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By a Tarski plane (resp. plane) I understand a mathematical structure $(X,B,\equiv)$ consisting of a set $X$, a ternary betweenness relation $B\subseteq X^3$ and the 4-ary congruence relation ${\equiv}\subseteq X^2\times X^2$ satisfying the Tarski's axioms minus the Euclides parallel postulate (in which the Axiom schema of continuity is replaced by two axioms of segment-circle continuity and circle-circle continuity).

Since axioms of a Tarski plane include the axioms of a plane, each Tarski plane is a plane. In a plane one can produce standard geometric constructions with a compass and ruler.

If a plane satisfies the Euclid parallel postulate, then it is called a Euclidean plane.

For distinct points $o,e\in X$ of a plane $(X,B,\equiv)$, the set $$L(o,e)=\{ x\in X:Bxoe\vee Boxe\vee Boex \}$$is called the line containing the points $o,e$.

The line $L(o,e)$ carries a unique structure of a linearly ordered commutative group $(L(o,e),+,\le)$ such that

$\bullet$ $o$ is zero of $L(o,e)$ and $o<e$,

$\bullet$ for any $x,y\in L(o,e)$ we have $x\le y$ iff $Bxyo\vee Bxoy\vee Boxy$,

$\bullet$ for any $x,y,z\in L(o,e)$ with $o\le x$ and $o\le y$ we have $z=x+y$ if and only if $Boxz$ and $xz\equiv oy$.

If the plane is Euclidean (and Tarski), then the line $L(o,e)$ carries a structure of a (real closed) ordered field such that $e$ is the unit of $L(o,e)$ and for any element $y\in L(o,e)$ with $o\le y$ there exists $x\in L(o,e)$ such that $x^2=y$.

Question. What is an algeraic structure of a line $L(o,e)$ in a (Tarski) plane? It should be something more general than the structure of a (real-closed) ordered field, closed under taking square roots.

Or more precisely:

Problem. Characterize ordered groups, which are isomorphic to the lines in (Tarski) planes.

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    $\begingroup$ I do not fathom the downvotes on this interesting question. $\endgroup$
    – Joel David Hamkins
    Commented Oct 21, 2022 at 14:51
  • $\begingroup$ Looking at the question from non-semantic POV, I doubt the premise that "curvature" should somehow affect algebraic structure of lines; geodesics in hyperbolic plane (or, fwiw, any reasonable Alexandrov plane-like space) are usual lines, curvature only comes into play when you consider triangles. That structure can be possibly encoded algebraically as an operation on set of lines, which are otherwise more or less indistinguishable. $\endgroup$
    – Denis T
    Commented Oct 21, 2022 at 15:33
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    $\begingroup$ @DenisT I had in mind that the line in an non-Euclidean absolute plane should be closed under certain (probably, hyperbolic) functions (corresponding to constructions with compass and ruler), which include some parameters which can be interpreted as the curvature of the absolute plane. A line in an elementary absolute plane can be countable, in fact, a minimal model of an elementary absolute plane is countable. Observe that instead of a full continuity axiom we have two axioms postulating the existence of points in the intersections of circles and lines. $\endgroup$
    – Taras Banakh
    Commented Oct 21, 2022 at 15:36
  • $\begingroup$ Aren’t these simply the euclidean fields? Can’t you construct a hyperbolic plane over any such field? (You can certainly construct a euclidean plane.) My memory may be rusty, but I believe such questions are answered in the Schwabhäuser–Szmielew–Tarski monograph. $\endgroup$
    – Emil Jeřábek
    Commented Oct 22, 2022 at 7:46
  • $\begingroup$ @EmilJeřábek Indeed, lines in Euclidean planes are isomorphic to Euclidean fields. But in hyperbolic planes this is not true anymore because the ratio (the longest length)/(side) in a right isosceles triangle in the hyperbolic plane does not belong to the smallest Euclidean field (I hope so). For a hyperbolic plane the lines should carry a structure of an ordered field, closed under some functions of many variables (corresponding to constructions with compass and ruler). So, my question is to determine those functions and the smallest number of such functions. $\endgroup$
    – Taras Banakh
    Commented Oct 22, 2022 at 8:10

1 Answer 1

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This question is addressed by W. Schwabhäuser on p. 156 of his paper Metamathematical methods in foundations of geometry. Logic, Methodology and Philosophy of Science (Proc. 1964 Internat. Congr.) North-Holland, Amsterdam, 1965, pp. 152–165. If the Tarski plane (as you have defined it) is hyperbolic (i.e. not Euclidean), then by a result of W. Szmielew (A new analytic approach to hyperbolic geometry. Fund. Math. 50 (1961/62), pp. 129–158) the plane is isomorphic to a Klein space over a Euclidean ordered field if and only if the plane satisfies Hilbert's hyperbolic axiom of parallels. If it does not satisfy Hilbert's axiom then the algebraic characterization is substantially more complicated and can be obtained from Pejas's classification of Hilbert planes, as discussed in W. Pejas, Die Modelle des Hilbertschen Axiomensystems der absoluten Geometrie, Math. Ann. 143 (1961), 212–235 and in F. Bachmann Zur Parallelenfrage Abh. Math. Sem. Univ. Hamburg 27 (1964), 173–192. Unfortunately, I no longer have these papers and it's been so long since I've studied the matter, I can't be of help with the details.

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  • $\begingroup$ Thank you for the answer and the references. $\endgroup$
    – Taras Banakh
    Commented Oct 23, 2022 at 8:41
  • $\begingroup$ So, at least for the Klein plane $K$ over a Euclidean field $E$ each line in $K$ is isomorphic (as an ordered group) to the additive group $\ln[E]:=\{t\in \mathbb R:e^t\in E\}$? And this is a desirable characterization of the lines in hyperbolic Tarski planes that satify the Hilbert hyperbolic axiom. Right? $\endgroup$
    – Taras Banakh
    Commented Oct 23, 2022 at 15:47
  • $\begingroup$ Are you assuming K satisfies the Archimedean axiom? Your definition of ln[E] seems to suggest this. $\endgroup$
    – Philip Ehrlich
    Commented Oct 24, 2022 at 0:26
  • $\begingroup$ Yes, for defining $\ln[E]$ in the real line I indeed need the Archimedean axiom. In the non-Archimedean case we can consider the Lobachevskyi function $L(\alpha)$ assigning to each acute angle $\alpha$ in the length of the unique bounded side a right idean triangle with an an angle $\alpha$ and two infinite sides. According to Bonola's consruction such a triangle can be constructed using the ruler and compass. Then each line in $K$ is isometrically isomorphic to the ordered group $\{L(\alpha):\cos(\alpha)\in E\}$, where we identify angles in $K$ with suitable angles in the completion of $E^2$. $\endgroup$
    – Taras Banakh
    Commented Oct 24, 2022 at 4:13
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    $\begingroup$ OK, yes, for the Archimedean case, that seems right. By the way, if I remember correctly, by a result of M. J. Greenberg (On J. Bolyai's parallel construction. J. Geom. 12 (1979), no. 1, pp. 45–64), Archimedean, non-Euclidean Tarski planes always satisfy Hilbert's hyperbolic parallel axiom. $\endgroup$
    – Philip Ehrlich
    Commented Oct 24, 2022 at 12:25

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