how do we prove that a sum of two periods is still a period?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?
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
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
New contributor
asked 4 hours ago
periodsperiods
411
411
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
periods is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326977%2fhow-do-we-prove-that-a-sum-of-two-periods-is-still-a-period%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 3 hours ago
GaussianGaussian
1788
1788
add a comment |
add a comment |
periods is a new contributor. Be nice, and check out our Code of Conduct.
periods is a new contributor. Be nice, and check out our Code of Conduct.
periods is a new contributor. Be nice, and check out our Code of Conduct.
periods is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326977%2fhow-do-we-prove-that-a-sum-of-two-periods-is-still-a-period%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown