Proof. Since the exponential map is well defined on the $\rho $ -ball centered at $0$ of $T_x\widehat N$ , the geodesic $s\mapsto \widehat \pi (g_sv)$ can be extended at least until it exits $\widehat B_{2r}$ ; thus, $\{s>0 \mid \widehat {\pi }({g}_sv)\not \in \widehat {B}_{r}\}$ and $\{s>0 \mid \widehat {\pi }({g}_sv)\not \in \widehat {B}_{2r}\}$ are non-empty. They are closed because the balls $\widehat {B}_{r}$ and $\widehat {B}_{2r}$ are open.

Let $s \in [s_1,s_2)$ . Then, $\widehat {\pi }({g}_sv)$ belongs to $\widehat B_{2r}$ by definition of $s_2$ . If it did belong to $\widehat {B}_{r}$ , then, by convexity of $\widehat {B}_{r}$ , there would exist a geodesic segment connecting $\widehat {\pi }(v)$ and $\widehat {\pi }({g}_sv)$ that is contained in $\widehat {B}_{r}$ . By the strong convexity of $\widehat {B}_{\rho }$ , this segment would have to be equal to $\{\widehat {\pi }({g}_tv) \mid 0\leq t\leq s\}$ , which is contradictory because the latter is not contained in $\widehat {B}_{r}$ . Hence, $\widehat {\pi }({g}_sv) \in \widehat {B}_{2r}\setminus \widehat {B}_{r}$ .

Moreover, by the strong convexity of $\widehat B_\rho $ , the segment $\{\widehat \pi (g_tv)\mid 0\le t\le s_1\}$ is a length-minimizing geodesic; thus, the triangle inequality yields $s_1\le 2r$ .

Finally, the reversed triangle inequality yields

$$ \begin{align*}s_2-s_1=d_{\widehat N}(\widehat\pi(g_{s_2}v),\widehat\pi(g_{s_1}v))\ge |d_{\widehat N}(x, \widehat\pi(g_{s_2}v)) - d_{\widehat N}(x, \widehat\pi(g_{s_1}v))| \ge 2r-r = r.\end{align*} $$

Hence, $s_1\le \tfrac 23 s_2$ .

Proof. If U is bounded, the limit of ${ \unicode{x3bb} ([0,T]\cap U)}/{T}$ is zero and the lemma is proved. Let us assume that U is unbounded. As a non-empty open subset of $\mathbb {R}$ , it is a finite or countable union of disjoint intervals. Let $(t_{-1},{t_1})$ be one of those intervals (with $t_1 \in (0,\infty ]$ ).

Let $\tilde E$ be a connected component of the unit tangent bundle of $\phi (\{x\}\times \tilde N)$ such that $E = dp(\tilde E)$ . The geodesic distance induced by the Sasaki metric on $T\tilde M$ is bounded below by the geodesic distance on $\tilde M$ . Hence, $\tilde \pi (\tilde B_r(\tilde E))$ is contained in the r-neighborhood of $\phi (\{x\}\times \tilde N)$ , and since $\widehat N\times \tilde N$ is equipped with the product metric, the following inclusion holds:

(*) $$\begin{align}\tilde \pi(\tilde B_r(\tilde E))\subset \phi(\widehat B_r\times \tilde N).\end{align} $$

Since $\tilde N$ is connected and geodesically complete, $\phi (\{x\}\times \tilde N)$ is a connected component of $p^{-1}(N)$ and since $(p\circ \phi )(\widehat N\times \tilde N)$ is a tubular neighborhood of N, then $\phi (\widehat N\times \tilde N)$ is a connected component of $p^{-1}((p\circ \phi )(\widehat N\times \tilde N))$ . Thus, any geodesic segment in $(p\circ \phi )(\widehat N\times \tilde N)$ admits a lift in $\phi (\widehat N\times \tilde N)$ . In particular, we can find a lift $\tilde u\in T^1\tilde M$ of u such that $g_{t}\tilde u$ belongs to $\tilde B_r(\tilde E)$ for all $t\in (t_{-1}, t_1)$ .

Let us pick $t_0 \in (t_{-1},{t_1})$ , and let $v\in T\widehat N$ and $w\in T\tilde N$ such that $g_{t_0}\tilde u=d\phi (v,w)$ . To prove by contradiction that $\|v\|$ is positive, let us assume that $\|v\|=0$ . Then, $\tilde u = d\phi (v, g_{-t_0}w)$ holds and one has $d_{T\widehat N}(v, 0_{T_x\widehat N}) = d_{\widehat N}(\widehat \pi (v), x)$ . Hence,

(**) $$\begin{align}d_{T\tilde M}(g_{t_0}\tilde u, d\phi(0_{T_x\widehat N}, w)) = d_{T\tilde M}(\tilde u, d\phi(0_{T_x\widehat N}, g_{-t_0}w)) = d_{\widehat N}(\widehat \pi(v), x).\end{align} $$

Since $g_{t_0}\tilde u\in \tilde B_r(\tilde E)$ , (*) shows that this distance is smaller than r by (*). Moreover, the vector $d\phi (0_{T_x\widehat N}, w)$ belongs to the unit tangent bundle of $\phi (\{x\}\times \tilde N)$ . The latter is either equal to $\tilde E$ (if $\dim N\ge 2$ ) or the union of $\tilde E$ and a lift of F (if $\dim N\le ~1$ ). According to (**), the vector $d\phi (0_{T_x\widehat N}, w)$ belongs also to the $2r$ -neighborhood of $\tilde E$ , which must be disjoint from any lift of F since $2r<d_{TM}(E, F)$ . In both cases, $d\phi (0_{T_x\widehat N}, w)$ must belong to $\tilde E$ . Since $\tilde N$ is geodesically complete, the vector $d\phi (0_{T_x\widehat N}, g_{-t_0}w)$ also belongs to $\tilde E$ , and (**) yields $\tilde u\in \tilde B_r(\tilde E)$ , which leads to a contradiction because $u\not \in B_r(E)$ . Hence, the norm of v must be positive.

By (*), since $g_{t_0}\tilde u\in \tilde B_r(\tilde E)$ , the point $\widehat \pi \big ({v}/{\|v\|}\big )$ belongs to $\widehat B_r$ ; thus, ${v}/{\|v\|}$ satisfies the assumptions of Lemma 1, and we can set $s_1$ and $s_2$ as in the statement of the lemma. Moreover, $g_s({v}/{\|v\|})$ is well defined for all $s\in [s_1, s_2)$ and the following holds:

$$ \begin{align*}\text{ for all } s \in [s_1, s_2), \quad g_{(s/{\|v\|})+t_0} \tilde u = {\|v\|}\,d\phi\bigg(g_s\frac v{\|v\|}, g_{s}\frac w{\|v\|}\bigg).\end{align*} $$

Hence, the geodesic arc $\{\pi (g_{(s\!{\|v\|})+t_0}\tilde u)\mid s_1\le s<s_2\}$ is contained in $\phi (\widehat N\times \tilde N)$ , which is a lift of a tubular neighborhood of N, and since the normal exponential map of $\phi (\{x\}\times \tilde N)$ coincides with he pull-back of the normal exponential map of N by $dp$ , one has

$$ \begin{align*} \begin{aligned} \text{for all } s \in [s_1, s_2), \quad d_M(\pi(g_{(s/{\|v\|})+t_0} u), N) &= d_{\tilde M}(\tilde\pi(g_{(s/{\|v\|})+t_0} \tilde u), \phi(\{x\}\times \tilde N))\\&\ge d_{\widehat N}\bigg(\widehat \pi\bigg(g_s\frac v{\|v\|}\bigg), x\bigg). \end{aligned} \end{align*} $$

This is larger than r by Lemma 1. Hence, the interval $[{s_1}/{\|v\|}+t_0, {s_2}/{\|v\|}+t_0)$ is disjoint from U, and the inequalities ${s_1}/{\|v\|}+t_0 \ge t_1$ and ${s_2}/{\|v\|}+t_0\le t_2$ hold, with $t_2=\inf (U\cap (t_1, +\infty ))$ . Notice that $t_2$ exists because U is unbounded and $t_1$ is not $+\infty $ , since it is bounded by ${s_1}/{\|v\|}+t_0$ . Additionally, the conclusion of Lemma 1 yields

$$ \begin{align*}t_1-t_0 \le \frac{s_1}{\|v\|} \le \frac{2s_2}{3\|v\|}\le \frac23 (t_2 - t_0).\end{align*} $$

Since $t_1-t_{-1}=\sup \{t_1-t_0 \mid t_0\in (t_{-1},t_1)\}$ , we also get the inequality:

$$ \begin{align*}t_1-t_{-1} \le \frac23 (t_2 - t_{-1}).\end{align*} $$

This argument can be applied to any maximal open interval in U. In particular, every such interval is bounded. Since U is unbounded, its partition in open intervals must be countable, which means that there exist two sequences $(t_{-1}^{(n)})_{n\in {\mathbb {N}}}$ and $(t_1^{(n)})_{n\in {\mathbb {N}}}$ such that

$$ \begin{align*}U = \bigsqcup_{n\in{\mathbb{N}}} (t_{-1}^{(n)},t_1^{(n)})\quad\text{and}\quad\text{for all } n \in {\mathbb{N}}, \quad\begin{cases}0<t_{-1}^{(n)}<t_1^{(n)}<t_{-1}^{(n+1)},\\ t_1^{(n)}-t_{-1}^{(n)}\le\frac23(t_{-1}^{(n+1)}-t_{-1}^{(n)}). \end{cases}\end{align*} $$

Hence,

$$ \begin{align*}\text{for all } n \ge1, \quad \frac{\unicode{x3bb}\big([0,t_{-1}^{(n)}] \cap U\big)}{t_{-1}^{(n)}} \leq \frac2{3t_{-1}^{(n)}}\sum_{k=0}^{n-1}\bigg(t_{-1}^{(k+1)} - t_{-1}^{(k)}\bigg)\leq \frac23.\end{align*} $$

The sequence $t_{-1}^{(n)}$ tends to infinity because U is unbounded, which proves Lemma 2.

Proof. Under those assumptions, there exists an open isometric embedding $\phi $ of $\widehat N\times \tilde N$ into $\tilde M$ , where $\tilde M=M$ , $\tilde N=N$ , and the cover $p: \tilde M \to M$ is the identity map. Let $x\in \widehat N$ and let us prove that the submanifold $\phi (\{x\}\times N)$ satisfies Assumption 1 of Proposition 1.

Restricting, if necessary, $\widehat N$ to an open geodesic ball centered at x, we can assume that $\phi (\widehat N\times \tilde N)$ is a tubular neighborhood of $\phi (\{x\}\times N)$ . Moreover, $\phi (\{x\}\times N)$ is geodesically complete and contains a closed geodesic since it is isometric to N. Finally, the set $\widehat N\setminus \{x\}$ is non-empty since $\dim \widehat N\ge 1$ and for all $y\in \widehat N\setminus \{x\}$ , the submanifold $\phi (\{y\}\times N)\subset M\setminus \overline {\phi (\{x\}\times N)}$ also contains a closed geodesic.

Proof. The inclusion map $\Sigma _{\mathrm {half}}\times S^1 \rightarrow M$ is an embedding that satisfies the assumptions of Corollary 2, with $S^1$ being a geodesically complete manifold that contains at least one closed geodesic.