The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent bundle.


This means that a framed manifold (one whose tangent bundle is trivial, e.g. a Lie group) represents an element of the stable homotopy groups of spheres.


So some elements are represented by honestly framed manifolds (as opposed to stably framed).


What is known about such elements? Is every element of the stable homotopy groups of spheres represented by an honestly framed manifold (i.e. with a trivial tangent bundle)?



Which stable homotopy groups are represented by parallelizable manifolds?
I think all elements are representable by honestly framed manifolds.


Let $M$ be a closed $d$-manifold with a stable framing, and consider the obstructions to destabilising a stable framing. Asumng $M$ is connected, which we can arrange by stably-framed surgery, there is a single obstruction, lying in $H^d(M ; \pi\_d(SO/SO(d)))$.


If $d$ is even then $\pi\_d(SO/SO(d)) = \mathbb{Z}$ and this obstruction may be identified with half the Euler characteristic of $M$. (As $M$ is stably framed, its top Stiefel--Whitney class vanishes and so its Euler characteristic is even.) We can change $M$ to $M \# S^p \times S^{2n-p}$ by doing a trivial surgery in a ball, and the stable framing extends over the trace of such a surgery. By taking $p$ to be 1 or 2 we can therefore change the Euler characteristic by $\mp 2$: thus we can change $M$ by stably framed cobordism until its Euler characteristic is 0, whence the stable framing destabilises to an actual framing.


If $d$ is odd then then $\pi\_d(SO/SO(d)) = \mathbb{Z}/2$ and the obstruction is obscure to me (it is realised by the stable framing induced by $S^d \subset \mathbb{R}^{d+1}$, and is non-trivial even in Hopf invariant 1 dimensions where $S^d$ does admit a framing). I can't see an elementary argument for $d$ odd, but I think it is nontheless true by the following.


Let $d=2n+1$ with $d \geq 7$ (lower dimensions can be handled manually). Consider the manifold
$$W\_g^{2n} = \#g S^n \times S^n.$$
This has a stable framing by viewing it as the boundary of a handlebody in $\mathbb{R}^{2n+1}$. By doing some trivial stably-framed surgeries as above (with $p=2,3$ say, to keep it simply-connected), we can change it by a cobordism to a manifold $X$ having an honest framing $\xi$. I wish to apply [Corollary 1.8 of Galatius, Randal-Williams, ``Homological stability for moduli spaces of high dimensional manifolds. II"], to $(X, \xi)$. There is a map
$$B\mathrm{Diff}^{fr}(X, \xi) \to \Omega^{\infty+2n} \mathbf{S}$$
given by a parameterised Pontrjagin--Thom construction. Now there is a step that I would have to think about carefully, but I think that the choices made can be arranged so that $(X,\xi)$ has genus $g$ in the sense of that paper, and so taking $g$ large enough the map above is an isomorphism on first homology. But this has the following consequence: any element $x \in \pi\_{2n+1}(\mathbf{S})$ is represented by the total space of a fibre bundle
$$X \to E^{2n+1} \overset{\pi}\to S^1$$
with a framing of the vertical tangent bundle (and the Lie framing of $S^1$).


(Again, I'm sure there must be a more elementary way of seeing this.)


