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If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree. (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}

Concerning, "What are the edges?": They are geodesics, such that there are two distinct shortest paths from $x$ to every interior point of an edge of the cut locus. Points of the cut locus of degree $k$ have $k$ distinct shortest paths from $x$.

If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree. (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}

Concerning, "What are the edges?": They are geodesics, such that there are two distinct shortest paths from $x$ to every interior point of an edge of the cut locus. Points of the cut locus of degree $k$ have $k$ distinct shortest paths from $x$.

If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree. (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}

If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree. (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}

Concerning, "What are the edges?": They are geodesics, such that there are two distinct shortest paths from $x$ to every interior point of an edge of the cut locus. Points of the cut locus of degree $k$ have $k$ distinct shortest paths from $x$.

If you will excuse me substituting a polyhedron for the Riemannian manifold, this figure shows how the cut locus (red) from source $x$ is a tree.  (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}

If you will excuse me substituting a polyhedron for the Riemannian manifold (imagine rounding the vertices), this figure shows how the cut locus (red) from source $x$ is a tree.  (The green arcs are equidistant from $x$.)
     
     _{(Figure from Discrete and Computational Geometry)}