A Lambert quadrilateral is a quadrilateral three of whose angles are right angles. And in 2-d hyperbolic space $\mathbb H^2$, we have nice formulas for the fourth angle.

If $AOBF$ is a Lambert quadrilateral in $\mathbb H^2$, with $\angle A=\angle O=\angle B=\pi/2$, then $$\cos\angle F=\sinh |OA|\cdot \sinh |OB|.$$ My question is, do we have a some kind of comparison result for $\angle F$ of Lambert quadrilaterals with the lengths $|OA|$ and $|OB|$ fixed? (maybe as corollary of some classical comparison theorems?)

By this I mean that if we consider another Lambert quadrilateral $A'O'B'F'$ with $|O'A'|=|OA|$ and $|O'B'|=|OB|$ on a compact surface $(M^2,g)$ with curvature $\kappa<-1$, can we compare the two acute angles $\angle F$ and $\angle F'$? In this case it seems natural to expect $\angle F'\le \angle F$.

  • $\begingroup$ In order to have a hope of proving something like this, you are going to need to assume that the sides of the quadrilateral form a contractible loop in the surface. On a compact surface with $K\equiv -1$, you can have 'squares', i.e., a periodic sequence of 4 congruent geodesic segments such that each successive pair meet at right angles at their endpoints. (Of course, this can't happen in the hyperbolic plane, but the hyperbolic plane is simply-connected.) $\endgroup$ – Robert Bryant Feb 23 '16 at 18:05
  • $\begingroup$ @RobertBryant Thanks, you are right. For the application I have in mind, the two manifolds would be simply-connected. (In fact, I am considering $M$ here to be a universal cover of some compact surface with negative curvature bounded from above, so it's diffeomorphic to $\mathbb R^2$.) $\endgroup$ – forevenone Feb 23 '16 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.