at 18:55
  • $\begingroup$ The "$\int_\sigma f$" being used as a definition of a period is presumably an ordinary volume integral - no surface integrals allowed. $\omega$ has zero volume, so the integral is zero. $\endgroup$ – Dap Apr 11 at 6:48

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