# Gauss Bonnet theorem validation with hyperbolic circles

How to verify the theorem in case of a hyperbolic circle radius $r$ for constant negative Gauss curvature $K=-1/a^2$ and constant geodesic curvature $k_g$ ... e.g., like here...

$$k_g=1/r \tag1$$

$$\int k_g ds + \int\int K dA = 2 \pi \tag2$$

(Geodesic polar coordinates) Perimeter and Area plugged in from :

$$Perimeter = 2 \pi a \sinh (r/a) ,\, Area= 4 \pi a^2 \sinh^2(r/2a) \tag3$$

as these reduce to $( 2\pi r, \pi r^2)$ when $a\rightarrow \infty$

$$\frac{\pi a \sinh (r/a) }{ r } -{ 4 K \pi a^2 \sinh^2 (r/2a)} = 2 \pi$$

does not tally in general. Also when $r\rightarrow 0, \pi + \pi (r/a)^2 \ne 2 \pi$

Clearly (1) is assumed intentionally wrong, but then what is correct ? It needs to be defined properly in the tangent plane or in the hyperbolic plane which is not known to me.

Since a direct calculation (like for case $K=+1/a^2$) appears elusive.. an indirect back -calculation is the only resort . For this case we have after some simplification:

$$\boxed{a \, k_g= \coth (r/a ), R_g=a \tanh (r/a)}\tag4$$

verified by plugging this into GB thm (1). When

$$r\rightarrow \infty,\,k_g \rightarrow \,(1/a),\, R_g \rightarrow \,a \tag5$$

$$r=0 ,k_{g} = \infty \tag6$$

as expected for a point circle.

(4) is a new result (for me) surprising and apparently anomalous because it is bounded. There is no way to check it directly or conceptually reinforcing it by other theorems or better known results in hyperbolic geometry.

And that is the motivation of this posting...

Show how theses hyperbolic circles are/can be placed geometrically in the hyperbolic models available (Poincaré half plane, Poincaré disc and Klein's )

• What value are you using for $k_g$, and how do you justify that value? – Lee Mosher Jul 31 '17 at 23:03
• I am unable to calculate here. For sphere $K=+1/a^2$ it is easy – Narasimham Aug 1 '17 at 7:04
• That's a source of error, then. You cannot just pick the formula $k_g=1/r$ out of a hat and hope that it is the correct formula; although it may be correct in the Euclidean plane, it is wrong in the hyperbolic plane. On the other hand, you could use the Gauss-Bonnet theorem to calculate a formula for $k_g$ (which should depend on $r$ and on $a$). – Lee Mosher Aug 1 '17 at 11:19
• I known this. I wish to be able to directly evaluate it. Just as in the case with $K=+1/a^2$ where $k_g$ is directly calculated and plugged into GB thm to verify it successfully, it is not possible to do so for $K=-1/a^2$. Right? – Narasimham Aug 1 '17 at 15:59

Fix $a > 0$. As you've found, in the hyperbolic plane with curvature $-\frac{1}{a^{2}}$, a circle of hyperbolic radius $r$ has geodesic curvature $$k_{g} = \tfrac{1}{a} \coth \tfrac{r}{a}. \tag{*}$$ The integral is $$\int k_{g}\, ds = k_{g} \times \text{perimeter} = 2\pi \cosh \tfrac{r}{a};$$ since the enclosed total Gaussian curvature is $$4\pi \sinh^{2}\tfrac{r}{2a} = 2\pi(\cosh \tfrac{r}{a} - 1),$$ the Gauss-Bonnet theorem yields the expected result.

Here are a couple of ways to see why (*) is true.

First, the geodesic curvature of a circle of intrinsic radius $r$ on a sphere of radius $a$ is $\frac{1}{a} \cot \frac{r}{a}$. On the principle that "corresponding" spherical and hyperbolic quantities are given by corresponding formulas with circular functions replaced by hyperbolic functions, we'd expect the geodesic curvature of a hyperbolic circle of radius $r$ to be $\frac{1}{a} \coth \frac{r}{a}$.

Second (and less magically), model the hyperbolic plane as the open unit disk in the Cartesian plane, equipped with the metric $$g = \frac{4a^{2}(dx^{2} + dy^{2})}{(1 - (x^{2} + y^{2}))^{2}}.$$ If $0 \leq R < 1$, the radial segment $(t, 0)$ with $0 \leq t \leq R$ has hyperbolic length $$r := \int_{0}^{R} \frac{2\, dt}{a^{2} - t^{2}} = a\log \frac{a + R}{a - R};$$ that is, the circle of hyperbolic radius $r$ centered at the origin has Euclidean radius $$R = \tanh \tfrac{ar}{2}, \tag{1}$$ and hyperbolic circumference $$\frac{2a(2\pi R)}{1 - R^{2}} = 2\pi a \sinh \tfrac{r}{a}.$$

To find the geodesic curvature of a circle $C$ of hyperbolic radius $r$, draw the circle of Euclidean radius $R$ centered at the origin, fix a point $p$ of this circle, and draw the hyperbolic line $\ell$ tangent to the circle at $p$. Because $\ell$ is the arc of a Euclidean circle crossing the unit circle at right angles, a bit of geometry shows $\ell$ is part of the Euclidean circle of center $(\frac{1 + R^{2}}{2R}, 0)$ and radius $R' := \frac{1 - R^{2}}{2R}$.

Let $d\theta$ be a small angle at the center of $C$, subtending an arc of hyperbolic length $$ds = \frac{2aR}{1 - R^{2}}\, d\theta = a(\sinh \tfrac{r}{a})\, d\theta. \tag{2}$$ If $d\phi$ is the indicated Euclidean angle subtended at the Euclidean center of the circle representing $\ell$, then (up to first order in $ds$) $$R\, d\theta = R'\, d\phi,\qquad\text{i.e.,}\quad \frac{d\phi}{d\theta} = \frac{R}{R'} = \frac{2R^{2}}{1 - R^{2}}.$$ By (1), we have $$1 + \frac{d\phi}{d\theta} = \frac{1 + R^{2}}{1 - R^{2}} = \frac{1 + \tanh^{2} \frac{r}{2a}}{1 - \tanh^{2} \frac{r}{2a}} = \cosh \tfrac{r}{a}. \tag{3}$$

Parallel transport along $C$ of the boldface tangent vector at $p$ is the vector labeled $v$. The relative angle of turning of the tangent $v'$ to the circle is $d\alpha = d\theta + d\phi$. The geodesic curvature at $p$ is the rate at which the tangent to $C$ rotates per unit length. By (2) and (3), this is equal to $$k_{g} = \frac{d\alpha}{ds} = \frac{d\theta + d\phi}{ds} = \left(1 + \frac{d\phi}{d\theta}\right) \frac{d\theta}{ds} = \frac{\cosh \frac{r}{a}}{a \sinh \frac{r}{a}} = \tfrac{1}{a} \coth \tfrac{r}{a}.$$

• At first for physical dimension tally, can I suggest $R=a \tanh \dfrac{r}{2a}, \, (1-R^2) \rightarrow (a^2-R^2)\, etc.$ in the disc model ? – Narasimham Aug 2 '17 at 21:44
• Yes, the same hyperbolic plane can be modeled using the disk of radius $a$ with the metric$$g = \frac{4a^{2}(du^{2}+ dv^{2})}{(a^{2} - (u^{2} + v^{2}))^{2}}.$$(The mapping $(u, v) = (ax, ay)$ is an isometry with the metric I used.) – Andrew D. Hwang Aug 2 '17 at 22:19
• Then are we having ( in the fraction numerator as well as denominator of first metric $g$ the same $a,1$ representing the same radius? and could it perhaps be changed throughout ? – Narasimham Aug 3 '17 at 4:10
• The entire calculation can be expressed in terms of the metric on the disk of radius $a$: hyperbolic quantities won't change (because the two metrics are isometric); "my" Euclidean radius $R$ could be written $R/a$ with $0 \leq R < a$. (In fact, I initially tried to work things out (and type up the results) in the disk of radius $a$. It appears some formulas didn't get converted to the unit disk, thinking specifically of the display before (1).) – Andrew D. Hwang Aug 3 '17 at 10:03
• I used symbols $T$ for center circle radius and $a$ for various hyp. geodesic radii. FullHyperbolicGeodesics – Narasimham Aug 3 '17 at 16:59