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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:

enter image description here

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

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