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It is often said that he discovered non-Euclidean geometry. But in which sense?

I am reading the book 'geometry' by Brannan et al. They use the disk model as an example of hyperbolic geometry. Did Lobachevsky have a similar model?

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Yes, Lobachevsky was a discoverer of non-Euclidean (more precisely, hyperbolic) geometry. (It was independently discovered by few other people, but Lobachevsky's treatment was by far the most complete). More precisely, he just accepted as an axiom the negation of the "5-th postulate", and developed a comprehensive geometry based on these new axioms.

But he did not have any model. It was developed axiomatically, in the same way as Euclid. The role of the later models (Beltrami, Poincare, Klein) was to show rigorously that IF there is no contradiction in Euclidean geometry, THEN these is no contradiction in the Lobachevsky's one. So they are equally "true".

(The question whether there is a contradition in Euclidean geometry, or in the rest of mathematics, does not belong to mathematics itself: this is unprovable by the usual mathematical methods).

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    $\begingroup$ Parenthesis needs reference... The first-order theory of Euclidean geometry (equivalenly the first-order theory of the real numbers, real-closed fields) is o-minimal. That theory is much easier than the first-order theory of $\mathbb N$, where G\"odel's incompleteness theorem applies. $\endgroup$ – Gerald Edgar Sep 1 '17 at 18:24
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When developing his geometry, Lobachevsky worked exclusively in the (Lobachevskian) plane. It was the Italian mathematician Beltrami who first showed that the geometry of (part of) the Lobachevskian plane coincided with the geometry of a certain surface - namely the pseudosphere. Beltrami's work came some forty-two years after Lobachevsky first formulated and completed his ideas: 1868 vs 1826.

Beltrami's interpretation shows that to every statement of Lobachevskian geometry referring to part of the plane there corresponds an immediate fact about the intrinsic geometry of the psuedosphere.

However, not all of the Lobachevskian plane is realised on the pseudosphere, but only part of it. It was Klein, in 1870, who first gave an actual interpretation of Lobachevskian geometry on the whole plane and, more generally, his geometry in space. (See, for example, Klein's interpretation in the circle and the sphere.)

Source: MATHEMATICS Its Contents, Methods, and Meaning by Aleksandrov, Kolmogorov, and Lavrent'ev.

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  • $\begingroup$ Has it been later shown that Lobachevskian plane corresponds to intrinsic geometry of hyper and hypo pseudospheres as well? Both have an extra non-unity constant describing them, unity for Beltrami pseudosphere. $\endgroup$ – Narasimham Jan 4 '18 at 15:37
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Another important contributions of Lobachevsky was the development of hyperbolic trigonometry (from the axioms: unlike the modern treatments, he did not use any models since he did not have any). An interesting, and often overlooked, application of this development is the uniqueness of the Lobachevskian geometry:

Theorem. Suppose that $X_1, X_2$ are spaces satisfying Lobachevsky axioms with the same "curvature normalization": Each ideal triangle has area $c$ in both geometries. (You can take $c=\pi$ if you like, this corresponds to the curvature $-1$.) Then $X_1$ is isometric to $X_2$.

I can explain how it follows if you are interested.

Edit:

Proof. Pick a base point $o_1\in X_1$ and a reference ray $\rho_1$ emanating from $o$. This defines the "polar coordinates" on $X_1$, namely $P(r,\theta)$, where $r$ is the distance from $P$ to $o_1$ and $\theta\in [0,2\pi)$, the (oriented) angle between $o_1P$ and $\rho_1$. Now, do the same in the second space $X_2$: Pick a base-point $o_2$, a ray $\rho_2$, etc.

Then, define a map $f: X_1\to X_2$ by sending $P(r,\theta)\in X_1$ to $Q(r,\theta)\in X_2$. It is clearly a bijection, let us prove that it is an isometry. Take two points $A, B\in X_1$. Then their distance in $X_1$ is given by the hyperbolic cosine law (due to Lobachevsky: I did not check, but it is possible that Bolyai also proved it):
$$ cosh(|AB|)= cosh(|o_1A|)cosh(|o_1B|) - sinh(|o_1A|)sinh(|o_1B|) cos(\angle(Ao_1B)). $$
By the construction, the map $f$ preserves all the quantities on the right-hand side of this equation (the distance to the origin and the angle). Since the hyperbolic cosine formula also holds in $X_2$, it follows that $f$ is an isometry. qed

This proof uses the notion of an oriented angle which can be easily avoided by working in one half-plane in $X_i$ (defined by the unique line through $\rho_i$ $i=1,2$) at a time.

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