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There are familiar analytic equiconsistency proofs for Euclidean and hyperbolic geometry.  Those proofs are so robustly geometric that it seems like they must have synthetic analogues.

Looking into the literature, though, I wonder if I am too optimistic about this.  The most common rigorous axiomatizations for synthetic Euclidean and hyperbolic geometry, so far as I can tell, are Hilbert's from his Foundations of Geometry, with two changes.  They omit the non-elementary axiom of continuity, and they add something to assure that circles will actually have points of intersections with lines and other circles that they cross.  The only difference, in these axiomatizations, between Euclidean and hyperbolic geometry is in the axiom of parallels. But this synthetic hyperbolic geometry is incapable of some constructions that we take pretty much for granted both in Euclidean geometry and also (more to my point) in analytic hyperbolic geometry.   Notably $n$-section of lines, see trisection-of-a-hyperbolic-line-segment.  Maybe the analytic equiconsistency proofs really do not transfer well to synthetic geometry.

I know Greenberg's discussion of axiomatic issues in Greenberg. But I do not have all his references at hand.  Judging from the ones I do have (including Hartshorne) it seems likely that  his discussion of equiconsistency for the two geometries (pp.213-214) refers to analytic presentations of geometry.

Can I find mutually interpretable axioms for synthetic Euclidean and hyperbolic geometry?

Edit: Erik Walsberg's comment about Tarski's axioms answers my title question, even though not in the way I had in mind when I wrote the text.  My text was ambiguous about "synthetic" methods, in just the way that Tarski What is elementary geometry? means when he says   "In colloquial language the term elementary geometry is used loosely [...with] no well determined meaning."   

Hilbert, Tarski, and Greenberg all show that the important logical distinction characterizing elementary methods is not between using or not using coordinates in some field.  It is between using or not using higher order notions like point-set continuity and limits.  First order algebraic considerations on fields (notably pythagorean fields) are already implicit in Euclid and  central to successful first-order elementary geometry.  

The work that led me to this question is about interpretation in first order logic, and not about compass (or horocompass) and straightedge or other such construction methods, and really not about avoiding coordinates.  So my title was true to my actual concern.  Some of my text concerning synthetic methods was less relevant (though those questions too intrigue me).  Walsberg's comment is really an answer both to the question as titled and to my actual concern, though he correctly saw I had another issue about methods also in mind. 

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  • $\begingroup$ I remember years ago discussing axiomatic hyperbolic geometric with Daniel Allcock. If you run into him, you could ask him about it, but I don't remember the subject appearing in any of his publications. $\endgroup$
    – Ben McKay
    Nov 28, 2021 at 9:51
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    $\begingroup$ In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. The Tarski axioms for Euclidean geometry are pretty well known (they are on wikipedia), and of course you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,\times,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof. $\endgroup$ Nov 29, 2021 at 20:41
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    $\begingroup$ But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane. You should be able to do this by fixing a line $L$ in the hyperbolic plane $\mathbb{H}$, then $L^2 \subseteq \mathbb{H}^2$ should be a copy of the Eucliean plane. This gives two interpretations and I think it's pretty clear that this is a bi-interpretation. Of course there are things to check here. $\endgroup$ Nov 29, 2021 at 20:46
  • $\begingroup$ @ErikWalsberg Yes, you can do something like this in the way that Greenberg cites, from A. Ramsey and R. D. Richtmyer, using a grid of "equidistant curves (in the hyperbolic sense of distance)" to put real coordinates on the hyperbolic plane, and then defining a Euclidean geometry treating those equidistant curves in the hyperbolic plane as a grid of (Euclidean) parallel lines. But I do not know enough about axiomatizing equidistant curves in hyperbolic geometry to know if this works well in the synthetic setting, $\endgroup$ Nov 29, 2021 at 20:54
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    $\begingroup$ Don't know if this is useful, but $PU(2,R)$ is the group of elliptic isometries when $R = \mathbb C$, the group of Euclidean isometries when $R = \mathbb R[\varepsilon]/(\varepsilon^2)$, and the group of hyperbolic isometries when $R = \mathbb R \oplus \mathbb R$. This fact might help you. Godspeed. $\endgroup$
    – wlad
    Nov 30, 2021 at 17:01

2 Answers 2

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Here is an expanded version of my previous comments. There are a lot of things to check here and I haven't.

In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. Tarski works in a system where the domain is $\mathbb{R}^2$ and you have two relations, a ternary betweenness relation and an equidistance relation $E(x,y;x',y')$ which says that the line segments $\overline{xy}$ and $\overline{x',y'}$ are congruent. In this system all the axioms are phrased in terms of points, not in terms of both points and lines like in Euclid. Lines emerge as definable sets.

The Tarski axioms for Euclidean geometry are on Wikipedia, and you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,×,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof.

But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane.

Edit: I think that my initial attempt at the Hyperbolic model of the Euclidean plane fails, because equidistance isn't definable. Following a suggestion of @ColinMcLarty, I will describe a different model that I think works. This model is from Greenberg's book "Euclidean and Non-euclidean geometries".

Let $\mathbb{H}$ be the Hyperbolic plane. In this model the points are just the points in $\mathbb{H}$ and the lines all all lines in $\mathbb{H}$ through the origin together with the curves in $\mathbb{H}$ that are equidistant from a line through the origin.

To get a model of Euclidean geometry we need a betweenness and equidistance relationship, and these relationships need to be definable in $\mathbb{H}$. The betweenness relationship is easy, you just need to observe that the lines form a uniformly definable family of sets. Equidistance is more complicated.

Greenberg describes a map $\rho : \mathbb{H} \to \mathbb{R}^2$ and says that the equidistance relation on $\mathbb{H}$ is the pull-back of the equidistance relation on $\mathbb{R}^2$ by $\rho$. We can put polar coordinates on $\mathbb{H}$ in the same way as on the Euclidean plane. We fix a ray $\ell$ through the origin and let the polar coordinates of $p \in \mathbb{H}$ be $(r,\theta)$ where $r$ is the Hyperbolic distance from the origin to $p$ and $\theta$ is the angle that $p$ makes with $\ell$. If $p \in \mathbb{H}$ is the point with polar coordinates $(r,\theta)$ then $\rho(p)$ is the point $$ \rho(p) = (\sinh r \sin \theta, \sinh r \cos \theta) = \sinh r (\cos \theta, \sin \theta). $$ So $\rho(p)$ is the point in $\mathbb{R}^2$ whose (euclidean) polar coordinates are $(\sinh r, \theta)$.

Now let's suppose that $\mathbb{H}$ is the Poincare disc model. I think it is enough to show that $\rho$ is semialgebraic, i.e. definable in $(\mathbb{R},+,\times,<)$. The Hyperbolic distance $r$ of $p$ from the origin is not a semialgebraic function of $p$, but it is the arcsinh of a semialgebraic functions, see Wikipedia. (On Wikipedia there is a factor of $2$ in from the arcsinh, but that is just a scaling factor, so we can drop it). I think it should be pretty clear that $\rho$ is semialgebraic.

Once we know that $\rho$ is semialgebraic we know that the equidistance relation on our Hyperbolic model of Euclidean geometry is semialgebraic. This should imply that it is definable in Hyperbolic geometry, but one would need extra arguments to see it directly.

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  • $\begingroup$ How do we define equidistance in this hyperbolic model of the Euclidean plane? For a Euclidean metric, $E(a,b,c,d)$ should be equivalent to $$(a_x-b_x)^2+(a_y-b_y)^2=(c_x-d_x)^2+(c_y-d_y)^2$$ but I don’t know how to get the squaring in hyperbolic geometry: I don’t think the usual Euclidean constructions via similar triangles will work straightforwardly. $\endgroup$
    – user44143
    Dec 2, 2021 at 2:12
  • $\begingroup$ I think this is the only technically possible answer to my question. Greenberg well says "Geometrically, Tarski-elementary plane geometry certainly seems mysterious." It is very hard to prove anything from the axioms until you develop a lot of lemmas. But metamathematically they are the right axioms. They transparently agree with intuitive Euclidean (resp. hyperbolic) geometry. And because both axiom sets are complete (as first order theories) they are full analyses of the elementary content of those intuitions. $\endgroup$ Dec 2, 2021 at 3:49
  • $\begingroup$ @MattF. This is the Klein-Beltrami plane model. The analytic distance function is given for example in en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model $\endgroup$ Dec 2, 2021 at 3:52
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    $\begingroup$ @ColinMcLarty, I know that direction; my question is how to define the Euclidean equidistance on $L^2\subset \mathbb{H}^2$, as in the fourth paragraph of the answer. Another way of asking this is: how do you interpret multiplication in hyperbolic geometry? If Szmiliew indeed wrote this down, I don’t have an easy reference for it. $\endgroup$
    – user44143
    Dec 2, 2021 at 7:31
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    $\begingroup$ @MattF. For multiplication in hyperbolic geometry see Hilbert's arithmetic of ends. But Greenberg seems right: No one put this into the form of a hyperbolic plane model of the Euclidean plane until 1995 by A. Ramsey and R. D. Richtmyer, and even that had to be explicated in Greenberg's book Euclidean and Non-Euclidean Geometries (2007) pp. 514–515. Hartshorne's book is a good source on the arithmetic of ends, and for a fuller account see "horocycles" in Borsuk and Szmielew Foundations of Geometry: Euclidean, Bolyal-Lobachevskian, Projective. $\endgroup$ Dec 3, 2021 at 1:10
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One should not forget Arthur Cayley's claim "And thus projective geometry is all geometry".

The treatise of Oswald Veblen and John Wesley Young, "projective geometry", follows that path. The first volume is dedicated to synthetic projective geometry. The second one builds both euclidean and non-euclidean geometry in that context, the first one by choosing a line (to be the line at infinity) and an involution on that line (to be the orthogonal relation on the set of directions), the second one by choosing a conic (to be the ideal points of the absolute circle at infinity). Then Veblen studies the group of transformations leaving invariant these datas, by pure synthetic methods. This is Klein original idea. Both geometries are consistent because they are derived from projective geometry, and Veblen provides a set of axioms for both.

Nowadays, projective geometry is often built from affine geometry (the set of lines through a point), and that closes the loop if really you want to go from euclidean to non-euclidean geometry. But really, projective geometry should not be set aside, at least that's what the ancients say.

From the introduction of the second volume of Veblen and Young,

The ideal of such books should be not merely to prove every theorem rigorously but to prove it in such a fashion as to show in which spaces it is true and to which geometries it belongs.

The treatise is 990+ pages long and certainly achieves its goal. From the synthetic viewpoint, modern treatises such as Hartshorne and Greenberg certainly pale in comparison to the works of geometers from the beginning of the XX century.

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  • $\begingroup$ Sometimes geometries are interpreted in projective geometry using recursively constructed nets, so the results in Veblen’s treatise might not be first-order. $\endgroup$
    – user44143
    Dec 2, 2021 at 13:29
  • $\begingroup$ @Matt Indeed there are many results in the book of Veblen that are not of first order. For example, results on polygons, since the notion of a polygon with an arbitrary number of sides is not first order. $\endgroup$
    – coudy
    Dec 2, 2021 at 14:45
  • $\begingroup$ This book is a Landmark in the history of mathematics, for both geometry and logic and axiomatics. And don't forget co-author Young. I hope Greenberg and Hartshorne never meant to rival this book for what it does. Tarski's work is also a landmark, on the more purely logical side, for making complete (and therefore decidable) axiomatic first order theories a serious, productive subject. Reading Veblen and Young was a huge step in my own education but Tarski's axioms seem to me better for the purpose of this question. $\endgroup$ Dec 2, 2021 at 16:42

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