For an algebraic number $\alpha$ one can define its "height" in many ways. Informally, you could use its minimal polynomial over $\mathbf{Q}$ and consider the maximum of the heights of its coefficients. Or consider all the valuations of $\alpha$, etc. In this context, the height is supposed to be some kind of measure of complexity.

Question. Is there a reasonable definition of the "height" of a transcendental number.

I'm not sure what such a height would mean though in this context.

If there isn't any reasonable definition, is there any reasonable explanation for why this isn't possible?

  • 2
    $\begingroup$ For what it's worth - the Kolmogorov complexity of the number in question. Does it help in any way ? probably not. $\endgroup$ Oct 12, 2011 at 8:25
  • 2
    $\begingroup$ There is Mahler's classification en.wikipedia.org/wiki/… which roughly speaking distinguishes in terms of approximation properties by algebraic numbers. That's possibly rather coarser than what you are asking for though. $\endgroup$
    – dke
    Oct 12, 2011 at 10:21
  • $\begingroup$ The (logarithmic) height of a rational number $x$ is roughly the number of digits needed to write $x$. Given an arbitrary real number $x$, one could try to define the ``height'' of $x$ as the minimal number of symbols needed to write to $x$. There are two problems with this definition : 1) it is not precise - what expressions are allowed ? 2) it is very ineffective... $\endgroup$ Oct 12, 2011 at 17:56
  • $\begingroup$ In my answer mathoverflow.net/questions/53724/… to a similar question, I describe several commonly used hierarchies for measuring the complexity of transcendental real numbers. $\endgroup$ Oct 13, 2011 at 22:38

2 Answers 2


It does'nt quite answer your question, but maybe there is a resaonable definition of the height of a period. The ring of period is a countable over-ring of the field of algebraic numbers, sharing (at least conjecturally) many properties with them. This ring contains most of interesting transcendental numbers.

  • $\begingroup$ For example, you could define the "dimension" of a period to be the minimal dimension of an algebraic set one can integrate over to obtain your period. This isn't so much a "height" in the usual sense of Diophantine analysis, but at least (possibly) filters the periods according to some measure of complexity. $\endgroup$ Oct 13, 2011 at 20:04
  • $\begingroup$ One could make Cam's suggestion compatible with traditional heights by allowing height functions to take values in a higher-dimensional group. One coordinate would indicate dimension, and another would indicate some kind of traditional height information about the defining ideals of the algebraic set. $\endgroup$
    – S. Carnahan
    Oct 14, 2011 at 2:22
  • $\begingroup$ From Kontsevich and Zagier's article Periods (p. 8) : The "simplicity" — the analogue of the height in the case of algebraic numbers — should be measured in terms of the dimension of the integral defining the period and the complexity of the polynomials occurring in the description of the integrand and domain of integration (or, if one wishes, simply by the amount of ink or the number of \TeX keystrokes required to write down the integral). $\endgroup$ Oct 17, 2011 at 8:58

Here is one potentially reasonable explanation for why such an invariant shouldn't exist. One property of height that can be useful is that the height of an algebraic number is invariant under all automorphisms of all rings that contain that number. For any algebraically independent pair of transcendental complex numbers, one may choose a ring-theoretic automorphism of $\mathbb{C}$ that exchanges them. If we want the same sort of invariance as in the algebraic setting, then all heights of transcendental numbers must be equal.

I think the existence of a single correct answer to your question would require a specific application.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.