has zero, one or infinitely many homologically visible closed geodesics.

Similar results can also be obtained when $M\simeq{{\mathbb{R}}}^2$ is a complete plane:

\[thm:plane\] Let $M$ be a complete Riemannian plane with isolated closed geodesics and assume one of the following hypothesis:

1.  \[it:plane:selfintersecting\] there exists a self-intersecting closed geodesic,

2.  \[it:plane:intersecting\] there exist two distinct closed geodesics that intersect,

3.  \[it:plane:mind\] there exists a closed geodesic of non-zero average index,

4.  \[it:plane:homological\] there exists a homologically visible closed geodesic.

Then $M$ contains infinitely many geometrically distinct homologically visible closed geodesics.

Any complete Riemannian plane with isolated closed geodesics has zero or infinitely many geometrically distinct homologically visible closed geodesics.

It is easy to give counter-examples to Theorem \[thm:main\] when none of the assumptions \[it:contractible\]–\[it:homological\] hold by considering embedded cylinders of revolution $$\label{eq:revolution}
    (\theta,z)\mapsto (r(z)\cos\theta,r(z)\sin\theta,z),$$ for well-chosen smooth maps $r:{{\mathbb{R}}}\to (0,+\infty)$. A complete cylinder may have no closed geodesic at all: take $r'>0$. It can have an arbitrary large finite number $k\in{{\mathbb{N}}}$ of homologically invisible closed geodesics: take $r'(z)>0$ for all $z\in{{\mathbb{R}}}\setminus\{z_1,\ldots,z_k\}$ and $r'(z_i)=0$. It can also have a unique visible closed geodesic: take $r'<0$ on $(-\infty,0)$, $r'(0)=0$ and $r'>0$ on $(0,+\infty)$ (one can as well add to this cylinder an 