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We know that for algebraic $e^{it\pi}$, $t$ can not be algebraic irrational by Baker's theorem, but his proof is analytic; is there some algebraic understanding for such fact? If $t$ is transcendent, can $t$ be a period? We say $t$ is period, iff $t$ can be expressed as the volume of domain in $R^{n} defined by polynomial inequalities with algebraic coefficients.

edited Mar 11 at 6:45

unknown (google)
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asked Mar 11 at 1:39

connes
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What do you mean by "can $t$ be [a] period?" – Noah S Mar 11 at 2:09

(I fixed some grammar/spelling; I hope you don't mind.) – Noah S Mar 11 at 2:09

Noah S: Presumably this means "Can t be an element of the Ring of Periods?" I think the question is clear as written. See en.wikipedia.org/wiki/Periods_%28ring%29 . – Steven Landsburg Mar 11 at 4:01

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The number $\pi$, the exponential function, a period... are not objects from Algebra, but from Analysis. So an algebraic proof would require to extract from these beasts their relevant properties and use them in an algebraic manner. However, the conjectures about periods by Grothendieck and Kontsevich-Zagier give a motivic interpretation of Baker's theorem and of (almost) all conjectural statements in Transcendental Number Theory. – ACL Mar 11 at 6:58

GH · Answer 1 · 2013-03-11 02:49:32Z UTC

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The heart of all transcendence proofs is the analytic fact that there is no integer between $0$ and $1$. This is one of the things I learned from Baker's book "Transcendental Number Theory".

answered Mar 11 at 2:49

GH
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How can we understand Baker’s theorem about transcendence ?

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