In graph theory, a ball homogeneous graph is one where, for any two vertices, there is an automorphism of the graph that maps one vertex to the other while preserving the structure of the graph. Such graphs are often studied in the context of their symmetry properties. A vertex transitive graph is a graph in which, for any two vertices, there exists an automorphism of the graph that maps one vertex to the other. This means that all vertices are structurally identical in terms of their connections to other vertices. To address the question: Yes, ball homogeneous graphs are indeed vertex transitive. The reason is that the definition of ball homogeneity implies that for any two vertices, the neighborhoods (or "balls") of these vertices can be transformed into each other through an automorphism. Since this transformation can be extended to the vertices themselves, it follows that ball homogeneous graphs must also exhibit vertex transitivity. In summary, all ball homogeneous graphs are vertex transitive due to their inherent symmetry properties.

A Riemannian manifold is referred to as ball homogeneous if the volume of any geodesic ball with radius \( r \) centered at a point \( m \) is solely a function of \( r \) and is independent of the choice of \( m \). It is evident that homogeneous Riemannian manifolds are inherently ball homogeneous. The reverse implication is a conjecture in the field of Riemannian geometry, questioning whether every ball homogeneous Riemannian manifold is also homogeneous.

To investigate whether a finite graph \( G \) with the property that the number of vertices within distance \( r \) from any vertex \( v \) depends only on \( r \) and not on \( v \) must be vertex transitive, we can consider a specific counterexample. Consider the graph \( G \) that is a union of two disjoint complete graphs, \( K_3 \) and \( K_2 \). Let's denote the vertices of \( K_3 \) as \( a_1, a_2, a_3 \) and the vertices of \( K_2 \) as \( b_1, b_2 \). The graph \( G \) looks like this: ``` a1 | a2 -- a3 b1 -- b2 ``` Now, let's analyze the distances from any vertex in \( G \): - For any vertex \( a_i \) (where \( i = 1, 2, 3 \)): - Distance 0: \( 1 \) vertex (itself) - Distance 1: \( 2 \) vertices (\( a_{j} \) for \( j \neq i \)) - Distance 2: \( 3 \) vertices (\( a_1, a_2, a_3 \)) - For any vertex \( b_j \) (where \( j = 1, 2 \)): - Distance 0: \( 1 \) vertex (itself) - Distance 1: \( 1 \) vertex (the other vertex in \( K_2 \)) - Distance 2: \( 5 \) vertices (all vertices in \( K_3 \) plus the other vertex in \( K_2 \)) Now, let's summarize the number of vertices within distance \( r \): - For \( r = 0 \): \( 1 \) vertex (both \( K_3 \) and \( K_2 \)) - For \( r = 1 \): \( 2 \) vertices for \( K_3 \) and \( 1 \) vertex for \( K_2 \) - For \( r = 2 \): \( 3 \) vertices for \( K_3 \) and \( 5 \) vertices for \( K_2 \) Notice that the counts depend on which vertex we start from, indicating that the property does not hold uniformly across all vertices. Thus, while it may seem that we could find a graph where the number of vertices at distance \( r \) is independent of the choice of vertex, this particular graph \( G \) does not satisfy the vertex transitive property, as the vertices in \( K_3 \) and \( K_2 \) are not equivalent in terms of their distances to the other vertices in the graph. Therefore, we conclude that it is indeed possible for a graph to satisfy the condition about the number of vertices within distance \( r \) from \( v \) without being vertex transitive. The answer is no, \( G \) does not have to be vertex transitive, and the example of the union of \( K_3 \) and \( K_2 \) serves as a counterexample.

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