Fibrewise homotopy-equivalence of unit sphere bundles vs isomorphism of tangent bundles Let $M$ be a smooth $m$-dimensional manifold, $TM$ its tangent bundle and $SM$ its unit sphere bundle. 
Are there some simple examples where $SM$ is fibrewise homotopy-equivalent to the trivial bundle $S^{m-1} \times M$, yet $TM$ is not trivial as a vector bundle?  Does it ever happen for $M$ a sphere?
Via classifying space machinery this amounts to comparing the orthogonal group $O_m$ to the space of homotopy-equivalences of $S^{m-1}$, $HomEq(S^{m-1})$, in particular its asking for tangent bundle classifying maps $M \to BO_m$ such that the composite $M \to BO_m \to BHomEq(S^{m-1})$ is null.  
As far as I know I've never come across examples of this sort, but then again I haven't studied the homotopy-properties of the map $O_m \to HomEq(S^{m-1})$ in much detail.  Are there many canonical references on this topic? 
This is related to a math.stackexchange question: https://math.stackexchange.com/questions/16779/conditions-for-a-smooth-manifold-to-admit-the-structure-of-a-lie-group
 A: It is proved in [Kaminker, J., Proc. Amer. Math. Soc. 41 (1973), 305–308] that the tangent sphere bundle of a closed smooth H-manifold is (unstably) fibre homotopy trivial. On other other hand, surgery theory allows to construct H-manifolds with non-trivial rational Pontrjagin classes, see e.g. [Victor Belfi, Pacific J. Math. vol 36, Number 3 (1971), 615-621] but this is a standard surgery-theoretic argument. 
Finally, in [Milnor-Spanier, Pacific J. Math. vol 10, Number 2 (1960), 585-590] it is shown that the tangent bundle to $S^n$ is fiber homotopy trivial if and only if $n=1,3,7$,
which are exactly the parallelizable spheres (by Bott-Milnor). 
Thus the above is a complete answer to your question. This phenomenon you ask for never occus for spheres but occurs for lots of H-manifolds.
A: The rational homotopy groups of $HomEq(S^{m-1})$ can be calculated via (Sullivan) minimal models (refer page 314 of D. Sullivan's Infinitesimal computations in topology). In short, if I'm not mistaken one can show that $\pi_i(HomEq(S^{2n})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,4n-1$ and $\pi_i(HomEq(S^{2n+1})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,2n+1$. One the other hand, $Spin(2n+2)$ is rational homotopy equivalent to $S^3\times S^7\times\cdots\times S^{4n-1}\times S^{2n+1}$ while $Spin(2n+1)$ is rationally $S^3\times S^7\times\cdots\times S^{4n-1}$. This should imply at least that, even rationally, the map $O_m\to HomEq(S^{m-1})$ is not a homotopy equivalence in general. May be more is known about the specifics of this map.
