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Is there a first-order formula ϕ(x) with exactly one free variable $x$ in the language of fields together with the unary function symbol $\exp$ such that in the standard interpretation of this language in $\Bbb C$ (where $\exp$ is interpreted as the exponential function $x \mapsto e^x$), $ϕ(x)$ holds iff $x \in \Bbb R$?

My thoughts: If one takes the algebraic theory of the complex field without the exponential function, then $\Bbb R$ is not definable because there are isomorphisms of $\Bbb C$ mapping reals to non-real numbers. Does a similar approach work in this case? In particular, is there any other isomorphism of the exponential $\Bbb C$ structure besides the identity and complex conjugation?

Dominik
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    At least one can get $\mathbb Q$:

    $$z\in\mathbb Q\iff\exists a\colon\exp(a)=1\land a\ne 0\land \exp(za)=1$$ This is because this implies that $a$ and $za$ are both integer multiples of $2\pi i$.

     – Hagen von Eitzen Jun 12 '13 at 16:16
  • @HagenvonEitzen: Unfortunately, I don't see any possibility to extract topological information from \exp. – Dominik Jun 12 '13 at 16:20
  • I think that this problem is equivalent to defining complex conjugation. If you can, then you can define norm, which is always real; hence you can define $\mathbb{R}$. And if you can't, then how do you distinguish $\mathbb{C}$ from its image under conjugation? – vadim123 Jun 12 '13 at 16:55
  • By the axiom of choice, there is an isomorphism of $(\Bbb C,0,1,+,\times,\exp)$ that does not fix the reals. – GEdgar Jun 12 '13 at 17:14
  • @GEdgar: Could you be a bit more explicit? – Dominik Jun 12 '13 at 17:16
  • @Dominik: see this answer http://math.stackexchange.com/a/412034/442 – GEdgar Jun 12 '13 at 17:27
  • That doesn't seem to include exp, as far as I can tell. – Harry Altman Jun 12 '13 at 17:28
  • You are right, it does not include exp. – GEdgar Jun 12 '13 at 17:32
  • Maybe one could argument in this way: Let $pi$ be a real transcendental number, there are only countably many numbers standing in a nontrivial exponential-polynomial relation with $pi$. Thus there is a $z$ which is exponentially independent of $\pi$ and we map $\pi$ to $z$. Now hopefully some compactness argument may allow us to extend this to a whole $\Bbb C_{exp}$ isomorphism. – Dominik Jun 12 '13 at 17:34
  • $x$ is real if and only if ${ \exp(i n x) : n \in \mathbb Z}$ is bounced. Now $\mathbb Z$ is definable, but is boundedness? – GEdgar Jun 12 '13 at 17:36

1 Answers1

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This is a major open problem in model theory. Recent attempts to study ${\mathbb C}_{\exp}$ have been centred around Zilber's pseudo-exponentiation, which is a nice structure and module some (very serious) algebraic conjectures coincides with ${\mathbb C}_{\exp}$. Zilber's pseudo-exponential field is quasiminimal (i.e. every definable set is countable or co-countable) and hence $\mathbb R$ is not definable there. Have a look at Marker's paper for an introduction.

Levon Haykazyan
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