# Hyperbolic geometry when the curvature is constant and negative but not -1

Help I am getting completely confused

Hyperbolic geometry is the geometry of surfaces of a constant negative Gaussian curvature, in most formula's it is almost assumed this constant negative curvature is $-1$.

But what when the constant negative curvature is $-2$ or $- \frac{1}{3}$ ?

For example is the area of a triangle the defect divided or multiplied by the (absolute value of the) Gaussian curvature.

Wikipedia ( https://en.wikipedia.org/wiki/Hyperbolic_triangle )says you have to divide the defect by the curvature.

$(\pi-A-B-C) R^2.$ where $R = \frac{1}{\sqrt{-K}}$

Sommerville "The elements of non euclidean geometry" page 81 gives $k^2 ( \pi -A -B -C)$

and I fear that when i reach for another book i will even get another formula.

Who is right and can you give a canonical proof?

• There are various elementary proofs, but the result follows easily from the Gauss Bonnet theorem for surfaces. – Ted Shifrin Feb 22 '15 at 14:05

The quick way to get the correct answer is dimensional analysis: to get an area from a dimensionless angle defect you need to multiply by something with units length$^2$, so Wikipedia's formula $A = (\pi - \sum\theta)R^2$ is correct.

Sommerville is a very old text and uses $k$ for the negative inverse curvature - see e.g. page 75 where it states "The formulae of hyperbolic trigonometry become of those of euclidean plane geometry as $k \to \infty$."

The proof is going to depend on how you are constructing things - from the Riemannian perspective you just calculate how curvatures and areas behave under multiplication of the metric by a constant: covariant derivatives are unchanged, so Ricci curvature is unchanged, so scalar curvature scales as the inverse metric. Area scales as the metric. Thus area is inversely proportional to curvature under scaling.

Spherical geometry

The Gaussian curvature of a real sphere of radius $R$ is $$K=\frac{1}{R^2},$$ or $$R^2=\frac{1}{K}$$ where $K$ is a positive number. So the angle excess of a spherical triangle, expressed by the Gaussian curvature, is $$\Delta_{Spherical}=\frac{1}{K}(\alpha+\beta+\gamma-\pi)$$

Hyperbolic geometry

In hyperbolic geometry the so called space constant, $k$, is defined as follows. Let $A_1B_1$ and $A_2B_2$ two horocyclic segments lying on two parallel horocycles and lying between the same two radii of the horocycle. Let the hyperbolic distance between these horocycles be $x$. It was discovered by Lobatchevsky and Bolyai that the ratio of the two segments is an exponential function of $x$. Bolyai introduced the constant at stake as $$\frac{A_1B_1}{A_2B_2}=e^{\frac{x}{k}}.$$ Lobatchevsky chose $k=1$ for his investigations.

The angle defect of a triangle in hyperbolic geometry, expressed by the space constant, is $$\Delta_{Hyperbolic}=k^2(\pi-\alpha-\beta-\gamma).$$ The pseudosphere

The pseudosphere is a sphere of imaginary radius, $iR$. The Gaussian curvature is then $$K=-\frac{1}{R^2},$$ a negative number. One has then $$R=\frac{1}{\sqrt{-K}}.$$ The angle excess of a triangle on the pseudosphere is $$\Delta_{PSphere}=(iR)^2(\alpha+\beta+\gamma-\pi)=R^2(\pi-\alpha-\beta-\gamma).$$

The pseudosphere as a model of hyperbolic geometry

The pseudosphere is known to be a model of hyperbolic geometry. Comparing the angle excess formulae we find that $$k^2=R^2 \text{ and } R=\frac{1}{\sqrt{-K}}.$$