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In their 2001 paper defining periods, Kontsevich and Zagier (pdf) without further comment state that $e$ is conjecturally not a period while many other numbers showing up naturally (conjecturally) are. The former claim is repeated at many other internet sources including Wikipedia but nowhere could I find a heuristic making the conjecture that $e$ is not a period more plausible than its negation. Does anyone here know of such an argument?

EDIT: I figured it would be good measure (i.e. 'shows research effort') to write what was the best I could come up with myself. I don't find it very convincing however so feel free to ignore. The number $e$ is more or less defined as the value at a rational number (1) of a function that is a solution to a ordinary differential equation ($y' = y$) with rational boundary condition ($y(0) = 1$). Now K & Z point out that all periods arise in this way (replace a rational number in the defining integral with a parameter and it will satisfy an ODE). However they also warn us that the differential equations are really special and (conjecturally) satisfy a lot of criteria among which having at most regular singularities.

Now the singularity at infinity of $y'= y$ is not regular as it has order 2 (while the equation is of order 1) but of course this proves nothing since nothing is stopping $e$ from being the value at some rational number of a solution to a much more complicated differential equation which might be of the right class. So what is missing from an argument along these lines is some way of making precise that $y'= y$ really is the simplest equation which produces $e$ and that 'therefore' more complicated equations can be 'reduced' to it by a series of simplifications innocent enough to preserve the regularity of the singularities if it exists (quod non). Now personally I would not buy such a claim if it wasn't for the fact that it is a bit akin to conjecture 1 from K & Z. However this line of reasoning requires a lot of 'making precise' and perhaps is an entirely wrong way of looking at it, so better ideas are welcome!

Vincent
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    I would think that the lower bound on the subword complexity of $e$ derived by Adamczewski gives some ground for the belief that $e$ is only an exponential period. http://adamczewski.perso.math.cnrs.fr/Complexity_Periods.pdf – Carlo Beenakker Sep 04 '14 at 10:46
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    The article is certainly interesting, but could you elaborate a bit on how it gives ground for the believe that $e$ is only an exponential period? Aren't periods like $\pi$ expected to satisfy similar lower bounds, even if the method of Adamczewski cannot prove that? – Vincent Sep 04 '14 at 13:54
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    $e\ $ is not a period... of what? in what sense? – Włodzimierz Holsztyński Sep 06 '14 at 19:24
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    @Wlodzimierz: http://en.wikipedia.org/wiki/Ring_of_periods – Qiaochu Yuan Sep 06 '14 at 19:54
  • I'd guess the question about $e$ not being a period is of the same difficulty as the one about $1/\pi$ not being a period. Is that your feeling, too? – Wolfgang Sep 11 '14 at 14:54
  • What does it mean, singularity having order 2? – Anixx Jun 23 '21 at 18:50
  • @Annix The notation $y' = y$ obscures with respect to what variable the derivative is taken, but let's say it is with respect to a variable $x$. Now to see what happens with solution to this equation 'at infinity' we look at what happens at $u = 0$ for a variable $u$ defined as $u = 1/x$. Suppose that $y$ is a function of $x$ that satisfies $y' = y$ and we define $g(u) = y(1/u)$ then what differential equation does $g$ satisfy? By the chain rule $g$ is a solution to $g'(u) = (1/u^2)g(u)$. (ctd in next comment) – Vincent Jun 26 '21 at 13:11
  • Generally speaking we say that a differential equation $g'(u) = g(u)/f(u)$ for some fixed polynomial $f$ has a singularity at $u = 0$ of order equal to the deg $f$. Here this order is 2. And the singularity takes place at $u = 0$ or equivalently at $x$ equals infinity. – Vincent Jun 26 '21 at 13:12
  • Singularities are called regular if they are of order less than or equal to the degree of the equation itself – Vincent Jun 26 '21 at 13:13

4 Answers4

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Periods arise from the comparison between Betti and de Rham cohomology for an algebraic variety. The Period Conjecture, due to Grothendieck, is a transcendence conjecture for periods which says that every algebraic relation between periods arises from geometry (in a certain precise sense).

More generally, there is a wider class of complex numbers called exponential periods arising from the comparison of rapid decay cohomology and de Rham cohomology. The number $e$ is an example of an exponential period. There is an analogue of the Period Conjecture in the setting of exponential periods, and the truth of this conjecture would imply that $e$ is transcendental over the ring of ordinary periods (see Proposition 10.1.5 of the paper Exponential Motives by Javier Fresán and Peter Jossen). So the exponential Period Conjecture provides a heuristic coming from algebraic geometry that $e$ is not a period.

Julian Rosen
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To my understanding, the reason is simple: in the almost 300 years since $e$ was discovered, no representation of it as a period has been found. I think this is quite a strong evidence.

Remark. To those who think that periods were introduced by Kontsevich and Zagier, I recommend the paper of Euler, On highly transcendental quantities which cannot be expressed by integral formulas English translation. (Strange that he does not mention his own $e$ as a candidate. Perhaps he was still looking for an integral that equals $e$ when he wrote this paper.)

Alexandre Eremenko
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    How many (wo)man hours were spent on this project in the 300 years? I view this argument as weak, at best. – Igor Rivin Sep 04 '14 at 17:29
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    I've always imagined that in the old days mathematicians spent a lot of time sitting around compiling integral tables and inventing special functions. They never got $e$ as an answer. Maybe no hours with $e$ as the direct goal, but surely quite a few indirectly! –  Sep 04 '14 at 17:55
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    @Mark: How can we be so sure they never got it? Mathematical literature is vast. Imagine someone during these 300 years encountered an obscure integral that evaluated to a messy formula that happened to involve $e$, in the middle of an otherwise unremarkable proof of a minor result in a long forgotten publication. They would have no idea that years later someone will invent the notion of a period which will make the result interesting. – Emil Jeřábek Sep 04 '14 at 19:02
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    @Igor Rivin: I am sure, many thousands. And some of the best mathematicians. – Alexandre Eremenko Sep 05 '14 at 18:46
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    @Emil Jerabek: The problem is famous enough and there are people who know and search old literature. It is actually not so large as one can imagine: most of the math publiucations were published in the recent years, after the invention of TeX:-) – Alexandre Eremenko Sep 05 '14 at 18:48
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    @AlexandreEremenko I am, of course, just a caveman, but I have never heard of this question before, so I think characterizing it as famous is a little extreme. Of course, if you exhibit other places (than Kontsevich-Zagier) where it is mentioned over the last 300 years, I would be happy to change my mind. – Igor Rivin Sep 05 '14 at 19:25
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    @Igor Rivin: Look in the last AMS Notices, http://www.ams.org/notices/201408/rnoti-p898.pdf – Alexandre Eremenko Sep 05 '14 at 22:44
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    @AlexandreEremenko I did not know about the (very nice) article, but it sort of confirms my suspicion that the work on the subject has taken place in (to be generous) the last twenty years, not three hundred. In particular, it does not answer the OP's question: why did K&Z believe $e$ was not a period? – Igor Rivin Sep 05 '14 at 23:54
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    @Igor Rivin: indeed the formal definition of a period is due to Kontsevich and Zagier. When I said 300 years I meant of course the nature of the $e$ constant. I am ready to bet that no integral representation of the required type has been found. And they were searching for ALL kinds of representation. – Alexandre Eremenko Sep 06 '14 at 00:37
  • @AlexandreEremenko,excellent answer,but could you reformulate it in a strict way? – XL _At_Here_There Sep 11 '14 at 01:05
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    @XL_at_china: I am not sure what a "strict way" means. It is not a theorem. It is a statement about real world:-) – Alexandre Eremenko Sep 11 '14 at 02:06
  • @AlexandreEremenko,yes.So a relevant question is why we find ,discover or prove mathematical theorems or even physical law in such a order but other order?Could we reformulate the order of discovery?I think this may be reformulated by computational complexity and Kolmogorov Complexity.Since the distance between periods and $e$ is infinite ,although mathematicians have found periods(integral representation of number,not the idea by Kontsevich) and $e$ almost at the same time,there is no way or order between them?Of course this is just a feelings or guess. – XL _At_Here_There Sep 11 '14 at 02:25
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    @XL_at_China: You are raising complicated issues which would require long discussion, possibly out of scope of this forum. – Alexandre Eremenko Sep 11 '14 at 18:11
  • @AlexandreEremenko,yes,then let's stop discussion of the question. – XL _At_Here_There Sep 11 '14 at 22:32
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    Ok, it's been a month, I think I should accept this answer. I do agree that this is strong evidence even if somewhat anticlimactic. – Vincent Oct 04 '14 at 22:50
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One very weak heuristic comes from the continued fraction expansion $$e=[2;1,2,1,1,4,1,1,6,1,1,\dots],$$ which exhibits a very simple pattern. If you exclude rational numbers and quadratic irrationals, which have finite and periodic continued fractions respectively, the rest are expected to have "generic" continued fractions. In other words you expect their continued fractions to exhibit properties of "almost all" real numbers. One such property is the convergence of the partial geometric means to Khinchin's constant. For example this property seems to be satisfied numerically by $\pi$.

I guess this is in the same vein as the heuristics that non-rational algebraic numbers (or even periods) are normal in every base.

Gjergji Zaimi
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    I think this is indeed a rather weak argument. Take the generalized CF's of $\pi$ as in http://en.wikipedia.org/wiki/Pi#Continued_fractions, also very simple patterns. Moreover, the reciprocal of a number has essentially the same CF, but reciprocality seems (almost) completely unrelated for periods. – Wolfgang Sep 10 '14 at 13:18
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    @Wolfgang,regular or simple continued fraction is very much different from generalized CF.So your refutation is rather much more weaker. – XL _At_Here_There Sep 11 '14 at 01:18
  • "the rest are expected to have generic continued fractions." I'm confused here, as to what "the rest" refers to. The rest of what? The rest of the periods? – Gerry Myerson Oct 27 '17 at 21:20
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    @GerryMyerson, yes the rest of periods. – Gjergji Zaimi Oct 27 '17 at 21:58
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$e$ is the value at a rational number of a single-valued meaningful holomorphic function whereas periods are values of multivalued functions (they have "conjugates"). It would take some work to make this statement really meaningful, though...

Some further intuition may be found in Yves André's paper "Galois theory, motives and transcendental numbers".

Community
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L.E.
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    And $\pi$ is the value of the constant function $\pi$ at $0$, yet it's a period. Honestly, I don't understand your intuition here. – Donu Arapura Sep 06 '14 at 21:32
  • This doesn't seem to be an answer to the question. – S. Carnahan Sep 06 '14 at 23:44
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    A highly reputable user has opined that this answer deserves to remain open, because it expresses the intuition of the first 3 pages of Yves André's paper "Galois theory, motives and transcendental numbers" arxiv.org/abs/0805.2569. He believes the answer could be very useful for this reason, although even more useful if work were done to flesh out what is meant by "meaningful" holomorphic function. – Todd Trimble Oct 27 '17 at 19:09
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    More like the last 2 pages imho. As I understand the argument: we consider holomorphic functions arising as solutions of algebraic regular differential equations associated with a flat connection. For a smooth proj. family of varieties the Gauss-Manin connection on $H^*_{dR}$ identifies the periods of fibers with a multi-valued solution of a diff. eq. on the base, the periods are values of these functions at rational points. The motivic Galois group associates nontrivial multivalued functions to any period. If one could show that $e$ isn't a value of such functions, we'd have a contradiction. – Anton Fetisov Oct 28 '17 at 04:42