how do we prove that a sum of two periods is still a period?
$begingroup$
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...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
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add a comment |
$begingroup$
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...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
$endgroup$
add a comment |
$begingroup$
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...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
$endgroup$
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...
ag.algebraic-geometry nt.number-theory
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
asked Apr 2 at 14:27
periodsperiods
492
asked Apr 2 at 14:27
periodsperiods
492
asked Apr 2 at 14:27
periodsperiods
492
asked Apr 2 at 14:27
periodsperiods
492
492
add a comment |
add a comment |
1 Answer
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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 $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, 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.
answered Apr 2 at 15:15
GaussianGaussian
38118
$endgroup$
2
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
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– 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
add a comment |
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1 Answer
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$begingroup$
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 $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, 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.
answered Apr 2 at 15:15
GaussianGaussian
38118
$endgroup$
2
$begingroup$
@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
add a comment |
$begingroup$
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 $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, 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.
answered Apr 2 at 15:15
GaussianGaussian
38118
$endgroup$
2
$begingroup$
@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
add a comment |
$begingroup$
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 $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, 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.
answered Apr 2 at 15:15
GaussianGaussian
38118
$endgroup$
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 $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, 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.
answered Apr 2 at 15:15
GaussianGaussian
38118
answered Apr 2 at 15:15
GaussianGaussian
38118
answered Apr 2 at 15:15
GaussianGaussian
38118
answered Apr 2 at 15:15
GaussianGaussian
38118
38118
2
$begingroup$
@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
add a comment |
2
$begingroup$
@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
2
2
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
$begingroup$
@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
$begingroup$
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
add a comment |
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