7

Hilbert's axioms provide a synthetic system for Euclidean geometry. Is it possible to do the same thing for the Taxicab plane? It would seem that one would only need to alter the axioms for congruence, since all the other properties are the same as in the Euclidean plane, and the congruence axioms are the ones that determine the metric properties of the plane. If so, how? Note that if we remove SAS and leave all the other axioms in place, Taxicab geometry becomes a model of the Hilbert axioms, but can we add some more axioms in place of it that make it the unique model? If so, which ones, and if not, why not?

The_Sympathizer
  • 19,651
  • The question seems to depend on what one uses as lines -- the same lines as those of the Euclidean plane, or the geodesics of the taxicab metric. Most sources seem to assume the former (although the latter seems more natural to me). Anyway, the Wikipedia article for taxicab geometry claims that https://en.wikipedia.org/wiki/Taxicab_geometry#Properties taxicab geometry is a model of all of the Hilbert axioms except for the SAS postulate (the last of the congruence axioms). This view also seems to be supported by the book by Millman and Parker, Geometry: A Metric Approach with Models. – Chill2Macht Jul 03 '17 at 15:40
  • So in terms of making it the unique model, one would only need to remove the SAS axiom (not any other congruence axioms). As for what to replace it with, I have no idea. Especially since the answer would depend seemingly on what we choose as our primitive notion of lines for the geometry, Euclidean lines, or the geodesics of the $L^1$ metric. EDIT: this page http://www.ams.org/samplings/feature-column/fcarc-taxi also claims that taxicab geometry satisfies all of the congruence axioms except for SAS, but again they also claim that lines should be the same thing as Euclidean lines. – Chill2Macht Jul 03 '17 at 15:42
  • 1
    @Chill2Macht: The difficulty is that the taxicab geodesics are not unique between two points -- in fact, there are infinitely many of them. So if we were to use them, even more of Hilbert's axioms would have to give way. FWIW, Euclidean lines ARE taxicab geodesics -- but essentially "curves" in some fashion, when they are measured by suitable measure of "taxicab arc length" as seen here: https://math.stackexchange.com/questions/12906/is-value-of-pi-4 to "fallaciously" measure "pi" as being "4". Also note the top answer, it is more directly relevant. (cont'd) – The_Sympathizer Jul 04 '17 at 09:01
  • 1
    (cont'd) The method there used to claim "$\sqrt{2} = 2$" is the way to measure the length of a Euclidean line with the taxicab metric and in taxicab metric legitimately assigns the diagonal of the square the length of 2 in that setting. It is easy to see from this example that the length of a line is the same as the length of any other geodesic, thus it must be a geodesic as well. – The_Sympathizer Jul 04 '17 at 09:03
  • That does make sense -- also I really appreciate you sharing the link of that question with me, because I actually had that exact same question and was debating whether or not to ask it -- now I know the answer and get to avoid the embarrassment of accidentally asking a duplicate of a question with 537 upvotes. Anyway that does make more sense -- restrict lines to be a subclass of geodesics, rather than make lines something completely unrelated. I figured that the geodesics weren't unique, but that the non-uniqueness didn't matter -- I guess not though. – Chill2Macht Jul 04 '17 at 09:52
  • If you want, you could post your above comments as an answer to my recent question here: https://math.stackexchange.com/questions/2345213/why-arent-the-lines-in-taxicab-geometry-the-geodesics-of-the-l1-metric -- I would upvote it and accept it. Even though personally I think that non-uniqueness of geodesics might be something possible to work around, it is a legitimate concern, especially since there are infinitely many geodesics between two points, as opposed to, say, two in elliptic geometry. – Chill2Macht Jul 04 '17 at 09:54

1 Answers1

2

Note:
This is not even remotely a definitive answer to this question, so please don't accept it. What follows is too long for a comment, but may help with understanding the problem. (Also, it may not help with understanding the problem, which is another reason to not accept it. In any case, it would be better to leave the question open, i.e. with no accepted answer, in order to attract a definitive answer which could then be accepted. I am also very interested/invested in this question too and would like it to one day receive a definitive answer, which this answer most certainly is not. Since accepting this answer might discourage receiving a definitive one, please do not accept it.)

BEGINNING OF UNINTERESTING INTRO

I assume you already know most of what will be in this introduction, I am just stating it so as to organize my thoughts and to write as cogent of an answer as possible.

  • A system of axioms is "consistent" if there exists at least one model satisfying the axioms.
  • A system of axioms is "complete" if there exists only one model (up to model isomorphism) satisfying the axioms.
  • What we know already is that the axioms for Euclidean geometry are complete.
  • What we also know is that the axioms for Euclidean geometry $-$ SAS are consistent, but not complete, because both taxicab and Euclidean geometry satisfy them.
  • Thus our goal is to find a complete set of axioms such that taxicab geometry is the unique (up to isomorphism) model. Since the axioms for Euclidean geometry $-$ SAS are consistent, and taxicab geometry satisfies them, we can assume that this hypothetical complete set of axioms will contain the axioms for Euclidean geometry as a subset.

One thing to consider first would be the following question:

Are there other geometries, besides taxicab geometry, which satisfy all of Hilbert's axioms for Euclidean geometry except for perhaps the SAS postulate?

In other words, how "close" is this consistent set of axioms close to being complete?

If we can characterize all, or at least most, of such geometries which satisfy all of the axioms of Euclidean geometry except SAS, we will have a better idea of how to add additional axioms to restrict the resulting geometry to being taxicab geometry.

This is essentially the analog of the approach taken historically first with the parallel postulate -- we ask if there are other models of geometry which satisfy the axioms of neutral geometry (Euclidean minus parallel postulate), we find one, hyperbolic geometry.

Then we ask how close to being "complete" is neutral geometry -- well, it turns out, fairly complete, since hyperbolic geometry is actually a model of neutral geometry plus the negation of the parallel postulate, and not only that, but upon further investigation it turned out to be the only model of neutral geometry plus the negation of the parallel postulate.

Since the parallel postulate is either true or false (assuming the law of excluded middle), this shows that there are only two models of neutral geometry up to isomorphism.

END OF UNINTERESTING INTRO

Since removing the SAS postulate from Hilbert's axioms does not prevent any model of the resulting geometry from still being isomorphic (as a model) to the set $\mathbb{R}^2$ (with some metric), as the example of taxicab geometry itself shows,

A good place to look would be $\mathbb{R}^2$ with various metrics besides the Euclidean metric.

In particular, we know that the Euclidean axioms with SAS are complete, that taxicab geometry satisfies the negation of SAS, so really our question reduces to:

We know that (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) is consistent, since taxicab geometry satisfies it -- how close is it to being complete?

In particular, if taxicab geometry were the only model of this axiom system (up to isomorphism), we would be done. But if it is not the only model, then we would want to characterize the entire class of models satisfying the axiom system [(Euclidean geometry) $-$ (SAS) $+$ (negation of SAS)] as thoroughly as possible, because we could then basically just add on as axioms whatever makes taxicab geometry unique among this class, guaranteeing a complete axiom system

Our question has become: what makes taxicab geometry unique among the class of models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) ?

Now the reason why this answer is not a complete answer to your question is because I don't have an exhaustive characterization of the class of all models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) -- in fact, I only have what seems to be a possible subset, but I don't have any rigorous proof that in fact (1) these are all models for this axiom system, and (2) these are the only models for this axiom system. In fact, I only conjecture that (1) is true:

Conjecture: ($\mathbb{R}^2, d_p$), with $d_p$ the metric induced by the $p$-norm, is a model for (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) for every $p \not=2$.

Basis for the conjecture: As far as I can tell, the SAS postulate seems to be directly connected with isotropy of norms, i.e. whether or not the norms are "rotationally invariant", or whether or not there are a set of preferred directions, how strongly the norm is determined by its isometry group, how many "rotations" said isometry group has, or whether or not the norm is induced from an inner product.

A $p$-norm on $\mathbb{R}^2$ is isotropic, has no preferred directions, is "rotationally invariant", is completely determined up to scalar-multiple by its isometry group, and is induced by an inner product, if and only if $p = 2$.

You mentioned in the comments how the geodesics are not unique for the $L^1$ metric -- in fact, it seems to be the case that the geodesics are unique (and correspond exactly to straight lines, rather than just having them as a subset) for a metric induced by an $L^p$ norm if and only if $p = 2$. See this question. So uniqueness of geodesics may also be related to isotropy somehow.

See (1)(2)(3)(4)(5)(6)(7)(8) -- not all of these are directly relevant, but basically my point is that it is widely believed (rightly or wrongly) that having an inner product is necessary "to be able to talk about angles in a sensible way", and that my conjecture is that the SAS postulate is the synthetic geometry axiom which says that "we can talk about angles in a sensible way".

In other words, the negation of SAS may be equivalent to "metric is not induced by an inner product". It may be the case that (Euclidean geometry) $-$ (SAS) also somehow implies "space if "flat" (i.e. isomorphic to $\mathbb{R}^2$ as set), and the metric is induced by a norm".

If that is so, then we would even have that the class of all models satisfying (Euclidean geometry) $-$ (SAS) $+$ (negation of SAS) is equivalent to "$\mathbb{R}^2$ with some norm besides the Euclidean norm". That would mean that the only additional axiom to add, besides (negation of SAS), would just be (norm is the $L^1$ norm) (although that raises the open question of how to phrase that in terms of synthetic geometry -- I don't know, I'm not even 100% certain how to phrase the existence of an inner product in terms of synthetic or metric geometry, much less a specific $p$-norm for $p \not=2$).

To summarize, there are several levels of uncertainty for me regarding this conjecture:

  • Are all of the proposed models ($\mathbb{R}^2$ with $p$-norm for $p \not = 2$) actually models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS)?

  • What is the entire class of models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS)?

  • Does every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) consist of $\mathbb{R}^2$ with some metric? Or do there exist models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) which are "not flat", i.e. not $\mathbb{R}^2$?

(At the very least, Hilbert's dimension axioms and second-order continuity schema should most likely ensure that any model is at the very least a 2-dimensional metrizable manifold, although I'm not even 100% certain of that. Still, I think we don't have to worry about things which look locally like $\mathbb{Q}^2$ or other oddities like that.)

  • Assuming that every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is "flat" (and metrizable), do the axioms of (Euclidean geometry) $-$ SAS $+$ (negation of SAS) guarantee us at the very least that the metric is translation-invariant and homogeneous, i.e. induced by a norm?

  • Assuming that every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with a norm-induced metric, does the norm have to be a $p$-norm?

Certainly, any norm which isn't the Euclidean norm isn't isotropic, not just $p$-norms for $p \not = 2$, which is why I only dare to conjecture that every $p$-norm for $p \not= 2$ satisfies (Euclidean geometry) $-$ SAS $+$ (negation of SAS), but that the converse is not true, i.e. that there may exist models for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) which are not $\mathbb{R}^2$ with a $p$-norm for $p \not=2$.

Still, the latter might not matter if you can devise a synthetic geometry axiom which reduces to "the norm-induced metric is induced by a $p$-norm" and then a third synthetic geometry axiom "$p = 1$".

I don't know if the above actually makes any sense as written -- it makes sense in my head, but that's not a very good criterion for communicability.

To conclude:

(Really bold) Conjecture: Any and every model of (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with some translation-invariant and homogeneous but non-isotropic metric.

I.e., any and every model for (Euclidean geometry) $-$ SAS $+$ (negation of SAS) is isomorphic to $\mathbb{R}^2$ with any norm-induced metric (not even necessarily $p$-norm induced) besides the Euclidean metric.

If really bold conjecture is true, then as I tried to explain above, then theoretically at most two more synthetic axioms may be needed to add on to (Euclidean geometry) $-$ SAS $+$ (negation of SAS) in order to get a complete axiom system describing taxicab geometry.

Community
  • 1
Chill2Macht
  • 20,920
  • 1
    This just struck me while reading your post. You asked how to specify that the norm should be $p = 1$ synthetically. What if you introduce axiom: "There exists a square which is also a circle"? (That is, EG - SAS + $\neg SAS$ + SqrCirc = Taxicab?) ("Is a circle" can be defined as saying that the lines from the center of the square to any point on the bounding lines are all congruent.) – The_Sympathizer Jul 05 '17 at 09:22
  • 1
    ADD: Although that doesn't seem to distinguish $p = 1$ from "$p = \infty$" or Chebyshev metric, but that at least makes it down to 2 if this works. Note that no Euclidean square is a circle. Last axiom must then be the tiebreaker between these two. – The_Sympathizer Jul 05 '17 at 09:29
  • 1
    (ADD 2: Or we should write it even better: "There exists a square circle." :) ) – The_Sympathizer Jul 05 '17 at 09:32
  • 1
    Though perhaps maybe defining "square" is tricky as if you've said "angle doesn't exist" although then again Hilbert's axioms do use angles as a primitive... but can you still say what constitutes a right angle so as to define a square without SAS? I haven't seen how Hilbert proves "All right angles are equal" from Euclid... What if we weaken to "there is a circle which is also an equilateral parallelogram", as the definition of a parallelogram does not require angles but only parallelism relations between the sides? – The_Sympathizer Jul 05 '17 at 09:37
  • @mike4ty4 Well I think angle exists, but it's not well-behaved basically. Like the definition of angle in terms of radians works, but SAS doesn't, so you could have two triangles with the same side lengths or different angles or something -- honestly I'm not sure, since there doesn't seem to be any consistent definition of "angle" outside of Euclidean spaces and Riemannian manifolds -- there doesn't seem to be a general definition for Finslier manifolds or general metric spaces, although I have found some papers which attempt to make definitions. – Chill2Macht Jul 05 '17 at 10:41
  • In particular there's the difficulty of distinguishing between "angles" (e.g. often triples of points, or a triple of a point and two incident line segments) and "angle measure" (the device/function we use to decide whether two angles are congruent) -- it often seems to be easier to define a notion of angle than to define a notion of angle measure, so to interpret SAS one needs both I think (or else one can't say whether the two angles are congruent) but one also needs definitions of triangles -- so does one use the lines or the geodesics for the triangles? It's highly ambiguous to me. – Chill2Macht Jul 05 '17 at 10:43
  • Also I read somewhere that the equivalence between the balls in the $L^1$ and $L^{\infty}$ norms hold only for dimension $d =2$, but for higher dimensions it's no longer a problem -- I forgot where I read it though, that's the problem – Chill2Macht Jul 05 '17 at 10:44
  • 1
    That is rather easy. For $L^\infty$ the unit sphere is $1 = \max { |x|, |y|, |z| }$. That is easily seen to be a cube -- the extreme points all must happen when one or more of x, y, and z are $\pm1$ (and they cannot be bigger than that). Driving them all that way gives 8 vertices, so a cube. For $L^1$, the unit sphere is $1 = |x| + |y| + |z|$. This is an octahedron: it is bound up by the 8 planes $1 = x + y + z$, $1 = x + y - z$, ... $1 = -x - y - z$ for each of the 8 choices of sign on $x$, $y$, $z$ (note how the number 8 is relevant in both, but in one counts vertices, the other, faces). – The_Sympathizer Jul 05 '17 at 11:23
  • 1
    more specifically the parts of those planes where that the applied signs convert the x, y, z to positives as in the original absolute values. 8 planes = 8 faces = octahedron != cube. :) Yeah there's probably more details to fill in but oh well :) – The_Sympathizer Jul 05 '17 at 11:25