# Deduce the curvature of the manifold by probing it with geodesics

Suppose in order to probe the curvature of a manifold in $$\Bbb R^2$$, you decide to shoot off geodesics from the four corners of the square.

Consider a set of geodesics, $$K,$$ each symmetrical about the line $$y=1-x,$$ embedded in the unit square, that pass through the points $$(0,0)$$ and $$(1,1):$$

$$K=\{s_1,s_2,s_3,...\},$$

such that, $$s_1>s_2>s_3>...$$

Furthermore, at the boundary of the square, the distance between successive geodesics goes to zero.

Define a new set of geodesics, $$G,$$ each symmetrical about the line $$y=x,$$ embedded in the unit square, that pass through the points, $$(0,0)$$ and $$(1,1):$$

$$G=\{t_1,t_2,t_3,...\},$$

such that,

$$...>t_2>t_1>s_1>s_2>...$$

At the boundary, the distance between successive geodesics goes to zero.

Define a new set $$R=\{K,G\},$$ that is the union of the sets $$K$$ and $$G.$$

Define a new set $$B$$ as the reflection of $$R$$ about the line $$x=1/2,$$ where all curves in $$B$$ are also geodesics.

Suppose each geodesic runs across a manifold, $$M,$$ in $$\Bbb R^2.$$

What is the curvature of the manifold?

What other information can one recover about the manifold?

This is the result of probing a particular manifold with geodesics:

• At least for me, some extra context would be helpful here. Is $M$ any submanifold in $\mathbb{R}^2$? Is $M$ topologically $[0,1]\times[0,1]$? – Santana Afton Feb 16 at 0:17
• Yes $M$ is any submanifold and $M$ is in $\Bbb R^2$, and $M$ is topologically $[0,1]\times[0,1]$ – Ultradark Feb 16 at 2:00