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Let $M$ be a complete Riemannian manifold and $a$, $b$ two different points on it. We define a set $A =\{x\in M | \ d(x,a)=d(x,b)\}$ where $d$ is the distance induced by the metric of $M$. My question is: is $A$ necessarily a measure $0$ set with respect to the volume form $dV$?

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up vote 2 down vote accepted

Let $C_a$, $C_b$ denote the cut loci of $a,b$ respectively. A result of Itoh and Tanaka gives that $C_a$ and $C_b$ has at most $n-1$ Hausdorff dimension, and hence has $n$ Hausdorff measure (and hence Lebesgue measure) 0.

By definition the cut locus is a closed set.

Let $C = C_a \cup C_b \cup \{a,b\}$. Outside of these points, it is well known that the Riemannian distance function from a point is smooth, and has non-zero gradient.

Since $C$ has measure 0, it suffices to consider $M\setminus C$. Let $q\in A\cap (M\setminus C)$. By definition there exists a unique geodesic connecting $q$ and $a$, and also $q$ and $b$. Let $v_a,v_b$ be the corresponding tangent vectors in $T_qM$. If $v_a \neq v_b$, the function $d(a,\cdot) - d(b,\cdot)$ has non-zero gradient and hence locally in a neighborhood of $q$ we have that $A\cap (M\setminus C)$ is a smooth submanifold with codimension 1 by the Implicit Function Theorem.

However, $v_a$ cannot be equal to $v_b$: this follows from existence and uniqueness of ordinary differential equations (if the two unit tangents are equal, then $a = b$ as $d(a,q) = d(b,q)$ and this contradicts our assumption).

Thus $A\cap (M\setminus C)$ is a codimension 1 submanifold of $M\setminus C$, an open $n$ dimensional manifold. Hence $A\cap (M\setminus C)$ has measure 0 also, proving the claim.

Note that the main difficulty posed by the manifold case as opposed to the $\mathbb{R}^n$ case is the cut locus. The rest follows from elementary differential geometric considerations. Hence the case for Hadamard manifolds (and in fact any complete manifold with infinite radius of injectivity) can also be done in the "simple" manner. Only at the cut loci do we need to appeal to more sophisticated machinery.

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