If $M$ is a closed simply connected manifold, is it possible for the space of free loops of Sobolev class $H^1$ not to be connected?

Question

If a closed manifold $M$ is simply connected, then is it possible for the space of free loops $\Lambda M$ of Sobolev class $H^1=W^{1,2}$ on $M$ not to be connected? More precisely, does this follow from the canonical isomorphisms $\pi_k(\Lambda M)\cong\pi_k(\Omega M)\rtimes\pi_k(M)\cong\pi_{k+1}(M)\rtimes\pi_k(M)$?

Thanks in advance!

Answer 1

The isomorphisms involving homotopy groups follow from the existence of a section (namely, the map sending a point on $M$ to the constant loop at $M$) in the fibration $\Omega M \to \Lambda M \to M$; in particular, you know that $\Lambda M$ is the total space of a fibration with connected fiber and base. That is enough to conclude that $\Lambda M$ is connected (use the long exact sequence in homotopy).