# Verification of Gauss Bonnet theorem in vicinity of pseudosphere cuspidal geodesic equator $K=-1$

In the Gauss-Bonnet theorem with Euler characteristic $\chi=1$

$$\int k_gds + \int K dA + \Sigma \psi_i = 2 \pi \tag1$$

$$0 - Area+ (\pi- A)+ (\pi- B)+ (\pi- C) = 2 \pi \tag2$$ giving area as pseudospherical defect

$$Area =\pi-(A+B+C) \tag3$$

Let us consider a particular curvilinear right triangle of finite area bounded by three geodesic arcs on a hyper pseudo-sphere $K=-1$ . One (green) is a meridian between cuspidal equators, one (blue) arc is along a cuspidal equator and one (red) arc is an arbitrary geodesic transversely cutting two cuspidal equators, alternate angles are equal.

Sum of angles on one side of these "parallels" is $\pi$ and for the right triangle sum we can add and remove $\pi/2$ leaving the sum still as $\pi$.

Although pseudo-spherical defect of such a right triangle vanishes, a finite area is enclosed that appears to be a violation of the theorem in this particular case

Next let us consider one half longitudinal section of any (or all.. by virtue of isometry) of the series of hyperbolic pseudospheres shown with "rectangular" patches of $K=-1$ surface but involving the edges this time.

Please allow me for the time being to call the edges as cuspidal equators. $(radius,\psi=\pi/2 )$ are constant, the slope $\phi =0$ at cusp, so they are geodesics. The demo plays to have invariant surface area $-4 \pi$ for the full segments and $-2 \pi$ for a half segment, as a consequence of the same metric/isometry a consequence of same Metric/ First Fundamental form/isometry.

Area of a segment is found to be constant.

$$dA=2\pi r dl=2\pi\frac{r}{\cos\phi}\cos\phi\frac{dl}{d\phi}d\phi =2 \pi \frac{\cos \phi \,d \phi}{\kappa_1\kappa_2}= =2 \pi \frac{\cos \phi \,d \phi}{K} \tag4$$

Integrating between $\phi$ limits $\pm \pi/2$ we get

$$A =\frac{4\pi}{K} \tag5$$

which is constant for all isometrically equivalent segment patches.

The Gauss-Bonnet equation (1) can be written for special case $K=-1$

$$0- 2\pi + 4\cdot \pi/2 = 2 \pi \tag 6$$

which relation does not agree. Even in case $K=+1$ we have ( ignore non-parallels, consider only zero longitudinal curvature cusp boundaries)

$$0+2\pi + 4\cdot \pi/2 = 2 \pi \tag7$$

which also does not agree. It would appear that the theorem does not hold good in the vicinity of cuspidal boundaries.

It can be noted that the above violation occurs even for the constant $K=+1$ spheres between two cuspidal equators where area expression holds good except for sign.

Where am I in error?

Are there extra considerations as regards smoothness, compactness etc. that need to be brought in to validate or properly understand of the elegant theorem? Right now it appears to me that the theorem is not valid for closed areas at least near such singularities.

Best Regards.

• Could you provide more details on why your picture represents a hyperbolic surface? The only pseudosphere im familiar with looks quite different. en.wikipedia.org/wiki/Pseudosphere – Deane Yang Aug 19 '18 at 14:53
• You are referring to the central or Beltrami pseudosphere which is among three types which you can see for example in Palais/Xah Lee 3D surface gallery.or by googling images "constant gaussian curvature surfaces revolution." – Narasimham Aug 19 '18 at 18:16
• The three types are diffeomorphic or isometric equivalents. One variety has cuspidal boundary and another a conical vertex apart from the central cases, We refer here to the former. – Narasimham Aug 19 '18 at 18:33
• There are no special considerations needed here. You just need to do your calculations more carefully, checking every assumption and step. – Deane Yang Aug 23 '18 at 3:36
• Why is the blue curve a geodesic? – Deane Yang Aug 23 '18 at 15:54