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For algebraic curves over a nice enough field $k$, we have a notion of what it means to be hyperbolic: If $\overline{C}$ is a smooth projective curve of genus $g$ and $P_1,\dots,P_n$ are closed points, then the open curve $C=\overline{C}\setminus\{P_1,\dots,P_n\}$ is hyperbolic if $2-2g-\sum \deg P_i<0$.

The general function-field/number-field analogy suggests the following question:

Is there a notion of hyperbolicity for rings of $S$-integers in number fields (possibly along the lines of some inequality appearing in arithmetic intersection theory)?

If there is such a notion of hyperbolicity, are most $S$-integer rings hyperbolic? Is it also true that number rings with trivial or finite étale fundamental group (as in this MO-question) are not hyperbolic?

Here is some of the background which motivated the question: In the function field case, I think I have convinced myself that the slightly stronger inequality $2-2g-n<-1$ implies that there will be cuspidal cohomology for the group $SL_2(k[C])$ (in the sense that the quotient of the product of Bruhat-Tits trees for the points $P_1,\dots,P_n$ is rationally non-contractible). I am trying to understand if similar statements can be made in the number-field situation. Therefore, the ideal answer to the question would be some numerical inequality which relates to Harder's Gauß-Bonnet formula for cohomology of arithmetic groups, implying that hyperbolicity forces non-trivial rational cohomology for the arithmetic group $SL_2(\mathcal{O}_{K,S})$.

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    $\begingroup$ There is an analogy between number fields and 3-manifolds proposed by Barry Mazur where primes in the ring of integers correspond to prime knots, cf. math.harvard.edu/~mazur/papers/alexander_polynomial.pdf The field of rational numbers is supposed to correspond to the 3-sphere. So presumably the field of rational numbers is not hyperbolic. $\endgroup$ Commented Dec 8, 2016 at 10:51
  • $\begingroup$ Do we ask that an extension (cover) of a hyperbolic number ring should be hyperbolic? $\endgroup$ Commented Dec 10, 2016 at 19:10

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My intuition is that there will not be a precise definition of a hyperbolic number field. However, there there may be some number fields you can confidently say are hyperbolic.

Consider the following characterization of a hyperbolic complete curve in the function field case - a degree $d$ cover of $\mathbb P^1$ is hyperbolic if its discriminant $\Delta$ is greater than $q^{2d}$.

This is equivalent to $g>1$ by the Riemann-Hurwitz formula $2-2g = 2d - \log_q (\Delta)$.

So it seems like a number field of degree $d$ should be hyperbolic if the discriminant is greater than $C^{d}$ for some constant $C$. If you want to prove arithmetic analogues of a certain consequence of hyperbolicity, you could try to prove it as a consequence of large discriminant. Is your desired result true for quadratic fields of large discriminant, say?

However for a full definition of hyperbolicity you would need to find a single value of $C$ that is the cutoff for all properties of hyperbolic curves. I don't think a plausible value of $C$ is known - I don't know if one is expected to exist.

One property $C$ should have is that every number field of discriminant $<C^g$, being parabolic, should have finite etale fundamental group. One knows this is true for $C = 22.3$ and false for $C=296.277$ but not a precise optimal value of $C$.

The converse (finite fundamental group implies not hyperbolic) may not actually be true - for many purposes, the etale fundamental group of a number ring is not a perfect analogue of the etale fundamental group of the function field. Instead, to model the etale fundamental group of a number ring, one can use the quotient of the etale fundamental group of a function field by the Frobenius element of a point at $\infty$. The reason is that the decomposition groups at infinite places over number fields are small or even trivial, and so one has to force the same thing to hold in the number field world to get a proper analogue. I'm not sure whether, for function fields, this group is always infinite for hyperbolic curves, but it is certainly true that the pro$-p$ quotients can be finite for hyperbolic curves, whereas even the pro$-p$ quotients of the geometric etale fundamental group are infinite.

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  • $\begingroup$ Thanks, that's an interesting suggestion. It is in fact true for imaginary quadratic fields that there are only finitely many where the rational cohomology of $SL_2(\mathcal{O}_{K,S})$ is trivial. $\endgroup$ Commented Dec 9, 2016 at 9:45
  • $\begingroup$ @MatthiasWendt Cool. I am becoming increasingly convinced that one is supposed to investigate this via a Gauss-Bonnet type formula (or trace formula, or Kusnetsov formula). These should express the dimension of the cuspidal cohomology as a main term (the volume) plus secondary contributions - one from the cusps, one from each element of finite order in $SL_2(\mathcal O_{K})$, and one from the trivial representation. One expresses these all in terms of $\Delta_K$ and then tries to show that, for $\Delta_K$ sufficiently large, the main term dominates the error terms. $\endgroup$
    – Will Sawin
    Commented Dec 9, 2016 at 10:33
  • $\begingroup$ Is there a reference for the sentence "One property 𝐶 should have is that every number field of discriminant <𝐶𝑔, being parabolic, should have finite etale fundamental group. One knows this is true for 𝐶=22.3 and false for 𝐶=296.277 but not a precise optimal value of 𝐶."? Thanks! $\endgroup$
    – dummy
    Commented Mar 12 at 23:42
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    $\begingroup$ @dummy The results I was mentioning are summarized at en.wikipedia.org/wiki/… For the $296.277$ statement, the claim is that $\mathbb Q(\sqrt{-3 \cdot 5 \cdot 7 \cdot 11 \cdot 19})$ has infinite fundamental group. In the other direction, the theorem is that there are only finitely many fields with discriminant less than $(22.3-\epsilon)^n$, which implies that all such fields have finite fundamental group. All of this is discussed in Tours le corps de classes et estimation de discriminants by Martinet. $\endgroup$
    – Will Sawin
    Commented Mar 13 at 1:23

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