3
$\begingroup$

I was introduced the Poincaré Disc model of hyperbolic geometry. The concept idea was clear but I had some questions about it that I could not figure out myself.

I understand that the geometric properties come from the metric. But it seems to me that we are studying it as a subset of the extended comlex plane, and we use the natural Euclidean coordinate system to calculate our formula. The professor told me that we are using the differential structure but not the Euclidean geometry to help us with the calculations. So in some sense we are studying hyperbolic objects (by considering a different metric) on the "euclidean structure".

I don't know if this question makes any sense, but is there a natural coordinate system in which every straight line (in the Euclidean sense) is itself a hyperbolic straight line? If not, why do we still need objects from Euclidean space to study non-euclidean geometry?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ The horosphere in hyperbolic three dimensional space can be turned into a model of Euclidean plane geometry. Would this mean we're studying Euclidean objects by considering a different metric on the hyperbolic structure? ;-) $\endgroup$ – egreg Jul 22 '16 at 17:04
  • $\begingroup$ Historically, there was doubt that other geometries were consistent. If you build a model of a non-Euclicean geometry on top of a model of Euclidean geometry. Then it must be as consistent as the Euclidean model that it was built upon. $\endgroup$ – steven gregory Jun 1 '18 at 21:49
3
$\begingroup$

Some comments in the form of an answer

  1. The Poincaré disk model is only a model of the hyperbolic geometry. There are other models as well. For example the Klein disk model uses euclidean straits falling within an open disk to model the hyperbolic straights. (if this answers your question.)

  2. We do not necessarily need euclidean objects if we want to study hyperbolic geometry. There is the synthetic treatment just like in the case of the euclidean geometry.

  3. We use euclidean models only because some are convinced that it is easier to understand the hyperbolic concepts based on certain artificial euclidean objects.

  4. The point is that there is a geometry (there are geometries...) in which the intuition is completely different from the euclidean one. To some extent -- in my opinion -- it is misleading to make the students believe that hyperbolic geometry can be understood based on strange formulas and enforced euclidean concepts.

  5. Perhaps most importantly: The usual (ruler and compass based) euclidean geometry is only a model of the real euclidean geometry -- we are used to it though.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your reply. It certainly makes me understand more about the topic, as I am new to the subject. I am not sure about what you mean by the last point though, do you mean that the geometry that Euclid and greek geometers did was just a model of Euclidean geometry as an abstract concept (which is obviously justified because it is the distance in the real-life sence)? $\endgroup$ – lEm Jul 22 '16 at 16:30
  • $\begingroup$ Yes, I do mean that the ancient Greek concept of Euclidean geometry is a model of the modern axiomatic concept of geometry. For example: if we lived in a hyperbolic world then we could create a model of the Euclidean geometry without unferstanding the Euclidean intuition. We could fool ourselves (hyp creatures) with formulas and hyp concepts. Only the greatest geniuses would really understand that the Euclidean world is another logically possibke world. $\endgroup$ – zoli Jul 22 '16 at 23:00
  • $\begingroup$ One thing which is lost in this answer is that the Euclidean model proves the existence of hyperbolic geometry, i.e. the existence of a geometry satisfying all of Euclid's axioms except for the parallel postulate. $\endgroup$ – Lee Mosher Nov 16 '16 at 3:40
  • $\begingroup$ @LeeMosher, yes you are right, I appreciate your remark. However, I tried to answer only the question(s). $\endgroup$ – zoli Nov 21 '16 at 14:01
5
$\begingroup$

The usefulness of the Poincaré model (or of Klein's) is that it shows that if Euclidean geometry is consistent, then also hyperbolic geometry is. This is because the hyperbolic axioms are true in the model, so a contradiction in hyperbolic geometry would yield a contradiction in Euclidean geometry.

It would be wrong to prove theorems in hyperbolic geometry using the Euclidean properties of the model, because we might be using facts that only pertain to the model. Just to make a simple example: an Euclidean segment included in the Poincaré model is generally not a segment in hyperbolic geometry, so the fact that three points of the model are “Euclidean aligned” does not mean they are as “hyperbolic points”.

On the other hand, the model can give us a better insight of what's happening in hyperbolic geometry, because we can draw “hyperbolic figures” in the model; for instance, we get a grasp of “hyperbolic parallels”: two lines (seen as orthogonal circles to the boundary circle or diameters) are parallel (in hyperbolic sense) if they have a common point in the boundary circle. Similarly if we use the half-plane variant model.

Another interesting exercise is to see where Saccheri went wrong when he concluded that “the hypothesis of the acute angle is absolutely false; because it is repugnant to the nature of straight lines”

PROPOSITIO XXXIII
Hypothesis anguli acuti est absolute falsa; quia repugnans naturae lineae rectae

enter image description here

Full scanned version

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your answer! After some thinking, I think I know why I got confused. I thought that cartesian coordinates and euclidean space are strongly and naturally related in the followinf sense: if we start with a real line, fix an origin O, then define the coordinate of a point $P= d(O,P)$. It looks natural to me because it is like a coordinate defined by metric. So basically I was making too much assumption, but in fact, cartesian coordinate is only one of the many ways of describing it. So when I study hyperbolic geometry I assume that a similar coordinates can be defined. $\endgroup$ – lEm Jul 23 '16 at 11:27
  • $\begingroup$ But of course it wouldn't work because the space should be curved. So any planar representation would be like a projection, and some information about distance and angle would be distorted. Anyway, thank you for clearing some ideas up, it helps me a lot ;) $\endgroup$ – lEm Jul 23 '16 at 11:29
0
$\begingroup$

The Euclidean metric on $\mathbb{R}^2$ has Euclidean geometry. It does not follow that no continuous metrics on $\mathbb{R}^2$ are hyperbolic. Some of you might be thinking Euclidean geometry is obviously true. If you say that by definition, the Euclidean metric on $\mathbb{R}^2$ is how distance is defined on $\mathbb{R}^2$, then you're right. According to that definition, that just means that if somebody defines a continuous hyperbolic metric of $\mathbb{R}^2$, they're lying about what the distance between any two points in $\mathbb{R}^2$ are. It doesn't mean that a continuous hyperbolic metric on $\mathbb{R}^2$ doesn't exist.

I think at a very young age, I think it was under the age of 5, I knew quite well about the existence of 3-dimensional spheres and figured out on my own that the surface of a sphere does not have Euclidean geometry and that's how I made sense of non-Euclidean geometry from a young age. As a result, I also thought of the concept of hyperbolic geometry on my own. I think I was in Grade 1 at that time. I think that's the reason I never got confused about it when I heard about that concept later.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.