how do we prove that a sum of two periods is still a period? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?optimizing Frobenius instance solutionsDoes “all points rational” imply “constant” for this “cubic” curve over an arbitrary field?What is the relationship between these two notions of “period”?Computer software for periodsIs special value of Epstein zeta function in 3 variables a period?Property of a derivative in global fieldWhy are Green functions involved in intersection theory?Waldspurger Formula as a Torus IntegralHow does $zeta^mathfrakm(2)$ and relate to $zeta(2)$?Is there an algorithm for numerical approximation of (naive) period integrals
how do we prove that a sum of two periods is still a period?
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?optimizing Frobenius instance solutionsDoes “all points rational” imply “constant” for this “cubic” curve over an arbitrary field?What is the relationship between these two notions of “period”?Computer software for periodsIs special value of Epstein zeta function in 3 variables a period?Property of a derivative in global fieldWhy are Green functions involved in intersection theory?Waldspurger Formula as a Torus IntegralHow does $zeta^mathfrakm(2)$ and relate to $zeta(2)$?Is there an algorithm for numerical approximation of (naive) period integrals
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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 $mathbbR^n$. 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
543
$endgroup$
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 $mathbbR^n$. 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
543
$endgroup$
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 $mathbbR^n$. 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
543
$endgroup$
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 $mathbbR^n$. 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
543
asked Apr 2 at 14:27
periodsperiods
543
asked Apr 2 at 14:27
periodsperiods
543
asked Apr 2 at 14:27
periodsperiods
543
asked Apr 2 at 14:27
periodsperiods
543
543
add a comment |
add a comment |
1 Answer
1
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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
39818
$endgroup$
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 |
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1 Answer
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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
39818
$endgroup$
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
39818
$endgroup$
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
39818
$endgroup$
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
39818
answered Apr 2 at 15:15
GaussianGaussian
39818
answered Apr 2 at 15:15
GaussianGaussian
39818
answered Apr 2 at 15:15
GaussianGaussian
39818
39818
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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