1
$\begingroup$

M is an n-dim manifold. $\pi :\tilde M \to M$ the universal cover of M. $\tilde p \in \tilde M$ a lift of p. We choose a measurable section $j:{B_1}\left( p \right) \to {B_1}\left( {\tilde p} \right)$, i.e. $\pi \left( {j\left( x \right)} \right) = x$ for any $x \in {B_1}\left( p \right)$. Let $T = j\left( {{B_1}\left( p \right)} \right)$. Let S be the union of g(T) over all $g \in {\pi _1}\left( M \right)$ such that $g\left( T \right) \cap {B_1}\left( {\tilde p} \right) \ne \emptyset $. Is then ${B_1}\left( {\tilde p} \right) \subset S$?

$\endgroup$
1
  • $\begingroup$ Yes, assuming that your manifolds are Riemannian and $\pi$ is an isometric covering map; you do not even need "measurable" assumption, it is just a set-theoretic statement. $\endgroup$
    – Misha
    Jun 5, 2013 at 4:55

1 Answer 1

1
$\begingroup$

I assume you are taking $M$ to be Riemannian and $\tilde{M}$ to be endowed with the induced metric?

Let $\tilde{q}\in B_1(\tilde{p})$. We want to show that $\tilde{q}\in S$. Let $q=\pi(\tilde{q})$. There exists a covering transformation $g\in \pi_1M$ such that $\tilde{q}=gj(q)$, so we just need to prove that $q\in B_1(p)$.

Let $\tilde{\gamma}$ be a smooth path from $\tilde{p}$ to $\tilde{q}$ of length less than one. Then $\gamma=\pi\circ\tilde{\gamma}$ is a smooth path from $p$ to $q$ of the same length, because $\pi$ is a local isometry. This finishes the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.