Let (M,g) be a compact Riemannian manifold. It is well known that there always exists a nontrivial closed geodesic in M, which is the so-called Lusternik-Fet theorem.

But such a geodesic could well self-intersect (in a transversal way).What is known about the existence of geodesics which do not self-intersect?

(I do not know whether ‘simple’ is the right term for describing this)Do they always exist? Or are there counterexamples in which all closed geodesics are self-intersecting?

**Answer**

We can easily find a simple closed geodesic if M is not simply connected:

*Proof*: Let ˜M be the universal cover of M. Then one can pull back the metric to ˜M, as π:˜M→M is a local diffeomorphism. Thus ˜M is also a Riemannian manifold and π is an local isometry. As M is compact, it’s complete, and one can check that ˜M is also complete.

Fix p∈M and let ˜p∈π−1(p) be fixed. Define

C(p)=minq∈π−1 (p)∖{˜p}d(˜p,q).

Note that C(p)>0 and is realized by some q∈π−1(p), as π−1(p) is a discrete set in ˜M. One can check C:M→R is continuous. Let p0∈M be its minimum. Let ˜p∈π−1(p) and ˜q∈π−1(p)∖{˜p} such that C(p)=d(˜p,˜q). Let η:[0,d]→˜M be a shortest geodesic joining ˜p and ˜q. Such a geodesic exists as ˜M is complete. η is obviously simple as it’s length minimizing. Let γ=π∘η. γ is a simple geodesic loop (γ(0)=γ(d)), which might not be smooth at p.

However, consider p1=γ(d/2)∈M. As C(p1)≥C(p), the same γ is a curve which is the shortest curve along all curves homotopic to γ with the same base point p1. Thus γ must be smooth at p and so γ is simple closed geodesic in M.

**Remark** When M is simply connected, the general question seems to be an open problem. For S2, the answer is affirmative. I am no expert in this question so I think I cannot say more.

**Attribution***Source : Link , Question Author : Mizar , Answer Author : Mizar*