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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!

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    $\begingroup$ 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. adamczewski.perso.math.cnrs.fr/Complexity_Periods.pdf $\endgroup$ Commented Sep 4, 2014 at 10:46
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    $\begingroup$ 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? $\endgroup$
    – Vincent
    Commented Sep 4, 2014 at 13:54
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    $\begingroup$ $e\ $ is not a period... of what? in what sense? $\endgroup$ Commented Sep 6, 2014 at 19:24
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    $\begingroup$ @Wlodzimierz: en.wikipedia.org/wiki/Ring_of_periods $\endgroup$ Commented Sep 6, 2014 at 19:54
  • $\begingroup$ 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? $\endgroup$
    – Wolfgang
    Commented Sep 11, 2014 at 14:54

4 Answers 4

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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.)

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    $\begingroup$ How many (wo)man hours were spent on this project in the 300 years? I view this argument as weak, at best. $\endgroup$
    – Igor Rivin
    Commented Sep 4, 2014 at 17:29
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    $\begingroup$ 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! $\endgroup$
    – user47305
    Commented Sep 4, 2014 at 17:55
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    $\begingroup$ @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. $\endgroup$ Commented Sep 4, 2014 at 19:02
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    $\begingroup$ @Igor Rivin: I am sure, many thousands. And some of the best mathematicians. $\endgroup$ Commented Sep 5, 2014 at 18:46
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    $\begingroup$ @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. $\endgroup$
    – Igor Rivin
    Commented Sep 5, 2014 at 19:25
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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.

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    $\begingroup$ Very interesting! Do you know if anybody did play this kind of game with any other kinds of cohomologies? Like, Faddeev cochains for example - what kind of "periods" can they correspond to? $\endgroup$ Commented Sep 30, 2017 at 5:57
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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.

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    $\begingroup$ I think this is indeed a rather weak argument. Take the generalized CF's of $\pi$ as in 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. $\endgroup$
    – Wolfgang
    Commented Sep 10, 2014 at 13:18
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    $\begingroup$ @Wolfgang,regular or simple continued fraction is very much different from generalized CF.So your refutation is rather much more weaker. $\endgroup$ Commented Sep 11, 2014 at 1:18
  • $\begingroup$ "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? $\endgroup$ Commented Oct 27, 2017 at 21:20
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    $\begingroup$ @GerryMyerson, yes the rest of periods. $\endgroup$ Commented Oct 27, 2017 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".

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    $\begingroup$ 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. $\endgroup$ Commented Sep 6, 2014 at 21:32
  • $\begingroup$ This doesn't seem to be an answer to the question. $\endgroup$
    – S. Carnahan
    Commented Sep 6, 2014 at 23:44
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    $\begingroup$ 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. $\endgroup$
    – Todd Trimble
    Commented Oct 27, 2017 at 19:09
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    $\begingroup$ 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. $\endgroup$ Commented Oct 28, 2017 at 4:42

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