4

In this article, the following is said:

On the other hand, there is no absolute standard of length in Euclidean geometry. If one wants to construct a segment of any particular length, one requires a ruler of some type. If you constructed a segment and somewhat arbitrarily defined it as your "unit length," then called a friend, you would be unable to describe to your friend how to construct a segment congruent to yours with straightedge and compass without making some reference to a ruler.

Big Idea: In hyperbolic geometry, it is possible to define an absolute standard of length, as well as an absolute standard of angle measure

I understand there is a prescription to construct angles without a pre graded scale but how is it possible to construct a length without a pre graded scale? An explicit presentation of how this could be done would be nice.

Secondly, Could I measure this absolute unit in hyperbolic geometry using the same ruler I use to measure lengths in Euclidean geometry? What would be my ruler's reading? (Suppose I fix K in the equation relating angular excess and area of triangle)

$$ \epsilon(\Delta) = K A(\Delta)$$

tryst with freedom
  • 11,538
  • 2
    It's a sloppily written article. The "big idea" is that in hyperbolic geometry similar triangles are congruent. – Moishe Kohan Nov 28 '21 at 14:47
  • 1
    I do not understand your second question: How are you planning to "import" a ruler from one geometry to another? Of course, you can do so if you already have a fixed notion of metric length in Euclidean geometry, but how would you do it otherwise? – Moishe Kohan Nov 28 '21 at 20:17
  • Well I mean, I give you the physical ruler that we can buy in stores, now somehow you take it to the hyperbolic world and measure a length in there using it. @MoisheKohan – tryst with freedom Nov 29 '21 at 02:43
  • 1
    "Somehow take it" is simply impossible. One world is the physical world you live in, the other is the world of mathematical abstractions. I suspect, what you mean to ask is: Is there a map $f$ from the Euclidean plane $E^2$ into the hyperbolic plane $H^2$ such that for some segment of positive length $s$ in $E^2$, whenever $s'$ is congruent to $s$, $f(s')$ is congruent to $f(s)$. This question will have negative answer. – Moishe Kohan Nov 29 '21 at 17:16

1 Answers1

2

I will answer the first question, since the 2nd question is unclear.

Now, to the question itself. You should read the continuation of the quoted passage, as it clarifies what is said in your quote:

This fact is a direct result of

Big Idea: In hyperbolic geometry, if two triangles are similar, then they are congruent.

For instance, if I want to communicate to a friend what segment I am planning to use as a unit, I can say:

Consider an equilateral triangle in the hyperbolic plane with angles equal $30^\circ$. Its edges are congruent to each other and will serve as the unit of measurement from now on.

Or, I can say: consider an isosceles triangle where one of the angles if the right angle and the other two have the angle $22.5^\circ$.

As long as your friend resides in the same hyperbolic plane as you do (same curvature), they will understand what you mean unambiguously and, thus, will use the same unit of measurement. That's all what the quoted passage is saying.

Edit. Both triangles I have mentioned are constructible in hyperbolic plane (using the hyperbolic compass and ruler). One uses the following criterion, due to Mordukhai-Boltovskoy:

  1. A number $x$ is constructible, as a length, in hyperbolic geometry if and only if $\cosh(x)$ is constructible in Euclidean geometry.
  2. A number $\alpha$ is constructible as an angle-measure in hyperbolic geometry if and only if it is constructible as an angle-measure in Euclidean geometry.

See for instance this answer. Or, you can find a self-contained (and rather long) proof of this theorem in the Bachelor's Thesis

Ruben de Vries, Compass and Straightedge Constructions in the Hyperbolic Plane, Utrecht University, 2021.

In fact, you do not need the full power of this result: In the proof of (1) they first establish the result for right-angled triangles, which clearly suffices in both (equilateral and right-angled) examples.

  1. In the case of the equilateral triangle I mentioned, by the dual hyperbolic law of cosines its side-length $a$ satisfies: $$ \cosh (a)= (\cos(\pi/6) + \cos^2(\pi/6) )/\sin^2(\pi/6) = 2\sqrt{3} + 6 $$ and the right-hand-side is Euclidean-constructible.

  2. The hypotenuse $c$ the right-angled triangle satisfies $$ \cosh(c)= 3+2\sqrt{2} $$ which is also Euclidean-constructible.

More generally, if $\alpha, \beta, \gamma$ are angles constructible in hyperbolic, equivalently, in Euclidean, geometry, then the side-lengths of the corresponding hyperbolic triangle are constructible in hyperbolic geometry (again, by the dual hyperbolic cosine law). Thus, in this situation, the entire triangle is constructible in hyperbolic plane.

Moishe Kohan
  • 97,719
  • 1
    @Blue: Oh, right... – Moishe Kohan Nov 28 '21 at 23:27
  • "Consider an equilateral triangle in the hyperbolic plane with angles equal $30^\circ$∘ Its edges are congruent to each other and will serve as the unit of measurement from now on." This raises the question: How can I duplicate your proposed unit for my own use? Off the top of my head, I can't recall an in-universe construction of an equilateral triangle with a given angle. The only thing that comes to mind is perhaps leveraging the angle-of-parallelism configuration, but even that requires dropping a perpendicular from an ideal point, which cheats a bit. Am I forgetting something obvious? – Blue Nov 28 '21 at 23:36
  • @Blue: I was just explaining what was written in the article. As I said in the first comment, it's sloppily written, confusing/conflating constructibility and existence of a universal measuring stick. – Moishe Kohan Nov 29 '21 at 00:01
  • @Blue, exactly my question after reading. How would the second person know how to repliace the equilateral triangle? – tryst with freedom Nov 29 '21 at 02:42
  • @User688539: "How would the second person know how to repliace the equilateral triangle?" This may be worth posting a separate question. That said, I'm reminded by Wikipedia's "angle of parallelism" entry of Bolyai's construction of the AoP for a given distance; constructing the distance for a given angle should also be possible by "finite" means. This would be enough to "encode" a ruler by an angle (we often say that "the" fundamental unit is the one with AoP $\pi/4$), but I'd like to see an equilateral triangle construction, myself. :) – Blue Nov 29 '21 at 02:55
  • 1
    Interesting issue raised by @Blue : How do I, a hyperbolic-space dweller, actually produce any triangle with the three angles given? Easy enough to construct a triangle from three given lengths, but off the top of my head, I don’t see how to do the given-angle construction. – Lubin Nov 29 '21 at 04:06
  • @Blue: See the edit. – Moishe Kohan Nov 29 '21 at 09:03
  • @Lubin: yes, as long as the angles are constructible. – Moishe Kohan Nov 29 '21 at 14:32
  • I guess, @MoisheKohan , that I was asking for a stricter definition of constructibility. I was hoping for a straightedge-and-compass construction of the triangle in question, using only a hyperbolic straightedge and compass. Can you specify one? – Lubin Nov 29 '21 at 23:26
  • @Lubin: Yes, it is a construction using hyperbolic compass and straightedge, so is in the strict sense. But the proof is indirect, via a reference to a theorem. – Moishe Kohan Nov 29 '21 at 23:29
  • Suppose I had a physical realization of the hyperbolic plane and I tried to find shortest path between two points by considering the path which string is taut, would it suddenly become like arc of circle rather than straight line? – tryst with freedom Nov 30 '21 at 03:08
  • @666User666 You mean of a hyperbolic circle? Then absolutely not. – Moishe Kohan Nov 30 '21 at 03:53