TITLE: The free loop space of spheres
QUESTION [21 upvotes]: Let $n>1$. The homology of the free loop space $\Lambda S^n$ of the sphere $S^n$ contains two torsion if $n$ is even. Thus the fibration
$$
\Omega S^n\rightarrow \Lambda S^n\rightarrow S^n
$$
is not trivial if $n$ is even (Here $\Omega S^n$ denotes the based loop space.).
The odd dimensional spheres do not have torsion in the homology of the free loop space and one can compute that $H_*(\Lambda S^{2k+1};\mathbb Z)\cong H_*(\Omega S^{2k+1}\times S^{2k+1};\mathbb Z)$. Hence the homology does not obstruct the existence of a trivialization of the free loop fibration. Indeed $\Lambda S^3$ and $\Omega S^3\times S^3$ are homeomorphic, which can be proven using the group structure on $S^3$. 
Is the free loop space fibration always trivial for odd dimensional spheres ? My guess would be that this is not the case, with a possible exception of $S^7$.

REPLY [16 votes]: I am grateful to Tobias Barthel, who sent me the following paper of J. Aguadé:
"On the space of free loops of an odd sphere". Pub. Mat. UAB No 25, June 1981. 
Aguadé proves the following theorem
Theorem: The following are equivalent:

$\Lambda S^{2n+1}\simeq S^{2n+1}\times \Omega S^{2n+1}$;
The free loop space fibration is homotopically trivial
$n=0,1,3$.

That 3 implies 2 is due to the fact that these spheres are $H$-spaces, which was already noted. 2 implies 1 is trivial. Hence the only thing Aguadé shows is that 1 implies 3. 
For this it is assumed that there is a map $f$ inducing a homotopy equivalence. From this there is an induced map $h:S^1\times S^{2n+1}\times \Omega S^{2n+1}\rightarrow S^{2n+1}$. Aguadé applies the Hopf construction to this map to get a map 
$$
\tilde g:S^{2n+1}*(S^1\times S^{2n})\rightarrow S^{2n+2}.
$$
The space on the left is a wedge of spheres $S^{2n+3}\vee S^{4n+2}\vee S^{4n+3}$, thus there is an induced map $g:S^{4n+3}\rightarrow S^{2n+2}$. Aguadé shows that this map has Hopf invariant one, hence the result follows.