The intersection rule is trivial. For any two points $p$ and $q$ in the intersection, they are both in each set, and therefore the unique geodesic $ gamma$ between them must have been in each set. But if the geodesic is in each set it is in the intersection. Therefore the definition is satisfied.

• Riemannian Geometry Lecture Notes
• Introduction To Riemannian Geometry

Riemannian Geometry Lecture Notes

Introduction To Riemannian Geometry

Contractibility seems harder. Let $p$ be a point in $M$. For $q$ in $M$ and $t$ in $0, 1$ define $f(t, q)$ to be the point a portion $t$ of the way along the unique geodesic $ gamma$ from $q$ to $p$. This is well-defined. Also $f(0, q) = q$ and $f(1, q) = p$.

That has to be the contraction. The question is whether this map is continuous.

The challenge is this. If you have a triangle between two 'nearby' points and a distant one, need the edges remain close? Well, in a general manifold, no. The 'unique geodesic' condition has to come in somehow. But I can't see how to do it.

As for the last part of your question, the hyperbolic plane does not fit in space and is strongly convex.

By Sakai, Takashi This quantity is an English translation of Sakai's textbook on Riemannian geometry which was once initially written in eastern and released in 1992. The author's motive in the back of the unique publication was once to supply to complex undergraduate and graduate scholars an advent to trendy Riemannian geometry which may additionally function a reference. The e-book starts with an evidence of the elemental idea of Riemannian geometry.

Precise emphasis is put on understandability and clarity, to lead scholars who're new to this zone. The rest chapters care for quite a few themes in Riemannian geometry, with the focus on comparability equipment and their functions. Read or Download Riemannian Geometry PDF Similar geometry & topology books.