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It is known that homotopy classes of free loops in a topological space are in bijection with conjugacy classes of the the spaces fundamental group, that every group is the fundamental group of some space, and that there exists an infinite (finitely generated!) group with exactly two conjugacy classes. This has the interesting consequence that there is some space with infinitely many distinct based loops but only two unbased loops (up to homotopy).

I'm wondering if there is any manifold that has this property. It is known that every compact manifold has finitely presented fundamental group (and that every finitely presented group is a fundamental group of a manifold), so showing there is a infinite finitely presented group with two conjugacy classes would suffice, although I believe this is unknown. But if we allow noncompact manifolds, perhaps things get easier.

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    Every countable group is the fundamental group of a manifold. There is even finitely generated infinite group with two conjugacy classes. For finitely presented groups, I think, it is an open problem. – Moishe Kohan Jan 28 '20 at 20:38

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  1. For every countable group $G$ there exists a (smooth) connected manifold $M$ with $\pi_1(M)\cong G$, see this MathOverflow discussion.

  2. There exist a countable infinite group with exactly two conjugacy classes: This is a classical result due to Higman, Neuman and Neuman ("Embedding Theorems for Groups", 1949). See this MSE post.

Hence, there exists a connected manifold with infinite fundamental group but only two free homotopy classes of loops.

The existence of a compact manifold with this property is equivalent to the existence of an infinite finitely presented group with exactly two conjugacy classes. The latter, I believe, is an open problem. A construction of finitely generated groups with this property is highly nontrivial and is due to D.Osin:

D. Osin, Small Cancellations over Relatively Hyperbolic Groups and Embedding Theorems, Ann Math. 172, no. 1 (2010), 1-39.

Moishe Kohan
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