Skip to main content
MathOverflow

Return to Question

removed capitals
Source Link
YCor
YCor
  • 63.9k
  • 5
  • 187
  • 286

Bisector of two points in a Riemannian Manifoldmanifold has measure $0$

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian Manifoldmanifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?

I was thinking about this problem with a colleague and we believe we can prove it when $M$ is a closed manifold, but even in that case the proof was much more complicated than expected. I encountered this problem while thinking about this question.

Bisector of two points in a Riemannian Manifold has measure $0$

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian Manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?

I was thinking about this problem with a colleague and we believe we can prove it when $M$ is a closed manifold, but even in that case the proof was much more complicated than expected. I encountered this problem while thinking about this question.

Bisector of two points in a Riemannian manifold has measure $0$

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?

I was thinking about this problem with a colleague and we believe we can prove it when $M$ is a closed manifold, but even in that case the proof was much more complicated than expected. I encountered this problem while thinking about this question.

Source Link
Saúl RM
Saúl RM
  • 10.6k
  • 2
  • 28
  • 48

Bisector of two points in a Riemannian Manifold has measure $0$

Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian Manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?

I was thinking about this problem with a colleague and we believe we can prove it when $M$ is a closed manifold, but even in that case the proof was much more complicated than expected. I encountered this problem while thinking about this question.